Diagonalization and Self-Reference RAYMOND M. SMULLYAN Department of Philosophy Indiana University
CLARENDON PRESS 1994
OXFORD
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Oxford is a trade mark of Oxford University Press Published in the United States by Oxford University Press Inc., New York © Raymond M. Smullyan, 1994
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All rights reserved. No part of this publication maybe reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press. Within the UK, exceptions are allowed in respect of any fair dealing for the purpose of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act, 1988, or in the case of reprographic reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms and in other countries should be sent to the Rights Department, Oxford University Press, at the address above. chi
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ISBN 0 19 853450 7 Typeset by J. M. Ashley
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Printed in Great Britain by Bookcraft (Bath) Ltd Midsomer Norton, Avon
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Preface
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This volume is written for the beginner and expert alike, since it is both an introduction to self-reference, diagonalization, and fixed points, as they occur in Godel's incompleteness proofs, recursion theory, combinatory logic, car
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semantics, and metamathematics, and a presentation of new results car
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partly in these areas, but mostly in their synthesis, as developed in the last nine chapters. No prior knowledge of these fields is presupposed, except that for the applications to mathematical logic in Chapters 5 and 9, a little acquaintance with the logical connectives and quantifiers, though not strictly necessary, is desirable. The rest of the book, however, makes no use of the symbolism of mathematical logic, since I wish the results to be as general and independent of syntax as possible. The main purpose of this study is to present a unified treatment of fixed points in the aforementioned areas. Originally, this book was planned as a companion volume to my previous books: G.I.T. (Godel's Incompleteness Theorems)[32] and R.M. (Recursion Theory for Metamathematics)[33] but it was finally decided to make it independent of the other two volumes, and thus available to a much wider class of readers. The first three chapters are of an introductory nature and consist mainly of exercises (sometimes called `problems') with solutions given to most of them. Despite their elementary nature, there are things there that should interest the expert as well. Chapter 1 introduces the subject of self-reference via quotation, and here, even the expert not familiar with my paper `Quotation and self-reference' should find it of interest that crosscar
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achieve using one-sided quotation instead of the more usual two-sided quotation. Chapter 2 is an updated version of my paper `Some unifying fixed point principles' and contains the germ of an idea developed in later chapters. Chapter 3 contains abstract versions of halting problems, and should be of interest to the expert mainly for pedagogical reasons. Chapter 4 is essential to several subsequent chapters and is a greatly expanded treatment of representation systems as defined in T.F.S. (Theory
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of Formal Systems). It consists of abstract versions of classical results of Tarski, Godel, and Rosser, as well as an account of Rowitt Parikh's demonstration that there are sentences in systems such as Peano Arithmetic car
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Chapter 5, in which mathematical logic first appears, can be read any time after Chapter 1, but its first part is a prerequisite to Chapter 9. Actually, Chapters 5 and 9 can be thought of as supplementary to the rest of the volume (and, if desired, could be read after all the other chapters). In Part I of Chapter 5, it is first shown how the method of normalization explained in Chapter 1 can be carried out for first-order arithmetic with class abstracts, as in my paper `Languages in which self-reference is possible'. This is followed by an account of Tarski's method of achieving self-reference in the more standard formulation of first-order arithmetic in which we don't have class abstract, and this is also compared to Godel's method. Part II of this chapter contains some particularly curious material which, though easily understood by the beginner who has at least a nodtin
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ding acquaintance with the logical connectives and quantifiers, should both 'C3
surprise and amuse the expert. It begins by showing how a minor modification of normalization works for the more usual first-order formulations of arithmetic in which there are not any class abstracts, and furthermore, that normalization itself works if the system is in Polish notation! It then goes on to simultaneously generalize normalization, diagonalization and Cantor's famous construction. Chapters 6, 7, and 8 constitute an introduction to recursion theory from the viewpoint of elementary formal systems, as originally presented in T.F.S.. I have devoted considerable space to their treatment for several reasons. First, these systems appear to be coming into vogue these days partly due to computer science, and partly due to their interesting generalization by Melvin Fitting in his study: `Elements of Recursion Theory.' Secondly, the original T.F.S. is out of print (though the Russian translation might still be available). Thirdly, I make it quite clear here (which I apparently failed to do to some readers of T.F.S.) that in application to incompleteness proofs for systems such as Peano Arithmetic, elementary formal systems not only show the existence of undecidable sentences, but can conveniently be used to find them. In Chapter 9 (using Part I of Chapter 8) it is shown quite explicitly how from an elementary formal system for an incomplete axiomatizable theory, an undecidable sentence can be found. We carry this out in particular for Peano Arithmetic. The approach is quite neat, in that Godel numbering comes in only at the last moment, so to speak; most of the work deals directly with the expressions car
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Preface
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Chapters 9 and 10 constitute an upgraded account of many of the results established in T.F.S. and R.M.. The present account is improved in several respects. For one thing, the theorems given in R.M. for first-order systems are shown here to generalize to representation systems. Secondly, the double indexing of r.e. relations introduced in Part II of Chapter 8, and carried out in the more general context of Chapter 10, makes the proofs simpler and more elegant. Also, many of the results for recursion theory proved '1y
in R.M. are established here in the more general context of generalized (D0
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recursion theory, in the sense of [5]. Also, as promised in R.M., Shepherd((D
son's exact separation theorem and some results in recursion theory are derived from a common construction. Furthermore, the strengthened form of Shepherdson's theorem that was stated and proved in the last chapter of R.M. is shown here to be a corollary of a result in pure recursion theory (Theorem 8.1, Chapter 11) which we believe to be of interest in its own s0.
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Chapters 12-20 fulfil the main purpose of this study, which is the unification of fixed point theorems in recursion theory (in fact in generalized recursion theory), combinatory logic, and metamathematics. These chapters are logically independent of the preceding chapters, but their motivation would be lost to those readers without a background comparable to that given in Chapters 5-8. [The expert could, of course, read Chapters 12-20 from scratch.] Chapter 12 introduces and begins the study of abstract structures called sequential systems, each fixed point theorem of which simultaneously yields a fixed point theorem for recursion theory, one for combinatory logic, and one for metamathematics. The results on sequential systems are also applicable to the uniform reflexive structures of Erik Wagner[37], but these applications are not given here. I have devoted considerable space to combinatory logic since some of its fixed point properties can be established as corollaries of fixed point theorems for sequential systems, and others, being of a highly specialized nature, are not even statable in terms of sequential systems. Chapters 17 and 18 constitute an introduction to combinatory logic, with special emphasis on fixed point properties and various ways of constructing fixed point combinators and related entities. Chapter 19 is devoted to a generalization and extensions of the result known as the second fixed point theorem of combinatory logic. This is usually stated in terms of numeral systems together with a Godel numbering, but we take a more general approach, using what we call an admissible naming relation between combinators. A specific such relation is given, which does not employ either a numeral system or a Godel numbering, but our more general results are applicable to the more usual naming relation as well, as is also explained. The fixed point results of Chapter 19 all have further generalizations to sequential ((DD
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Preface
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Fig. 1. Interdependence of chapter's
systems with some added structure, which constitutes the closing chapter of this study.
Contents I - Introduction to Self-Reference and Recursion Introduction to self-reference
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Part II
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Quotation and self-reference 2 2 Use and mention Self-reference using diagonalization 3 Normalization 4 One-sided quotation 7 9 Cross-reference using one-sided quotation Some general principles concerning designational systems 12 A pseudo-quotational system 14 Some other methods of self-reference 15 Self-reference in a more general setting 16 A more general setting 16 17 Applications to quotational systems Near Diagonalization 18 Cross-reference 19 Application to one-sided quotational systems 20 cad
Part I
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2 Some classical fixed point arguments compared Part I
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Five fixed point arguments §1 An argument from combinatory logic §2 Godel sentences §3 An argument from recursion theory §4 Another argument from recursion theory §5 Self-reproducing machines Part II A unification §6 A relational fixed point theorem Part III Quasi-diagonalization §7 Some stronger results
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3 How to silence a universal machine ix
25 25 25 26 27 28 29 31 31 34 34 C.0
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Contents
Part I
Silencing a universal machine A universal machine Part II Some related problems Part III Solutions to problems
39 39 43 44
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4 Some general incompleteness theorems Part I §1
§7
Part II §8
Self-reference in arithmetic
75
Part I
75
Part III Generalizations
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6 Introduction to formal systems and recursion Part I
84 85 85
87 88
90
93 97
97 97 105 110 113 113 118 122 126 129 irk
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Formal representability §1 Elementary formal systems §2 Some basic closure properties §3 Solvability over K Part II Recursive enumerability §4 Recursively enumerable relations §5 Recursive functions Finite quantifications and constructive definability §6 Recursive pairing functions §7 Dyadic Godel numbering §8
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Godelizing operations Generalized diagonalization .'3
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Normalization and Tarskification §1 Normalization in arithmetic §2 Tarski's method of achieving self-reference The more general situation §3 Part II Some special devices §4 Near normalization §5 Situation in Polish notation §6 More on quotational systems §7 Some variants
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Incompleteness 50 Diagonalization 50 An incompleteness argument using a truth set 52 Godel's argument and some variations 56 Rosser's argument 59 Complete representability and definability 63 Admissible functions 67 Kleene's symmetric incompleteness theorem generalized 68 Provability by stages 69 Provability by stages 69
Contents
§9
xi
Lexicographical Godel numbering and n-adic notation 132
7 A universal system and its applications Part I
135 135 139 141 141
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A universal system §1 The universal system (U) §2 The recursive unsolvability of (U) Part II Enumeration and iteration theorems §3 The enumeration theorem Iteration theorem
144
II - Systems with Effective Properties
Part II A concrete incompleteness proof §2 Peano arithmetic §3 Peano arithmetic is a formal system §4
The Godel and Rosser proofs
10 Doubly indexed relational systems Part I
Indexed relational systems §0 Definitions §1 Universal sets §2 Generative sets
Part II Double indexg
11 Effective representation systems Part I Effective representation
176 176 176 178 179 182 182 r-1
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Double enumeration and double universality Doubly generative pairs Semi-D.G. pairs Closing the circle Summary of main results
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165 166 166 169 169 170 174
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Truth in CO cannot be formalized The non-formalizability of arithmetic truth
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9 Elementary formal systems and incompleteness proofs
149 149 150 154 156 156 159 159 161
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Arithmetization of elementary formal systems §1 Some facts about Eo-relations §2 Arithmetization of elementary formal systems Part II Double enumeration §3 Some applications Part III More on arithmetization §4 All r.e. relations are E1 §5 More on arithmetization (optional)
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Part I
187 190
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8 Arithmetization of formal systems
195 198
200 200
xii
Contents +--F
Undecidability Generative systems Doubly generative systems Rosser systems Extensions Part II Exact and effective Rosser systems Exact Rosser systems §6 §7 Variations on a theme by Shepherdson' §8 Effective Rosser systems §9 Some simpler related results §10 Unconquerable systems and omniscient functions
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III - Fixed Point Theorems in a General Setting
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Part III Some fixed point properties §2 The weak fixed point property Universal systems and the weak Kleene property §3 §4 Integrated systems Part IV Summary and applications
13 Strong fixed point properties Part I
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14 Multiple fixed point properties Part I §1
Double fixed points The weak double fixed point property
245 245 245 249 251 252 253 255 256 257 258 259 260 261 262
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The recursion and extended recursion properties §1 The recursion property §2 A type 2 approach The extended recursion property §3 Part II The Myhill property and related properties §4 Strongly connected systems §5 The Myhill and recursion properties compared §6 Very strong connectivity §7 Universal integrated systems §8 The strong Kleene property Part III Summary and applications I Indexed relational systems II Sentential systems III Applicative systems
228 228 233 233 236 236 239 241 242
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12 Sequential systems Part I Definitions and purpose Part II Preliminary definitions
263 263 263
Contents
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xiii
Part II Double recursion properties Double recursion A type 2 approach
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Part III Double Myhill properties and related results §4 The double Myhill property The properties DMk §5 §6 Very Strongly Connected Systems Integrated systems of type 1 §7 Part IV Symmetric functions and nice functions §8 Symmetric functions coy
Nice functions Universal systems of type 2 Multiple fixed point properties
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Part I
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15 Synchronization and pairing functions coy
Synchronized fixed point properties §0 Synchronized sequences of functions §1 Synchronized fixed point theorems §2 Relation to multiple fixed point properties Part II Pairing functions §3 Exact and quasi-exact functions §4 Some consequences §5 Strongly connected systems of type 1* Part III Synchronized recursion resumed §6 The property DR* §7 The property DM* §8 Integrated systems of type 1* CG)
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16 Some further relations between fixed point properties Part I
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Single and double fixed point properties compared §1 Recursion and double fixed points §2 Recursion and double recursion §3 Systems of type 1* §4 Myhill and double Myhill properties Part II More on systetns of type 1* §5 Bar fixed points §6 Further topics
267 267 269 273 273 277 278 278 279 279 281 283 284 286 286 286 288 290 292 293 296 297 299 299 301 302
304 304 304 306 307 308 309 309 311
IV - Combinators and Sequential Systems 17 Fixed point properties of combinatory logic Part I §1
Interdefinability of some combinators The combinator B
315 315 316
Contents
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We add T We add M Two special combinators The combinator I A basis equivalent to {B, T, 1} .`n
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Some curiosities We now consider K Idempotent and voracious elements Part II Fixed points §10 Some sufficient conditions §11
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317 320 324 325 326 326 327 329 329 329
Fixed points in relation to idempotent and voracious
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Sages from B and W
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elements A fixed point principle Some fixed point combinators Turing sages Some special properties of 0
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Part III Multiple fixed points §18 coy coy coy
§19 §20 §21
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Some more useful combinators The cross-point property The weak double fixed point property The property DM Nice combinators and symmetric combinators
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Admissible name functions Admissible maps via numeral systems Part II Fixed point theorems of the second type §3 The second fixed point theorem §4 Applications §5 Other fixed point theorems of the second variety §1 §2
20 Extended sequential systems M-diagonalizers
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336 336
337 338 339
357 357 357 360 361
366 366 366 371 373 373 374 378
380 381 r-1
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19 A second variety of fixed point theorems Part I Self-reference in CLo
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Formal combinatory logic Consistency and models Extensional combinatory logic
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18 Formal combinatory logic Part I The axiom systems CL and CLo
330
Contents
§2 §3 §4 §5
M-fixed points The double M-fixed point theorem Strong M-fixed point properties Uniform double M-fixed point properties M-nice and M-symmetric functions
382 383 384 387 388 000
§1
xv
References
389
Index
392
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I
Introduction to Self-Reference and Recursion
Chapter 1
Introduction to self-reference Self-reference plays a crucial role in the famous incompleteness theorem of G6del[6]. What he did can be roughly described as follows. He showed that for a large class of mathematical systems, one can assign to each sentence car
a number called the Godel number of the sentence, and then construct a sentence X asserting that a certain number n has the property of being the Godel number of a sentence that is not provable in the system, but this number n is the Godel number of the very sentence X itself! And so X is true if and only if its Godel number is not the Godel number of a sentence provable in the system - in other words, X is true if and only if it is not provable in the system. This means that either X is true but not provable in the system, or X is false (not true) but provable in the system. Under the assumption that the system is correct in that only true sentences are provable, the second alternative is out, hence the sentence is true but not provable in the system. This is an extremely rough sketch of what Godel cad
did. Now, how did Godel manage to construct a sentence that says something about its own Godel number? Such a sentence is, in a sense, self-referential. There are many methods now known of achieving self-reference, as we will
see in the course of this book.
Indexicals
Suppose we wish to construct a sentence that ascribes to itself a certain property - say the property of being read by an individual named John. The simplest way to do so is to construct the following sentence. John is reading this sentence.
Chapter 1. Introduction to self-reference
2
Obviously, the above sentence is true if and only if John is reading it. Now,
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that sentence contains the indexical word this. (An indexical is a word or phrase whose denotation depends on its context, as for example, the denotation of the word "I", which depends on who is uttering it, or again, the denotation of "now" depends on the time it is uttered.) Now, although it is possible to formalize the use of indexicals (as in [29]), indexicals do not occur in the type of systems studied by Godel, and so we must turn to other
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methods. In Part I of this chapter, we study self-reference by quotation. This strikes us as a nice introductory approach, since quotation is a well known device for naming expressions, and its formalization leads to some intriguing problems that should interest the beginner and expert alike. We conclude Part I with a brief mention of other methods of achieving selfreference - including that of Godel numbering. A more general approach to self-reference is considered in Part II. Not all sections of this chapter are needed for subsequent chapters; most of them are of purely independent interest. The sections that are most relevant for this volume as a whole are §§1-3, 8, 9, and also exercises 33 and 34.
Part I Quotation and self-reference §1 Use and mention The following sentence is true. (1)
Ice is frozen water.
What about the following sentence? (2)
Ice has three letters.
Strictly speaking, sentence (2) is false. The substance ice has no letters
at all; it is the word "ice" that has three letters. And so the correct rendition of (2) is (2)'
"Ice" has three letters.
When one talks about a word, rather than that which is denoted by the word, one encloses the word in quotation marks. There is a difference between using a word and mentioning the word (which is talking about the word, instead of the denotation of the word). It can happen that a word is both used and mentioned in the same sentence - witness (3)
"Ice" is the name of ice.
§2 Self-reference using diagonalization
3
Or, a more startling illustration of how a phrase can be both used and mentioned in the same sentence is (4)
It takes longer to read the bible than to read "the bible."
Yes, it takes much longer! Here is another illustration. (5)
This sentence is longer than "this sentence."
On first encountering the distinction between use and mention, confusions often occur. For example, is the following sentence true of false? " "Ice" " has two pairs of quotation marks.
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Many would say that (6) is true; actually it is not! What is written has two pairs of quotation marks, but what is talked about has only one pair. And so it is rather the following sentence that is correct. " "Ice" " has one pair of quotation marks.
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When no ambiguity can arise, words and phrases are sometimes used autonomously, i.e., as names of themselves. For example, in a treatise on language, sentence (2) (ice has three letters) would be acceptable. Also, under autonomous use, sentence (6) would be acceptable. In textbooks and articles on symbolic logic, formulas are used as names of themselves. [For example, one would write "P D (P V Q) is a tautology" instead of " "P D (P V Q)" is a tautology."]. The context will usually make it clear when autonomous use is permissible.
§2 Self-reference using diagonalization
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Godel achieved self-reference by the method known as diagonalization, that we now illustrate using quotation marks instead of Godel numbering. We use the symbol "x" as a variable ranging over expressions of the English language. By the diagonalization of an expression, we mean the result of substituting the quotation of the expression for every occurrence of the variable "x" in the expression. For example, consider the following
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John is reading x
The expression (1) is not a sentence, true or false, but becomes a sentence (true or false) upon substituting the quotation of any expression for "x". If we substitute the quotation of (1) itself for "x", we obtain the diagonalization of (1), which is (2)
John is reading "John is reading x"
Chapter 1. Introduction to self-reference
4
Now, (2) is a sentence, and it asserts that John is reading (1). However,
(2) is not self-referential; it does not assert that John is reading (2); it asserts that John is reading (1). Let us consider the following expression. (3)
John is reading the diagonalization of x.
The diagonalization of (3) is the following (4)
John is reading the diagonalization of "John is reading the diagonalization of x."
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Sentence (4) asserts that John is reading the diagonalization of (3), but the diagonalization of (3) is (4) itself. Thus (4) asserts that John is reading the very sentence (4)! Thus sentence (4) is self-referential. This, in essence, is how Godel achieves self-reference. It might be easier to understand this if we use the following abbrevi-
ations. Let us use "J" to abbreviate "John is reading," and "D" to abbreviate "the diagonalization of." Then (3) and (4) assume the following abbreviated forms: (3)' (4)'
JDx JD"JDx"
The sentence (4)' asserts that John is reading the diagonalization of (3)', but the diagonalization of (3)' is (4)' itself.
§3 Normalization To carry through the idea of diagonalization for arithmetical systems is relatively complicated, since it involves the operation of substitution, which is difficult to arithmetize. In my paper Languages in which self-reference is possible[27], I showed another method of self-reference that does not involve substitution, nor does it use variables. To this we now turn. By the norm of an expression we shall mean the expression followed by its own quotation. For example, consider the following expression. (1)
John is reading
The norm of (1) is the following. (2)
John is reading "John is reading"
The sentence (2) is not self-referential; it doesn't assert that John is reading (2); it asserts that John is reading (1). But now consider the following.
§3 Normalization (3)
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John is reading the norm of
Its norm is the following sentence. (4)
John is reading the norm of "John is reading the norm OF
Sentence (4) asserts that John is reading the norm of (3), but the norm of (3) is (4) itself. And so (4) asserts that John is reading (4). Thus (4) is self-referential. -t4
Let us look at an abbreviated version. Again, we abbreviate "John is reading" by the letter "J." And we shall use the letter "N" to abbreviate "the norm of." Then (3) and (4) assume the following symbolic forms. (3)' (4)'
JN JN"JN" sue.
(4)' asserts that John is reading the norm of (3)', but the norm of (3)' is (4)' itself. Thus (4)' asserts that John is reading (4)'. This construction is like one due to Quine, whose version[18] of the semantical paradox of the liar is the sentence: "Yields falsehood when appended to its own quotation" yields falsehood when appended to its own quotation. This sentence asserts its own falsehood. Alternatively, using our notion of norm, the following is a version of the paradox. The set of false sentences contains the norm of "the set of false sentences contains the norm of" Using diagonalization instead of normalization, the paradox would be stated thus. cad
The set of false sentences contains the diagonalization of "the set of false sentences contains the diagonalization of x"
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Let us note in passing that, using diagonalization, the following is an expression that designates itself: the diagonalization of "the diagonalization of x." Or symbolically: D "Dx" . Using normalization, the following expression designates itself: the norm of "the norm of." Or symbolically: N "N". C.))
Exercise 1 Using the symbols J and N as well as opening quotes and closing quotes, construct a sentence X that asserts, not that John is reading X, but that John is reading the norm of X.
Exercise 2 Using the same symbolism, find an expression that names its own norm.
Chapter 1. Introduction to self-reference (FD
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Exercise 3 Now let us add the symbol "Q" to abbreviate "the quotation of." Construct an expression that names its own quotation.
Exercise 4 In this and the remaining exercises of this section (through Exercise 7) we continue to use "Q" for "the quotation of."
Find an X that names the norm of its own quotation and an X that names the quotation of its norm.
Exercise 5 Find a sentence X that asserts that John is reading the quotation of X.
Exercise 6 Find a sentence X that asserts that John is reading the quotation of the norm of X, and another sentence X that asserts that John is reading the norm of its quotation. Exercise 7 Find two distinct expressions X and Y such that each names the other.
Solutions
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1 Take X = JNN "JNN" . The sentence X asserts that John is reading the norm of the norm of JNN. Now, the norm of JNN is X, hence the norm of the norm of JNN is the norm of X. Thus X asserts that John is
reading the norm of X.
2 NN"NN". 3 QN"QN". It names the quotation of the norm of QN, which is the quotation of QN"QN" (since the norm of QN is QN"QN").
4 An expression that names the norm of its quotation is NQN"NQN". An expression that names the quotation of its norm is QNN "QNN" . 5 Take X = JQN "JQN" . This sentence X asserts that John is reading the quotation of the norm of JQN, which is the quotation of X (since the norm of JQN is X).
6 A sentence that asserts that John is reading the quotation of its norm
is JQNN"JQNN". cad
A sentence that asserts that John is reading the norm of its quotation
§4 One-sided quotation
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is JNQN"JNQN". (In unabbreviated English, this would be: John is reading the norm of the quotation of the norm of "John is reading the norm of the quotation of the norm of".)
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7 In Exercise 3, we found an X that names its own quotation which in turn names X. So we take X = QN"QN", and Y ="QN"QN" ".
§4 One-sided quotation
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To explore an interesting side stream, I would like to say something about one-sided quotation. Here we use only one quotation symbol - which we will take to be the symbol "*". For any expression X, we call *X the (onesided) quotation of X. One-sided quotation is used in certain programming languages such as LISP. For our purposes, one sided quotation has certain advantages over two-sided quotation - in particular it allows not only selfreference, but also easily allows cross-reference (pairs of sentences, each of ^r'
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which ascribe a certain property to the other) in a manner that we will shortly explain.
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We now define the associate of an expression X to be the expression X*X - in other words, X followed by its own one-sided quotation. We shall use "A" to abbreviate "the associate of" and we shall continue to use "J" to abbreviate "John is reading." Then a sentence that asserts that John is reading it is JA*JA. It asserts that John is reading the associate of JA, which is JA* JA. A conceptually simpler operation than association, which also accomcoo
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plishes self-reference, is that of repetition, where by the repeat of an expresI+.
sion X is meant the expression XX, i.e., X followed by itself. Let us use "R" to abbreviate "the repeat of." Then a sentence that asserts that John is reading it is JR*JR*. It asserts that John is reading the repeat of JR*, which is the very sentence JR*JR*. (The sentence JR*JR won't work; it asserts that John is reading JRJR, not that John is reading JR*JR.) These two schemes of self-reference - the one using association; the other using repetition - were discussed in further depth in my paper Quotation and self-reference [31]. For now, let me remark that an expression that is its own name is A*A (it names the associate of A, which is A*A), and another is R*R* (it designates the repeat of R*, which is R*R*). Before proceeding further, let us compare the four ways we have so far of constructing a sentence that asserts that John is reading it. L''
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Chapter 1. Introduction to self-reference
8
JD"JDx" JN"JN" JA*JA JR*JR*
(1) (2) (3) (4)
Let us also compare the four ways we have of constructing an expression that is its own name. (1) (2) (3) (4)
D "Dx" N "N" A*A R*R* car
In what follows, we consider a language containing at least the symbols J, A, R, * and we assume that if X is any expression built from the symbols
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of the language, the expression *X denotes (or is a name of) X, and that if X denotes Y, then AX denotes the associate of Y and RX denotes the repeat of Y. (For example, A*Y denotes the associate of Y; R*Y denotes the repeat of Y; AA*Y denotes the associate of the associate of Y; RA*Y denotes the repeat of the associate of Y.) Also, for any expressions X, Y of the language, if X denotes Y, then JX asserts that John is reading Y. (For example, JA*Y asserts that John is reading the associate of Y, since A*Y denotes the associate of Y.)
Exercise 8 Using just the two symbols A and *, find an expression that names (denotes) its own (one-sided) quotation. (Note that unlike the case of two-sided quotation, we get this result "free," as it were, that is, we do not have to add a symbol such as "Q" to abbreviate "the quotation of.") Do the same using R in place of A.
Exercise 9 Using any or all of the three symbols A, R, * find .-1
(1) An X that names its own repeat. (There are two solutions.) (2) An X that names its own associate. (There are two solutions.)
(3) An X that names the repeat of its associate. (There are two solutions.)
(4) An X that names the associate of its repeat. (There are two solutions.)
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Exercise 10 Using any or all of the same three symbols, together with "J" for "John is reading," find a sentence X that asserts that John is reading the repeat of X. (There are two solutions.)
§5 Cross-reference using one-sided quotation
9
Solutions 8 Using "A", a solution is A**A. It denotes the associate of *A, which is *A**A. Using "R", a solution is R**R*, which denotes the repeat of *R*, which is *R**R*. car
9 (1) RR*RR* is one solution. Another is RA*RA. (2) AA*AA is one solution. Another is AR*AR*.
(3) RAA*RAA is one solution. Another is RAR*RAR*. (4) ARA*ARA is one solution. Another is ARR*ARR*. 10
One such sentence is JRA*JRA. Another is JRR*JRR*.
§5 Cross-reference using one-sided quotation
er-
Suppose now we consider a second individual named Paul, and we wish to construct two sentences X and Y such that X asserts that John is reading
Y and Y asserts that Paul is reading X. Using one-sided quotation, this is relatively easy, using either association or repetition. There are two solutions in each case. We add "P" to our language to abbreviate "Paul is reading."
Exercise 11 Using the four symbols J, P, A, and *, construct such sentences X and Y. There are two solutions. Can you find them both? Exercise 12 Do the same, using R instead of A.
Exercise 13 Using the two symbols A and *, find two distinct expressions X and Y such that each is the name of the other. Do the same with R and *.
Exercise 14 [A fixed point principle] Show that for any expression E of the language, an expression X can be found that names EX (that is, it names the expression consisting of E followed by X. For example, taking E to be ARA, there is some X that names ARAX.) There are two ways of finding X, given E; one way uses A and the other uses R. Find both ways.
Chapter 1. Introduction to self-reference
10
Exercise 15 Show that for any expression E there is an X that denotes CAD
the associate of EX (there are two ways of getting such an X) and there is an X that denotes the repeat of EX (again there are two solutions).
Exercise 16 Find expressions X and Y such that X names the repeat of Y and Y names the associate of X (there are four solutions).
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Exercise 17 Given an expression E, find a sentence X that asserts that John is reading the repeat of EX (there are two solutions). Then show CAD
that there is a sentence X that asserts that John is reading the associate of EX (there are two solutions).
Exercise 18 Find sentences X and Y such that X asserts that John is cad
reading the repeat of Y and Y asserts that Paul is reading the associate of X (there are four solutions).
Exercise 19 Show that for any two expressions El and E2 there are expressions X and Y such that X names E1Y and Y names E2X (there are four solutions).
Solutions 11 One solution is X = J*PA*J*PA; Y = PA*J*PA. Another is X = JA*P*JA; Y = P*JA*P*JA.
12 One solution: X = J*PR*J*PR*; Y = PR*J*PR*. Another solution: X = JR*P*JR*; Y = P*JR*P*JR*.
13 Using A, take X = A**A and Y = *A**A. Using R, take X = R**R* and Y = *R**R*. 14 Using A, take X = A*EA. It denotes the associate of EA, which is
EA*EA, which is EX.
Using R, take X = R*ER*. It denotes the repeat of ER*, which is ER*ER*, which is EX.
15 An X that denotes the associate of EX is AA*EAA. Another is AR*EAR*. 4..
An X that denotes the repeat of EX is RR*ERR*. Another is RA*ERA.
§5 Cross-reference using one-sided quotation
11
16 Two solutions are obtained by finding an X that denotes the repeat of A*X and then taking Y = A*X. By the last exercise, there are two ways of finding such an X (taking A* for E), and so we have the following two solutions: (1) (2)
X = RR*A*RR* Y = A*RR*A*RR* X = RA*A*RA
Y = A*RA*A*RA
Two other solutions can be obtained by taking some Y that denotes the
associate of R*Y and then taking X = R*Y. We thus get the following two solutions: (3) (4)
X = R*AA*R*AA
Y = AA*R*AA
X = R*AR*R*AR* Y = AR*R*AR*
17 An X that asserts that John is reading the repeat of EX is JRA*EJRA. Another is JRR*EJRR*.
An X that asserts that John is reading the associate of EX is JAA*EJAA. Another is JAR*EJAR*. d
18 Two solutions are obtained by finding an X that asserts that John is reading the repeat of PA*X and then taking Y = PA*X (which asserts
that Paul is reading the associate of X). By the last exercise (taking PA* for E), we have the two solutions X = JRA*PA*JRA, or X =
o00 ,;4
JRR*PA*JRR*. In either solution, we take Y = PA*X. Two other solutions are obtained by taking some Y that asserts that Paul is reading the associate of JR*Y, and then taking X = JR*Y. One such Y is JAA*JR*JAA. Another is JAR*JR*AR*.
19 Two solutions are obtained by finding an X that denotes E1*E2X and taking Y = *E2X. These two solutions are: X = A*E1*E2A
Y = *E2A*E1*E2A
X = R*E1*E2R* Y = *E2R*E1*E2R*
*
(1) (2)
Two other solutions are obtained by finding some Y that denotes E2*E1Y and taking X = *E1Y. We thus get: X = *E1A*E2*E1A
Y = A*E2*E1A
X = *E1R*E2*E1R* Y = R*E2*E1R*
*
(3) (4)
Let us now add a third person named William and add to our symbolic language the symbol "W" to abbreviate "William is reading."
Chapter 1. Introduction to self-reference
12
Exercise 20 Find sentences X, Y, and Z such that X asserts that John is reading Y, Y asserts that Paul is reading Z, and Z asserts that William is reading X.
Exercise 21 Find sentences X, Y, and Z such that X asserts that John is reading Y, Y asserts that Paul is reading the associate of Z, and Z asserts that William is reading the repeat of the associate of X.
§6 Some general principles concerning designational systems
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Suppose now that we are given a language L in which there are rules determining that certain expressions of the language denote or designate others. Consider now a function f that assigns to every expression X of the language an expression f (X) of the language. Let us say that an expression F defines the function f in the language L if for any expressions X and Y such that X designates Y, the expression FX designates f (Y). (For example, in the one-sided quotation system that we have been considering, the symbol A defines the association operation and R defines the repetition operation. Also RA defines the function that assigns to each X the repeat of the associate of X. Any string of R's and A's defines some function or other.) And we say that f is definable in L if there is some expression F that defines it. Now, let us say that L has the designational fixed point property if for any expression E (of the language L) there is some X that designates EX. Let us say that L has the functional fixed point property if for any function f definable in L, there is some expression X that designates f (X).
Exercise 22 Prove that if L has the designational fixed point property then it also has the functional fixed point property.
Exercise 23 Let us say that L has the designational cross-point property if for any expressions El and E2 there are expressions X and Y such that X denotes E1Y and Y denotes E2X. Let us say that L has the functional cross-point property if for any functions f and g definable in L, there are expressions X and Y such that X denotes f (Y) and Y denotes g(X). Prove that the designational cross-point property implies the functional cross-point property.
6 Some general principles concerning designational systems
13
Solutions 22
Suppose F defines the function f. By the hypothesis, there is some Y
that designates FY. Then FY designates f (FY). We then take X = FY and so X designates f (X). 23 Suppose F defines f and G defines g. Then by hypothesis, there are
expressions Zl and Z2 such that Zl denotes GZ2 and Z2 denotes FZl. Then FZl denotes f (GZ2) and GZ2 denotes g(FZ1). And so we take X = FZl and Y = GZ2. Then X denotes f (Y) and Y denotes g(X). Self-referential and cross- referential languages
P..
Suppose now that certain expressions of L are called sentences and certain of the sentences are called true sentences. We shall say that a property P of expressions of L is defined by an expression H if for any expressions X and Y such that X designates Y, HX is a sentence and is a true sentence if and only if Y has the property P. (For example, in the language we have been studying, J defines the property of being read by John, since if X designates Y, then JX asserts that John is reading Y, and accordingly is true if and only if Y has the property of being read by John.) Now, let us say that L is self-referential if for any property P definable in L, there is a sentence X such that X is true if and only if it has the property P. (In a purely extensional sense, such a sentence X can be thought of as asserting that it has property P.) CAD
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Exercise 24 Prove that if L has the designational fixed point property then L is self-referential.
Let us now say that L is cross-referential, if for any two properties Pl and P2 definable in the language, there are sentences X and Y such that X is true if and only if Y has property Pl and Y is true if and only if X has property P2.
Exercise 25 Prove that if L has the designational cross-point property, then L is cross-referential. A fact about one-sided quotational systems
Call L a one-sided quotational system if there is an expression - which such that for any expression X, the we will take to be the symbol expression *X designates X.
Chapter 1. Introduction to self-reference
14
Exercise 26 Suppose L is a one-sided quotational system having the designational fixed point property. Prove that L has the designational cross point property.
Solutions 24 Suppose H defines property P in L. By hypothesis, there is some Y that designates HY. Then HY is true if and only if HY has property P. We thus take X to be HY. 25 Suppose H1 defines P1 and H2 defines P2 in L. By hypothesis there are expressions Z1 and Z2 such that Z1 designates H2Z2 and Z2 designates H1 Z1. Then H1 Z1 is true if and only if H2 Z2 has property P1 and H2Z2 is true if and only if H1Z1 has property P2. And so we take X = HI ZI and Y = H2Z2. c",
26 Assume the hypothesis. Take any expressions E1 and E2. By hypothesis there is some X that designates EI*E2X. Then take Y = *E2X. Alternatively, there is some Y that designates E2*E1Y. Then we can take X = *E1Y.
§7 A pseudo-quotational system Cam
Before leaving the subject of quotational systems, I would like to briefly
mention a streamlined method of achieving self-reference and crossreference which does so in a particularly simple manner. The essential feature of a quotational system is that a named expression
actually appears as a part of its name. The system to which I now turn also has this feature, but neither two-sided nor one-sided quotation is used. Accordingly, the system might aptly be called pseudo-quotational.
Let us now take the six symbols J, P, W, J, P, and W. For any expression X, we now boldly write JX to express the proposition that John is reading X. (We don't use any name of X, unless we regard X as its own name.) And we write JX to mean that John is reading XX (the repeat of X). And, of course, we use P and W similarly (for Paul and William).
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Self-reference now is completely trivial: an expression X that asserts that John is reading X is obviously JJ. Cross-reference, though not quite so trivial, is simple, compared to what we have done before. We want rte,
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sentences X and Y such that X asserts that John is reading Y and Y asserts that Paul is reading X. One solution is X = JPJ; Y = PJPJ. Another is X = JPJP; Y = PJP.
§8 Some other methods of self-reference
15
Exercise 27 In this system, find sentences X, Y, and Z such that X asserts that John is reading Y, Y asserts that Paul is reading Z, and Z asserts that William is reading X. coy
§8 Some other methods of self-reference
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Let us now briefly consider some other methods of achieving self-reference. In both quotational and pseudo-quotational systems, the expression named actually appears as part of its name. We now consider two other setups in which this is not the case. One method is what Quine [16] calls designation by spelling. In this method, there is a finite or denumerable, ordered alphabet of symbols including two special symbols, which we will currently assume to be S and the subscript. We then use the expressions Si, S1i, Sill, ... as respective
names of the first, second, third, ... symbols respectively. A compound expression is "spelled out" (named) by replacing each symbol by its name. For example, let us consider the following four symbols:
JNS1 "`S
The names of these four symbols are S1, Si1, Sill, and Sill,. (Note that S itself has a name, namely Sill, and the name of the subscript is 51111.) Then the name of the compound expression SNJ is SiiiSiSii; the
name of S11N is SlllSllilSliiiSll
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We now redefine the norm of an expression X as X followed by its own name, and we let N abbreviate "the norm of." That is, N followed by the name of an expression X (not by X itself!) designates the norm of X (e.g., since Si is the name of the symbol J, then NS1 designates the norm of J, which is JSi). Again, we let J abbreviate "John is reading," and if we want to write a sentence that asserts that John is reading a given expression X, we write down J followed, not by X itself, but by an expression that designates X. [A sentence that asserts that John is reading JJ is not JJJ, but JS1S1.] What sentence X asserts that John is reading the very sentence X? JNS1S11 is such a sentence. [Compare with JN"JN".] Now let's consider self-reference via Godel numbering. There are many useful ways to assign Godel numbers to expressions, and we now consider a way that I have termed lexicographical Godel numbering. Given a finite ordered alphabet al, a2, ... , an of symbols, we order all expressions in these symbols lexicographically i.e., according to length, and then alphabetically within each group of the same length. For example, if n = 3, our ordering of all expressions in the symbols al, a2, a3 begins as follows: al, a2, a3, a1a1, a1a2, a1a3, a2a1, a2a2, a2a3, a3a1, a3a2, a3a3, alaial, a1a1a2, aiala3,
ala2al, ...
.
16
Chapter 1. Introduction to self-reference
We then take the Godel number of an expression to be the position it occupies in the sequence, for example, for n = 3, the (lexicographical) Godel number of alas is 4; the Godel number of alala2 is 14. Let us now take the following three symbols.
JN1 We consider the lexicographical Godel numbering of all words (expressions) in these three symbols (the symbols being ordered in the given order-
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ing J, N, 1, so that, e.g., the Godel number of NN is 8. For any positive integer n, we let n consist of a string of is of length n (e.g., 3 ="111"). By the Godel numeral of an expression X, we mean n where n is the Godel number of X (e.g., the Godel numeral of NN is 11111111). And by the norm of an expression we now mean the expression followed by its Godel numeral (e.g., the norm of 1J is 1J1111111111). By a sentence we mean an expression of one of the two forms Jn (n is any positive integer) or JNn. We interpret Jn to mean that John is reading that expression whose Godel number is n, and we interpret JNn to mean that John is reading the norm of the expression whose Godel number is n. Then a sentence that asserts that John is reading it is JN11111 (since 5 is the Godel number of JN). Self-reference via Godel numbering will play a crucial role in later chapters.
Part II Self-reference in a more general setting §9 A more general setting We now consider self-reference in a far more general setting than that of quotational systems, or Godel numbering, and show how most of the results of Part I are but special cases of some very simple but general principles to which we now turn. We consider a language L in which certain expressions are called predicates and certain expressions are called sentences, and we are given a rule that assigns to each predicate H and any expression X a sentence denoted H(X), which we call the result of applying H to X. (Informally, predicates are names of properties, and the sentence H(X) is thought of as asserting that X has the property named by H.) How a predicate H is applied to an expression X to yield a sentence H(X) varies from system to system. For two-sided quotational systems, we take H(X) to be H "X" . Thus, for a predicate H (such as "J", for "John is reading"), H "X" can be thought
§10 Applications to quotational systems
17
of as asserting that X has the property named by H. For one-sided quotational systems, we take H(X) to be H*X. Thus, for a predicate H, the sentence H*X asserts that X has the property named by H. For systems in which Godel numbering takes the place of quotation, H(X) will be the result of performing a certain operation on H and the Godel number of X - as we will see in a later chapter. Next, we are given some equivalence relation 1 - between sentences. What this relation is varies from system to system. (For example, we car
might have some sentences determined as true, by some definition or other of truth, and we might then define two sentences to be equivalent if they are either both true or both not true. Or we might have an axiom system in which certain sentences are provable in the system, in which case we might call two sentences equivalent if they are either both provable or both
non-provable.) We write X - Y to mean that X is equivalent to Y. And now we shall use the term diagonalization in a far more general sense than was used in Part I. We shall say that a predicate H# diagonalizes a predicate H, or that H# is a diagonalizer of H, if for every predicate K,
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H#(K) is equivalent to H(K(K)). We call a sentence X a fixed point of a predicate H if X is equivalent to H(X). The following theorem, despite its simplicity, generalizes many fixed point theorems that occur in metamathematics.
Theorem 1 If H has a diagonalizer, then there is a fixed point for H. Proof Suppose H# diagonalizes H. Then for every predicate K, the sentence H#(K) is equivalent to H(K(K)). We take H# for K, and so H#(H#) is equivalent to H(H#(H#)). Thus H#(H#) is a fixed point of H. Remark
If we think of H as the name of a property of expressions, then a fixed point X of H can be thought of as asserting that X itself has the property named by H, and is in that sense self-referential.
§10 Applications to quotational systems Let us see how Theorem 1 generalizes some of the results of Part I. We now turn back to the study of quotational systems. In the sentences considered
in Part I, each sentence was given an interpretation (e.g., for two-sided 1That is, a relation = such that for any sentences X, Y, and Z: (1) X - X; (2) If
X -Y thenY-X; (3) IfX=_Y andY- ZthenX- Z.
Chapter 1. Introduction to self-reference
18
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quotational systems, J"X" was interpreted to mean that John is reading
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X, and in one-sided quotational systems, J*X was also interpreted to mean that John was reading X), and we will call two sentences equivalent if the truth of either one implies the truth of the other (e.g., the sentence JN"X" and J"X "X" " are equivalent). By an N-system let us mean a two-sided quotational system in which s-.
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there is an expression - which we will take to be the letter "N" - such that for any predicate H, HN is a predicate, and for any expression X, the sentence HN"X" is equivalent to H"X "X" ". (CD
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By an A-system let us mean a one-sided quotational system in which there is an expression - which we will take to be "A" - such that for any predicate H, the expression HA is a predicate, and for every expression X, the sentence HA*X is equivalent to H*X*X. By an R-system let us mean a one-sided quotational system in which there is an expression - which we will take to be "R" - such that for any predicate H, the expression HR is a predicate and for any expression X, the sentence HR*X is equivalent to H*XX. (Miniature A-systems and miniature R-systems were informally presented in §§4 and 5.) In an N-system, for any predicate H and any predicate K, the sentence HN "K" is equivalent to H "K "K" ", which means that HN diagonalizes H. Therefore, by Theorem 1, H has a fixed point, which by the proof of Theorem 1 is seen to be HN "HN" . In an A-system, for any predicates H and K, HA*K is equivalent to H*K*K, hence HA diagonalizes H, and so by Theorem 1 and its proof, HA*HA is a fixed point of H. In an R-system, the situation is different. Although HR does not diagonalize H (HR*K is equivalent to H*KK, not to H*K*K), H nevertheless does have a fixed point, namely HR*HR*. This, however, is not a consequence of Theorem 1, but is a consequence of the stronger Theorem 14
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§11 Near Diagonalization ,t."
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We shall say that a predicate H° is a near diagonalizer of a predicate H if for every predicate K there is at least one expression X (but not necessarily a predicate) such that H°(X) is equivalent to H(K(X)). Of .N.
course, a diagonalizer H# of H is also a near diagonalizer of H, since for any predicate K there is an expression X - namely K itself - such that v02
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H#(X) - H(K(X)). Therefore the following theorem is stronger than Theorem 1.
Theorem 1# If H has a near diagonalizer H°, then H has a fixed point.
§12 Cross-reference
19
Proof Suppose that H° is a near diagonalizer of H. Since for every predicate K there is some expression X such that H°(X) -_ H (K (X)), we can take H° for K, and hence there is some expression X (not necessarily a predicate) such that H°(X) - H(H° (X)). Thus H° (X) is a fixed point of H. An application
Let us see how this applies to 1Z-systems. Although HR is not a diagonalizer of H, it is a near diagonalizer of H, since for any predicate K, HR*K* - H*K*K*, and so if we take K* for X, then HR*X = H*K*X, and so H°(X) - H(K(X)) where H° is HR. And so HR is a near diagonalizer of H. Then by Theorem 1# and its proof, HR*HR* is a fixed point of H, which is as it should be.
§12 Cross-reference Let us return to the study of systems not necessarily quotational. We will call a pair (X1, X2) of sentences a cross-point for a pair (H1, H2) of predicates if X1 - H, (X2) and X2 - H2(X1). (If we interpret H1, H2 as names of properties, X1 asserts that X2 has the property named by H1 and X2 asserts that X1 has the property named by H2. Thus each of the two sentences ascribe a property to the other.)
Theorem 2 A sufficient condition for (H1, H2) to have a cross-point is that there exists a predicate L such that for every predicate K, the sentence L(H2(K)) is equivalent to H1 (H2(K(H2(K)))).
We will prove a strengthening of Theorem 2, but we first remark that although Theorem 2 has an application to A-systems, it does not have any application to R-systems, whereas the following strengthening of Theorem 2 does.
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Theorem 2# A sufficient condition for (H1, H2) to have a cross-point is that there exists a predicate L such that for every predicate K there is an expression X such that L(H2(X)) - Hl(H2(K(H2(X)))). car
Proof For then, taking L for K, there is an expression X such that
L(H2(X)) - H1(H2(L(H2(X)))). We then take X1 to be L(H2(X)) and H1(X2), and of course X2
III
X2 to be H2(X1), and so X1 X2 is H2(X1).
H2(X1), since
20
Chapter 1. Introduction to self-reference
Of course the hypothesis of Theorem 2 implies the hypothesis of Theorem 2# (taking X = L), and so Theorem 2 follows from Theorem 2#. Also, if the stronger hypothesis of Theorem 2 holds, then L(H2(L)) - H1(H2(L(H2(L)))), and so if we take X1 = L(H2(L)) and X2 = H2(L(H2(L))), then (X1, X2) is a cross-point of (Hi, 112).
§13 Application to one-sided quotational systems Let us first consider A-systems. Given two predicates H1 and H2, take L to be H1A. Now, for any predicate K, H1A*H2*K is equivalent to H1*H2*K*H2*K, which means that L(H2(K)) - HI(H2(K(H2(K)))), and so the hypothesis of Theorem 2 holds. Therefore by Theorem 2 and its proof, if we take X1 = H1A*H2*H1A and X2 = H2*H1A*H2*H1A, then (XI, X2) is a cross-point of (H1i H2).
For R-systems, given two predicates H1 and H2, we take L to
be H1R, and for any predicate K, we take X to be K*.
Then
L(H2(X)) - H1(H2(K(H2(X)))) (since H1R*H2*K* is equivalent to HI*H2*K*H2*K*), and so the hypothesis of Theorem 2# is satisfied. gin'
Then by Theorem 2# and its proof, a cross-point (X1, X2) of (H1, H2) is obtained by taking X1 = H1R*H2*H1R* and X2 = H2*H1R*H2*H1R*.
Exercise 28 In an A-system, a pair (H1, H2) of distinct predicates has two distinct cross-points. Why is this a consequence of Theorem 2? Also, what are the two cross-points? The same is true for R-systems, using Theorem 2# in place of Theorem 2. What are the two cross-points?
Exercise 29 In the pseudo-quotational system of §7, JJ is a fixed point of J. Why is this a consequence of Theorem 1? Also, (JPJ, PJPJ) is a cross-point of (J, P). Why is this a consequence of Theorem 2?
Exercise 30 Prove that a sufficient condition for (HI, H2) to have a cross-point is that there is a predicate L such that for every predicate K, L(K) - HI(H2(K(K))). Exercise 31 Suppose that there is a predicate L such that for every predicate K there is an expression X such that L(X) - HI(H2(K(X))). Does (HI, H2) necessarily have a cross-point?
Exercise 32 Suppose that every predicate has a fixed point and that for any predicate H1 and H2 there is a predicate H such that for every
§13 Application to one-sided quotational systems
21
expression X, H(X) - Hl(H2(X)). Prove that every pair (Hl, H2) of predicates has a cross-point.
.,.
Exercise 33 [ A Miniature Incompleteness Theorem] Suppose that we now have an axiom system A in which certain sentences of the language L are proved. To each sentence X is assigned a sentence -X called the negation of X, and a sentence X is called refutable (in A) if its negation -X is provable (in A). The system A is called consistent if no sentence is both provable and refutable in A. The system A is called complete if every sentence is either provable or refutable in A; otherwise A is called incomplete. A sentence X is called decidable in A if it is either provable or refutable in A; otherwise undecidable in A. Suppose now we are given the following three conditions.
Gl To each predicate H is assigned a diagonalizer H#, i.e., a predicate such that for every predicate K, the sentence H# (K) is provable if and only if H(K(K)) is provable.
G2 To each predicate H is assigned a predicate H' such that for every expression X, H'(X) is provable in A if and only if H(X) is refutable in A.
G3 There is a predicate P such that for every sentence X, P(X) is provable in A if and only if X is provable in A. Prove that if the system is consistent, then it must be incomplete!
Exercise 34 Suppose that in the preceding exercise we weaken condition Gl to the following condition Gi: every predicate H has a fixed point (a sentence X that is provable if and only if H(X) is provable). (This is a weakening of condition Gl by virtue of Theorem 1.) Now, suppose that A is consistent and satisfies conditions Gi, G2, and G3. Is A necessarily incomplete?
Exercise 35 (For those who have read the sections on one-sided quotation.) Let us consider a computing machine that prints out various expressions built from the following four symbols.
,.,PR* Let us call an expression printable if the computer can print it. We assume the computer is programmed such that every expression that the machine can print will be printed sooner or later (allowing an infinite future).
Chapter 1. Introduction to self-reference
22
By a sentence we shall mean any expression of one of the following four forms (where X is any expression build from the four symbols).
(1) P*X (2) PR*X
(3) -P*X (4) PR*X This language is build on one-sided quotation with * as the quotation symbol, and is an R-system, with the symbol R standing for "the repeat of." The symbol P stands for "printable." Thus truth of sentences is defined as follows.
(1) P*X is true if X is printable. (2) PR*X is true if XX (the repeat of X) is printable.
(3) -P*X is true if X is not printable. (4) -PR*X is true if XX is not printable. We have given a perfectly precise definition of what it means for a sentence to be true, and we have here an interesting case of self-reference: G1.
the computer is printing out various sentences that assert what it can or cannot print, and so it is describing its own behavior! (It somewhat resembles a self-conscious organism, and such computers are accordingly of interest to those working in artificial intelligence.) (1)
We are given that the machine is totally accurate in that all sentences printed by the machine are true. And so, for example, if the machine ever prints P*X then X really is printable (X will be printed by the machine sooner or later). Also if PR*X is printable, it is true, which means that XX must be printable. Now, suppose that X is printable; does it necessarily follow that P*X is printable? No; if X is printable then P*X is certainly true, but we are not given that the machine is capable of printing all true sentences, but only that the machine never prints any false ones. Is it possible that the machine can print all false sentences? The answer ..O
is no, and the problem for the reader is to find a true sentence that the machine cannot print. G].
Exercise 36 There also exists a curious pair (X, Y) of sentences such that it can be shown from the given conditions of Exercise 35 that one of the two must be true but unprintable, but there is no way to determine
§13 Application to one-sided quotational systems
23
which one it is! Can the reader find such a pair? (Hint: construct sentences
X and Y such that X asserts that Y is not printable and Y asserts that X is printable. There are two different ways of doing this!)
Solutions 33 Assuming the given conditions, the sentence P# (P#') is undecidable
in A. Reason: P#(P#') is provable if P(P#'(P#')) is provable (by G1, taking P for H), iff P#' (P#') is provable (by G3, taking P#' (P#') for X) if and only if P#(P#') is refutable (by G2, taking P# for H and P#' for X). Thus P#(P#') is provable iff it is refutable. This means that it is
.'3
either both provable and refutable or neither provable nor refutable. By the assumption of consistency, it is not both provable and refutable, hence it is neither - it is undecidable in the system A. Another undecidable sentence in A is P(P #(P'#)). Can the reader see why? (If not, see the solution to Exercise 34.)
ova
34 Yes; if we replace G1 by Gi, it still follows that A is incomplete, for suppose X is a fixed point of the predicate P'. Thus X is provable iff P' (X ) is provable, if P(X) is refutable. Thus X is provable iff P(X) is refutable. MSS
0.1
But also, by G3i X is provable if P(X) is provable. Therefore P(X) is provable if P(X) is refutable, and so by the assumption of consistency,
P(X) is neither provable nor refutable. Another point of interest: let us go back to the last exercise, in which we are given G1 instead of the weaker condition G. We have just seen Biz
(using just G2 and G3 and the assumption of consistency) that if X is any fixed point of P', then P(X) is an undecidable sentence of A. But with G1, a fixed point of P' is P #(P'#) (as we saw in the proof of Theorem 1), and so P(P'#(P'#)) is therefore undecidable in A (as stated at the end of the solution of the last exercise). Of course we can see this directly: P(P'#(P #)) is provable iff P'# (P'#) is provable if P'(P'#(P'#)) is provable if P(P'#(P'#)) is refutable. Thus the sentence P(P'#(Pt#))
°-o
is provable if it is refutable, and so by consistency it is neither.
air.
The sentence -PR*-PR* is true if the repeat of -PR* is not printable, but this repeat is the very sentence -PR*-PR*. Thus -PR*-PR* 35
is true if it is not printable, hence by the assumption that no false sentence is printable, the sentence must be true, but the machine cannot print it. 36
Let X be the sentence -P*PR*-P*PR* and Y the sentence
PR*-P*PR*. Thus X is the sentence -P*Y, hence is true iff Y is not
24
Chapter 1. Introduction to self-reference
printable. And Y is true if the repeat of -P*PR* - which is X - is printable. Thus X is true if Y is not printable, and Y is true if X is printable. Now, if X is not true, then Y is printable, hence true, which means that X is printable, and this is contrary to the given condition that only true sentences are printable. Therefore X must be true, and therefore Y is not printable. Next, if Y is true, then Y is a sentence that is true but not printable. If Y is not true, then X is not printable (since Y is true if X is printable), in which case X is true but not printable. ova
In summary, if Y is true, then Y is true but not printable, and if Y is not true, then X is true but not printable. Thus at least one of the sentences X, Y is true but not printable, but there is no way to tell which one it is. Another pair of sentences that works is X = -PR*P*-PR* and Y =
P*-PR*P*-PR*.
Chapter 2
Some classical fixed point arguments compared
,.y
In the last chapter we saw how fixed points are related to self-reference. We shall now look at fixed points in a larger perspective. Although no prior knowledge of mathematical logic, recursion theory, or combinatory logic is presupposed, Part I of this chapter consists of abstract versions of classical results in these three areas. We present them as problems, which the reader might have fun trying to solve. Solutions are given at the end of Part I. In Part II, we show how all five results of Part I are but special cases of one very general fixed point theorem, and then in Part III we show that all the preceding results go through under a curious weakening of the
xe.
hypotheses.
Part I Five fixed point arguments §1 An argument from combinatory logic By an applicative system is meant a set N together with an operation that assigns to each ordered pair (x, y) of elements of N an element of N that we denote xy. We call xy the result of applying x to y. The applicative system that will interest us most in this study is the system of combinatory logic that we will study in later chapters. It is closely related to the A-calculus, which can serve as a basis for the field known as recursion theory - some of which we will study later on. Combinatory logic and the A-calculus are playing an increasingly important role these days in computer science.
Chapter 2. Some classical fixed point arguments compared
26
For now, we will not consider anything as specific as combinatory logic,
but will consider a fixed point result in applicative systems of a much broader sort. In an applicative system, an element b (of N) is called a y,,
fixed point of an element a if ab = b. A solution of the following problem generalizes a famous result in combinatory logic. u-,
Problem 1 Suppose we are given the following two conditions. (1) There is an element m such that for every element x, mx = xx. [Such an element m might be called a duplicator.]
(2) For any elements a and b there is an element c such that for every element x, cx = a(bx). [This condition might be called compositional closure.]
The problem is to prove that every element a has a fixed point.
§2 Godel sentences
Dar
We now turn to a generalization of a famous problem from metamathematics (the theory of mathematical systems). Throughout this chapter, the word "number" shall mean a positive integer. We consider a denumerable collection C of sets of positive integers arranged in a denumerable sequence A1, A2, ...,An ..... We consider all sentences i c A; . We call the sentences i E A; true if i is a member of the set Aj. There are only denumerably many of these sentences (since there are only denumerably many ordered pairs (i, j) of positive integers). We arrange all these sentences in some 1-1 sequence S1, S2, ... , Sn..... Thus for each n, there is some i and j such that Sn is the sentence i E A2. We call n the Godel number of Sn. And for any set A of numbers, we call a sentence Sn a Godel sentence for A if the following condition holds.
Snistrue f-+nEA car
[In other words, either Sn is true and its Godel number is in A, or Sn is not true and its Godel number is not in A. In a purely extensional sense, a Godel sentence for A can be thought of as asserting that its own Godel number is in A.] For any number i, let d(i) be the Godel number of the sentence i c A. [Thus Sd(i) is the sentence i c Ai.] For any set A, by d-1(A) is meant the d(i) E A. set of all numbers i such that d(i) E A. Thus i E d-1(A)
Problem 2 Suppose that we are given that for every set A in C, the set d-1(A) is also in C. Prove that for every A E C there is a Godel sentence for A.
--1
§3 An argument from recursion theory
27
Problem 2a To highlight the significance of the above problem, suppose we have a mathematical system (M) in which certain sentences of the ."j
(D(
above problem are provable. And let us assume that the system is correct in that only true sentences are provable - in other words, for any numbers i and j, if the sentence i E Aj is provable in the system, then i really is a member of A;. We continue to assume, as in Problem 2, that for every set A in C, the set d-' (A) is in C. In addition, suppose that the following two conditions hold. a-,
..O
(1) For every set A in C, its complement A is in C. [By the complement A of A we mean the set of all numbers not in A.]
(2) The set P of Godel numbers of the provable sentences is one of the sets in C.
The problem now is to show that some true sentence is not provable in 'nom
the system (M) - in other words that for some numbers i and j, i is a member of A;, but the sentence i c Aj asserting this is not provable in the system.
§3 An argument from recursion theory We let N be the set of numbers (positive integers) and w, W2, ... , Wn,.. . be a denumerable sequence of subsets of N. We let E be a set of functions from N into N.
Problem 3 Suppose that there is a function F(x, y) from ordered pairs S."
of numbers to numbers and a function d(x) from numbers to numbers such that the following three conditions hold. (1) For every number n, Wd(n) = WF(n,n)
(2) For every number a there is a number b such that for all numbers x, WF(b,x) = WF(a,d(x))
(3) For every function f (x) in E there is a number a such that for all numbers x, WF(a,x) = Wf(x).
The problem is to prove that for every function f in E, there is at least one number n such that W f(n) = W. Remark
This is an abstract version of a famous result in recursion theory known as the recursion theorem (in one of several forms). We will later be studying
28
Chapter 2. Some classical fixed point arguments compared
sets of numbers known as recursively enumerable sets. [Informally, a recursively enumerable set is one that can be generated by a purely mechanical process - as, for example, the set of even numbers, which can be generated RR.
by starting with 2 and adding 2 to any number already obtained - thus generating the sequence 2,4,6,... , 2n, 2n+2,....] We will also be studying a class of functions known as recursive functions which, informally speaking, are functions that can be computed by a purely mechanical process. A precise definition of recursively enumerable sets and recursive functions will be given in a later chapter. Meanwhile, I wish to point out a fact that we will later verify in detail - namely, that it is possible to arrange all the recursively enumerable sets in a sequence W1, W2, ... ,Wn, ... and to find recursive functions F(x, y) and d(x) such that the given conditions of the previous problem do hold (taking E to be the class of recursive functions of one argument). The surprising conclusion, then, is that for any recursive function f (x) there must be at least a number n such that the sets Wn and W f(n) are identical!
As I said, this result is one form of what is known as the recursion theorem, which has numerous application. Many other forms and variants of this form will be studied in this volume - some of them being of a more complex nature. We now turn to another relatively simple form.
§4 Another argument from recursion theory CAD
In addition to the sequence w1, W2, ... , Wn, of sets, we have an infinite sequence R1 (x, y), R2 (x, y), ... , Rn (x, y), ... of binary relations of positive integers. We let C be the collection of all the relations Rn(x,y). con
At this point it will be useful to use the notation of class abstracts: Given any property P(x) of positive integers, by {x : P(x)} is meant the set of all numbers x such that x has the property P.
Problem 4 Suppose there is a function d(x) from numbers to numbers that satisfies the following two conditions.
(1) For every number n, Wd(n) = {x : Rn(x, n)} (in other words, Wd(n) is the set of all numbers x such that Rn (x, n) - or equivalently, for any numbers n and x, x E Wd(n) Rn(x, n). (CD
(2) For every relation R(x, y), in C, the relation R(x, d(y)) is in C (in other words, for every relation R in C there is a relation R' in C such that for all numbers x and y, R'(x, y) R(x, d(y)).
The problem is to prove that there exists at least one number n such that wn = {x : R(x, n)}.
§5 Self-reproducing machines
29
(gyp
CJ'
car
s."
(z7
009
.°.
This problem is again taken from recursion theory. In addition to the recursively enumerable sets, we have recursively enumerable binary relations of positive integers. [A binary relation is a set of ordered pairs, and a recursively enumerable relation R(x, y) can be informally described as one whose members (the ordered pairs) can be generated by a purely mechanical process. A simple example is the set of all ordered pairs (x, y) such that 5x = y. We start with the pair (1, 5) and given any pair (x, y) already obtained, we can form the new pair (x + 1, y + 5). We thus generate the sequence (1, 5), (2, 10),... , (n, 5n), ....] Now the recursively enumerable sets can be arranged in a sequence W1, W2, , Wn) ... and the recursively enumerable relations can be arranged in a sequence Ri(x, y), R2 (X, y), ..., Rn(x, y), ... and a recursive function d(x) exists such that conditions (1) and (2) of the above problem hold. A number n is called an index of a recursively enumerable set A if A = Wn. The amazing thing, then, is that for any recursively enumerable relation R(x, y) there exists a number n such that the set of x's for which R(x, n) holds - this set is a recursively enumerable set having the very number n as an index! This is another form of the result known as the recursion theorem. In later chapters, after giving a precise definition of recursive enumerability, v..
(Dc
O,!
(0D
...
cam
we will see that there is a recursive function d(x) such that (1) and (2) do hold. Meanwhile, it should be of interest to note that without knowing what recursively enumerable sets and relations are, knowing just (1) and (2), one can know that the conclusion follows. coy
r+.
§5 Self-reproducing machines '.3
Can a machine reproduce itself? Early results along these line were obtained by von Neumann. Hartley Rogers[20] uses the recursion theo-
car
car
'.7
rem (that we will study later) to obtain such a result. I will now show you a construction along these lines that does not depend on the recursion theorem, but uses a principle similar to that used in the proof of the recursion theorem. What now follows is completely self-contained. We are given a denumerable sequence P1, P2, ... , Pn,... for programs for constructing robots. The robots construct other robots and each robot bears a program which instructs it as to what robot it should create and what program to give it. A robot X will be called self-reproducing if the robot that it creates has the same program as X. For any two numbers x car
'Z3
7-+
cod
and y we will say that Robot x creates Robot y to mean that any robot .,,
0.H
with program PP will create a robot with program Py. We are given that there is a function d(x) from numbers to numbers and a function F(x, y) from ordered pairs of numbers to numbers such that
30
Chapter 2. Some classical fixed point arguments compared
the following two conditions hold.
(1) For any number x, Robot d(x) creates Robot F(x, x).
(2) There is a number b such that for all numbers x and y, if Robot y creates Robot F(b, x) then Robot d(x) creates Robot y.
Problem 5 The problem is to show that under these conditions there must exist at least one number n such that Robot n creates Robot n (and thus any robot with program Pn is self-reproducing).
Solutions to problems 1-5 1
By (1) there is an element m such that for all elements x, mx = xx.
Given a, it follows from (2) (taking m for b) that there is an element c such
that for all x, cx = a(mx), hence cx = a(xx). Taking c for x, we have cc = a(cc), and so cc is a fixed point of a. 2 Take any set A in C. Then d-1(A) is in C, hence for some number i, Ai = d-1(A). Then for all x, x E Ai +-4 d(x) E A. Taking i for x, i c Ai H d(i) E A. But d(i) is the Godel number of the sentence i E A.
And so the sentence i E Ai is a Godel sentence for A (it is true if and only if its Godel number d(i) is in A).
2a By (1) and (2) it follows that the complement P of the set P is in car
C, hence for some number i, Ai = P. Then, as_ we saw in the solution of Problem 2, the sentence i E Ai is true if d(i) E P, but d(i) E P if d(i) ¢ P, if d(i) is not the Godel number of a provable sentence of (M). However, d(i) is the Godel number of the sentence i E Ai, and so i E Ai is true if it is not provable in (M). This means that either the sentence is true but not provable in (M), or not true but provable in (M). The latter alternative is contrary to the assumption that only true sentences are provable in (M). And so the sentence is true but not provable in (M). Take any function f (x) in E. By (3) there is a number a such that for all x, WF(a,x) = Wf(x). Then by (2) there is a number b such that for all 3
x, WF(b,x) = WF(a,d(x)) Hence WF(b,x) = W f(d(x)) Taking b for x, WF(b,b) = W f (d(b)). But also by (1), Wd(b) = WF(b,b). Therefore Wd(b) = W f (d(b)), and so
Wn = Wf(n) for n = d(b).
§6 A relational fixed point theorem
31
cry
°:v
4 Given R(x, y) in C, by (2) the relation R(x, d(y)) is Rb(x, y) for some b. Thus for all numbers x and y, Rb(x, Y) R(x, d(y)). Therefore Rb(x, b) H R(x, d(b)). But also by (1), x E wd(b) H Rb(x, b). Therefore x E wd(b) H R(x, d(b)). And so for n = d(b), wn, = {x : R(x, n)}.
Take a number b satisfying (2). Then take b for x and d(b) for y, and then by (2), if Robot d(b) creates Robot F(b, b), then Robot d(b) creates Robot d(b). But Robot d(b) does create Robot F(b, b) by (1), and so Robot d(b) does create Robot d(b). Thus Pd(b) is a program for a self-duplicating Q..
5
robot.
Part II A unification §6 A relational fixed point theorem What do the five preceding problems have in common? Theorem 6.1 below provides one such answer (and so does Theorem 6.2 below). But first let
us pause and consider why our previous results are called "fixed point" theorems. Well, this term is usually used for functions: given a function f (x) from a set N into N, an element a of N is called a fixed point of the function f if f (a) = a. We shall generalize this notion as follows: Given a binary relation R(x, y) on a set N, we shall call an element a of N a fixed point of the relation if R(a, a) holds. [Thus a is a fixed point of the function f (x) if a is a fixed point of the relation f (x) = y.] Under this slightly extended notion of "fixed point" the results of the problems of Part I are indeed fixed point theorems. In Problem 1 we were required to show that for any element a, the relation ax = y (as a relation between x and y) has a fixed point. For Problem 2, we were to show that for each expressible set A, the relation "Sx is true if y c A" has a fixed point. For Problem 3, it was to be shown that for each function f (x) in E, the relation "wx = w f(y)" has a fixed point. In Problem 4, we were to show that for any cad
relation R E C there is a fixed point of the relation "wx = {z : R(z, y)}." For Problem 5 we were to show that there is a fixed point for the relation "Robot x creates Robot y." Now for a general relational fixed point theorem, of which all our preceding results can be obtained as corollaries.
Theorem 6.1 A sufficient condition for a relation R(x, y) on a set N to have a fixed point is that there be a relation S(x, y, z) on N and a function d(x) from N to N such that (1) For every x in N, S(d(x), x, x).
Chapter 2. Some classical fixed point arguments compared
32
(2) There is at least one element b of N such that for all x and y in N, S(x, b, y) implies R(x, d(y)).
car
Proof Assume (1) and (2). Take b satisfying (2), then take d(b) for x and b for y and we have: S(d(b), b, b) = R(d(b), d(b)). But S(d(b), b, b) holds by (1), hence R(d(b), d(b)). Thus d(b) is a fixed point of R.
ado
:s'
The solution of Problem 5 is obviously that special case of Theorem 6.1 in which R(x, y) is the relation "Robot y creates Robot x" and "S(x, y, z)" is the relation "Robot x creates Robot F(y, z)." Theorem 6.1 has the following corollary.
Theorem 6.2 Suppose that N and W are sets and z('(x, y) is a mapping from ordered pairs (x, y) of elements of N to elements of W and that O(x) is a function from elements of N to elements of W. Suppose that there is a function d(x) from N into N such that the following conditions hold.
(1) For all x in N, 0(d(x)) = O(x, x). (2) For every element a of N there is an element b of N such that for all
y E N: '(b, y) = 0(a, d(y)) Then for every a c N there is some c c N such that O(c) = 0(a, c).
Proof Given a, let R(x, y) be the relation O(x) = '(a, y). Let S(x, y, z) be the relation O(x) = z'(y, z). Then by (1), S(d(x), x, x) holds for every x. Also, since for every y, 0(b, y) = 0(a, d(y)), then for every x, O(x) = 0(b, y) implies O(x) = t(a, d(y)), which means S(x, b, y) implies R(x, d(y)). Then
by Theorem 6.1, there is some c such that R(c, c), which means O(c) _ 0(a, c). Discussion
It is perhaps easier to prove Theorem 6.2 directly as follows. Given a, take b satisfying (2). Then, taking b for y, 0(b, b) ='(a, d(b)) But also by (1), O(d(b)) = 0(b, b). Therefore O(d(b)) = 0(a, d(b)), so O(c) = 0(a, c) for c = d(b). Though this proof is more direct, we find it of interest that Theorem 6.2 can be derived from Theorem 6.1 without any further diagonalization.
a..
The solution of Problem 4 is that special case of the above theorem in which N is the set of positive integers, W is the set of subsets of N, q(x) is w, and 0(x, y) is the set of all z such that Rx(z, y).
§6 A relational fixed point theorem
33
Theorem 6.2 in turn has the following corollary.
Theorem 6.3 Let - be an equivalence relation on a set N and F(x, y) be a function from ordered pairs of elements of N to elements of N. Suppose
that there is a function d(x) from N into N such that the following two conditions hold.
(1) F(x, x) - d(x) for all x E N.
(2) For every a c N there is some b c N such that for all x c N, F(b, x) - F(a, d(x)). Then for every a E N there is some c c N such that c - F(a, c).
III
Proof We let W be the set of all subsets of N. For every x E N, let Ixj be the set of all y E N such that x - y. Thus, for any x and y in N, x - y iff 1xI = M.
Now take O(x) = xj and zb(x, y) = IF(x, y)l. Then by (1),'(x, y) _ q5(d(x)) and by (2), for each a there is some b such that 4(b, x) = '(a, d(x)). Hence by Theorem 6.2, there is some c (viz d(b)) such that O(c) = '(a, c), III
which means that c - F(a, c). Remark S."
Again, a direct proof is simpler. In (2) take b for x and we have F(b, b) F(a, d(b)). Then by (1), F(b, b) - d(b). Hence d(b) - F(a, d(b)), so d(b) is a fixed point of the relation x - F(a, y). But again, it is of interest to see that Theorem 6.3 is derivable from Theorem 6.1 via Theorem 6.2. Now the solutions of Problems 1, 2, and 3 are all special cases of The-
orem 6.3. For Problem 1, take - to be identity and F(x, y) to be xy and d(x) = mx. We are given that mx = xx, and so d(x) = F(x, x). Then, given a, by the second hypothesis of the problem, there is an element b such
that for all x: bx = a(mx), hence F(b, x) = F(a, d(x)). Thus the hypotheses of Theorem 6.3 hold, hence there is an element c such that c = F(a, c), which means that c = ac. As to Problem 2, define two sentences to be equivalent if they are either
both true (both in T) or both not true. Then define x - y to mean that the sentences SS and Sy are equivalent. Take F(x, y) to be the Godel number of the sentence y c Ax, and take d(x) = F(x, x). Then of course d(x) - F(x, x). Now, the given condition of the problem is that for any element a there is some element b such that for all elements x, x E Ab if and
only if d(x) E Aa, which means that for all x, F(b, x) - F(a, d(x)). Then
34
Chapter 2. Some classical fixed point arguments compared
by Theorem 6.3, for any element a there is some c such that c - F(a, c), which means that S, is equivalent to the sentence c c Aa, and thus S, is a Godel sentence for the set Aa. As to Problem 3, define x - y to mean that wx = wy. Then the three given conditions can be rewritten:
(1) For all numbers x, d(x) - F(x, x). (2) For every a there is some b such that for all x, F(b, x) - F(a, d(x)).
(3) For every f c E, there is some a such that for all x, F(a, x) - f (x).
41v
Given f c E, take a such that for all x, F(a, x) - f (x). But by (1), (2), and Theorem 6.3, there is some c such that c - F(a, c), but F(a, c) f (c), so c - f (c), which means that w, = wf(c). We now see how the solutions of Problems 1-5 are ultimately derivable from Theorem 6.1. 4-'
Part III Quasi-diagonalizat ion §7 Some stronger results We will now show that all the previous results go through under a curious weakening of the hypotheses.
Let us reconsider Problem 1 on applicative systems. An element m
LTA
such that mx = xx for all x is sometimes called a duplicator. Let us call an element m a quasi-duplicator if for all x there is at least one element x' (not necessarily x itself) such that mx' = xx'. Now, the interesting thing is that in Problem 1, if we replace the hypothesis that there is a duplicator (the first hypothesis) by the weaker hypothesis that there is a quasi-duplicator, it will still follow that every element has a fixed point! For suppose m is a quasi-duplicator. Given any element a, there is an element b such that ,i.
for every element x, bx = a(mx) (this by the second hypothesis). Also, since m is a quasi-duplicator, there is an element b' such that mb' = W. Also, since bx = a(mx) for all x, then bb' = a(mb'). But since mb' = bb', it follows that bb' = a(bb'). And so bb' is now a fixed point of a. This curious phenomenon spreads to all the results we have proved so far. Let us begin by considering the following strengthening of Theorem 6.1.
Theorem 7.1 A sufficient condition for a relation R(x, y) on a set N to have a fixed point is that there be a relation S(x, y, z) on N and a function d(x) from N into N such that
§7 Some stronger results
35
(1)* For every element x there is an element x' such that S(d(x'), x, x').
(2)* There is an element b such that for all x and y: S(x, b, y) = [Z,
R(x, d(y)).
Note ..T
*
Hypothesis (1)* is a weakening of hypothesis (1) of Theorem 6.1. Hypothesis (2)* is the same as (2) of Theorem 6.1.
Proof Choose b so that (2)* holds. Then by (1)* there is an element b' such that S(d(b'), b, b'). But also by (2)*, taking d(b') for x and b' for y, we see that S(d(b'), b, b') implies R(d(b'), d(b')). Hence R(d(b'), d(b')).
Of course, both Theorems 6.2 and 6.3 have similar strengthenings.
Theorem 7.2 Same as Theorem 6.2, replacing hypothesis (1) by the weaker hypothesis: (1)* For all x in N there is some x' in N such that 0(d(x')) = O(x, x).
Theorem 7.3 Same as Theorem 6.3, replacing hypothesis (1) by the weaker hypothesis: (1)* For every x E N there is some x' in N such that F(x, x') - d(x'). Exercise 1 Prove Theorems 7.2 and 7.3. S."
We have already shown a strengthening of Problem 1. It can be derived as a consequence of Theorem 7.1 (or more easily from Theorem 7.3). The other four problems have similar strengthenings, as indicated in the next four exercises.
N-4
Exercise 2 Problem 2 can be strengthened in the following interesting
-N-0
manner. Again we are given a denumerable collection C of sets of positive integers, but we are not given an enumeration of C. To each number i and each set A in C is associated the sentence i c A, which is called true if and
only if i is a member of A. And these denumerably many sentences are arranged in some 1-1 sequence S1, S2, ..., Sn,, .... [Thus the sentences are cad
arranged in a given sequence, but the sets - the members of C - are not.] We are now given a function d(x) from numbers to numbers such that the following two conditions hold.
Chapter 2. Some classical fixed point arguments compared
36
(1) For every element A in C there is at least one number n such that Sd(,,,) is true if and only if n E A.
(2) [As before] For any set A in C, the set d-1(A) is in C. Prove that for every set A in C, there is a Godel sentence for A.
Exercise 3 Suppose that in Problem 3 we replace the first hypothesis by the weaker hypothesis: (1)* For every number n there is a number n' such that Wd(n') = WF(n,n') Show that the conclusion still follows.
Exercise 4 The result of Problem 4 can be strengthened as follows. We again consider a denumerable sequence wo, w1, ... , wn, ... Of subsets of the set N of natural numbers. But now we are given a collection C of binary relations on N (but not arranged in any given denumerable sequence). We are given a function d(x) from N into N satisfying the following two conditions.
(1) For every relation R E C there is at least one number n such that Wd(n) = {x : R(x, n)}.
(2) For every relation R E C there is a relation R' E C such that for all x, y in N: R'(x, y) = R(x, d(y)) Prove that for every R E C there is a number c such that w _
{x : R(x, c)}. How does this result strengthen the result of Problem 4?
Exercise 5 State and prove an analogous strengthening of the result of Problem 5.
Exercise 6 Here is a variant of Problem 5 that is apparently of incomparable strength to Problem 5. We dispense with the function d(x). We are now given that there are numbers a and b such that the following two conditions hold. (1) For every x, Robot F(a, x) creates Robot x.
(2) For every x and y, if Robot x creates Robot y, then Robot F(b, x) creates Robot F(y, F(a, y)). Prove that there is some x such that Robot x creates Robot x.
.U"
The following relational fixed point theorem, though its proof is laughably simple, is nevertheless strong enough to yield Theorem 7.1 (and hence all previous results) as a corollary.
§7 Some stronger results
37
Theorem R A sufficient condition for a relation R(x, y) on a set N to have a fixed point is that there be a relation R'(x, y) on N and a function d(x) from N into N such that (1) There is at least one element a of N such that R'(d(a), a). (2) For all elements x and y of N, R'(x, y) implies R(x, d(y)).
Exercise 7 Prove Theorem R and show how it yields Theorem 7.1 as a corollary.
The following exercise is based on a somewhat different fixed point principle.
Exercise 8 [Axioms that assert their own consistency] Let us consider
(D.
car
a setup in which we have a denumerable 1-1 sequence S1, S2,.. ., Sn,.. . of sentences and a denumerable sequence H1i H27 ... , Hn, ... of predicates and an operation assigning to each predicate HZ and each natural number n a sentence H2(n). We now imagine that we have some axiom system A in which certain sentences are provable. Let us call a system consistent if not every sentence is provable in it. And now let us say that a sentence X is consistent with A if the result of adding X as a new axiom to the axioms of A is a consistent system. The system is given an interpretation, according to which each sentence asserts a definite proposition (whether true or false). We are given a predicate C and functions d(x), e(x) from numbers to numbers such that for any numbers n and m, the following three conditions hold.
(1) C(n) asserts that Sn is consistent with A. (2) The proposition asserted by Hn(n) is also asserted by Sd(n).
(3) The proposition asserted by Hn(d(m)) is also asserted by He(n)(m). The problem is to prove that there is a sentence X that asserts its own consistency with A. Remark
In the systems studied by Godel, there are recursive functions d(x), e(x)
and a predicate C such that the conditions (1), (2), and (3) above are satisfied. Thus there is a sentence X that asserts that the adjunction of X to the axioms of the system produces a consistent system. It so happens
Chapter 2. Some classical fixed point arguments compared
38
that the sentence X is false! If it is adjoined to the axioms, the resulting system is inconsistent. This follows from Godel's second incompleteness theorem, which, roughly speaking, is to the effect that consistent systems .`3
of sufficient strength cannot prove their own consistency.
Solution to Exercise 8 We are to find a fixed point for the relation: "S., asserts that Sy is consistent with A." III
Call two sentences X and Y equivalent - in symbols, X - Y - if (CD
X and Y assert the same proposition. The predicate C is H,, for some number c. Now, III
Sd(e(c))
=
He(c)[e(c)]
III
Hc[d(e(c))]
(by (2))
(by (3), taking c for n and e(c) for m)
But Hc[d(e(c))] is the sentence C[d(e(c))], which asserts that Sd(e(c)) is consistent with A (by (1)), hence so is Sd(e(c)).
Exercise 9 The solution of Exercise 8 can also be obtained as a corollary of Theorem 6.3. How?
Chapter 33
How silence a universal machine How to to silence
Part Part I
i..
COO
"s.
As we now now turn turn to aa fascinating As further applications applications of diagonalization, diagonalization, we fascinating problems that generalize variety of problems generalize some results in recursion recursion theory theory known known as halting halting problems. problems. That they they really really do do generalize generalize theorems theorems in in recursion recursion theory will also will be be seen seen in in aa later later chapter. chapter. The results of this chapter are also incompleteness theorems, related to incompleteness theorems, as as will will be be seen seen in in the the next chapter.
Silencing Silencing aa universal machine
§§1 1 A universal machine
cam
cam/]
We consider consider aa denumerable denumerablesequence sequenceMMl, M2, ...,, M,,,, ... (allowing We Mn, ... (allowing rep1, M 2 , ... etitions) of mathematical machines, or computers, computers, that behave etitions) machines, or behave in in the the folfoloperate on on positive positive integers. integers. The The word word lowing lowing manner. manner. First of all they operate number in this this chapter will mean positive integer. Each machine number in positive integer. machine is, so to speak, in in charge of aa certain property of numbers, and its function speak, charge of numbers, and function is to try and determine determine which which numbers have have the property and which which ones don't. One feeds feeds in in aa number number (positive (positive integer) integer.)x xasasinput inputtoto aa machine machineMM and and One goes into into operation operation and one of three things happens. the machine machine goes (1) The machine machine eventually eventually halts and and flashes flashes a signal signal (say (say a green green light) (1) The signifying that the number have the the property property in in question. signifying that number xx does does have question. If this happens, happens, we say that the the machine machine affirms affirms x. (2) The machine machine eventually eventually halts and and flashes flashes a different different signal (say a red (2) The light) signifying signifying that light) that xx does does not have have the the property. property. If this happens, happens, we say say that the we the machine machine denies denies x. x. ..o
(3) The The machine machine can never determine (3) determine whether whether or or not not xx has the property, so runs runs on onforever. forever. In that case case we we say say that the the machine machine is is and so
Chapter 3. How to silence a universal machine
40 P..
silenced by x, or that x silences the machine. We will call a machine total if it is not silenced by any number.
Given two machines M and N and numbers x and y, we say that M behaves the same way towards x as N behaves towards y if either M affirms x and N affirms y, or M denies x and N denies y, or M is silenced by x and
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N is silenced by y. We say that M and N are similar if for every number x, they behave the same way towards x. We now need a 1-1 function that maps each ordered pair (x, y) of numbers to a number x*y. We could take x*y to be the number which when written in ordinary base 10 notation consists of a string of 1's of length x followed by a string of 0's of length y, e.g., 3*4 = 1110000. Actually our choice of functions doesn't really matter; we could alternatively take x*y to be 2x3y, or we could take the recursive pairing function J(x, y) that we define in Chapter 6. To each machine M is associated a machine M# called the diagonalizer of M, which behaves towards any number x as M behaves towards x*x. Next, to each machine M is associated a machine M' called the opposer of M, which affirms those numbers that M denies, and denies those numbers that M affirms, and is silenced by those numbers that silence M. Finally, one of the machines U is called a universal machine and behaves towards any number x*y as Mx behaves towards y. We might remark that the universal machine U is as valuable as all the S."
machines M1, M2, ..., M, ... put together because to find out how Mx would react to y, instead of feeding in y to M, we could just as well feed in x*y to U. Let us now review and label the three key facts.
P1: M# behaves towards x as M behaves towards x*x.
P2: M' affirms just those numbers that M denies and denies just those numbers that M affirms. P3: U behaves towards x*y as Mx behaves towards y. The stage is now set and the show can begin.
Problem 1 The universal machine U, despite all the clever things it can do, is not omniscient (as the philosopher Leibniz might have hoped). There is a number x that silences U. How is this proved?
Problem 2 To make the problem more concrete, suppose that we are given conditions P1, P2, P3 in the following more specific forms.
§1 A universal machine
41
P1' For any number n, M# = P2 For any odd number n, Mn = Mn+1. P3 M1 is universal.
The problem now is to find a number x that silences M1. `ti
Problem 3 Still assuming P1', P2, P3, how many distinct numbers can you find that will silence M1?
Problem 4 Suppose we change Pi, P2, P3 as follows. P1 For any number n, Mn = M5n.
P2 For any odd number n, M# = Mn+1. P3' Same as P3 (M1 is universal).
Now how many numbers are there that silence M?
Problem 5 We now retract the special conditions P,", P2", P3 as well as P1, P2, P3, and go back to the more general conditions P1, P2, and P3 For any machine M, M' O is the diagonalizer of the opposer M' of M, whereas M#' is the opposer of the diagonalizer M# of M. Are the machines M#' and M# necessarily similar?
Problem 6 (1) Prove that there exists a machine C that behaves towards any number x as Mx behaves towards x. (2) Prove there exists a machine D such that for any number x, D affirms x if and only if Mx denies x, and D denies x if and only if Mx affirms X.
'C,
Problem 7 [A key problem] Prove that for any machine M, there is at least one number n such that U and M behave the same way towards n. The solution of Problem 1 is an easy consequence of this fact. Why?
Chapter 3. How to silence a universal machine
42
Problem 8 [A question of omniscience] Given two machines M and N, let us say that M knows at least as much as N if M affirms all numbers Q..
affirmed by N, and denies all numbers denied by N (but it might also affirm or deny some numbers that silence M - indeed it might affirm some numbers that silence M and deny other numbers that silence M). Let us call M omniscient if it knows at least as much as the universal machine U and is furthermore total (not silenced by any number). Prove one of the following two statements. (1) One of the machines must be omniscient. (2) None of the machines can be omniscient.
Problem 9 [A halting problem] Let us say that a machine halts at x if it either affirms or denies x. Prove that there is no machine that halts at those and only those numbers that silence U. 'ti
Problem 10 [A fixed point problem] For any numbers x and y, let x(y) be one of three integers +1, 0, -1. °a.
Suppose we are given the following three facts.
Fl For any number a there is a number b such that b(x) = a(x*x).
F2 For any number a there is a number b such that for all x, b(x) = -a(x). F3 There is a number h such that for all x and y, h(x*y) = x(y). We let h be a number satisfying F3. (1) Prove that for any number a there is a number n such that h(n) = a(n).
(2) Show that for any number a there is a number n such that h(n) =
-a(n). .°a
(3) Show that there is a number n such that h(n) = 0.
Problem 11 (1) Why is the solution of Problem 1 an immediate consequence of (3) above?
(2) Why is the solution of the key Problem 7 an immediate consequence of (1) above?
§1 A universal machine
43
Problem 12 Part (1) of Problem 10 can be shown to be a corollary of (CD
one of the fixed point theorems of Chapter 2. How?
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Problem 13 Let ai be the set of all numbers affirmed by Mi and let E be the collection of all the sets ai. Give two examples of an element A of E whose complement A (the set of all numbers not in A) is not in E.
Problem 14 Let D be the machine U#'. Describe a function f (x) such that if Mi is any machine that knows at least as much as D, the number f (i) silences D.
Part II Some related problems We now turn to some related problems that will be seen in the next chapter to be more closely related to incompleteness theorems. We shall be considering a denumerable collection E of sets of positive integer and a 1-1 function x*y such that the following two conditions hold.
Q1 For any set A in E the set A# of all n such that n*n E A is also in E.
Q2 There exists an enumeration (al, 8i),
(al, 8i), ... of all
disjoint pairs of elements of E such that the set Ul of all numbers x*y such that y c ax is in E, and so is the set U2 of all numbers x*y such that y c ,3x. Note
An important example of a collection E satisfying the above two conditions is the collection of all recursively enumerable sets, as we will see in a later chapter. In what follows, it will be assumed that E satisfies conditions Q1 and Q2 OED
Problem 15 Show that E must contain a set whose complement is not in E. [This will be seen to generalize a basic theorem in recursion theory: there exists a recursively enumerable set whose complement is not recursively enumerable.]
Problem 16 Given two disjoint elements A and B of E, call the pair (A, B) inseparable in £ - or £-inseparable - if there exists no pair (ai,,3i) such that A C ai and B C 3i and 3i is the complement of ai.
Chapter 3. How to silence a universal machine
44
Prove that there are disjoint sets A, B in E such that (A, B) is inseparable in E. [This will be seen to generalize the important result in recursion theory that there exists a pair of recursively enumerable sets that is recursively inseparable.]
Problem 17 The solutions to Problems 15 and 16 are really corollaries of results proved in Part I of this chapter. How? Indeed, collections E satisfying conditions Q1 and Q2 are intimately connected with machines satisfying P1, P2, and P3. How?
rd)
Part III Solutions to problems 1 - 6 It will be simpler and more instructive to solve the first six problems in a different order than that in which they were given. We will begin with Problem 6. We let C be the machine U#, where U is any universal machine, and we will see that for any number x, U# behaves towards x as Mx behaves
4.'
towards x. Well, U# behaves towards x as U behaves towards x*x, but U behaves towards x*x as Mx behaves towards x. Therefore C behaves towards x as Mx behaves towards x. We then take D to be the opposer C' of C, and this machine - which is U#' - obviously works. The solution of Problem 1 now follows. Let M be either D itself, or any machine similar to D. Then let h be any index of M, i.e., any number such that Mh = M. Then for any number x, Mh affirms x if and only if Mx denies x. We take h for x and so Mh affirms h if and only if Mh denies h. Hence Mh must be silenced by h, and therefore U is silenced by h*h. In summary, if h is an index of U#', or of any machine similar to U# then h*h is a number that silences U. Now for Problem 2. We take U to be M1, in which case U# is M5 and U#' is M6. And so 6*6 silences U. We next do Problem 5. The answer is yes; the machines M' O and M#' ...
are similar, for M'# affirms (denies) x if M' affirms (denies) x*x, if M denies (affirms) x*x, iff M# denies (affirms) x, if M#' affirms (denies) x. Now for Problem 3. It follows from what we have just shown that the machine U'# is similar to U#'. Now, U#' is M6 (as we have seen in the solution of Problem 2), whereas U'# is the machine M10 (since U' is M2). Thus M10 is similar to M6. We have see in the solution to Problem 1 that if Mh is any machine similar to U#', then Mh is silenced by h and U is silenced by h*h. Well, M10 is similar to M6, and so M10 is silenced by
r-+
§ 1 A universal machine
45
G2'
s0"
4.(D
10, just as M6 is silenced by 6. But also, since M10 and M6 are similar, any number that silences one also silences the other, and hence M10 is also silenced by 6 and M6 is silenced by 10. Therefore U is silenced by the numbers 10*10,6*6, 10*6, and 6*10. And so we now have four numbers that silence U. Four more are obtained by noting that M10 is the opposer of M9 and M6 is the opposer of M5, and a number silences a machine if and only if it silences its opposer. Therefore the numbers 6 and 10 also silence both M9 and M5, and so we have four more numbers that silence U, namely, 9*6, 9*10, 5*6, and 5*10. And so there are at least eight numbers that ..a
silence U.
Q..
(79
Now for Problem 4. Obviously, for any machine M, the machine M" is similar to M. Also, in our present indexing, the machines are numbered in such a way that for every number n there is a greater number m (25n, in fact) such that M,,,, is similar to M, and therefore there are infinitely many numbers h such that Mh is similar to M6, and U is then silenced by h*h for every such h. Therefore infinitely many numbers silence U.
Given a machine M, let Mh be any machine similar to the diagonalizer M# of M. Then for any x, Mh behaves towards x as M behaves towards x*x. Hence Mh behaves towards h as M behaves towards h*h. But also U behaves towards h*h as Mh behaves towards h, and so U behaves towards h*h as M behaves towards h*h. We then take n to be h*h, and U and M behave alike towards n. The solution of Problem 1 is immediate from this by simply taking U' for M. Then U and U' behave alike towards some number n, hence they are both silenced by n. (CD
pr,
7
cad
car
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car
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More generally, let M be any machine that knows at least as much as U. Then by the last problem there is a number n towards which U and M' behave alike. If M affirms n then M' denies n, hence U denies n (since U and M' behave alike towards n), hence M denies n (since M knows at least as much as U), and we have a contradiction. If M denies n, M' affirms n, U affirms n, M affirms n and we again have a contradiction. Hence n silences M. Thus M is not omniscient. Of course, this again solves Problem 1, since U obviously knows at least as much as U. 8
9 We could prove this from scratch, but it is immediate from the result of Problem 7. Given a machine M, take a number n towards which M and U behave alike. Then they either both halt at n or are both silenced by n.
Chapter 3. How to silence a universal machine
46
10 (1) Take any number a. Let b be such that for all x, b(x) = a(x*x). Then b(b) = a(b*b). But also h(b*b) = b(b). Therefore h(b*b) _ a(b*b), and so we take b*b for n, and we have h(n) = a(n). Fly
(2) Given a, take b such that for all n, b(n) = -a(n). Then by (1) there is some n such that h(n) = b(n). Hence h(n) = -a(n). in'
(3) By (2), taking h for a, there is some n such that h(n) = -h(n). Thus h(n) = 0. 11
The last problem ties up with all the earlier problems as follows. For cm.
any numbers x and y define x(y) to be +1 if Mx affirms y, -1 if Mx
Chi
OAF
Wpm
denies y, and 0 if Mx is silenced by y. Then F1, F2, and F3 are immediate consequences of conditions P1, P2, and P3 (for F1, take b to be an index of Ma ; for F2, take b to be an index of Ma; and for F3, take h to be any index of a universal machine). (CD
(1) By (3) of the last problem, there is some n such that h(n) = 0. Then U is silenced by n. (2) Given a machine Ma, by (1) of the last problem there is some n such
that h(n) = a(n), which means that U behaves towards n as Ma behaves towards n. 12 Part (1) of Problem 10 follows from just F1 and F3. By F3, there is a number h such that for all x, h(x*x) = x(x). To obtain the conclusion from Theorem 6.2 of Chapter 2, we take W = {-1, 0, +1}, d(x) = x*x, O(x, y) = x(y), and O(x) = h(x).
I--
,O++
13 The set of all n affirmed by U is one such set, for given any machine M, there is some x such that U and M behave alike towards x (Problem 7). For such an x, it is not possible that M affirms x if and only if U doesn't affirm x. Therefore the set of all numbers affirmed by M cannot be the complement of the set of all numbers affirmed by U. [Another such set is the set of all numbers denied by U. Another is the set of all numbers affirmed by C; another is the set of all numbers denied by C.] 14 Take f to be the identity function. Suppose Mi knows as much as D. Then if D affirms i, so does Mi, and if D denies i, so does Mi. But
also (Problem 6) if Mi affirms i then D denies i, and if Mi denies i then D affirms i, from which it follows that Mi can neither affirm nor deny i.
§1 A universal machine
47 coo
15 through 17 We will solve these in a different order than that in which ((DD
they were given. Let us go back to the machines of Part I. The mathematically significant
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car
aspects of what we called a "machine" is not its physical behavior (such as flashing a green or red light), but the set of numbers it affirms and the set of numbers if denies. And so we can take a "machine" M to be simply a disjoint ordered pair (a,,3) of number sets and we can say that (a, Q) affirms a number x if x c a, and denies x if x c Q, and is silenced by x if x ¢ (a U )3). Well, by conditions Q1 and Q2, the set of all disjoint ordered pairs (a,)3) of elements of £ can be arranged in a sequence (a,, 01), (an, /3n), ... that has the properties P1, P2, and P3 because we can take (A, B)' to be (B, A), and (A, B)# to be (A#, B#), where A# is the set of all x such that x*x E A and B# is the set of all x such that x*x E B. Finally, our "universal" pair is the pair (U1, U2), where U1 is the set of all x*y such that y E ax and U2 is the set of all x*y such that y E 3x. It then follows from the solution of Problem 8 that the pair (U1, U2) is inseparable in £, for consider any disjoint pair (A, B) of members of £ such that U1 C A and U2 C_ B. Then the "machine" (A, B) "knows as much" as (U1, U2), hence is not "omniscient," which means that there is some number n that is neither in A nor in B, so A is not the complement of B. Thus (U1, U2) is inseparable in £, which solves Problem 16. It of course then follows that neither U1 nor U2 can have its complement in £, which solves Problem 15.
Chapter 4
Some general incompleteness theorems car
This chapter is basic to several later chapters of this study, and is devoted to a reconstruction, generalization, and extension of several classical results in mathematical logic concerning incompleteness, provability, and truth. S."
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We give abstract versions of the incompleteness proofs of Godel[6] and Rosser[22] as well as the lesser known incompleteness proof using Tarski's notion of truth[35]. We conclude the chapter with a fascinating result of Rowitt Parikh[11]. Our results are presented in a purely abstract form which does not employ any apparatus of mathematical logic; the more concrete applications are given in Chapters 5 and 9 of this study. The central definition of this chapter is that of a representation system, which allows us to study the metamathematical properties of mathematical systems of highly diverse structures. The type of systems studied by Godel and Rosser have at least the following features. First a denumerable set E of elements called expressions is given, each of which is assigned a Godel number. Then there is a subset S of E whose elements are called sentences; next a set P of sentences called provable sentences and a set R of sentences obi
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called refutable (or disprovable) sentences. The system is called consistent if no sentence is both provable and refutable in the system, and inconsistent otherwise. The system is called complete if every sentence is either provable or refutable, and incomplete otherwise. A sentence X is called decidable in the system if it is either provable or refutable, and undecidable otherwise. Thus an incomplete system is one in which there is at least one undecidable sentence. Next, there are certain expressions called (unary) predicates and there is an operation that assigns to each predicate H and each numbers n, a sen'In some applications, "number" will mean natural number, and in others, positive
Chapter 4. Some general incompleteness theorems
49
.`3
tence denoted H(n), and which may be referred to as the result of applying H to n. [Informally, predicates can be thought of as names of properties of numbers, and the sentence H(n) intuitively expresses the proposition
that n has the property named by H.] A predicate H is said to represent the set of all n such that H(n) is provable in the system. How a predicate H is applied to a number n to yield a sentence H(n) varies considerably from system to system, and our whole purpose of working in representation systems is to be independent of the formal peculiarities of individual systems, and thus to study representability in systems of highly diverse syntactical structures. In later chapters we will see how representation systems are applicable to first-order systems of arithmetic. In later chapters we will also need the notion of representing relations of numbers, as well as sets, and so for each positive n we then have a set of expressions called predicate of degree n, and there is an operation that assigns to every predicate L of degree n and every n-tuple (a1, . . . , an) of numbers, a sentence L(a1,... , an), called the result of applying L to (a1, ... , an). And we say that L represents the set of all n-tuples (al,.. . , an) such that L(a1,... , an) is provable in the system. For technical reasons it will be helpful to consider an operation that assigns to every expression E, whether a predicate or not, and every ntuple (al,.. . , an) of numbers, an expression denoted E(a1,... , an) such that if E is a predicate of degree n, then E(a1,... , an) is a sentence. Again, for technical reasons, it will be convenient to presently assume that our Godel numbering is onto the set N of positive integers, i.e., we will assume that every positive integer n is the Godel number of some expression that we denote En. We remark that if we start with the more S."
usual type of Godel numbering in which not every number is a Godel number, we can always arrange all the expressions in the infinite sequence E1i E2) ..., En, ... according to the magnitude of their Godel numbers, and then forget about the old Godel numbering and define the new Godel number of EZ to be i. The collection of all the items just described will be called a representation system. To review the definition, we now formally define a representation system Z to be a collection of the following items. O..
(1) A 1-1 denumerable sequence E1i E2, ..., En, ... of elements called expressions. We call i the Godel number of E. (2) A set of expressions called sentences. (3) For each positive n, a set of expressions called predicates of degree n, or n-ary predicates, or n-place predicates, or predicate of n arguments. natural number. The results of this chapter hold under either application.
Chapter 4. Some general incompleteness theorems
50
(4) A mapping that assigns to each expression E and each n-tuple (a1i ... , an,) of numbers an expression E(al,... an), subject to the ,
condition that if E is a predicate of degree n, then E(al,... , an) is a sentence.
(5) An ordered pair (P, R) of sets of sentences. The elements of P are called the provable sentences of the system, and the elements of R are called the refutable sentences of the system.
This concludes our definition of a representation system. The only predicates that will be considered in this chapter are predicates of degree one, and so for the rest of this chapter, predicate will mean predicate of degree 1.
Part I Incompleteness §1 Diagonalization
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For any expression Ei (i being the Godel number of Ei) by its diagonalization we mean the expression Ei(i). If Ei is a predicate (of degree 1) then its diagonalization Ei(i) is a sentence. We shall always use the letter "H" to stand for a predicate (of degree 1) and "h" to stand for its Godel number. Thus H(h) is the diagonalization of H and H(h) is a sentence. [Intuitively, H(h) is supposed to express the proposition that the Godel number of H lies in the set represented by H.] For any number i, we let d(i) be the Godel number of Ei(i), and we call d(x) the diagonal function of Z. For any set W of expressions we let WO be the set of its Godel numbers
- the set of all i such that Ei E W - and we let W* be the set of all i such that Ei(i) E W. Thus W* = d-'(Wo). For any number set A, we let A# = d-1(A) (the set of all i such that d(i) E A). Thus W* = WO W. For any number set A we let A be the complement of A. It is easily verified that Wo = (W)o (W being the complement of W with respect to the set of expressions) and that W* = W.
Exercise 1 Verify that WO = (W) o and W * = W * . First diagonal lemma
We recall that for any predicate H of degree 1 and any set A of numbers,
H is said to represent A if for all numbers n, n E A H H(n) E P. We shall have need to generalize this notion as follows: Given an arbitrary
§ 1 Diagonalization
51
set K of sentences, we will say that H represents A with respect to K if
for any number n, n E A H H(n) E K. And we will say that A is Krepresentable if there is a predicate H that represents A with respect to K. [Thus P-representable is what we have called "representable."]
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Lemma 1.0 (First diagonal lemma) If A# is K-representable then there is a sentence X such that X E K <-a Xo E A (Xo is the Godel number of X). More specifically, if H represents A# with respect to K, then its diagonalization H(h) is such a sentence.
Proof Suppose H represents A# with respect to K. Then H(h) E K h E A# <-a d(h) E A. But d(h) is the Godel number of H(h). Discussion
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If we were working with a Godel numbering which is not onto the set of positive integers, we would proceed as follows. We would define a function r(x, y) (from ordered pairs of numbers to numbers) to be a representation function for the system Z if for every predicate H with Godel number h, and every number n, r(h, n) is the Godel number of H(n). Then we would take d(x) to be r(x, x). We would then take A# to be d-1(A) and W* to be d-1(Wo) and Lemma 1.0 would hold, as is easily verified.
Exercise 2 Verify the above. As an obvious corollary of Lemma 1.0 we have
Proposition 1.1 If W* is K-representable then there is a sentence X such that X E K X E W. More specifically, if H represents W* with respect to K, then its diagonalization H(h) is such a sentence.
Since there cannot be a sentence X such that X E K " X E K, we have
Proposition 1.2 The set K* (or the set K*, which is equal to it) is not K-representable. Remark
If the Godel numbering is not onto the set of positive integers, we proceed a bit differently: for every i that is a Godel number, we let Xi be the
Chapter 4. Some general incompleteness theorems
52
expression whose Godel number is i. If i is not a Godel number, Xi is undefined. We then define W* as the set of all Godel numbers i such that Xi(i) E W. Then Lemma 1.0 and Proposition 1.1 still hold, as the reader can easily verify. Also, although the sets k* and K* are not the same, they have the same Godel numbers, and so neither K* or K* is K-representable. Normality We shall say that Z is normal with respect to K if for every K-representable
set A, the set A# is also K-representable. This of course implies that for any set W of expressions, if WO is K-representable, so is W*. We then have
Proposition 1.3 If Z is normal with respect to K, then (1) If WO is K-representable then there is a sentence X such that X E K H X E W; (2) The complement Ko of Ko is not K-representable. For later use we will need the following
Proposition 1.4 If Z is normal with respect to K, then for any predicate H, there is a sentence X such that X E K H H(Xo) E K. Proof Suppose Z is normal with respect to K. Take any predicate H. It represents some set A with respect to K. Then A# is also K-representable, so by Lemma 1.0 there is a sentence X such that X E K H Xo E A. But H(X) E K (since H represents A with respect to K). And so Xo E A
XEKH(Xo)EK.
Normality and fixed points
We call Z normal (without reference to any K) to mean that Z is normal with respect to P - in other words, for any set A, if A is representable, so is A#. We call a sentence X a fixed point of a predicate H if the following holds: X is provable - H(Xo) is provable. Taking P for K in Proposition 1.4 we have
Proposition 1.5 If Z is normal, then every predicate H has a fixed point.
§2 An incompleteness argument using a truth set In the systems studied by Godel and Rosser (such as Peano Arithmetic, which we will discuss later in this volume) there is, in addition to, the set P
§2 An incompleteness argument using a truth set
53
of provable sentences, a set T of sentences called true sentences (as defined later by Alfred Tarski[35]). The system is called correct if every provable sentence is true and no refutable sentence is true. Formally, given a set T of sentences, we will say that Z is correct with respect to T if P C_ T and R
is disjoint from T. Let us note that if there exists a set T with respect to which Z is correct, then Z is automatically consistent (P must be disjoint from R). And now we have the following result which yields a method of proving incompleteness that is less well known than it should be.
Theorem 2.1 Suppose that T is a set with respect to which Z is correct. Then if P* is T-representable then Z is incomplete. More specifically, if Z
is correct with respect to T and H is a predicate that represents P* with respect to T, then its diagonalization H(h), though in T, is not in P, nor is it in R. Proof Assume hypothesis. Then H represents P* with respect to T, and so by Proposition 1.1 (taking T for K) the sentence H(h) E T H H(h) E P.
We let X be the sentence H(h). Thus X E T H X
P. This means
that either X E T and X 0 P, or X E P and X
T. The second
alternative is ruled out by the correctness assumption that P C T, and so X E T but X ¢ P. Since T is disjoint from R (again by the assumption of "correctness") and X E T, then X ¢ R. Thus X ¢ P and X ¢ R, so X is undecidable in Z. Discussion
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For a system in which T is the set of "true" sentences of the system, the sentence X in the above proof, which is in T if it is not in P, is thus a sentence that is "true" if it is not provable in the system. Such a sentence can be thought of as asserting its own non-provability in the system, and might be thought of as saying: "I am not provable in the system." G6del[6] likens this to the paradox of the liar: "This sentence is not true." Actually, the paradox of the liar is more clearly related to Tarski's theorem (which we will state shortly). The following problems are more closely related to Godel's theorem. [Solutions to Problems 1-5 are given following Problem 5.]
Problem 1 Suppose we have a book in which every sentence in the book is true. Construct a sentence X that must be true but cannot be anywhere in the book.
Chapter 4. Some general incompleteness theorems
54
Problem 2 In a certain land, some of the natives are Athenian and some are Cretan. Every native of this land either always tells the truth or always lies. The Athenians always tell the truth and the Cretans always lie. What statement could a native make that would imply that he must be a truthteller but not an Athenian?
Problem 3 Here is a more interesting one: We now make the additional assumption about the land of the last problem that every native is either an Athenian or a Cretan. A perfectly accurate logician one day visits this land. He is perfectly accurate in the sense that everything he proves is true; he never proves anything false. A native makes a statement to him, from which it follows that the native must be an Athenian, but the logician can never prove that he is. What statement would work?
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Problem 4 Suppose I place a penny and a quarter on the table and ask you to make a statement, with the understanding that if the statement is true, I will give you one of the two coins, not saying which one, but if the statement is false, I give you neither coin. Assuming I keep my word, what statement could you make that would force me to give you the quarter. A dual of theorem 2.1 Theorem 2.1 has the following "dual." car
Theorem 2.1° Suppose that T is a set with respect to which Z is correct 0.0
and that R* is T-representable. Then Z is incomplete. More specifically, if Z is correct with respect to T and H represents R* with respect to T, then the sentence H(h) is undecidable in Z (and also is outside T).
Proof Since R* is T-representable, there is now a sentence Y - H(h) - such that Y E T +-4Y E R. By the assumption of T-correctness, R is disjoint from T, so Y is neither in R nor T. Since Y is not in T, and P C T, then Y ¢ P. Thus Y is neither in P nor R, hence is undecidable in Z. Discussion
Again, if we call the elements of T "true" sentences, then the sentence Y
c_°
of the above proof is true if it is refutable, and can be thought of, not as saying: "I am not provable" (as does the sentence X of Theorem 2.1), but rather as saying: "I am refutable." The sentence X is true but not provable in the system, whereas Y is false (not true) but not refutable in
§2 An incompleteness argument using a truth set
55
the system.
Problem 5 (1) In the setup of Problem 2, what statement could a native make that would imply that he is a liar but not a Cretan?
(2) In the setup of Problem 3, what statement could a native make to an accurate logician such that the native must be a Cretan, but the logician can never prove that he is?
Solutions to problems 1-5 1
A sentence that works is: "This sentence is not in the book."
2 A statement that works is: "I am not an Athenian." 3 A statement that would work is: "You can never prove that I am an Athenian." If the logician could prove that the speaker is an Athenian, that would falsify the statement, thus making the speaker in fact a Cretan, hence the logician would have proved something false, contrary to the given fact that the logician never proves false statements. Hence the logician can never prove that the speaker is Athenian. This makes the speaker's statement true, hence the speaker is in fact an Athenian. Thus the speaker is an Athenian, but the logician can never prove that he is.
4 A statement that works is: "You will not give me the penny." I leave the proof to the reader. [The relation to the Godel argument is that the penny corresponds to the provable statements and the quarter corresponds to the statements that are true but not provable in the system under consideration.]
5 (1) I am a Cretan. (2) You can prove that I am Cretan. A rudimentary form of Tarski's theorem
Theorem 2.2 [After Tarski] Suppose that T has the following properties. (1) The complement of every T-representable set is T-representable.
Chapter 4. Some general incompleteness theorems
56
(2) (T-normality) For every T-representable set A, the set A# is Trepresentable.
Conclusion The set To of Godel numbers of the elements of T is not T-representable.
Proof By Corollary 1.2, the set T* is not T-representable. Then by (1), the set T* is not T-representable, so the set To (which is T*) is not Trepresentable. Then by (2), the set To is not T-representable. Discussion
In the systems investigated by Godel, Rosser, and Tarski, the set T of "true" sentences does satisfy the hypothesis of the above theorem. The car
word "expressible" is sometimes used for T-representable. Thus a predicate P expresses the set of all numbers n such that H is true for n, i.e., such that H(n) is a true sentence. Thus for such systems, the set of Godel numbers
of the true sentences is not expressible - there is no predicate H that is true for those and only those numbers n that are Godel numbers of true
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sentences. This is Tarksi's theorem, and is sometimes paraphrased that for systems of sufficient strength, truth within the system is not expressible in the system. On the other hand, in these systems, the set Po of Godel numbers of To, the provable sentences is expressible, and since To isn't, then Po
hence P and T are difference sets. Thus some sentence X is either in P
but not in T, or in T but not in P. Assuming correctness, it must be that X is in T but not in P - it is true but not provable in the system. More constructively, if T satisfies the hypothesis of Theorem 2.2 and Po is expressible (T-representable) then by (1) and (2) of the hypothesis, the set P* is T-representable, and so by Theorem 2.1, the sentence H(h) is in T but not in P, where H is any predicate that represents P* with respect to T. We thus have
Corollary 2.3 Suppose that T satisfies the hypothesis of Theorem 2.2 and that P0 is T-representable. Then Z contains an undecidable sentence.
§3 Godel's argument and some variations Godel's original incompleteness proof did not use any truth set. He worked
purely with provability and refutability, and we now turn to an abstract
study of this. We no longer use any truth set T; we stick entirely to provability and refutability, and "representable" means P-representable.
§3 Godel's argument and some variations
57
Earlier in this volume we defined what it means for a sentence X to be a Godel sentence for a set A of numbers; we will now define X to be a Godel sentence for a set W of expressions if X E P H X E W (X is either provable and in W, or neither). By Proposition 1.1 (taking P for K) we know that if W* is representable (P-representable) then there is a Godel sentence for W - to wit, H(h), where H is any predicate that represents W*.
Now we shall start with a "dual" of Godel's argument.
Theorem 3.1 If R* is representable in the system Z then Z is either inconsistent or incomplete. More specifically, if Z is consistent and if H represents R*, then its diagonalization H(h) is an undecidable sentence of Z.
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Proof Assume hypothesis. Then by Proposition 1.1, there is a Godel sentence for R - in fact, H(h) is such a sentence. Thus H(h) is provable ¢+°
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if it is in R - it is provable if it is refutable. Thus it is either both provable and refutable, or neither provable nor refutable. If the system is consistent, then the first alternative cannot hold, hence H(h) is undecidable in Z.
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Exercise 3 Let us say that H almost represents A if H is provable for every n in A and not refutable for any n not in A (i.e., n E A H(n) E P, and n ¢ A = H(n) ¢ R). Prove that for a consistent system Z, if H represents A then H almost represents A. Then prove the following strengthening of Theorem 3.1: if Z is consistent and R* is almost representable in Z, then Z is incomplete. Contrarepresentability
We shall henceforth use the word "contrarepresentable" for R-representable. Thus H contrarepresents A if for all n, H(n) is refutable if n E A. By simply interchanging the roles of P and R in Theorem 3.1, we have
Theorem 3.1
[After Godel] If P* is contrarepresentable in Z, then Z is inconsistent or incomplete. More specifically, if Z is consistent and if H contrarepresents P*, then its diagonalization H(h) is undecidable in Z. Remark
Theorem 3.1 comes closer to what Godel did than does Theorem 2.1. For the system for which he gave his incompleteness proof he constructed a
58
Chapter 4. Some general incompleteness theorems
predicate that contrarepresents P*.
Exercise 4 Let us say that H almost contrarepresents A if H(n) is refutable whenever n is in A, and whenever n ¢ A then H(n) is not provable. Suppose Z is consistent and P* is almost contrarepresentable. Does it follow that Z is incomplete?
Exercise 5 Here is a more interesting one. Suppose that Z is consistent, P* is representable, and for any predicate H there is a predicate H' such
that for all n, H(n) E P - H(n) E R. Does it follow that Z must be incomplete? [The answer will emerge in a later theorem.] Normal systems
We recall that we are calling Z normal if it is normal with respect to P, as defined in §1 - which means that for any representable set A, the set A# is also representable (and hence for any set W of expressions, if Wo is representable, so is W*).
Theorem 3.2 If Z is normal and if PO is contrarepresentable, then Z, if consistent, is incomplete.
Proof Suppose Z is normal and H contrarepresents Po. Then by Proposition 1.4 (taking P for K) there is a sentence X such that X is provable H(Xo) is provable. But since H contrarepresents P, then X is provable if H(Xo) is refutable. Thus H(Xo) is provable H H(Xo) is refutable. Assuming consistency, H(Xo) is then undecidable in Z. The negation property
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We shall say that Z has the negation property if there is a mapping that assigns to every predicate H a predicate H', called. the negation of H such that for every n, H'(n) is provable if H(n) is refutable, and also, for any n, if H(n) is provable, so is H"(n) (and thus if H(n) is provable, H'(n) is c0.
refutable). If Z has the negation property, then every contrarepresentable set is representable, because if H contrarepresents A, then for all n, n E A H H(n)
is refutable F4 H'(n) is provable, and so H' then represents A. But the negation property does not appear to necessarily imply that every representable set is contrarepresentable. Let us say that Z has the double negation property if for every predi-
cate H and number n, H(n) is provable if and only if H"(n) is provable
§4 Rosser's argument
59
- and thus H(n) is provable if H'(n) is refutable. Now, representation systems associated with classical systems, in which every sentence is prov(ADD
ably equivalent to its double negation - these classical systems do have the double negation property (taking the negation of a formula F(x) to be -F(x)). But systems based on intuitionistic logic don't have the double negation property. Now, if Z has the double negation property, then every representable set is contrarepresentable (because if H represents A, then for every n, n E A F-, H(n) is provable H H'(n) is refutable, and so H' contrarepresents A). And so in a consistent system having the double negation principle, if P* is representable, the system has an undecidable sentence. However, the double negation property is a stronger hypothesis than we need; the negation property suffices by virtue of our next theorem (which is accordingly applicable to intuitionistic systems as well as classical ones).
In what follows we shall use the following notation for systems with the negation property: H is a predicate with Godel number h, H' is the negation of H, and h' is the Godel number of H'.
Theorem 3.3 Suppose Z is consistent and has the negation property and that P* is representable in Z. Then Z contains an undecidable sentence.
Proof Assume hypothesis and let H represent P*. It is now the sentence H(h') that is undecidable in Z, for H(h') is provable F-, h' E P* H H'(h') E P H(h') E R. Thus H(h') E P H H(h') E R, so by consistency, H(h') P and H(h') ¢ R.
Exercise 6 Suppose that Z is consistent and has the negation property and that H is a predicate whose negation H' represents P*. Is the sentence H(h) necessarily undecidable in Z?
Theorem 3.4 If Z is consistent, normal, and has the negation property, and if Po is representable in Z, then Z is incomplete. Exercise 7 Prove Theorem 3.4. Also, show that it is a consequence of the miniature incompleteness theorem stated in Exercise 33, Chapter 1.
§4 Rosser's argument Godel's method of proving incompleteness boils down to either representing R* or contrarepresenting P*. To do this, however, Godel had to make a certain assumption known as w-consistency, which we will discuss in Part
60
Chapter 4. Some general incompleteness theorems
II of this chapter. And so Godel did not prove outright that the systems under his investigation were incomplete; he proved that they were incomplete under this additional assumption of "w-consistency." Subsequently, J. Barkely Rosser[22] was able to prove these systems incomplete under the much weaker assumption of consistency. What Rosser did boils down, not to representing the set R* itself, but to representing some superset A of R* disjoint from P*, and as we will see, this is enough to give incompleteness. [We call B a superset of A if A is a subset of B.] Separability within Z
We shall say that a predicate H is provable for n if H(n) is provable, and refutable for n if H(n) is refutable. We now say that H weakly separates A from B if H is provable for every element of A and not provable for any element of B. Equivalently, H weakly separates A from B if H represents some superset of A that is disjoint from B. And we say that A is weakly separable from B - or that the ordered pair (A, B) is weakly separable in Z - if some predicate H weakly separates A from B in Z. We now need the following diagonal lemma.
Lemma 4.0 [Second diagonalization lemma] If H weakly separates W* from P*, then its diagonalization H(h) is outside both P and W.
Proof Suppose H represents A, W* C A, and A is disjoint from P*. Since H represents A, H(h) E P H h c A. Also, H(h) E P h c P*. Therefore h E P* H h c A. Since P* is disjoint from A, It P* and h ¢ A. Since h ¢ A and W* is a subset of A, then h V W*. Thus h V P* and h V W*, which means that H(h) V P and H(h) ¢ W. Taking R for W in the above lemma, we have the following strengthening of Theorem 3.1.
Theorem 4.1 If R* is weakly separable from P* then Z is incomplete. More specifically, if H weakly separates R* from P* in Z, then its diagonalization H(h) is outside both P and R, and is hence undecidable in Z.
Remark
(1) We note that the hypothesis of the above theorem did not involve the assumption that Z is consistent. Consistency enters Rosser's incom-
§4 Rosser's argument
61
HIV
pleteness proof in showing that some superset of R* disjoint from P* is representable in Z! We will say more about that shortly. (2) The hypothesis of Theorem 4.1 is genuinely weaker than the hypothesis of Theorem 3.1, because if Z is consistent and if H represents R*, then it of course represents some superset of R* disjoint from P*, namely, R* itself. Thus the above theorem is genuinely stronger than Theorem 3.1. The difference in strength may appear to be minuscule, but as we will see, in application it is not. [Because of the added strength of Theorem 4.1, Rosser was able to avoid Godel's assumption of w-consistency.] (3) For later use, let us note that the proof of Theorem 4.1 reveals the
o>,
fact that if A is the superset of W * that H represents, the Godel number h of H is not only outside both W* and P*, but is outside both A and P*. Theorem 4.1 says that if H represents some superset of R* disjoint from P*, then H(h) is an undecidable sentence. Obviously by interchanging the roles of P and R we have the following "dual" of Theorem 4.1.
(-+x
Theorem 4.1° If H contrarepresents some superset of P* disjoint from R*, then H(h) is undecidable in Z. We also have
Theorem 4.2 If Z has the negation property and P* is weakly separable from R* then Z is incomplete.
Proof Suppose H weakly separates P* from R*. We assert that H(h') is undecidable in Z (where h' is the Godel number of H').
(1) Suppose H(h') is provable. Then H'(h') is refutable, hence h' E R*, hence H(h') is not provable (since H is not provable for any element of R*). This shows that H(h') cannot be provable.
(2) Since H(h') is not provable, then h' ¢ P* (since H is provable for every element in P*), hence H'(h') is not provable, hence H(h') is not refutable.
Thus H(h') is neither provable not refutable in Z.
Exercise 8 For those who like curiosities, let us say that a predicate H' is a weak negation of H if for all n, H'(n) is provable if H(n) is refutable (but it is not required that if H(n) is provable then H'(n) is refutable). Now, suppose H weakly separates P* from R* and that there exists a weak negation H' of H. It can then be shown that Z is incomplete - in fact that
Chapter 4. Some general incompleteness theorems
62
either H'(h') is undecidable, or H(h') is undecidable, but there appears to be no way to tell which! How is it shown that one of those two sentences must be undecidable in Z? Strong separability
We shall say that H strongly separates A from B in Z if H is provable for all elements of A and refutable for all elements of B - in other words, H represents some superset of A and contrarepresents some superset of B. Thus if H strongly separates A from B, then for all numbers n,
(1) n c A = H(n) is provable,
(2) n c B = H(n) is refutable. Suppose now that H strongly separates A from B and that Z is consistent. Then for any n in B, since H(n) is refutable, H(n) is not provable (by consistency), and so H is then provable for all n c A, but for no n in B, which means that H weakly separates A from B. We thus have the obvious fact.
Fact 4.2 If H strongly separates A from B and Z is consistent, then H weakly separates A from B.
By virtue of the above fact and Theorem 4.1 we have
Theorem 4.3 [A rudimentary form of Rosser's incompleteness theorem] If H strongly separates R* from P* and Z is consistent, then H(h) is undecidable in Z.
Lemma 4.4 Suppose Z has the negation property. Then if H strongly separates A from B, its negation H' strongly separates B from A.
Proof Obvious. From the above Lemma and Theorem 4.3 we have
Theorem 4.5 If Z is consistent and has the negation property, then if P* is strongly separable from R*, the system Z is incomplete - moreover, if H effects the separation, then H'(h') is an undecidable sentence of Z.
§5 Complete representability and definability
63
Exercise 9 Suppose Z has the negation property. (a) Show that if H'(n) is undecidable, H(n) is also undecidable. [The converse doesn't appear to be true, unless, of course, Z has the double negation property.]
(b) Suppose that Z is consistent and that H strongly separates P* from R*. Is H(h') necessarily undecidable in Z?
Theorem 4.6 Suppose that Z is consistent, normal, and has the negation property. Then if Po is strongly separable from Ro, then Z is incomplete - in fact, if X is any fixed point of H', where H strongly separatesPo from Ro, the sentence X is undecidable in Z.
Proof That H' has a fixed point X is guaranteed by Proposition 1.5. Now, let X be any fixed point of H'. Thus X is provable " H'(Xo) is provable, and so
(1) X is provable F-, H(Xo) is refutable. Now, suppose H strongly separates Po from Ro. Then
(2) X is provable = H(Xo) is provable
(3) X is refutable = H(Xo) is refutable. If X is provable, then H(Xo) is provable (by (2)) and is also refutable (by (1)) and the system is then inconsistent. If X is refutable, then H(X0) is refutable (by (3)), hence X is provable (by (1)), and again the system is inconsistent. And so if Z is consistent, then X is undecidable in Z. Exact separability
In a later chapter we will need the notion of exact separability. Given two
disjoint sets A and B, we say that H exactly separates A from B if H represents A and contrarepresents B.
§5 Complete representability and definability We say that H defines A if H strongly separates A from its complement A
- in other words, for any number n: (1) n E A
H(n) is provable,
Chapter 4. Some general incompleteness theorems
64
(2) n ¢ A = H(n) is refutable. We say that H completely represents A if H represents A and contrarepresents A (H exactly separates A from A) - in other words, for any number n: (1)* n E A
H(n) is provable,
(2)* n ¢ A
H(n) is refutable.
Let us note that for a consistent system, complete representability -m°0
and definability are equivalent, complete representability obviously implies definability, and in the other direction, suppose that H defines A and that Z is consistent. Then for every n, (1) and (2) hold, and we must show that
the converse implications also hold. So suppose H(n) is provable. Then H(n) is not refutable (by consistency), hence by (2), n ¢ A cannot hold, so n E A. Thus H(n) is provable implies n E A. Similarly, H(n) is refutable H(n) is not provable = n E A cannot hold = n ¢ A. We thus have
Proposition 5.0 If Z is consistent then H defines A if H completely represents A.
Exercise 10 Suppose Z is consistent and has the negation property. Show that the complement of any definable set is representable. From Lemma 4.4 we have
Proposition 5.1 Suppose Z has the negation property. Then if H defines A then H' defines A. Next we need
Proposition 5.2 No superset of R* disjoint from P* is completely representable in Z.
Proof Suppose R* C A and A is disjoint from P* and H represents A. We will show that H fails to contrarepresent A, and hence that H doesn't completely represent A. Since H represents a superset of R* disjoint from P* (the set A), then by Theorem 4.1, the sentence H(h) is undecidable in Z. Since H represents
A and H(h) is not provable, then h ¢ A, and so h E A. Since h E A and
§ 5 Complete representability and definability
65
H(h) is not refutable, then H fails to contrarepresent A. Thus H does not completely represent A. From the above and Proposition 5.0 we have
Theorem 5.3 Suppose Z is consistent. Then (1) No superset of R* disjoint from P* is definable in Z.
(2) R* is not definable in Z. We also have BCD
N
Theorem 5.4 Suppose Z is consistent and has the negation property. Then
(1) No superset of P* disjoint from R* is definable in Z.
(2) P* is not definable in Z.
Proof Suppose Z is consistent and has the negation property. (1) Then if H defines a superset A of P* disjoint from R*, H' would define A, which is a superset of R* disjoint from P*, and this is contrary to Theorem 5.3. (2) This is immediate from (1).
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Exercise 11 Suppose Z is consistent and H strongly separates R* from P*. Let A be the set represented by H and B the set contrarepresented by H. Show that the Godel number h of H lies outside both A and B. This yields another proof of Proposition 5.2. How?
Theorem 5.5 A sufficient condition for a consistent system Z to be incomplete is that there be a set A that is representable in Z but not definable in Z.
Proof We will prove the equivalent result that if Z is both consistent and complete, then every set representable in Z is definable in Z. Well, suppose Z is consistent and complete and that H represents A in Z. Since H is provable for every element of A, it is not refutable for any element of A (by consistency), and so for every number n, H(n) is refutable
Chapter 4. Some general incompleteness theorems
66
n ¢ A. But also n ¢ A implies H(n) is refutable, because if n ¢ A, then H(n) is not provable, hence H(n) is refutable (by completeness). And so H contrarepresents A, hence H completely represents A, and therefore H defines A.
Another way of looking at it is this. Suppose that Z is consistent and that H represents A but fails to define A. Then H does not contrarepresent A, so either H is refutable for some n in A, or fails to be refutable for some n in A. The former alternative implies the inconsistency of Z (since H is provable for every n in A), hence by our assumption that Z is consistent, it must be that H fails to be refutable for some number in A. For such an n, H is not provable for n either (since H represents A) and so for such an n, the sentence H(n) is undecidable in Z. Remark 0
This proof is less constructive than our former incompleteness proofs, in that it reveals that for some number n in A, the sentence H(n) is undecidable, but it gives no indication how to find such an n. A more constructive version of this theorem will be provided in a later chapter. We might also note that the above theorem together with (2) of Theorem 5.4 provides an alternative proof of Theorem 3.3. Suppose Z is consistent and has the negation property and that P* is representable in Z. Since P* is not definable in Z (by Theorem 5.4) then by the above theorem, Z must be incomplete. But this proof is again non-constructive, in that it gives us no idea of what the undecidable sentence could be - all we know from this proof is that if H represents P*, then for some number n outside P*, the sentence H(n) is undecidable in Z. By contrast, the former proof of Theorem 3.3 reveals that H(h') is undecidable, where h' is the Godel number of H'. As a corollary of Theorem 5.5 we have
Theorem 5.6 Suppose Z is consistent and has the negation property. Then if there is a set A that is representable in Z but whose complement A is not representable in Z, then Z is incomplete.
Exercise 12 Prove Theorem 5.6.
§6 Admissible functions
67
§6 Admissible functions }j"
We will call two sentences X and Y equivalent - in symbols, X - Y - if they are either both provable, both refutable, or both undecidable. Thus if X = Y, then if either one is provable, so is the other, and if either one is refutable, so is the other. CFA
We now define a function f (x) from numbers to numbers to be admissible in Z if for any predicate H there is a predicate H1 such that for every
number n, Hi(n) - H(f (n)), i"1
Proposition 6.1 Suppose f (x) is admissible in Z. Then for any sets A and B:
(1) If A is representable (contrarepresentable) in Z, so is f -1(A).
(2) If A is definable in Z, so is f-1(A).
(3) If A is strongly (exactly) separable from B in Z, then f -1(A) is car
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strongly (exactly) separable from f-1(B) in Z.
Proof Obvious Hence we have ..o
Proposition 6.2 Suppose the diagonal function d(x) is admissible in Z. Then from any sets W, V of sentences,
(1) If Wo is representable (contrarepresentable) in Z, so is W*. (2) If Wo is definable in Z, so is W*. (3) If (Wo, Vo) is strongly (exactly) separable in Z, so is (W*, V*).
From Theorem 5.4 and Proposition 6.2, we have the following generalization and extension of a theorem of Tarski (cf. note below).
Theorem 6.3 Suppose Z is consistent and has the negation property ±o"
and that the diagonal function d(x) is admissible in Z. Then the set P0 of Godel numbers of the provable sentences Z is not definable in Z - in fact, no superset of P0 disjoint from Ro is definable in Z.
Chapter 4. Some general incompleteness theorems
68
Note
For arithmetical systems considered in the language of first-order logic, there is a notion of a function f (x) being definable in the system. This notion essentially involves the logical connectives and quantifiers, hence is not applicable to our more abstract representation systems. However, it implies admissibility in our sense, which is meaningful for representation systems. It was proved in Undecidable Theories[34] that Po and the diagonal function are not both definable in a consistent system. This is generalizable to representation systems, as we have just seen, and the added fact that the hypothesis implies that no superset of Po disjoint from Ro is definable is a further strengthening along the lines of Rosser.
§7 Kleene's symmetric incompleteness theorem generalized In Chapter 3 we stated that the results of Part II were related to incompleteness theorems. So let us now consider a collection E of sets of numbers
such that E satisfies conditions Q1 and Q2 defined in Chapter 3, Part II. Let us call Z a E-system if all representable sets of Z and all contrarepresentable sets of Z are members of E. The following theorem is suggested by Kleene's symmetric form of Godel's theorem[7].
Theorem K Suppose Z is a consistent E-system such that any disjoint pair (a, /3) of members of E is strongly separable in Z. Then Z is incomplete.
Proof Assume the given conditions. As we proved in Chapter 3 (Problem 16), the pair (U1, U2) is inseparable in E, yet the pair (U1, U2) is strongly !/)
separable in Z; let H be a predicate that effects the separation. Then H represents some set Al and contrarepresents some set A2 such that U1 C Al and U2 C_ A2, and Al is disjoint from A2 (by the assumption of consistency). Also Al and A2 are elements of E (since Z is a E-system), and since (U1, U2) is inseparable in E, then Al is not the complement of A2. Hence some number n is outside both Al and A2, which means that H(n) is neither provable nor refutable in Z. Remark
It is interesting that the Godel numbering of the expressions of Z plays no role in the above proof. In applications, however, in which E is the collection of all recursively enumerable sets, the Godel numbering comes in
§8 Provability by stages
69
in showing that Z is a E-system.
Part II Provability by stages We now go a bit more deeply into the Godel and Rosser incompleteness proofs and also consider an interesting related result of Rowitt Parikh[11]. For the remainder of this chapter, Z will be assumed to be normal and to have the negation property.
§8 Provability by stages t,"
In the type of system studied by Godel and Rosser (such as Peano Arithmetic) there is a well defined relation P(X, n) between sentences of Z and positive integers, which we read "X is provable at stage n." [In applications, one assigns Godel numbers not only to formulas, but also to proofs.
We then say that X is provable at stage n to mean that n is the Godel
((DD
number of a proof of X. In Chapter 9 we consider a closely related idea. But for now, we can think of the theorems of Z as generated by a computer that prints out the provable sentences in some sequence. We will say no more about this at present, since we want our relation P(X, n) to be quite general.] It is understood that the relation P(X, n) enumerates the set P, in that X is provable if and only if X is provable at some stage n. In the type of system amenable to Godel's incompleteness proof, there is associated with each positive integer n a predicate Bn, whose intuitive meaning is this: for any number i, Bn(i) expresses the proposition that i is the Godel number of a sentence that is provable at stage n. Thus for any sentence X, Bn(Xo) intuitively says that X is provable at stage n. And then there is in these systems a predicate B - a so-called provability predicate - whose intuitive meaning is this: for any i, B(i) intuitively says that i is the Godel number of a sentence that is provable (at some stage or other). Thus for any sentence X, the sentence B(Xo) can be read: "X is provable," or "There exists some n such that Bn(Xo)." We are now interested in systems that satisfy the following three conditions. fir
Gl: If X is provable at stage n, then Bn(Xo) is provable. G2: If X is not provable at stage n, then Bn(Xo) is refutable. G3: If for any n, Bn(Xo) is provable, then B(Xo) is provable.
Informally, the idea behind G1, G2, and G3 is that at any stage the system has memory, so to speak, of what has and what has not already been proved. And so if X is proved at stage n, then (perhaps at a later
Chapter 4. Some general incompleteness theorems
70
stage) B,,,(Xo) will be proved, and if X is not proved at stage n, then
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(perhaps at a later stage) B' (Xo) will be proved'. As for G3, since B(Xo) intuitively means that X is provable at some stage or other, it follows that if for any n, Bn(X0) is true, then B(X0) is true. The point now is that the system is strong enough so that if Bn(Xo) is provable, then B(Xo) is not only true, but actually provable in the system. It is obvious that if B(Xo) is true, then there must be at least one number n for which Bn(X0) is true, for if X is provable, then it must be provable at some stage n or other. And so if B(X0) is true, then it cannot be that all the infinitely many sentences B1(X3), B2(Xo), ... , Bn(X0), ... are false; at least one of them must be true. Now, suppose that B(X0) were provable, and at the same time, all the infinitely many sentences B1(Xo), B2(Xo), ... , Bn(X0), ... were refutable; would this mean that the system would necessarily be inconsistent? No; it would mean that there was at least one provable sentence that was false under the intended interpretation of the predicates B, 131, B21 ... , Bn, . .., but it wouldn't necessarily mean that some single sentence was both provable and refutable, which is what inconsistency means. And so it is possible for a consistent system to have the property that there is a sentence X such that B(X0) is provable, yet for every n, Bn(X0) is refutable. Such systems we will call unstable, and we will call Z stable if it is not unstable. Thus in a stable system, if B(Xo) is provable, then there is at least one n such that Bn(Xo) is not refutable. [What we are calling instability, is a special case of the condition known as w-inconsistency, which, informally speaking, is that, on the one hand, the system can prove that there exists at least one number having a certain property, and, on the other hand, for every particular number n, the system can prove that n does not have the property. We will be more precise about this in Chapter 9 of this study.] We now have the following result which is an abstract form of Godel's theorem that comes closer to Godel's original proof than any theorem of Part I. v0)
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Theorem G [After Godel] Suppose Z is normal and has the negation property and satisfies conditions G1, G2, and G3. Then
(1) If Z is consistent and stable, then Z is incomplete.
(2) More specifically, there is a fixed point X for the predicate B', and for any such X: (a) If Z is consistent then B'(Xo) is not provable in Z. (b) If Z is also stable, then also B(Xo) is not provable in Z. 2We are assuming the negation property.
§8 Provability by stages
71
Problem 6 Prove Theorem G. [Solutions to this and the remaining problems are given at the end of the chapter.]
Problem 7 Actually, (1) of Theorem G can be established as a corollary of Theorem 3.4. How?
More on Rosser's proof
In Rosser's version of Godel's theorem[22] it is not necessary to assume stability (w-consistency). We continue to assume conditions G1 and G2 but we now abandon G3.
In fact we now abandon altogether the predicate B used by Godel and cod
consider another predicate H used by Rosser, whose meaning will soon be explained. We must now assume that to each sentence X is associated a sentence X' (also written -X) called the negation of X, which is provable if and only if X is refutable. And we say that X is refutable at stage n to mean that X' is provable at stage n. And now we say that X is provable (refutable) by stage n to mean that
X is provable (refutable) at stage n or at some earlier stage. Thus X is provable (refutable) by stage n if there exists some m < n such that X is
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provable (refutable) at stage m. We now have for each positive integer n a predicate Sn, where Sn(Xo) intuitively means that X is provable by stage n. We still have, for each n, a predicate Bn, where Bn(Xo) means that X is provable at stage n. Coming back to Rosser's predicate H, the sentence H(Xo) is interpreted to mean that there is some n such that X is provable at stage n and is not refutable
at either stage n or at any earlier stage - in other words, X is provable at stage n and not refutable by stage n. Thus H(Xo) is true if and only if there is some n such that Bn(Xo) is true and Sn(Xo') is false. Also with a bit of argument it can be shown that if Bn(Xo) is true and Sn(Xo) is false, then H(Xo) must be false. Can the reader see why?
Problem 8 Why? We thus have the following four facts:
(1) If Bn(Xo) is true and m < n, then Sn(XO) is true. (2) If for every m < n, Bm(Xo) is false, then Sn(Xo) is false.
(3) If Bn(X) is true and Sn(Xo') is false, then H(Xo) is true. (4) If Bn(Xo') is true and Sn(Xo) is false, then H(Xo) is false.
Chapter 4. Some general incompleteness theorems
72
Now, Rosser showed that for the systems studied by Godel, the above
four facts were formalizable in the system, that is, they held replacing "true" by "provable" and "false" by "refutable." And so each such system satisfies conditions G1 and G2 as well as the following four conditions (for any sentence X).
R1: If B,,(Xo) is provable and m < n, then S,,(Xo) is provable. R2: If for every m < n, Bm(Xo) is refutable, then S,,(Xo) is refutable. R3: If Bn(X) is provable and Sn(Xo') is refutable, then H(X0) is provable.
R4: If Bn(Xo') is provable and S,,(Xo) is refutable, then H(X0) is refutable.
And now we have the following general form of Rosser's theorem.
Theorem R [After Rosser] If Z is a consistent normal system with the negation property and Z satisfies conditions G1, G2, and R1-R4, then Z is incomplete.
Problem 9 Prove Theorem R. Parikh sentences
C/]
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We now turn to a lovely observation of the logician Rowitt Parikh[11]. We consider a consistent normal system with the negation property and which satisfies conditions G1, G2, R1, and R2. [Rosser's predicate H is no longer relevant, and so we no longer need conditions G3 or G4.] Since Z is assumed normal, then by Proposition 1.5, for each n there is a fixed point
of S, - let Yn be such a fixed point.
Problem 10 What is the status of Yn? Is it provable, refutable, undecidable? Is it provable by stage n? Is it provable at a later stage? Is
it refutable by stage n? Is it refutable at a later stage? Just what is its status? [The answer has a very curious corollary which we discuss following the solution.]
Solutions to problems of Part II Since Z is assumed normal, then by Proposition 1.5 there is a fixed point for B'. Let X be any fixed point of B'. 6
§8 Provability by stages
73
'No
ate'
(1) Suppose B'(Xo) is provable. Then X is provable. Then X is provable at some stage n, hence Bn(Xo) is provable (by G1), hence B(X0) is provable (by G3), hence Z is then inconsistent (since B'(Xo) is also
provable). And so if Z is consistent, B'(Xo) is not provable in Z. Also, if Z is consistent, then X is not provable in Z (since X is provable if B'(Xo) is provable).
'"'
(2) Assuming consistency, X is not provable in Z, as we have just seen, hence X is not provable at any stage n, so for every n, the sentence Bn(Xo) is refutable (by G2). Therefore, if Z is also stable, then B(X0) is not provable in Z. And so if Z is both consistent and stable, then B(Xo) is also not provable in Z, and is hence undecidable in Z.
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It easily follows from Gl and G3 that if X is provable, so is B(Xo). Also from G2 it easily follows that if B(X0) is provable and Z is stable, then X must be provable. And so for a stable system satisfying G1, G2, and G3i the predicate B represents the set P0. And so Theorem 3.4 can then be invoked. 7
For any two sentences X and Y, let us say that X is provable before Y if there is some n such that X is provable at stage n and Y is not provable by stage n. [It is to be understood that if X is provable and Y is not provable at all, then X is provable before Y.] It is obviously impossible for X to be provable before Y and also for Y to be provable before X. To say that H(Xo) is true is to say that X is provable before X'. Now, if Bn(XX) is true and Sn(Xo) is false, then X' is provable before X, hence it is false that X is provable before X', which means that H(X) is false. 8
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9 Step 1 - It follows from G1, G2, R1, and R2 that if X is provable
by stage n, then Sn(Xo) is provable, and if X is not provable by stage n, then Sn(Xo) is refutable, for suppose X is provable by stage n. Then X is provable at stage m for some m less than or equal to n, hence B"(Xo) is provable (by G1), hence Sn(Xo) is provable (by R1). Now suppose X is not provable by stage n. Then for every m less than or equal to n, X is not provable at stage n, hence all the sentences B1(Xo),... , Bn(Xo) are refutable (by G3) and hence Sn(Xo) is refutable (by R2). Step 2 - We now show that it follows from G1, G2, and R1-R4 that for any sentence X, if X is provable before X', then H(Xo) is provable, and if X' is provable before X, then H(X0) is refutable. Well, suppose X is provable before X. Then for some n, X is provable at stage n and X' is not provable by stage n. Then Bn(Xo) is provable (by R1) and
Chapter 4. Some general incompleteness theorems cad
74
car
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Sn(Xo') is refutable (as shown in Step 1), and therefore H(X) is provable (by R3). Now suppose X' is provable before X. Then by a similar argument (interchanging X and X') there is some n such that Bn(Xo) is provable and Sn(Xo) is refutable, and so H(Xo) is refutable by R4. Step 3 - Now suppose Z is consistent and satisfies conditions G1i G2, and R1-R4. Since Z is consistent, then to say that X is provable before X' is equivalent to saying that X is provable, and to say that X' is provable before X is equivalent to saying that X is refutable. And so, by Step 2, if X is provable, so is H(X0), and if X is refutable, so is H(X0), which means that H strongly separates Po from Ro! And so, if Z is normal and has the negation property, then Z is incomplete by Theorem 4.6. [In fact any fixed point X of H' is undecidable in Z.] 10 As we saw in Step 1 of the solution of the last problem, if X is provable by stage n, then Sn(Xo) is provable, and if X is not provable by stage n, c''
then Sn(XO) is refutable. The converse of the second fact also holds, that is, if Sn(XO) is refutable, then X is not provable by stage n, for if it were, then Sn(Xo) would be provable, hence both refutable and provable, contrary to
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our assumption that the system is consistent. And so we see that Sn(Xo) is refutable if and only if X is not provable by stage n. Since Yn is a fixed point of Sam,, then Y, is provable F-, Sn((Yn)o) is refutable. But also, as we have just shown, Sn((Yn)o) is refutable if Yn is not provable by stage n. And so Yn is provable if and only if it is not provable by stage n. Hence if Yn were not provable, it would be provable by stage n, which is absurd. Therefore Yn must be provable, but not by stage n. And so the status of Yn is that it is provable, but only at some stage later than stage n.
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This result has a very curious corollary. Consider now a consistent system satisfying G3 as well as G1, G2 and R1, R2. Then, since Parikh's sentence Yn is in fact provable, so is B((Yn)o). Now, in the systems studied by Godel and Rosser (such as Peano Arithmetic) the fact is that for fairly large n, the sentence B((Yn)o) can be proved at a much earlier stage than cwt
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stage n. And so B((Yn)o) can be proved before Yn! We thus have the curious situation that there are sentences X such that the sentence B(Xo) - which asserts that X is provable - is provable before X is provable! We might say that X is provably provable before it is provable. Thus an observer watching the output of the system would realize that there was a proof of X before knowing what the proof was!
Chapter 5
Self-reference in arithmetic
0
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We now turn to the study of self-reference in arithmetic couched in the language of first-order logic. For this chapter and Chapter 9, a prior acquaintance with the logical connectives and quantifiers, though not strictly necessary, is certainly desirable. Part I of this chapter is necessary for the incompleteness proof of Peano Arithmetic given in Chapter 9. The remaining parts of this chapter are not needed for any of the subsequent chapters, but should be of interest to those who wish to see some CAD
new and unusual methods of achieving self-reference in arithmetic, as well as a generalization that also generalizes Cantor's famous proof that a set cannot be put into a 1-1 correspondence with its power set.
Part I Normalization and Tarskification
(1)
In §1, following [27], we show how the idea of normalization, as defined in Chapter 1, can be carried out for first-order arithmetic with class abstracts added. Without these abstracts, self-reference can be achieved by Godel's method of diagonalization (which involves substitution), or by a simpler method due to Alfred Tarski, which we explain in §2. A still simpler (though bizarre) method is explained in Part II of this chapter.
§1 Normalization in arithmetic We will first consider a language G1 for arithmetic based on plus, times, and power (exponentiation) couched in first-order logic with identity. We will take - (negation) and D (material implication) for our primitive logical connectives and the universal quantifier V. We use the usual symbols "+" for plus, for times, and "E" for the exponential symbol (xEy will mean
Chapter 5. Self-reference in arithmetic
76
xv). We use the accent symbol' for the successor operation (thus x' means the successor of x, which is x + 1). For variables we will take the letter v followed by one or more subscripts. We let v,, be the expression consisting of v followed by n subscripts (e.g., v3 = viii). And so we use a finite alphabet - the following thirteen symbols: 0
'
-
D O V v i=+ E
We inductively define the notions of numeral, variable, term, formula,
free and bound occurrences of variables, designation, and truth by the following rules. (1)
For any natural number n, by n we mean the symbol 0 followed by
n accents and we refer to n as the numeral designating n. [e.g., 0"" is the numeral designating the number 4.] The symbol 0 itself designates 0.
(2) v alone or followed by a string of subscripts is a variable.
(3) Every numeral and variable is a term.
,.+
(4) If tl and t2 are terms, so are (tl +t2), (tl t2), (t1Et2), and ti. Terms containing no variables are called constant terms. For any numeral n, the constant term n designates the number n (as already stated) and if t1, t2 are constant terms that respectively designate numbers ni and n2i then the terms (tl + t2), (tl t2), (t1Et2) respectively designate ni plus n2, ni times n2, and nit (nl raised to the power n2), and t' designates ni + 1. We sometimes write tit for (t1Et2).
(5) If ti and t2 are terms, then tl = t2 is a formula, called an atomic formula, and all occurrences of variables in it are free. If it contains no variables, then it is called an atomic sentence and is called true if and only if tl and t2 designate the same number.
(6) For any formula Fl and F2, the expressions -F1 and (F1 D F2) are formulas, and for any variable vi, its free occurrences in -F1 are those of F1, and its free occurrences in (Fl D F2) are those of Fl and those of F2. If there are no free occurrences of any variable, then -Fl, (F1 D F2) are sentences, and -F1 is called true if Fl is not true, and (Fl D F2) is called true if either Fl is not true or F2 is true.
(7) For any formula F and any variable vi, the expression dv2F is a formula. All occurrences of v2 are bound (not free) in it and for any variable vj different from vi, the free occurrences of vj in dv2F are those of F. If dv2F has no free variables, it is called a sentence and is
§1 Normalization in arithmetic
77
called true if for every number n, the sentence F(n) - the result of substituting n for all free occurrences of v2 in F - is a true sentence. By the degree of a formula is meant the number of occurrences of the logical connectives (- and D) and the universal quantifier V. Let us note that our definition of truth for sentences is inductive, in that the truth of a sentence is defined in terms of truths of sentences of lower degree. Abbreviations
We use the following standard abbreviations (F, G are formulas; vi is a variable).
(1) (F V G) for (- F
G),
(2) (F A G) for -(F
-G),
(3) 3v2F for -Vv2-F. Substitution By a formula in v1, ... , vn we mean a formula in which V 1 ,- .. , vn are the only free variables. W e usually denote such a formula as ( V i , . ... , vn),
and for any numbers al, ... , an, by F(al,... , an) is meant the result of substituting the numerals a1, ... , an for all free occurrences vi, of- .. , vn in F respectively. In particular, F(vl) is a formula in v1 and F(n) is the
cad
result of substituting n for all free occurrences of v1 in F. Also, by a term in vl, ... , vn is meant a term in which-vi, . , vn are the only variables. We denote such a term t(vl,... , vn), and by t(a1i... , an) is meant the result of substituting in t, the numerals d1 ,. .. , an for v1, ... , vn respectively. [Of course in a term, all occurrences of variables are free.] Expressibility
We say that n satisfies F(vi) if F(n) is a true sentence. And we say that F(vi) expresses the set of all n that satisfy F. Thus F(vl) expresses the number set A if for all n the following condition holds:
F(n) is true iffnEA More generally, we say that F(vl,... , vn) expresses the set of all ntuples (al,... , an) such that F(al,... , an) is a true sentence. Thus the t,"
formula F(vl,... , vn) expresses a relation R(xl,... , xn) if for all numbers a1, ... , an the following condition holds
F(a1i... , an) is true - R(al,... , an)
Chapter 5. Self-reference in arithmetic
78
CAD
Following the terminology of [32] we call a set or relation Arithmetic (note the capital A) if it is expressed by some formula, and arithmetic (note the small a) if it is expressed by a formula in which the exponential symbol does not appear. Actually all Arithmetic sets and relations are arithmetic (as we will later see), but for the time being we will make a distinction. 'F.
We call a function f (xl, ... , xn) from n-tuples of natural numbers to natural numbers Arithmetic if the relation f (xl,... , xn) = y is Arithmetic. W e shall call the function term definable if there is a term t(vl, ... , vn)
such that for all numbers al,... , an, the term t(al, ... , an) designates f (al, ... , an) - and such a term t will be said to define f . Of course if t(vl, ... , vn) defines f , then the formula t(vl, ... , vn) = vn+l expresses the relation f (xl, ... , xn) = y, and so every term definable function is expressible in L1 (but the converse is not true). Let us note that if A is an Arithmetic set and f (x) is an Arithmetic function, then f -1(A) (the set of all n such that f (n) (E A) is also Arithmetic,
.ova
CAD
for suppose F(vi, V2) is a formula that expresses the relation f (x) = y and A(vi) is a formula expressing the set A and is such that the variable v2 does not occur in it. Let A(v2) be the result of substituting v2 for all free occurrences of vi in A(vi). Then f -1(A) is expressed by the formula Vv2 (F(vl, V2) D A(v2)). [Alternatively, it is expressed by the formula 3v2(F(vl, v2) A A(v2)).] We might note that if A is Arithmetic and f (x) is not only Arithmetic,
(D,
O0'
but term definable, there is then a simpler method of getting a formula to express f -'(A), namely, take a formula A(vi) expressing A and a term t(vl) defining f, and f-l(A) is then expressed by the formula A(t(vl)), i.e., the formula obtained by substituting the term t(vl) for all free occurrences of vi in A(vi). At any rate, we see
Fact A If A is Arithmetic and f (x) is Arithmetic, then f -'(A) is Arithmetic. Gbdel numbering
We assign Godel numbers to the individual symbols as follows. 0
1
1
0
,., 2
D 3
(
)
bv
4
5
6
7
l 89
=
+
899
8999
E 89999
899999
Then Godel numbers are assigned to compound expressions by replacing each symbol above by the Arabic numeral written under it and then reading the resulting Arabic numeral in ordinary base 10 notation, e.g., the Godel
number of the expression "-V =" is 26899. By a proper expression we
§1 Normalization in arithmetic
79
shall mean one that does not begin with an accent, unless it consists of the accent alone. Proper expressions include all terms and formulas and no two proper expressions have the same Godel number. [This is not true for
improper expressions, e.g., '- and "" both have the same Godel number 2 (since 002 = 02 = 2).] We now consider the following question: We let T1 be the set of Godel
numbers of the true sentences of G1 and we ask whether or not the set T1 is Arithmetic. That is, does there exist a formula F(vi) such that for any number n, F(n) is true if and only if n is the Godel number of a true sentence? Tarkki's famous theorem - which we will prove by two different methods - is that the set T1 is not Arithmetic. We will first prove a similar result for a minor extension Gi of the language G1i since the proof is much simpler and a cute modification of it will yield a totally new and simplified proof of Tarksi's theorem for the langauge L. Class abstracts
By a class abstract we shall mean an expression of the form (F), where F is a formula with vl as the only free variable. And now we add to the class of sentences of G1 all the expressions of the form (F) ft, where (F) is a class
abstract and n is a numeral. We define (F)n to be a true sentence if n satisfies F (i.e., if the sentence F(n) is true). [Informally, we read (F) as
F".
"the set of numbers that satisfy F," and we read the sentence (F)n as "the set of numbers that satisfy F contains n." The more usual notation for the sentence (F)n is n E {vl : F}. But we will use the simpler notation.] We have thus extended our class of sentences and the resulting language °++
we call L'. We now aim to prove that the set T1' of Godel numbers of the true sentences of Gi is not Arithmetic. To this end we can use the normalization operation of Chapter 1, which we cannot quite do for the language Gl (though we will do something very close to it in Part II of this chapter). Self-reference in Gi (DC
For any expression E of L, by #E we shall mean the Godel number of E. We define a sentence X of Gi to be a Godel sentence for a set A of (natural) numbers if the following condition holds.
Xistrue +--+ #XEA In a purely extensional sense, a Godel sentence for A can be thought of as a sentence asserting that its own Godel number is in A. [If the sentence is true, then its Godel number is in A; if the sentence is false, then its Godel
Chapter 5. Self-reference in arithmetic
80
s0.
number is not in A.] We are aiming to show that for every Arithmetic set A there is a Godel sentence (in Gi) for A, from which it will easily follow that the set T1' is not Arithmetic. We call two sentences equivalent if they are either both true or both false (not true). Let us note that if F(vi) is a formula that expresses a set A, then a Godel sentence for A is nothing more nor less than sentence X ,a)
vii
that is equivalent to F(#X) - in other words a fixed point of F(vl), in the sense of Chapter 1. Our method of constructing Godel sentences in L' is that of normalization in the sense of Chapter 1. For any expression E, by its norm is meant Ee (E followed by the numeral e), where e is the Godel number
relatively easy.
(moo'
of E. By a norm function we shall mean a function N(x) from natural numbers to natural numbers such that for any E with Godel number e, the number N(e) is the Godel number of its norm EE. We now need to show the existence of a norm function N(x) which is Arithmetic. This is
o'=
We first note that for any number n, the numeral n, like every other expression, has a Godel number. Under the Godel numbering we have chosen, the Godel number of n is simply 1On (e.g., 5 is the expression 0""', so its Godel number is 100000, which is 105). For any n that is the tea)
Godel number of some expression, we let En be the expression whose Godel
number is n. The Godel number of Enn consists of En followed by the symbol 0 followed by n accents, and so its Godel number is n 10n+1 + ion (e.g., the norm of E5 is E50""', whose Godel number is 5100000, which is 5000000 + 100000, which is 5 106 + 105). We thus define N(x) = x 10x+1 + 10x. The function N(x) is not only Arithmetic but even term definable! Now we can prove tai
Theorem 1.1 If A is Arithmetic then there is a Godel sentence for A. 'C3
cam,
.,,
Proof Suppose A is Arithmetic and is expressed by A(vi). Since the .`3
am.
function N(x) is Arithmetic, then by Fact A the set N-1(A) is Arithmetic. Let F be a formula in v1 that expresses N-1(A) and let H be the abstract (F) and let h be the Godel number of H. Then Hh is true F-, F(h) is true
HhEN-1(A)HN(h)EA. ThusHh is true
N(h) e A, but N(h) is
the Godel number of Hh, and so Hh is a Godel sentence for A. Tarski's theorem
gad
For any number set A, by its complement A is meant the set of all numbers not in A. Obviously the complement of every Arithmetic set is Arithmetic,
§2 Tarski's method of achieving self-reference
81
CAD
°i'
because if F(vi) expresses A, then its negation -F(vi) expresses A. We are letting T1' be the set of Godel numbers of the true sentences of L'. There obviously cannot be a Godel sentence for the complement T1' of T, since such a sentence would be true if and only if it were not true (the Godel number of a sentence X is in T1' if X is not true). Therefore, by Theorem 1.1, the set T1' is not Arithmetic, and hence T1' cannot be Arithmetic. And so we have
Theorem 1.2 [After Tarski] The set T1' of Godel numbers of the true sentences of Gi is not Arithmetic.
§2 Tarski's method of achieving self-reference Now let's consider the language G1 in which we don't have class abstracts. Then the method of constructing Godel sentences by using the norm func-
tion N(x) won't work. One way of constructing Godel sentences is by Godel's diagonalization, which is this: call a function d(x) a diagonal func-
tion for G1 if for every n that is the Godel number of a formula F(vl), the number d(n) is the Godel number of F(n). Now, there is indeed an Arithmetic diagonal function d(x), but the proof of this is quite elaborate (it involves the Arithmetization of the operation of substituting numerals for variables in formulas). Anyway, once we get an Arithmetic diagonal function d(x), then for any Arithmetic set A, the set d-1(A) is then Arithmetic, and if F(vi) is a formula that expresses d-1(A), its diagonalization F(a) (where a is the Godel number of F) is a Godel sentence for A (since F(a) is true if a E d-1(A) H d(a) E A, and d(a) is the Godel number of F(a)). This method is perfectly feasible, but as I have indicated, it is far ?;.
p..
more complex than need be, due to the complexities involved in proving the existence of an Arithmetic diagonal function. Fortunately, Tarski devised a much simpler method, which is based on the following idea. Informally, to say that a given property P holds for a given number n is equivalent to .g2
IC'
saying that for every number x equal to n, P holds for x. Formally, given any formula F(vi) with vi as the only free variable, the sentence F(n) is equivalent to the sentence Vvl(vl = n F(v1)). [Incidentally, it is also equivalent to the sentence Ivl(vl = n n F(vl)).] The whole point now is that it is a relatively simple matter to show that the Godel number of the sentence Vvl(vl = n D F(vi)) is an Arithmetic function of the Godel number of F(vi) and the number n (far simpler than showing that the Godel number of F(n) is an Arithmetic function of the Godel number of F(vi) and the number n). For any formula F(vi) and any n, we henceforth write F[n] to mean
Chapter 5. Self-reference in arithmetic
82
car
::r
cad
the sentence Vvl(vl = n D F(n)). [Notice the square brackets!] To repeat an important point, the sentences F(n) and F[n], though clearly not the same, are equivalent sentences (both true or both false). As a matter of fact, for any expression E, whether a formula or not, the expression Vvl (vl = n D E) is a perfectly well defined expression (though a meaningless one, if E is not a formula), and we shall denote this (possibly
meaningless) expression E[n]. If E is a formula with vl as the only free variable, then E[n] is, of course, a sentence, but in all cases, E[n] is a 0..
(o+
well defined expression. We might define the Tarskification of E as E[e], coy
where e is the Godel number of E. Thus the Tarskification of E is the expression Vvl(vl = e D E). What is now needed is to show that there is ,..h
an Arithmetic function t(x) (which we might call a Tarskification function)
X.-
such that for any expression E with Godel number e, t(e) is the Godel .0,
number of its Tarskification E[E]. To prove this, several preliminaries are necessary. For any numbers m and n, we define m * n to be the number which when written in base 10 notation consists of n (in base 10 notation) followed by m (in base 10 notation). For example 53 * 792 = 53792. We note that 53792 = 53000 + 792, which is 53. 103 + 792. We also note that apt
a-0
3 is the number of digits of 792 (when written in base 10 notation), or as we'll say, 3 is the length of 792 (in base 10 notation). More generally, -x-
x * y = X. 10L(y) + y, where L(y) is the length of y (in base 10 notation).
Our present aim is to show that the function x * y is Arithmetic. [The c0+
significance of this function is that if x, y are respective Godel numbers of Ex and Ey, then x * y is the Godel number of ExEy, i.e., of Ex followed by Ey.] To show that the function x * y is Arithmetic we must first show
car
Q..
that the function 10L() is Arithmetic. Now, for any x 0, 10L() is the smallest power of 10 greater than x. We let powlo (x) be the condition that x is a power of 10. This condition is Arithmetic, because it holds if Iy(x = 10y). [More formally, the set of powers of 10 is expressed by the formula ]1v2(v1 = (10Ev2)). We shall not be this formal from now on!] Next, the relations x < y, x < y are Arithmetic, for the former is simply Iz(x + z = y) and the latter is x < y n ' (x = y). Now let s(x, y) be the relation that y is the smallest power of 10 greater than x. It is Arithmetic, since it can be written (powlo(y) Ax < y) A Vz ((powlo (z) Ax < z) y < z). Then the condition 10L(x) = y is equivalent to the condition
(x = 0 A y = 10) V (x # 0 A s(x, y)), hence is Arithmetic. Finally, the relation X. 10L(y) + y = z (which is x * y = z) is Arithmetic, for it is the condition 2z12z2(10L(y) = zl n x zl = z2 A z2 + y = z). And so we have proved
Proposition 2.1 The relation x * y = z is Arithmetic.
§2 Tarski's method of achieving self-reference
83
As a corollary we have
Proposition 2.2 For any n > 2, the relation x1 * ... * xn = y (as a relation of n + 1 arguments among x1, ... ) xn, y) is Arithmetic.
Proof By induction on n > 2. We have already proved this for n = 2. Now suppose n > 2 is a number such that the relation x1 * ... * xn = y is Arithmetic. Then the relation xl * ... * xn * xn+1 = y is Arithmetic, since it holds if 2z(xl * ... * xn = z A z * xn+1 = y).
Next, we wish to have an Arithmetic function r(x, y) such that for any expression Ex with Godel number x and any number y, r(x, y) is the
Godel number of Ex [y] - which is bvl (vl = y D Ex). Well, let k be the Godel number of the expression "bv1(vl =" (this number happens to be 67894899). Then, since the Godel number of y is 10y, and the Godel
number of ":)" is 3, and the Godel number of Ex is x, and the Godel number of the right parenthesis is 5, the Godel number of Vvl (vi = y D Ex)
isk*10y*3*x*5. And so we taker(x,y)tobek*10y*3*x*5. This function is Arithmetic, since r(x, y) = z if 2w(10Y = wAk*w*3*x*5 = z). We then take t(x) = r(x, x) and we have:
Proposition 2.3 (1) There is an Arithmetic function r(x, y) such that for any Godel number x and any number y, r(x,y) is the Godel number of Ex[y]. (2) There is an Arithmetic function t(x) such that for any Godel number x, t(x) is the Godel number of Ex [x].
Though the function t(x) is Arithmetic, it is not, like the norm function N(x), term definable. Nevertheless, by Fact A of §1, for any Arithmetic set A, the set t-1(A) is Arithmetic. So there is a formula F(vl) expressing t-1(A) and its Tarskification F[a] (a is the Godel number of F(vi)) is a Godel sentence for A, since F[a] is true H F(a) is true +-4a E t-1(A) H t(a) E A, and t(a) is the Godel number of F(a). And so we have
Theorem 2.4 For every Arithmetic set A there is a Godel sentence for A in the language L.
Chapter 5. Self-reference in arithmetic
84
Remark
A much simpler method of constructing Godel sentences for Gl will be presented in Part II of this chapter. It follows from Theorem 2.4 that the complement T1 of the set T1 of true sentences of G1 is not Arithmetic, and hence T1 is not Arithmetic, and so we have
Theorem T1
(Tarksi's theorem for G1) The set T1 of Godel numbers of the true sentences of G1 is not expressible in L.
§3 The more general situation To what extent does Tarski's theorem for G1 depend on the particular Godel numbering that we used? Well, the two properties of the Godel numbering
that played the crucial role in the proof are: (i) the fact that there is an Arithmetic function c(x, y) (in this case x * y) such that c(x, y) is the Godel number of ExEy; (ii) there is an Arithmetic function num(x) (in this case 10x) such that num(x) is the Godel number of x. For any other Godel numbering with those two properties, the proof would go through. A further generliazation is possible. To what extent does the success of the proof of Theorem T1 depend on our choice of plus, times, and power as our arithmetical primitives? Well, in the first place, we will see in Chapter 8 that the exponential relation xy = z is arithmetic (note the small a!), i.e., definable from plus and times, and so we can delete the exponential symbol
from G1 and the proof of Tarski's theorem for the resulting language -
00O
CAD
''Y
which we will call Lo - goes through. [Actually there is a simpler proof of this that we will give in Chapter 9.] There are also many other choices of arithmetic primitives that could be taken and the proof would go through. Also, there are many schemes other than 0, 0, 0",... of naming the natural numbers (for example, ordinary decimal notation, or binary notation) that would work. car
And so let us now turn to a very general situation: Let L be a first004
order language - a language in which we have adequate logical connectives (enough to do propositional logic) and one or both of the quantifiers b, ], ((D
and which contains instead of +, , and E, arbitrary function symbols, each of which is interpreted as some function from natural numbers, or n-tuples of natural numbers, to natural numbers. The language might also have relational symbols, each of which is interpreted as some relation of natural numbers. In addition, there is associated with each natural number n, a certain term n, called the numeral for n. Call this formalism,
§4 Near normalization
85
'C3
together with the interpretation of the function and relation symbols, G. Since the functions and relation symbols are interpreted, every sentence is either true or false under the intepretation, and we let T be the set of true sentences of G. Once we have the notion of truth in G, then we have the notion of expressiblity in L - a formula F(v1, . . . , vn) expresses in G the set of all n-tuples (al, ... , an) of numbers such that F(a1, ... , an) is true in G. And a function f (x1, ... , xn) is called expressible in G if the relation f (x1i ... , xn) = y is expressible. Fact A that we proved for G1 has the following obvious generalization: if A is a set expressible in G and f (x) is a function expressible in G, then the set f-1 (A) is expressible in G (and again, if A(vi) expresses A and F(v1, v2) expresses the relation f (x) = y, then the formula bv2(F(v1,v2) D A(v2)) expresses f-1(A)). Next, consider an arbitrary Godel numbering g that assigns to every expression X of G a number g(X) that we will call the g-number of X. We will say that the Godel numbering g is adequate (for L) if: (i) there is a function c(x, y) expressible in L such that for any expressions Xn, X,n with respective g-numbers n and m, the number c(n, m) is the g-number of XnX,; (ii) there is a function num(x) such that for every number x, the number num(x) is the g-number of the numeral x. Then Theorem T1 .'S
P''
SAC
c34
holds in the following more general form.
Theorem T [Tarski's theorem] If g is any adequate Godel numbering of the expressions of G, the set of g-numbers of the true sentences of G is not expressible in G.
The proof of Theorem T is an obvious generalization of the proof of Theorem T1 and is left as an exercise. [Actually, it is a consequence of Theorem 2.2, Chapter 4.]
Exercise 0 How is Theorem T above derivable from Theorem 2.2 of Chapter 4?
Part II Some special devices As we have already remarked, the remainder of this chapter is not necessary for any subsequent chapters; the results are purely of independent interest.
§4 Near normalization We now turn to a new method of constructing Godel sentences in the language G1, which, despite its bizarre character, is really quite workable.
Chapter 5. Self-reference in arithmetic
86
'.7
.'S
A"'
'-'
.,,
Tarski's method of achieving self-reference is simpler than Godel's method, which involves arithmetizing substitution, but it is less simple than the method of normalization. However, the method of normalization, though applicable to L', in which we have class abstracts, is not applicable to languages like L1, in which we don't have abstracts. However, we have hit on another method which is almost as simple as the method of normalization and is moreover applicable to languages like G1 which don't have class abstracts. We call this method near normalization. We recall that the norm of an expression E with Godel number e is Ee. We now define the near norm of E to be the norm of E followed by the right parenthesis. Thus the near norm of E is EE). The whole point now is that for any formula F(vi) there is an expression E (not a formula, to be sure, but an expression nevertheless) whose near norm EE) is a sentence equivalent to F(e) - in fact, for any number n, En) is a sentence equivalent to F(n). We simply take E to be the expression Vvi(-F(vl) D Vi =. Then En) is the sentence bvl(-F(vl) -vi = n), which is equivalent to Vvi(vi = n D F(vi)), which is the sentence F[ii], which is equivalent to F(n). In particular, if e is the Godel number of E, then the near norm Ee) of E is equivalent to F(e). And so we have C01
IC'
(Ds
Fact 1 For every formula F(vl) there is an expression E whose near norm is a sentence equivalent to F(e), where e is the Godel number of E.
LCD
.,,
.Or,
coo
We now call a function M(x) a near norm function if for every expression E with Godel number e, M(e) is the Godel number of the near norm Ee) of E. What we now need is to show the existence of an Arithmetic near norm function. This is considerably simpler than showing the existence of a Tarskification function, and is indeed about as simple as getting an Arithmetic norm function. Since 5 is the Godel number of the right parenthesis, then for any expression X with Godel number x, the Godel number of X) is 10x + 5. And so if N(x) is a norm function, then 10 N(x) + 5 is a near norm function. Now, x 10x+1 + 10x is a norm function, and so x . 10x+2 + 10x+1 + 5 is a near norm function. We thus take M(x) = x . 10x+2 + 10x+1 + 5, or 10x+1 (10x + 1) + 5.
Our near norm function M(x) is thus not only Arithmetic, but even cad
term definable (like our norm function, and unlike the Tarskification function), and so if t(vl) is a term defining it, then for any Arithmetic set A, if
F(vi) expresses A, then F(t(vi)) expresses M-1(A), and so a Godel sentence for A is Ee), where E is the expression Vvl(-F(t(vl)) D -vi = and e is the Godel number of E. [Reason: Ee) is equivalent to F(t(e)), which
'T1
§5 Situation in Polish notation
87
is true if e E M-1(A), if M(e) E A, but M(e) is the Godel number of EE).]
And so all the arithmetization necessary to obtain self-reference in L1 is given by the simple function 10'+1(10x + 1) + 5. [We don't even need to first show that the function x * y is Arithmetic!] Some observations
Let us say that an expression E covers a formula F(vi) if for every n, En) is a sentence equivalent to F(n). We have seen that one expression that covers F(vi) is Vv1(-F(v1) w1 =. Another is -Vv1(F(v1) w1 =. If instead of taking -, D, V as our logical primitives, we had taken ", n, and 2, then just using n and 3 we could construct an expression that covers F(vi), namely 2v1(F(v1) A v1 =. And so if A is Arithmetic and F(vi) is a formula that expresses the Arithmetic set M-1 (A), then a Godel sentence
for A would be 2v1(F(v1) A v1 = h), where h is the Godel number of 2v1(F(v1) A v1 =.
It appears that whether or not near normalization works depends on what logical primitives are taken. For example, if we took joint denial .. as a sole logical connective (pI q means that p and q are both false) and if we took either V or 3 as our quantifier, we see no way in which near normalization could give Godel sentences. The same appears to be true using Sheffer's stroke I instead of 1 (p I q means that p and q are not both true). However, we have checked that if we take - as one of our logical connectives, and if we take any one of the three standard connectives D, V, n as primitive, and either of the two quantifiers V, 3, then in each of these six
cases there is some E that covers F(v1), and so near normalization then yields Godel sentences.
Exercise 1 In each of these six cases, find an E that covers F(vi). [Solutions to exercises are given at the end of the chapter.]
§5 Situation in Polish notation If we had formulated the language L1 in Polish notation instead of the more usual notation using parentheses, we could have got by with the norm function instead of the near norm function!' In Polish notation, we delete the parentheses from our alphabet and the rules for constructing terms and formulas are the following.
(1) Every numeral and variable is a term. 'For this neat observation. the author is indebted to J. Michael Dunn.
Chapter 5. Self-reference in arithmetic
88
(2) If t1 and t2 are terms, so are +t1t2, t1t2, Et1t2, and ti. (3) For any terms ti and t2, = t1t2 is an atomic formula - also a formula.
(4) For any formulae F and G and any variable v2, the expressions -F, D FG and dv2F are formulae.
In Polish notation, as in standard notation, if G is a formula in just the free variable vi, by G(n) is meant the result of substituting n for all free occurrences of v1 in G. The point to observe now is that for any such formula G, there is an expression E such that for any number n, En is a sentence equivalent to G(n), namely, we take E to be Vv1 D -G- = v1. [We note that the translation of Vvi D -G- = vin into standard notation
is Vvi(-G' D -vi = n), where G' is the translation of G.] And so if A is an arithmetic set so is N-1(A) (N(x) is the norm function) and if G is a formula in Polish notation that expresses N-1(A), then a Godel sentence for A is Vv1 D -G- = vin, where n is the Godel number of Vv1 D -G- = v1. If we took -, n, and as our logical primitives, then the Godel sentence would be 3v1 n G = vin, where n is the Godel number of 3v1 n G = v1. The interesting thing, then, about Polish notation is that we have the power of class abstracts, and so normalization is applicable.
§6 More on quotational systems Next, I would like to look at Tarskification and near normalization in the context of quotational designation instead of Godel numbering. So let us go back to quotation as studied in Chapter 1. Tarskification
We shall use the eight symbols J, t, D, V, x, as well as opening and closing quotes. The letter x will be used as a variable and by the Tarksification of an expression E we shall mean the expression Vx(x = "E" D E). As in Chapter 1, we use the symbol "J" to abbreviate "John is reading" and now we use "t" to abbreviate "the Tarskification of." Let us now consider the following expression.
(1) Jx The Tarskification of (1) is
(2) Vx(x = "Jx" D Jx)
§6 More on quotational systems
89
The sentence (2) says that John is reading anything that is identical
to (1) - in other words, that John is reading (1). And so (2) is not self-referential. Now, instead of (1), consider
(3) Jtx Its Tarskification is (4) Vx(x = "Jtx" D Jtx)
Sentence (4) asserts that John is reading the Tarskification of anything that is identical to (3) - in other words, that John is reading the Tarskification of (3), but the Tarskification of (3) is (4). And so (4) asserts that John is reading (4), and is thus self-referential.
Exercise 2 Suppose we add to our eight symbols the symbol "A" (for conjunction) and "P" (for "Paul is reading"). Write down a sentence that asserts that both John and Paul are reading it.
Exercise 3 Do the same using the existential quantifier 2 in place of V, redefining the Tarskification of E to be 2vl(vl = "E" A E). Near normalization mot
Let us now look at near normalization in the context of quotation. Let us use the symbols J, P, M, 3, A, =,x, (1), and opening and closing quotation marks. The symbol "M" abbreviates "the near norm of," where by the near norm of an expression E we mean E "E") (the norm of E followed by the right parenthesis).
Exercise 4 In this notation, write a sentence that asserts that John is reading it. cad
Exercise 5 Do the same using V instead of I and 9 and - instead of A
and -. Polish notation, normalization, and quotation via
Let us use the symbols J, P, N, 2, A, =, x, as well as opening and closing quotation marks. The symbol "N" abbreviates "the norm of," as formerly. Of course a sentence that asserts that John is reading it is our old friend JN "JN". What about a sentence that asserts that John and Paul are both reading it? [The sentence would have to be in Polish notation, of course.
Chapter 5. Self-reference in arithmetic
90
Exercise 6 (a) Find such a sentence.
(b) Do the same, using the logical primitives ' , D, b instead of 11, A.
§7 Some variants Here is another method for achieving self-reference, which is a variant of near normalization, and is applicable to systems without class abstracts and with parentheses. We will first illustrate it at the quotational level. We use the symbols J, L, D, V, _, (, ), x (the same symbols we used for Tarskification, only with "L" instead of "t"). We define the display of an expression E to be Vx(x = "E"E. [Again, we are also using opening and closing quotes.] The symbol "L" is to abbreviate "the display of." In this language, there is a sentence that asserts that John is reading it.
Exercise 7 What expression E is such that its display is a sentence X that asserts that John is reading X? Now let's look at this method in terms of the arithmetical language L. For any Godel number n, define the display of En as the expression bvl (vl = nEn. The Godel number of this display is k * 10' * n, where k is again the Godel number of Vvl(vl =. We let L(x) = k * 10x * x, and so for any Godel number n, L(n) is the Godel number of the display of En. Of course the function L(x) is Arithmetic (though not term definable), and so for any Arithmetic set A, the set L-1(A) is Arithmetic.
Exercise 8 (a) Show that for every formula F in v1 there is an expression E whose display is a sentence equivalent to F(e), where e is the Godel number of E.
(b) Given an Arithmetic set A, and a formula F(vi) expressing A, how would you go about getting an expression E whose display is a Godel sentence for A?
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We remark that this method of constructing Godel sentences is less simple than near normalization (which doesn't even involve showing that the
§8 Godelizing operations
91
function x * y is Arithmetic!), and is only slightly simpler than Tarskification. Its interest lies in the fact that it has a certain feature in common with Tarskification, near normalization, normalization, and diagonalization that we will look at in Part III of this chapter. Two other methods for achieving self-reference that have this same common feature are presented in the two exercises below.
Exercise 9 (a) For quotational languages, define the splash of an expression E to be
the expression E 3 -x = "E"). Let "S" abbreviate "the splash of." Then, using the symbols J, S, 3, -, b, x, =, (,), and quotation marks, find an expression E whose splash is a sentence X that asserts that John is reading X. ti.
(b) For the Arithmetic language G1i define the splash of an expression E =-'
to be E D -vl = e), where e is the Godel number of E. Show that for any Arithmetic set A there is an expression E whose splash is a Godel sentence for A.
Exercise 10 For quotational languages, define the presentation of an ex(1.
pression E to be E = "E", and for L1i define the presentation of E to be E = e), where e is the Godel number of E.
(1)
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(a) Using "P" to abbreviate "the presentation of" and the symbols J, D , V, x, =, (, ), and the quotation marks, construct an expression E whose presentation is a sentence that asserts that John is reading it. cad
(3)
(b) Show that for any Arithmetic set A there is some expression E of G1 whose presentation is a Godel sentence for A.
Part III Generalizations §8 Godelizing operations Now, what significant features do all those methods of constructing Godel
sentences have in common? To begin with, instead of considering the particular language C1i let us consider an arbitrary language L of the type considered at the end of Part I, together with some Godel numbering g of the expressions of L. A Godel sentence for a set A will now be understood
as relative to g, i.e., a sentence that is true in L if its g-number is in
Chapter 5. Self-reference in arithmetic
92
`Or
`-'
'-'
A. Now let us consider an operation 4) that assigns to each expression E of G an expression 4)(E) of G. Let us call it a Godelizing operation (with respect to g understood) if for every expressible set A, there is an
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expression E such that 4D(E) is a Godel sentence for A. As we have seen, the operations of diagonalization, Tarskification, near normalization, as well as the operations of taking the display, splash, presentation of E are all Godelizing operations for G1 (and so is normalization for Gi, or for 'Cl in Polish notation). And so we wish to see what significant feature these operations have in common which makes them Godelizing operations. Let us say that D is pre-Godelizing if for every formula F(v1) there is an expression E with g-number e such that 4)(E) is a sentence equivalent to F(e). Stated otherwise, D is a Godelizing operation if for every set A expressible in G there is some expression E such that 4)(E) is a sentence, and is true if the g-number of E (not of 4)(E)!) is in A. cad
Next, let us say that a function f (x) from numbers to numbers is admissible in G if for every set A expressible in G, the set f -1(A) is expressible in G. [We saw in the last chapter that if the function f (x) is expressible in G, then it is admissible in G.] Now, let us say that a function O(x) from numbers to numbers is a representative of an operation 4) from expressions to expressions (a representative with respect to g understood), if for every
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expression E with Godel number e, O(e) is the Godel number of 1(E). Finally, let us say that an operator 1 is admissible in G if it has at least
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and admissible (as can be seen from inspection). The fact that they are all Godelizing operations is then a consequence of the following simple
Theorem 8.1 If 4D is pre-Godelizing and admissible, then 4D is Godelizing.
Although the above theorem is but a special case of something more general that we will shortly consider, let us look at its proof anyway. Let 0 be an admissible representative of T. Take any set A expressible in G. Then 0-1(A) is expressible in G. Then there is some expression E
with g-number e such that 1(E) is a sentence and is true if e E 0-1(A), if O(e) E A, but q(e) is the g-number of -D(E), and so -D(E) is a Godel sentence for A (with respect to g).
§9 Generalized diagonalization
93
§9 Generalized diagonalization Now we shall consider a far more general situation. We consider a set N, a subset G of N, a subset S of G, a subset T of S, and a collection C of subsets of N. Let S be the quintuple (N, G, S, T, C). A function '(x) from N into N will be called a generalized diagonal function for S if for every
`-'
set A in C there is at least one element e in G such that V(e) E S and also ?P(e) E T F-, e E A. A function f (x) from N into N will be called admissible (with respect to S) if for every set A in C, the set f -1(A) E C. --1
The following theorem generalizes Theorem 8.1 (as well as Cantor's theorem
that the power set of a set N cannot be put into a 1-11 correspondence with N).
Theorem 9.1 If 0 is an admissible generalized diagonal function for S, then for any set A in C there is an element e in G such that
O(e)ET"O(e)EA
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Proof Given A in C, the set-1(A) E C (since is admissible), hence there is some e in G such that ''(e) E S and also ''(e) E T " e E'-1(A).
But eEV-1(A)*V(e)EA,and so O(e)ETH?P(e)EA. That Theorem 8.1 is a special case of Theorem 9.1 can be seen as follows:
Given the language L and a Godel numbering g of the expressions of L, we let N be the set of natural numbers, G the set of Godel numbers of the expressions, S the set of Godel numbers of the sentences, T the set of Godel numbers of the true sentences and C the collection of sets expressible in L. Suppose now that' is a pre-Godelizing operation for L and that ?P(x) is a representative of T. Then the function qp(x) is a generalized diagonal function for the quintuple (N, G, S, T, C). If, furthermore, %F is admissible,
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then 'i is admissible and so by Theorem 9.1, for any set A in C, there is some element e of G such that O(e) E S, and also ?P(e) E T iff''(e) E A. Then e is the Godel number of some expression E and V(e) is the Godel number of W(E), and so W(E) is a sentence (since b(e) E S) and W(E) is true F-,''(e) E T H'' (e) E A. Thus %F (E) is true if its Godel number 0(e) is in A, which means that W(E) is a Godel sentence for A. Thus Theorem
-4.
8.1 is a special case of Theorem 9.1. Let us now make the following observations. Call a function f (x) a strict diagonal function for L if for every formula F(vi) with Godel number n, f (n) is the Godel number of F(n). [Godel's diagonal function is thus a strict diagonal function.] Call f (x) a near diagonal function if for every
formula F(vi) with Godel number n, f (n) is the Godel number of some
Chapter 5. Self-reference in arithmetic
94
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sentence equivalent to F(n) (but not necessarily F(n) itself). The Tarski function t(x) that we considered is an example of a near diagonal function that is not a strict diagonal function, since if n is the Godel number of F(vi), then t(n) is the Godel number, not of F(n), but of the equivalent sentence F[n] (the sentence Vvl(vl = n D F(vi))). Next, let us call a function q(x) a quasi-diagonal function for G if for every formula F(vi) there is some number n (but not necessarily the gnumber of F(vi)) such that q(n) is the Godel number of a sentence that is equivalent to F(n). The function M(x) (a representative of the near norm operation) is an example of a quasi-diagonal function that is not even a
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near diagonal function. Indeed, if %P is any pre-Godelizing operation, any representative 0 of T is a quasi-diagonal function. Of course every strict diagonal function is also a near diagonal function and every near diagonal function is also a quasi-diagonal function, but the converses of these two facts are not true. The interesting thing now is that for the construction of Godel sentences in G, we don't need an admissible strict diagonal function (which Godel used), nor even an admissible near diagonal function (which Tarski used), but an admissible quasi-diagonal function suffices.
Exercise 11 How is Cantor's theorem a special case of Theorem 9.1?
Solutions to exercises 1
Here are the cases (some of them have more than one solution). V) (Using (Using 3) n, V) (Using (Using n, 3) V, V) (Using (Using ' , V, 3)
VV,(-F(vi) I _vl =
3vi' (F(vi) j -vi = _Vvi-(F(vl n vl = 3vl(F(vl) n vl = Vvl (F(vl) V -v1 =
'vi'-(F(vi) V -vi =
2
Vx(x = "(Jtx n Ptx)" D (Jtx A Ptx))
3
2x(x = "(Jtx n Ptx)" A (Jtx A Ptx))
Such a sentence is the near norm of 2x(JMxnx =, which is 3x(JMxn x = " 9x(JMx n x ="). 4
§9 Generalized diagonalization
95
Such a sentence is the near norm of Vx(-JMx I -x =, which is b'x(-JMx ij 'x = "Vx(-JMx ID 'x ="). 5
6 (a) Such a sentence is the norm of ]x n AJNxPNx = x, which is ]x n AJNxPNx = x "2x n AJNxPNx = x".
(b) Take the norm of Vx » JNx-PNx- = x. 7 Take E to be ID JLx. Its display is Vx(x = "ID JLx)" ID JLx).
8 (a) Take E to be ID F).
(b) For any Godel number n, the display of En, is Vvi(vi = nEn), and the Godel number of this display is c * 10n * n, where c is the Godel number of Vv1(vl =. We let L(x) = c * 10x * x, and so for any Godel 4.1
number n, L(n) is the Godel number of the display of En. Clearly L(x) is an Arithmetic function, so given a formula F(vi) expressing an Arithmetic set A we can get a formula G in v1 that expresses L-1(A). Its display is then a Godel sentence for A.
9 (a) We take E to be Vx(-JSx. Its splash is the sentence Vx(-JSx ID
-x = "Vx(-JSx"). (b) We leave it to the reader to verify there is an Arithmetic function s(x)
such that for any expression En, s(n) is the Godel number of the splash of En. Then for any formula G in v1 that expresses s-1(A), the splash of G is a Godel sentence for A.
10 (a) Vx(-JPx ID -x = (b) Left to the reader.
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11 Given a set N and a function assigning to each element x of N some subset Ax of N, we are to show that some subset of N is different from each of the sets Ax. We let G = S = N, and we let T be the set of all x in N such that x E Ax and we let C be the collection of all the sets Ax. It is trivial that the identity function I (x) on N is an admissible generalized
-0-
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diagonal function for the quintuple (N, G, S, T, C), and so by Theorem 9.1, for any set A E C, there is some element x such that I (x) E T F-, I (x) E A,
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thus x E T H x E A, which is but another way of saying that T (the complement of T with respect to N) is not in C. Thus C lacks the set T. Of course this is a ridiculously roundabout way of proving Cantor's theorem, but it should be of some interest in that it and Theorem 7.1 are
96
Chapter 5. Self-reference in arithmetic
both obtainable as special cases of Theorem 9.1. It has long been felt that Godel's argument and Cantor's argument have a common source.
Chapter 6
Introduction to formal systems and recursion
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This and the next two chapters are written primarily, but not exclusively, for those readers with no prior knowledge of recursive function theory. We develop those elementary portions of the theory that will be needed for the remaining chapters. Our approach to the theory is via elementary formal systems, as defined in T.F.S. (Theory of Formal Systems). These chapters might also be of interest to readers who are familiar with recursion theory, but not with elementary formal systems, since these systems will be seen to facilitate a neat and simple proof of Godel's incompleteness theorem for first-order arithmetic. Also, readers who are familiar with T.F.S. might find Chapter 8 of interest in that it contains an improved method of arithmetizing metamathematics.
Part I Formal representability §1 Elementary formal systems Elementary formal systems, as we will define them, provide a direct characterization of recursive enumerability for sets (and relations) of formal expressions without recourse to Godel numbers. Then by fixing some notation for natural numbers, a definition of recursive enumerability for sets and relations of numbers follows. The purpose of elementary formal systems is to explicate the notion of "definability by recursion." We refer to this type of definition, frequent
in the literature, which instead of defining a set or relation W outright, defines it implicitly by first giving a set of axioms for membership in W,
Chapter 6. Introduction to formal systems and recursion
98
then certain axioms of the form "if such and such and so and so are in W, then such and such a combination of them is also in W." Then the
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final "recursion" clause: "Furthermore, nothing is in W unless its being so follows from the preceding axioms." Now, it is the last clause that must be made precise. One might ask: "follows from in what logic?" Elementary formal systems provide one such type of logic. Let us consider a simple example. Let K be the alphabet consisting of just two symbols, a and b. Suppose we wish to define the set S of all strings of the two signs a and b which alternate with a and b, i.e., which contain no two consecutive occurrences of a or b. [Examples of alternating strings are a, ab, aba, babab.] We can implicitly define the set S by stipulating the following axioms for membership in S. p.,
Axiom 1: a is an element of S. Axiom 2: b is an element of S. Axiom 3: ab is an element of S. Axiom 4: ba is an element of S. Axiom 5: if xa is an element of S, where x is any string in the symbols a and b, then xab is an element of S.
Axiom 6: If xb is an element of S, then xba is an element of S. --I
*Axiom 7: [The recursion clause] Furthermore, no element is in S unless its being so follows from Axioms 1-6. We now convert the above informal axiom system into a precise formal
system (E) - an example of an elementary formal' system. We first abbreviate any phrase such as "x is an element of S" by "x E S," or, more simply still, "Sx" (which we read "S contains x"). Then we take a symbol -* for material implication and so any such phrase as "if (-) then (. .)" is written "(-) --> (. . .)." We now rewrite Axioms 1-7 in the following symbolic form and take them as formal axioms of our system (E). Al - Sa A2 - Sb
A3 - Sab A4 - Sba
A5 - Sxa -* Sxab
§ 1 Elementary formal systems
99
A6 - Sxb -> Sxba Now, instead of Axiom 7 (the recursion clause) we shall give a precise definition of "follows from."
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By an instance of an axiom of (E) we shall mean the result of substituting for the variable x any string composed of the symbols a and b. [For example, Sbab -* Sbaba is an instance of A6, substituting ba for x. If the axiom does not contain the variable x (i.e., if it is not A1, A2, A3, or A4), then its only instance is itself.] And now we shall define an expression in the symbols a and b to be initially provable in (E) if it is an instance of an axiom of (E). Thus A1i A2, A3, and A4 are initially provable in (E), and so are all instances of A5 and A6, and these are the only initially provable expressions of (E). Finally, we define an expression to be provable in (E) if it is either initially provable, or is derivable from the set of initially provable expressions by a finite number of applications of the following rule:
Rule Do [A rule of detachment] From expressions X and X -* Y, the expression Y may be inferred. Thus our definition of "provable" has the following consequences.
(1) Every instance of an axiom is provable.
(2) If X and X -> Y are provable, so is Y.
(3) The only way that Y can be provable is that either Y is an instance of an axiom, or there is some X such that X and X -* Y are both provable.
S."
Let us look at some expressions that are provable in this system. Obviously, the four strings Sa, Sb, Sab, and Sba are provable (they are initially provable). Since Sab is provable and Sab -> Saba is provable (it is an instance of A5), then Saba is provable (by Rule Do). But also Saba -> Sabab is provable (it is an instance of A6), and so Sababa is provable (again by Rule Do). Then we can successively prove Sababa, Sababab, and so forth. Also, since Sba is provable, we can successively prove Sba, Sbab, Sbaba, and so forth. And since So and Sb are provable, we see that for any alternating string X, the expression SX is provable. [A more rigorous demonstration of this is by mathematical induction on the length (number of symbols) of "dam
X.]
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Next, we wish to show that, conversely, if SX is provable, where X is a string in the symbols a and b, then X must be alternating. To this end it will be useful to give an interpretation of the system in the following
100
Chapter 6. Introduction to formal systems and recursion
chi
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manner. By an atomic sentence we shall mean an expression of the form SX, where X is any string of the symbols a and b, and by a compound sentence we shall mean an expression of the form SX -* SY, where SX and SY are atomic sentences. Obviously every expression provable in (E) is a sentence (either atomic or compound). And now we interpret the symbol S to denote the property of being an alternating string, and we interpret -> as standing for material implication, and so we call an atomic sentence SX true (under this interpretation) if X is alternating, and we define a compound sentence SX -> SY to be true if X is alternating implies that Y is alternating (in other words, either X is not alternating, or X and Y are both alternating). Now, it is obvious that sentences Al, A2, A3, and A4 are true (under our interpretation), and that all instances of A5 and A6 are true (because if Xa is alternating, so is Xab, and if Xb is alternating, so is Xba). Thus all initially provable sentences of (E) are true. And also, for any atomic sentence X and Y, if X and X -> Y are both true, so is Y - which we paraphrased by saying that Rule Do preserves truth. Since all initially provable sentences are true, and Rule Do preserves truth, it then follows that all sentences provable in (E) are true. And so, if SX is provable, where X is a string in a and b, then SX must be true, which means that X must be alternating. We have now shown that for any string X in a and b, the sentence SX is provable in (E) if and only if X is alternating. We say that S represents the set of all strings X in a and b such that SX is provable in (E). And so we have shown that S represents the set of all alternating strings. Elementary formal systems also provide means of representing relations between strings, as well as sets of strings. Here is an example: let K be the 3-symbol alphabet {a, b, c}. By the reverse of a string of these three symbols we mean the string in which the symbols are in reverse order, for example the reverse of bcaac is caacb. We shall take the letter "R" to represent the relation "X is the reverse of Y," and we take a comma, and read "Rx, y" as "x is the reverse of y." We then take the following axioms.
(1) Ra, a (2) Rb, b
(3) Rc, c
(4) Rx, y -> Rxa, ay (5) Rx, y -> Rxb, by
(6) Rx, y -> Rxc, cy
§ 1 Elementary formal systems
101 car
Of course all these axioms are correct under the intended interpretation - as examples, (1) is correct, since a is its own reverse. Likewise (2) and (3) are correct. Axiom (4) is correct, since if x is the reverse of y, then xa is the reverse of ay. Likewise (5) and (6) are correct - all instances are true under the intended interpretation.
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The rule of derivations for this system is the same as for the first system - Rule Do. Then' it can be seen that for any strings X and Y in the symbols a, b, and c, the expression RX, Y is provable in the system if and only if X is in fact the reverse of Y. And thus we say that the symbol "R" represents the relation "X is the reverse of Y." In each of the two systems that we have considered, the axioms (and hence also all the provable sentences) involve at most one occurrence of the symbol However, in the more general case of elementary formal systems, the axioms may involve more than one arrow. And here we must carefully observe a crucial point: if a sentence contains more than one arrow, the arrow is interpreted as implication with association to the right, e.g., X -> Y -* X is read "X implies that Y implies Z," or alternatively, "If X and Y are both true, so is Z." Now let us consider another elementary formal system in which we again represent the relation "X is the reverse of Y," but by a different set of axioms (involving five axioms instead of six). To this end we need an additional predicate "S" to represent the set of symbols of K. Here are the axioms.
(1) Sa (2) Sb
(3) Sc (4) Sx --> Rx, x
(5) Rx, y -> Sz -> Rxz, zy Axiom (1) says that a is a symbol. Axiom (4) says that if x is a symbol, then x is its own reverse. Axiom (5) says that if x is the reverse of y and if z is a symbol, then xz is the reverse of zy. And now we must modify rule Do! The idea is this: since the arrow is interpreted as implication with association to the right, then given three
sentences X, Y, and Z, we do not wish to infer Z from X --> Y and X -* Y --> Z, because if X implies Y is true and if X implies that Y implies Z, it does not follow that Z is true! What is the case is that if X is true and X implies that Y implies Z, then Y implies Z is true. And so Do must be replaced by the following more restrictive rule.
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102
Rule D From two strings X and X -* Y, the string Y can be derived provided that the arrow does not occur in X! Rule D is really the rule that will be used in elementary formal systems
in general. And now in the above system, it can be shown that for any string X and Y of the symbols a, b, and c, the sentence SX is provable if and only if X is one of the three symbols, and RX, Y is provable if and only if X is the reverse of Y. Thus R represents the relation: "x is the reverse of y." Here is another elementary formal system in which "R" again represents the relation "x is the reverse of y," in which we do not also need to represent
the set of symbols, and which cuts down the required number of axioms from five to four (but again the system uses expressions with more than one arrow). The axioms are the following
(1) Ra, a (2) Rb, b
(3) Rc, c (4) Rx, y --> Rz, w --> Rxz, wy
[Axiom 4, in the intended interpretation, tells us that if x is the reverse of y and z is the reverse of w , then xz is the reverse of wy, which is indeed true.] Elementary formal systems in general
Now that we have considered a few examples, let us turn to a precise and general definition of an elementary formal system.
We let K be a finite set of elements called symbols. By a word, or string, or expression in K, we mean any finite (non-empty) sequence of the symbols of K. For any words X and Y, by XY is meant the word consisting of X followed by Y, e.g., if X is the word abc and Y is the word de, then XY is the word abcde. By an elementary formal system (E) over an alphabet K, we mean a collection of the following items.
(1) The alphabet K. (2) Another alphabet of symbols called variables.
(3) Another alphabet of symbols called predicates, each of which is assigned a unique positive integer called its degree.
§1 Elementary formal systems
103
(4) Two more symbols called the implication sign and the punctuation sign (or comma). CDR:
(5) A finite sequence A1,. .. , An of strings which are well formed formulas, according to the definition given below; these strings are called the axioms of the system (E).
Dpi
car
The alphabets (1)-(4) are to be mutually disjoint. The symbols of K Q':
.,,
"(y
will usually be denoted by "a", "b", "c", or "d", with or without subscripts. The variables will be denoted by "x", "y", "z", "w", "v", with or without subscripts. We will usually use capital letters for predicates. The implication sign (also called the arrow) will usually be the symbol The punctuation sign will usually be the ordinary comma.
Let al,... , an be the symbols of K. By a term t of (E) we mean '.Y
any string composed of the ai and variables (or either one alone), e.g., a1xya2ya1 is a term. By an atomic formula we mean an expression of the
form Pt1, ... , tn, where P is a predicate of degree n and t i ,. .. , to are terms. [If n = 1, then Pt is an atomic formula, where P is a predicate of degree 1 and t is a term.] By a formula F of (E) we mean either an atomic formula or an expression of the form F1 --> F2 -> ... -> Fn, where ..j
"'_
each Fi is an atomic formula. As previously indicated, in a compound formula F1 -* F2... --> Fn, the arrow is to be interpreted as implication with association to the right, and F1 --f F2 --> Fn is to be read: "Fn is implied by the conjunction of the premises F1, F2, ..., Fn_1." In fact, we refer to F 1 ,.. . , Fn-1 as the premises and Fn as the conclusion. (DD
By an instance of a formula X is meant any expression attained from X by substituting words in K for all variables. [Of course the same word must be substituted for different occurrences of the same variable.] By a sentence of (E) is meant a formula with no variables. Obviously every instance of a formula is a sentence. Also, if X is a sentence, the only instance of X is X itself. We say that a sentence X is initially provable in (E) if it is an instance ..4r
of an axiom of (E). And now we define a sentence X to be provable in (E), or to be a theorem of (E), if it is either initially provable in (E), or is derivable from the set of initially provable sentences by a finite number of applications of the following inference rule, which we restate:
Rule D [Rule of detachment or modus ponens] From sentences X and X -> Y, the sentence Y may be inferred, provided X is atomic. Thus a sentence X is a theorem of (E) if and only if there is a finite . , Xn of sentences such that Xn = X and for each i < n, either Xi is an instance of an axiom of (E), or there exist earlier members Xj, Xk of the sequence such that Xi is derivable from Xj and Xk by the sequence X1, . .
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Chapter 6. Introduction to formal systems and recursion
rule of detachment (which means that Xk is atomic and Xk is the sentence X j -4X. And such a sequence X 1 ,.. . , Xn is called a proof of X (in the system (E)). Representability
And now we say that a predicate P of degree n represents in (E) the set of all n-tuples (X1, ... , Xn) of words in K such that the sentence PX1i ... , Xn is provable in (E). [For P of degree 1, P represents the set of all words X
in K such that PX is provable in (E).] A set or relation W of words in K is said to be formally representable over K if there exists an elementary formal system (E) over K in which W is represented by some predicate. CAD
J-,
Exercise 1 Suppose W1, W2 are sets of words in K and are formally representable over K. By W1 U W2 (called the union of W1 and W2) is meant the set of all x that belong to at least one of the sets W1, W2, and by the intersection W1 fl W2 of W1 and W2 is meant the set of all x that belong to both W1 and W2. Prove that W1 U W2 and W1 fl W2 are both f.r. (formally representable) over K.
Exercise 2 For any relation R(x, y), by its domain is meant the set of all x such that R(x, y) holds for at least one y. Show that if R(x, y) is f.r. over K, so is its domain.
Exercise 3 Suppose R(x,y) is f.r. over K. By R-1 is meant the set of cad
all ordered pairs (x, y) such that R(y, x) holds. Show that R-1 is f.r. over K.
III
Exercise 4 Suppose R(x, y) is f.r. over K. Define S(x, y, z) to mean that R(z, x) holds. Show that the relation S(x, y, z) is f.r. over K. [In other words, show that the set S of all triples (x, y, z) such that R(z, x) - this set S is formally representable over K.]
Exercise 5 Suppose f and g are both functions from words in K to words in K. For any word X in K let h(x) = f (g(x)). Suppose that the relations "f (x) = y" and "g(x) = y" are both f.r. over K. Show that the relation "h(x) = y" is f.r. over K.
Exercise 6 If S is a finite set of words in K, is S necessarily formally representable over K?
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105
Exercise 7 Suppose K is a sub-alphabet of L and that W is a set of words in K. Prove that if W is f.r. over K then W is f.r. over L. [The converse is also true, but its proof is trickier - a partial result along those line is the subject of the next two exercises.]
Exercise 8 Let K be a finite alphabet whose symbols are arranged in some definite order k1, k2, ... , kn, where n > 2. Let CI, C2.... , c,,,, ... be the respective words klk2, klklk2, ... , klkl ... k1 k2, i.e., for each i < n, ci n
consists of i occurrences of kl followed by k2. For any word X in K, define its code word to be the result of replacing each symbol ki by ci (e.g, the
code word of k3klk2 is klklklk2klk2klklk2). For any set W of words in K, let WO be the set of its code words. (a) Show that if W is f.r. over K then WO is f.r. over the two-sign alphabet {kl, k2}.
(b) Let Cx, y be the relation "the code word of x is y." Prove that this relation is f.r. over K. (c) Using (b) and the last exercise, show that if WO is f.r. over {k1, k2} then W is f.r. over K.
Exercise 9 Suppose K is a sub-alphabet of L and that K contains at least two symbols. Using the last two exercises, show that any set W of words in K that is f.r. over L is f.r. over K. [This result is also true if K contains only one symbol, but the proof is more elaborate.] Formal mathematical systems Elementary formal systems provide one way of characterizing what it means for a mathematical system to be a formal system. We will define a math-
ematical system in a finite alphabet K to be a formal system if its set of provable sentences is formally representable over K.
§2 Some basic closure properties For any n > 2, by a relation of degree n, or an n-place relation, or a relation of n arguments on a set S is meant a set of n-tuples of elements of S. We
regard subsets of S as special cases of relations on S - relations of one argument. We let R(K) be the collection of all relations (including sets) of words in K that are formally representable over K. We wish to show that R(K) has some basic closure properties.
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106
Preparatory to this, we observe the following. Let (El) and (E2) be elementary formal systems over a common alphabet K. We shall call them 'C7
independent if they contain no common predicates. By (E1) U (E2) we mean the E.F.S. over K whose axioms are those of (E1) together with those of (E2). Clearly every theorem of (E1) and of (E2) is again a theorem of (E1) U (E2). And it is obvious that if (E1) and (E2) are independent, then every provable formula of (El) U (E2) is provable in
(p,
ca"
pi'
U,_
either (E1) or (E2) alone, and hence the representable sets and relations of (El)U(E2) are those representable in (E1) alone or in (E2) alone. Similarly, if (El), (E2), ... , (En) are mutually independent (no two have a common predicate), then the representable relations of (El) U (E2) U ... U (En) are those of each of (E1), (E2), ... , (En) alone. Now, suppose W1, W2,..., Wn are each formally representable over K in systems (E1), (E2), ... , (En). Since we are assuming an unlimited stock '.s
of symbols at our disposal that we can use as predicates, we can construct (E1), (E2),. - -, (En) so as to contain no common predicates, and then W1,... , Wn are all represented in the common system (El) U ... U (En). We thus have
Proposition 2.1 If W1,. .. , Wn are each f.r. over K, then they can all be represented in a common E.F.S. over K. Now for our closure properties.
(a) Unions and intersections For any two relations Rl (xi.... , xn), R2(xl, ... , xn) of the same number of arguments, by their union R1 U R2 - also written Rl (xl, ... , xn) V R2(x1 i ... , xn) - is meant the set of all n-tuples ( X1 ,--- , xn) that are in R1 or in R2 (or both), 110,
and by their intersection R1 n R2 - also written R1(x1, ... , xn) A R2(x1i ... , xn) - is meant the set of all n-tuples (x1, ... , xn) that are in both R1 and R2.
(b) Existential quantifications If R(xl,... , xn, y) is a relation of two or more arguments, by its existential quantification ayR(x1i ... , xn, y) - also written EIR - we shall mean the set of all n-tuples (x1,.. . , xn) such that R(x1i ... , xn, y) holds for at least one element y.
(c) Explicit transformations For any relation R(xl,... , Xk) of k arguments and for any positive n and any positive integers i1 ,-- . , ik each
< n, by )xi ... xnR(xil,... )xik) is meant the set of/ all n-tuples (x1, ... , xn) such that R(xii,... , xik) [e.g., )xix2x3x4R(x3, x4, x1) is the set of all n-tuples (x1, x2, x3, x4) such that R(x3i x4, xl) holds.
§2 Some basic closure properties
107
-Ti
We could also write this relation as AxyzwR(z, w, x).] We say that the relation Ax1 ... xnR(xi, , ... , xij is explicitly definable from R(xl,... , xk). Thus a relation S of n arguments is explicitly definable from a relation R of k arguments if there are positive integers il, ... , ik such that for all elements x1, ... ) xn, S(xi, ... , xn) holds if and only if R(xi17 ... , xij holds 1. A.,
(d) Introduction of constants For any relation R(x, yl,... , yn) of two or more arguments, and any fixed element c, by R, we mean the set of all n-tuples (yl, ... , yn) such that R(c, yl, ... , yn). Thus Rr(yi, ... , yam.) - R(c, yl, ... , yn).
We recall that we are letting R(K) be the collection of all sets and relations that are formally representable over K.
Theorem 2.2 The collection R(K) of all relations f.r. over K is closed under unions, intersections, existential quantifications, explicit transformations, and introduction of constants, i.e., (a) For any relations W1, W2 of R(K) of the same number of arguments, W1 U W2 and W1 fl W2 are f.r. over K.
(b) If R(xl,... , xn, y) is f.r. over K, so is zlyR(xi, ... , xn, y). (c) If R(xl,... , xk) is f.r. over K, so is Ax1 ... xnR(xi17... , xik) (i1, ... , ii. each < n).
(d) If R(x, yl,... , yn) is f.r. over K then for any word c in K, the relation R(c, yl,... , yn) (as a relation among yl, ... , yn) is f.r. over K.
Proof C!1
(a) Union and intersection Suppose W1, W2 of degree n are f.r. over K. By Proposition 2.1, there is a single E.F.S. (E) over K in which W1 and W2 are both represented - say by predicates P1 and P2. To represent the union, we take a new predicate P and add to (E) the "SIN
(1)
following two axioms.
Pixi,...,xn
->
Pxl.... ,xn
P2x1i...,xn -> Px1,...,xn SAC
In this extended system, P represents W1 U W2. 'This definition is narrower than the one I gave in Theory of Formal Systems, in that I do not allow the introduction of constants, which I treat separately in (d).
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108
As for the intersection, instead of the above two axioms we add the following single axiom. Plx1i...,xn
->
P2xl,...,xn - Pxl,...,xn
(b) Existential quantification Let P represent the relation R(xi, ... , xn, y) in (E). Take a new predicate Q (of degree n) and add the axiom.
Pxl,...,xn,y
--+
Qxl,...,xn
Then Q represents the relation zlyR(xl, ... ) xn, y).
(c) Explicit transformations Suppose P represents R(xl,... , xk) in (E). Take a new predicate Q and add the following axiom. Pxi1,...,xik
-f Qx1,...,xn
Then Q represents Axl,... , xnR(xi, , .... xik ).
(d) Introduction of constants Suppose P represents R(x, yl,... , yn) in (E) and c is any fixed word in K. We take a new predicate Q and add the axiom.
PC,x1,...,xn
Qx1,...,xn
Then Q represents the relation R(c, yi, ... , yn) (as a relation among yl) ... , yn), i.e., Q represents the relation Ay1, ... , yn . R(c, y1,
...)yn).
We note that it is obvious from (d) that if a relation R(xl,... , x.,,,,, , yn) is f.r. over K, then for any words cl, ... , c, in K, the relation R(cl,... , C,n, yl, ... , yn) (as a relation among yl, ... , yn) is f.r. over K. y1,
We say that a function f (x1, ... , xn) is explicitly definable from a func-
tion g(x1i ... , xk) if there are numbers il, ... , ik, all < n, such that for all elements X1, ... , xn, f (xl, ... , xn) = g(xi1, ... xik) In A-notation, by ... xn g(xi17 ... , xik) is meant the function f such that for all x1, ... , xn, f (x1, ... , xn) = g(xi...... xin ) . Obviously if f (x1, ... , xn) is explicitly definable from 9(X1, ... , xk), then the relation f (x1, ... , xn) = y is explicitly definable from the relation g(xl, ... , xk) = y. )
Ax1
We say that a function f (xl, ... , xn) is f.r. over K if the relation f (x1, ... , xn) = y is f.r. over K.
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109
Theorem 2.3 (a) The class of functions f.r. over K is closed under composition, i.e., for any functions f1(x1, , xn), ... , fk(xl, .. , xn), g(xl, , xk) that are f.r. over K, the function g(f 1(x1, ... - , X . ) ,- . , fk (xl, ... , xn)) is f.r. over K.
(b) For any functions fl (xl,... , xn), ... , fk (xl, ... , xn) f.r. over K and any relation R(xl, ... , xk) f.r. over K, the relation R(fl (xl, ... , xn), ..., fk(x1i...,xn)) is f.r over K. (c) The class of functions f.r. over K is closed under explicit definability.
(d) The class of functions f.r. over K is closed under introduction of constants, i.e., if f (y, x1i ... , xn) is f.r. over K, then for any word c in K, the function f (c, xl,... , xn) (as a function of xl,... , xn) is f.r. over K.
0
Proof
G".
(a) g(fl (xl, ... , xn), ... , fk (xl, ... , xn)) = y if and only if 2yl ... ]yk (fl (x1, ... , xn) = yl A ... A f k (xl,... , xn) = yk A g(yl, ... , yk) = y). The result then follows since the class of relations f.r. over K is closed under existential quantifications and conjunctions.
(b) R(fl (xl, ... , xn), ... , fk (xl, ... , xn)) if and only if 3yl ... 3yk (f l (xl, ... , xn) = yl A ... A fk (xl, ... , xn) = yk A R(yl, ... , yk)). Again, the result follows by closure under existential quantifications and conjunctions. (c) Suppose that f (X 1, ... , xn) is explicitly definable from g(xl, ... , xk) and that the latter is f.r. over K. Then the relation f (xl, ... , xn) = y is explicitly definable from the relation g(xl, ... , xk) = y, and since the latter is f.r. over K, so is the former (by (c) of Theorem 2.2.), which means that function f (xl, ... , xn) is f.r. over K. (d) This follows easily from (d) of Theorem 2.2.
Let us note that (b) of the above theorem also holds when R is a set A, i.e., if A is a set of words in K that is f.r. over K, then for any function f (xl, ... , xn) f.r. over K, so is the set of all n-tuples (x1, ... , xn) such that f (xl, ... , xn) E A, i.e., so is the set ) x1 ... xn(f (xl, ... , xn) (E A).
In particular, this also holds when f is a function of one argument - in
110
Chapter 6. Introduction to formal systems and recursion
other words, if f (x) and A are f.r. over K, so is the set f -1(A) (the set of all x such that f (x) E A). More generally, for any relation R(xl,... , xn) and any function f (x), by f -1(R) we shall mean the set of all n-tuples (x1, ... , xn) such that R(f (xl),... , f (xn)). And so we have
Theorem 2.4 If R and f are formally representable over K, so is f -1(R). Exercise 10 (a) For any set W and function f (x), by f"(W) is meant the set of all elements f (x) such that x E W. Show that if f and W are f.r. over K, so is f"(W). (b) For any sets W1 and W2, by W1 x W2 (the Cartesian Product of W1 and W2) is meant the set of all ordered pairs (x, y) such that x E W1 and y E W2. Show that if W1 and W2 are f.r. over K, so is W1 x W2.
§3 Solvability over K For any relation W(xl, ... , xn) on a set S, by W (the complement of W with respect to the set of all n-tuple elements of S) is meant the set of all n-tuples.. (x1, . , xn) of elements of S that are not elements of W. [Thus W (X 1, ... , xn) if -W (X 1, ... , xn).] We presume the reader to be familiar with the fact that if W = W1 U W2, then W = W1 fl W2. A relation W of words in K is said to be solvable over K if W and its complement W are both f.r. over K. Given any collection R of relations on a set S, we let RX be the colBCD
lection of all relations W such that W and W are both in S. We write RX (K) for (R(K))x - thus RX (K) is the collection of all relations that are solvable over K.
Theorem 3.1 The collection RX (K) of all relations solvable over K is closed under union, intersection, explicit transformations, and introduction of constants.
Proof More generally, we will show that for any collection R of relations on a set S, if R is closed under union, intersection, explicit transformations, and introduction of constants, then RX has these same closure properties and is also closed under complementation. And so suppose R has these closure properties.
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111
Complementation is trivial - If W E RX, then W and W are both in R. Hence Wis in R (since W = W), so W and W are both in R, which means that W E RX. As for union, suppose W1 and W2 are elements of RX of the same degree. Then W1, W2, W1, W2 are all in R. Let W = W1 U W2. Then of course W is in R (by hypothesis). Now, W = W1 n W2, and since W1 and W2 are both in R, so is W1 nW2, thus W is in R. Since W and W are both in R, then W E RX . The proof for intersection is obviously dual (or indeed follows from the fact that RX is closed under union and complementation).
As for explicit transformations, we note that if S is explicitly definable from R, then S is explicitly definable from R, because if S(xl, .... xn) is equivalent to R(x2,, ... , xi. ), then S(xl, ... , x,,,) is equivalent to R(x2,, ... Xik), and so if R E RX and S is explicitly definable from R, then S is explicitly definable from R, and since R and R are both in R then S and S are both in R, so S E RX. The proof for introduction of constants is similar and is left to the )
reader.
Theorem 3.2 If R(xl,... , x,,) is solvable over K and if fl (xl, ... , xn), .... f k (x1, ... , xn) are functions that are solvable over K then the relation R(f l (xl, ... , xn), ... , fk (xl, ... , xn)) is solvable over K.
Proof Let S(xl,... , xn) be the relation R(fl (xi, ... , xn), ... , fk (xl, ,xn)). Then S(xl,...,xn) is the relation R(f1(_x1,...,xn),...,fk(xl) xn)). If now R is solvable over K, then R and R are both f.r. over K, hence S and S are both f.r. over K (by (b) of Theorem 2.3), hence S is ,
solvable over K.
Exercise 11 Show that the relation x = y is solvable over K. [It is easy to represent the relation x = y, but it is trickier to represent the relation x
y!]
Extension of alphabets
We call an alphabet L an extension of an alphabet K if every symbol of K is also a symbol of L. We wish to show that any relation of strings in K which is formally representable over K is formally representable over any extension L of K. Suppose now that (E) is an elementary formal system over K. For any extension L of K, let (E)L be that elementary formal system over L whose
Chapter 6. Introduction to formal systems and recursion
112
.j'
W'S
axioms are the same as those of (E). [In particular, (E)K = (E).] In general, a predicate P of (E) does not represent the same relation in (E)L as it does in (E). [As a trivial example, suppose (E) has the sole axiom Px. Then in (E)K, P represents the set of all strings in K, whereas in (E)L, P represents the set of all strings in L.] However, given any elementary formal system (E) over K, we can construct an elementary formal system (E*) over K whose predicates include all predicates of K and which is such that for any extension L of K, each predicate of (E) represents the same relation in (E*)L as it represents in (E). We do this as follows. Let al, ... , a,, be the symbols of K and let Xl,... , X,,, be the axioms of (E). We take a new predicate Q (which is to represent the set of strings in K) and for each axiom X of (E), let X* be the string Qx1 -> ... -> Qxt -> X, where x1, ... , xt are the variables that occur in X. We then define (E*) to be that elementary formal system whose axioms are: Qal
QaT,
Qx -*Qy -*Qxy
Xi X*
Now let L be any extension of K. It is easily seen that in (E*)L, the predicate Q represents the set of all strings in K, and that each of the other predicates represents the same relation as it does in the original system (E). We have thus given a solution to Exercise 7 and have proved
Theorem 3.3 Any relation formally representable over K is formally representable over any extension L of K. The converse of Theorem 3.3 is also true, as we shall see later on. As an obvious corollary of Theorem 3.3 we have:
Theorem 3.5 Any relation solvable over K is solvable over any extension L of K.
§4 Recursively enumerable relations
113
Part II Recursive enumerability §4 Recursively enumerable relations Just as any non-negative integer is uniquely expressible as a polynomial in powers of 2 with coefficients 0 and 1, so is any positive integer uniquely expressible as a polynomial in powers of 2 with coefficients 1 and 2. We let D2 be the two-sign alphabet {1, 2}; these two symbols we call dyadic digits; any string dndn_1 ... dldo of dyadic digits is called a dyadic numeral and denotes the positive integer do + 2d1 + 4d2 + ... + 2ndn. This way of naming
?.'
positive integers, which we call dyadic, has certain technical advantages over the better known binary notation (using 0 and 1). Until further notice we shall identify the positive integers with the dyadic numerals that denote them. We note that if we order the dyadic numerals lexicographically, i.e., according to length and alphabetically within each group of the same length, then the position that any numeral has, in this sequence, is the same as the number which it denotes. Any elementary formal system over the two-sign alphabet D2 will be called an elementary dyadic arithmetic - or a dyadic arithmetic for short. We now define a relation W of positive integer to be recursively enumerable
- abbreviated "r.e." - if it is representable (in dyadic notation) in some dyadic arithmetic, i.e., formally representable over {1, 2}, and recursive if W and W are both recursively enumerable. Having defined recursive enumerability for relations of positive integers, we can now define recursive enumerability for relations of natural numbers as follows. For any relation W (x1, ... , xn) of natural numbers, let W+ be
cam.
the set of all n-tuples (x1, ... , xn) of positive integers such that W(xi 1, ... , xn - 1). Then we define W to be a recursively enumerable relation of natural numbers if and only if W+ is a recursively enumerable relation of positive integers.
Until further notice we will be concerned only with positive integers, and the word "number" shall mean positive integer, unless specified to the contrary. Remark
In our definition of recursive enumerability, our choice of 2 as a base was arbitrary, but will prove to be convenient. Whatever base b we chose, the definition of "recursively enumerable" would be equivalent, as we will see.
By a numerical notation - more briefly, a notation - we shall mean a mapping A that assigns to each positive integer n an expression An -
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114
,.o
also written n when A is fixed for the discussion - called the A-numeral for n. By the alphabet of A we mean the set of symbols from which the A-numerals are formed. We shall say that A is an admissible notation if the alphabet K of A is finite and the set of all ordered pairs (On, An+1) is formally representable over K. Given an admissible A and a relation R(x1, ... , xn) of positive integers, by RA we shall mean the set of all nsue.
-a)
tuples. (Dy...... Ax.) such that R(xl,... , xn) holds. And now we shall Eau
say that R is formally representable in A-notation if R. is formally representable over the alphabet K of A. By the dyadic notation we of course mean the mapping that assigns
to each number n its dyadic numeral. And we are defining a relation of numbers to be recursively enumerable if it is f.r. in dyadic notation.
,1'
[Actually the choice of notation doesn't matter, so long as the notation is admissible, but the dyadic notation has certain particular advantages.] The dyadic notation is admissible, because the successor relation is represented (in dyadic notation) by "S" in the dyadic arithmetic whose axioms are
`pa
Si, 2 S2,11 Sxl, x2 Sx, y -* Sx2, yl Eau
More generally, let us consider n-adic notation for n > 2. In this notation, we take n symbols 61, 62, ... , 6n called, n-adic digits, as names of the positive integers 1, 2, ... , n respectively, and any string 6i,.Si,._1 ... 6o of these n-adic digits denotes the positive integer io + iln + i2n2 + ... + irn''. Again, if we order all the n-adic numerals lexicographically, the position a numeral occupies in the lexicographical sequence is the number it designates. We let Dn be the alphabet {61i ... , 6n}. For any n > 2, n-adic notation is admissible, since in n-adic notation the successor relation is represented by "P" in that elementary formal systems over Dn whose axioms are the following :.a
61,62
6n-1, 6n, P6n, 6161
Px61i x62
§4 Recursively enumerable relations
115
Px8,,,_1, x6",
Px, y --* Px&",, y51
We thus have
Theorem 4.1 For any n > 2, n-adic notation is admissible. Unary and Peano notations
In unary notation we take the numeral n to be a string of is of length '.3
.'3
cal
n. Obviously unary notation is admissible, since the successor relation is represented by S in that E.F.S over the alphabet {i} whose sole axiom is Sx, xi. In Chapter 5 we used what might be called Peano notation, in which An (usually written n) is taken to be the symbol "0" followed by n accents. [Thus the Peano positive numerals are 0', 0", 0"', ....]
Theorem 4.2 Peano notation is admissible. Proof The alphabet of the Peano notation is {0,' }. We construct the following elementary formal system over {0,' }.
We first take a unary predicate N to represent the set of (positive) Peano numerals and we take the following axioms. NO'
Nx -fNx' BCD
Then we take a predicate S to represent (in Peano notation) the successor relation and we add the axiom
Nx-->Sx,x' Theorem 4.3 For any admissible notation 0, the following relations are f.r. in A-notation.
(a) The set of 0-numerals. (b) The relations x < y, x < y, x = y, x
(c) x+y=z,xxy=z,xy=z.
Y.
116
Chapter 6. Introduction to formal systems and recursion
Proof We let K be the alphabet of A and we let (E) be an elementary formal system over K in which S is a predicate that represents the successor relation in A-notation.
(a) To represent the set of A-numerals, we take a new predicate N and add the axiom Sx, y --> Nx
(b) To represent the relation x < y (in A-notation), we take a new predicate L and add the axioms Sx, y --> Lx, y
Lx, y -> Ly, z -* Lx, z
(c) To represent the relation x < y we take a new predicate L' and add the axioms Lx, y --> L'x, y Nx --> L'x, x
To represent the relation x = y we take a new predicate I and add the axiom
Nx -->Ix,x To represent the relation x # y we take a new predicate I' and add the axioms
Lx, y -' I'x, y Ly, x --> I'x, y
(d) To represent the relation x+y = z (in A-notation) we add a predicate A and the axioms
Sx,y-*Ax,A,,y Ax, y, z --> SY, yi --> Sz, zi -4 Ax, yi, zl
Next we take a new predicate M to represent the multiplication relation x x y = z and we add the following axioms
Nx-> Mx,A,,x Mx,y,z-+Sy,yl
Az,y,w->Mx,yl,w
§4 Recursively enumerable relations
117
Next we take a predicate E to represent the exponential relation xy = z and add the following axioms
Nx --fEx,Al,x Ex, y, z --> Sy, yl --> Mz, x, w --> Ex, yl, w
Dyadic concatentation
For any numbers x and y, by x * y we mean that number whose dyadic numeral is the dyadic numeral of x followed by the dyadic numeral for y (e.g., in dyadic notation, 12112 * 212 = 12112212). We call the relation x * y = z the relation of dyadic concatenation. This relation satisfies the following three conditions and is uniquely determined by them.
(1) x*1=2x+1 (2) x*2=2x+2 (3) x*(y*z)=(x*y)*z From this we have
Theorem 4.4 For any admissible notation A, the relation of dyadic concatenation is f.r. in A-notation.
Proof Suppose A is admissible and K is its alphabet. Let (E) be an elementary formal system over K in which "A" represents the addition relation and "M" the multiplication relation (in A-notation). To represent the dyadic concatenation relation we take a new predicate "C" and add the following axioms.
MA2,x,y -* Ay,A,,z -* Cx,A,,z M1 2,x,y--'Ay,J2,z--->Cx,02,z Cx,y,v --> Cv,z,t -> Cy,z,w --> Cx,w,t From Theorems 4.1, 4.3, and 4.4 we have
Theorem 4.5 The following relations are recursively enumerable (a) x+ = y [By x+ is meant the successor x + 1 of x.]
(b) x
Chapter 6. Introduction to formal systems and recursion
118 coy
§5 Recursive functions
A function f (xl,... , x,,,) from n-tuples of numbers to numbers is called recursive if the relation f (xl, . . . , xn) is recursive.
Theorem 5.1 If the relation f (xl, ... , xn) = y is recursively enumerable then it is recursive.
Proof Suppose the relation f (xl, ... , xn) = y is r.e. Then the relation f (xl, ... , xn) 54 y is also r.e., since it can be written z1z (f (xl .... xn) _ zAz54 y).
Theorem 5.2 (a) The functions x+, x + y, x x y, xy, x * y are recursive.
(b) The relations x < y, x < y, x = y are recursive.
0
Proof (a) This follows from Theorem 4.4 and 5.1. sue.
(b) Since the relations x = y and x y are both r.e. (by Theorem 4.4), then the relation x = y is recursive.
Ins
The relations x < y, x < y, are both r.e. (by Theorem 4.4) and the complement of the relation x < y is the r.e. relation y < x, and the complement of x < y is y < x, and so the relations x < y, x < y are both recursive.
CD'
We will henceforth be working mainly in dyadic notation and we write xy to mean x * y. No confusion with multiplication will result since we denote x times y by x x y, or x y.
Exercise 12 We say that x is part of y (in dyadic notation) if x = y, or xz = y for some z, or zx = y for some z, or z1xz2 = y for some zl and z2. Prove that the relation "x is part of y" is r.e. For Theorems 2.3, 3.2, and 5.1 we have
§5 Recursive functions
119
Theorem 5.3 (a) The class of recursive functions is closed under composition, explicit transformations, and introduction of constants.
(b) If f 1(x1, ... , xn), ... , fk (x1, ... , xn) are recursive and if R(xi, ... , xk) is r.e. (recursive) then the relation R(fl (xl, ... , xn),
.... fk (xl, .
. .
,
xn)) is respectively r.e. (recursive).
Exercise 13 Show that if A is recursive and f (x) is a recursive function, then the set f -1(A) is recursive. .s:
Theorem 5.4 If R(xl,... , xn) is r.e. and f (x1i ... , xn) is recursive, then the set A of all numbers f (XI, ... , xn) such that R(xl, ... , xn) - this set A is r.e.
Proof x E A F-, 2x1 ... 3xn(f (x1, ... , xn) = x A R(xl, ... , xn))Primitive recursive functions Although primitive recursive functions (which we will shortly define) play no significant role in this study, they are of importance generally and we shall say something about them. Given a function h(x, y) and a number c, a function f (x) is said to be definable by recursion from h and c if for every number x, the following conditions hold: (1) f (1) = c (2) f(x + 1) = h(x,f(x))
The idea is that f (1) is given outright, and for any x, once f (x) has
,yam
been computed, then f (x + 1) can be computed using (2) (assuming h(x, y) can be computed). More generally, given functions g(x1 ... ) xn), h(xl,... ) xn, y, z), a function f (x1, ... , xn, y) is said to be definable by recursion from the functions g and h if for all X 1 ,- .. , xn, y the following conditions hold:
(1) f(x1,...,xn,1) =g(xl,...,xn) F-3
(2) f(x1,...,xn,y+1) = h(x1i...,xn,y,f(x1,...,xn,y)) Again, assuming we know how to compute g(x1i ... , xn) and h(xl, .... xn, y, z), we can successively compute f (xl, ... , xn,1), f (X1, ... , xn, 2), . .., f (x1i ... , xn, y), f (x1, ... , xn, y + 1), .... These informal remarks can be made precise by:
Chapter 6. Introduction to formal systems and recursion `CS
120
Theorem 5.5 (a) If h(x, y) is recursive and if f (x) is definable by recursion from h(x, y) and c, then the function f (x) is recursive.
(b) If g(x1i ... , xn) and h(xl,... , xn, y, z) are recursive functions and if f (x1i ... , xn, y) is definable from them by recursion, then f (xl, ... , xn, y) is a recursive function.
Proof (a) Let (E) be a dyadic arithmetic in which H represents the relation h(x, y) = z and S represents the relation x + 1 = y. Take a new predicate F to represent the relation f (x) = y and add the following axioms.
F(1, c) F(x, y) --> H(x, y, z) --> Sx, x' --> Fx', z
(b) Let (E) be a dyadic arithmetic in which S represents the successor relation x + 1 = y, G represents the relation g(x1i ... , xn) = y, and H represents the relation h(xl, ... , xn, y, z) = w. Then take a new predicate F and add the axioms
Gx1,...,xn,y Fx1,...,xn,y,z
Fx1,...,xn,1,y Hx1i...,xn,y,z,w Sy,y' --> Fx1i...,xn,y',w
Then F represents the relation f (x1, ... , xn, y) = Z. And now for a definition of primitive recursive functions. First, the projection function Ui (x1, ... , xn) (i < n) is defined by the condition: Ui (x1, ... , xn) = xi. Of course projection functions are recursive (the relation Ui (x1, ... , xn) = xi is represented by P in that dyadic arithmetic whose only axiom is Px1,... , xn, xi). Well, the class of primitive recursive functions is defined as the smallest class containing the successor function S(x) = x + 1, all projection functions, and closed under composition and definability by recursion. And so by Theorem 5.5, and the facts that the successor function and the projection functions are recursive, and that the recursive functions are closed under composition, we have:
Theorem 5.6 Every primitive recursive function is recursive.
§5 Recursive functions
121
Minimization
Let us call a numerical function f (xl, ... , xn, y) regular if for every X1, ... , xn there is at least one number y such that f (x1 .... ) xn, y) = 1. If now f is regular, then we define µy f (xl, ... , xn, y) = 1 to be the smallest
number y such that f (xl, ... , xn, y) = 1. We then say that the function ,ay f (xl, ... , xn, y) = 1 (which is a function of X1.... , xn) is the minimization of the function f.
Theorem 5.7 The minimization of any regular recursive function is recursive.
Proof Exercise. Another characterization of recursive functions We now see that any class of functions that contains all primitive recursive functions and is closed under minimization of regular functions contains all recursive functions. Actually, the class of recursive functions happens to be the smallest class of functions that contains all primitive recursive functions and is closed under minimization of regular functions. Indeed, some treatments of recursion theory define the recursive functions to be the members of this class. *Partial recursive functions
."5
.-y
CDR
Let q(xl,... , xn) be a function defined on some but not necessarily all n-tuples of numbers. We write q(x17 ... , xn) -- y to mean that q is defined on (x1, ... , xn) and its value on x1i ... , xn) is y. The function q is called a partial recursive function if the relation q(xl,... , xn) ^_ y is recursively enumerable. In general, this does not imply that the relation q(x1i ... , xn) ^ y is recursive, but we have the following theorem of which .s:
Theorem 5.1 is a special case.
*Theorem 5.8 If q(xl,... , xn) is partial recursive and if the set of all n-tuples (x 1, ... , xn) on which q is defined is recursive, then the relation q(x1i ... , xn) ^_ y is recursive.
Proof Exercise. One also defines the minimization py(q(xl,... , xn, y) ^_ 1) of a partial recursive function q(xl, .... xn, y) as follows. If there is some y such that
Chapter 6. Introduction to formal systems and recursion
122
q(xl,... , xn, y) = 1, then we take pyq(xl, ... , xn, y) to be the smallest such
inn
'0-h
.'3
car
coo
y; otherwise 1cy(q(x1 i ... , xn, y)) - 1 is undefined. It is easy to prove that for any partial recursive function q, its minimalization is a partial recursive function. An alternative characterization of the class of partial recursive functions is the smallest class of partial functions that contains all primitive recursive functions and is closed under minimalization. This result, though well known, is not needed for this study. Although many treatments of recursion theory take partial recursive functions as the fundamental objects, we prefer the approach of Post[13] which takes recursively enumerable sets and relations as the fundamental objects. And now we turn to something that will play a fundamental role in the approach that we will be taking.
§6 Finite quantifications and constructive definability For any relation R(xi, ... , xn, y) of numbers, by the relation (Vz < y)R(xl,... , xn, z) we mean the set of all (n + 1)-tuples (x1, ... , xn, y) such that R(x1, ... , xn, z) for all z less than n equal to y. By the relation (]z < y)R(xl,... , xn, z) (read "There exists a number z less than or equal to y such that R(xl, ... , xn, z)") is meant the set of all (n + 1)-tuples (x1, .... xn, y) such that R(x1, ... , xn, z) holds for at least one z less than or equal to y. We abbreviate the relation (Vz < y)R(xl,... , xn, z) by VF(R), and (1]z < y)R(xl, .... xn, z) by IF (R), and we say that VF(R) is .s:
the finite universal quantification of R, and that 3F(R) is the finite existential quantification of R. We also say that VF(R) arises from R by finite universal quantification and that ]F(R) arises from R by finite existential quantification. As previously stated, we abbreviate "recursively enumerable" by "r.e."
Theorem 6.1 (a) If R(xl, ... , xn, y) is r.e., so are the relations IF (R) and VF(R). (b) If R is recursive, so are ZIF(R) and VF(R).
Proof (a) Suppose R(xl,... , xn, y) is r.e. Then the relation (3z < y)R(xi, ... , xn, z) is equivalent to ]z(z < y A R(x1,... , xn, z)), which is r.e., hence ZIF(R) is r.e.
§6 Finite quantifications and constructive definability
123
As to VF(R), let (E) be a dyadic arithmetic in which P represents R and S represents the relation "the successor of x is y." To represent VF(R), take a new predicate Q and add the following axioms.
Pxl,...,xn, 1 - Qxl,...,xn, 1 Qxl, ... xn7 y ` Sy, Y
Pxl,... , xn, y
Qxl,... , xn, y1
(b) This follows from (a), since the complement of EIF(R) is VF(R), and the complement of VF(R) is EIF(R).
Theorem 6.2
(a) If R(xl,... , xn, z, y)
is
r.e. so are the relations (3y < z)R(xl,
.... xn, z, y) and (Vy < z)R(xl,... , xn, Z, y) (b) Same with "recursive" instead of "r.e." Proof
(a) Suppose R(xl,... , xn, z, y) is r.e. Then the relation (3y < z)R(xl, ... , xn, z, y) is equivalent to 3y(y < z A R(xl,... , xn, z, y)), which is
.-:
r.e. As to (Vy < z)R(xl,... , xn, z, y), let S(xl,... , xn, z, w) be the relation (Vy < w)R(xl,... , xn, z, y). By (a) of the preceding theorem, the relation S(xl,... , xn, z, w) is r.e. Then the relation S(xl, ... , xn, z, z), which is explicitly definable from S(xl,... , xn, z, w), is r.e., and this is the relation (Vy < z)R(xl, .... xn, z, y). cad
ova
(b) The proof of this, using (a) is similar to the proof of (b) of Theorem 6.1.
For any relation R(xl, ... , xn, y, z) and function f (x), we write (Ely < f (z))R(xl,... , xn, y, z) to mean zly(y _< f (z) A R(xl,... , xn, y, z)), and (Vy < f , xn, y, z) to mean Vy(y < f (z) D R(xl,... , xn, y, z)). (z))R(xl,...
Theorem 6.3 Suppose R(xl,... , xn, y, z) is r.e. and f (x) is a recursive function. Then
(a) The relations (3y < f (z))R(xl,... , xn, y, z) and (Vy < f (z))R(xl, ... , xn, y, z) are r.e.
(b) If R is recursive, so are the relations (ly < f (z))R(xl,... , xn, y, z) and (by <_ f(z))R(xl, ... , xn,Y, z)
Chapter 6. Introduction to formal systems and recursion
124
Proof IA
(a) (3y < f (z)) is the relation 3y(y < f (z) A R(xl,... , xn, y, z)), which is r.e. (under the hypothesis that R is r.e. and f is recursive). As to (Vy < f (z))R(xl,... , xn, z), the relation Vy < w)R(xl,... ) xn, y, z)) is an r.e. relation among x1, ... , xn, z, w (by Theorem 6.1), and the relation f (z) = w is r.e., hence so is the relation 11w(f (z) _ w/ A (by < w)R(xl, ... , xn, y, z)), and this is equivalent to (Vy
f(z))R(xl,...,xn,y,z) (b) This follows from (a), since the complement of (Ily < f(z))R(xl, ... ) xn, y, Z) is (Vy C f lz))R(xi, .... Xn, y, z). Constructive definability
We say that a relation W (of positive integers) is constructively definable from relations W1,. .. , Wn if it is obtainable from them by a finite number of applications of the following operations: (1)
Unions and intersections,
(2) Explicit transformations, (3)
Introduction of constants,
(4)
Finite quantifications (both existential and universal),
(5) Complementation. [We remark that if we started with the relations W1, ... , Wn and their complements W1i ... , Wn, then we would not need scheme (5). This can be verified by an argument similar to the "prenex normal form construction," but this fact will not be needed.] If instead of (4) we allowed unbounded existential quantifications, then
the relations thus obtained are the ones known as first-order definable from W1,... , Wn. As stated in Chapter 5, a relation is called arithmetic (after Godel) if it is first-order definable from plus and times. If W is constructively definable from plus and times, then W is called constructive
arithmetic - abbreviated C.A., and also called E0. By a E1-relation also called an E.C.A. relation - we mean a relation that is the existential quantification of a E0-relation - a relation of the form 3yR(x1, ... , xn, y), where R is E0. From our summary we immediately obtain
§6 Finite quantification and constructive definability
125
Theorem 6.4 (a) Every Eo-relation is recursive.
(b) Every E1-relation is r.e.
In Chapter 8 we will show that all r.e. relations are in fact El, and hence that the E1-relations coincide with the r.e. relations. Let us now observe the following simple proposition.
Proposition 6.5 The relations x = y, x < y, x < y are Eo.
Proof (a)
(b) x
(x
The following principle will prove quite useful.
:-0
In".
Proposition 6.6 Given any condition (-), the condition 3x]ly(-) is equivalent to 11w(Ix < w)(11y < w)(-).
'"'
-In
Proof Obviously ]w(]1x < w)(zly < w)(-) implies ]1x]1y(-). But also im,
the latter implies the former, for if x, y are numbers satisfying the condition
(-), we can take w to be the larger of x and y, and then we have (Ilx <
w)Py < w)(-).
As a consequence we have
Theorem 6.7 The existential quantification of a E1-relation is E1.
'---
Proof Suppose R(xl, ... , xn, y) is El. Then for some Eo-relation S(xi, ... , xn, y, z) the relation R(xi,... , xn, Y) is I zS(xl, ... , xn, y, z). Then the relation z1yR(xl,... , xn, y) is the relation ]y]zS(xl.... , xn, y, z). But this is equivalent to zlw(ly < w)Pz < w)S(xl, ... , xn, y, z) (by the above proposition). Therefore I]yR(x1, . . . , xn, y) is the existential quantification of the Eo-relation (Ely < w) (Ely < w)S(xl, ... , xn, y, z), and hence is
El.
Chapter 6. Introduction to formal systems and recursion
126
Exercise 14 Show that if R(xl,... (Ely < z)R(xl,...,xn,y,z)
,x
y, z) is Eo, so is the relation
Exercise 15 Show that the collection of E1-relations is closed under unions and intersections.
Exercise 16 [Tricky!] Show that the class of E1-relations is closed under finite universal quantifications. [Hint: show that (Vx < y)Elz(-) is equivalent to ]w(dx < y)(Iz < w)(-).]
§7 Recursive pairing functions
v-.
It will be handy to have a 1-1 recursive function J(x, y) from the set of ordered pairs of numbers (positive integers) onto the set of numbers. [To say that J is 1-1 means that if J(x, y) = J(z, w) then x = z and y = w. To say that J is onto the set N of positive integers is to say that for every z in N, there are elements x, y of N such that J(x, y) = z.] There are many ways of constructing such a pairing function J(x, y). The following is one way. We start with (1, 1), followed by the three pairs whose highest number is 2, in the order (1, 2), (2, 2), (2, 1), followed CDR
by the five pairs whose highest number is 3, in the order (1, 3), (2, 3), (3,3),(3,1),(3,2).... followed by the 2n - 1 pairs whose highest number is n, in the order (1, n), (2, n), ... , (n, n), (n, 1), (n, 2).... , (n, n - 1), and so forth. We then take J(x, y) to be the position that (x, y) occupies in this enumeration - in other words, the number of pairs that precede (x, y) in the ordering plus one. Thus J(1,1) = 1, J(1, 2) = 2, J(2, 2) = 3, J(2, 1) = 4, J(1, 3) = 5, J(2, 3) = 6, J(3, 3) = 7, J(3, 1) = 8, J(3, 2) = 9, J(1, 4) = 10, and so forth. We will now show:
Theorem 7.1 (a) If m < n then J(m, n) = (n - 1)2 + M. (b) If m > n then J(m, n) = (m - 1)2 + m + n.
"'O
Proof For any n, the pairs (x, y) that precede (1, n) are those in which x and y are both less than n, and there are (n - 1)2 such pairs. Therefore J(1, n) = (n-1)2+1. Hence J(2, n) = (n-1)2+2, ... , J(n, n) = (n-1)2+n. Thus for m < n, J(m, n) _ (n - 1)2 + in. This proves (a).
Now, J(m, m) = (m - 1)2 + m, hence J(m,1) = (m - 1)2 + m +
1,J(m,2) = (m-1)2+m+2,...,J(m,m-1) = (m-1)2+m+(m-1), and so for n < in, J(m, n) = (m - 1)2 + m + n. This proves (b).
§7 Recursive pairing functions
127
It then easily follows that the function J(x, y) is recursive - in fact the relation J(x, y) = z is Eo, because for any numbers m, n, and p
J(m, n) = p
if
(m
nA(m-1)2+m+n=p)
tea.
c,,
'c3
[The minus sign shouldn't bother us; we can write (n - 1)2 + m = p in the equivalent form (zlx < p) (x + 1 = n A x2 + m = p).] We leave it to the reader to supply the remaining details that prove that the relation J(x, y) = z is Eo, and hence is recursive. Remark -IN
An alternative pairing function that works is J(x, y) = z (x + y - 1)(x + y - 2) + x (see [8], p. 14). If J(x, y) is a pairing function for the positive integers, then it is easily seen that the function J(x+1, y+1) -1 is a pairing function for the natural numbers.
Exercise 17 Prove the last statement above. The inverse functions K and L
Each number x is J(xl, x2) for exactly one pair (xl, x2), and we define
K(x) = xl and L(x) = X2. Then J(K(x), L(x)) = J(xl, x2) = x. Thus J(Kx, Lx) = x. [Given a function f (x) of one argument, we often write f x for f (x).]
Also, for any numbers x and y, if we let z = J(x, y), then Kz = x and Lz = y - in other words K(J(x, y)) = x and L(J(x, y)) = y. The functions K(x), L(x) are recursive (in fact constructive arithmetic) '+1
since K(z) = x if (Ely < z)(J(x,y) = z), and L(z) = y if (Ix < z) (J(x, y) = z). We thus have
Theorem 7.2 There is a 1-1 recursive (in fact constructive arithmetic) function J(x, y) and recursive (in fact constructive arithmetic) functions K(x), L(x) such that for all numbers x and y.
(1) J(Kx, Lx) = x (2) K(J(x, y)) = x and L(J(x, y)) = y
Chapter 6. Introduction to formal systems and recursion
128
Exercise 18 Show that for any recursive function f (x, y) there is a recursive function z/' (x) such that for all x and y, f (x, y) = 0(J(x, y)). The functions Jn(x1, ... , xn)
For each n > 2 we define the function Jn(x1i ... , xn) by the following inductive scheme J2(xl, x2)
J3(xl,x2,x3)
= J(xl, x2) = J(J2(xl,x2),x3)
Jn+l(xl,...,xn+l) = J(Jn(xlo...,xn),Xn+l) An obvious induction argument shows that for each n > 2, the function Jn(x1, ... , xn) is a 1-1 recursive function onto the set N of positive integers.
Theorem 7.3 (a) For any r.e. set A and any n > 2, the relation Jn(x1i ... , xn) E A is r.e.
(b) For any r.e. relation R(xl, ... , xn), if A is the set of all numbers J(xl, ... , xn) such that R(x1,... , xn), then A is r.e. a16
Proof (a) follows from Theorem 2.3 (b), and (b) follows from Theorem 5.4.
The inverse functions Kin
For each n > 2 and i < n we define the functions Ki, K2, ..., KZ, ..., Kn by the following scheme. We take K1(x) = K(x) and K2 (x) = L(x). Now suppose n > 2 and the functions Ki , ... , Kn are defined. Then we
take Ki +l(x) = Ki (K(x)), for i < n, and we take Kn+1(x) = L(x). [For example, Ki (x) = K(K(x)); K2 (x) = L(K(x)); K3 (x) = L(x).] An obvious inductive argument yields.
Theorem 7.4 For any n > 2 and i < n: Ki (Jn(xl, ... , xn)) = xi. From which easily follows
Theorem 7.5 Jn(Ki (x), ... , Kn(x)) = x. Later we will need
§8 Dyadic Godel numbering
129
Theorem 7.6 For any r.e. relation R(xi,... , xn, yl, ... , ym) there is an r.e. relation M(x, y) such that for all -Ti, ... , xn, yl, .... ym: R(xl, ... , xn, yl, ... , ym) H M(Jn(xl, ... , xn), Jm(yl, ... , ym))
i
Proof Take M(x,y) = R(Ki (x), ... , Kn (x), Km (y), ... , K,, (y))
§8 Dyadic Godel numbering We now turn to a very useful method of assigning Godel numbers to expressions.
For any positive n, we let IIn be the number 2n+1, which in dyadic notation consists of a string of is of length n followed by 2 [e.g., 113 = 24, which in dyadic notation is "1112"]. Now let K be a finite ordered alphabet (k1, k 2 ,-- . , kn). For any string X in K we will take its (dyadic) Godel number to be the number which in dyadic notation is the result of replacing each symbol ki by the dyadic numeral Ili (e.g., the Godel number of k3klk2 is 111212112 (when written in dyadic notation)). We let g(X) (also written Xo) be the Godel number of X, and for any relation W
of words in K, we let g(W) - also written Wo - be the corresponding relation of Godel numbers, i.e., the set of all n-tuples (g(Xi),... , g(Xn))
-M'
such that (X1, ... , Xn) E W. We note that this dyadic Godel numbering has the nice technical advantage of being an isomorphism with respect to concatenation, i.e., if x is the Godel number of X and y is the Godel number of Y, then xy (i.e., x followed by y) is the Godel number of XY. [As previously stated, by xy ,we always mean x followed by y not x times y, which is written x y.] It sometimes happens that 1 and 2 are themselves symbols of K, in which case we will order K so that 1,2 are the beginning symbols (thus K is in some order 1, 2, k 3 ,-- . , kn), and so each dyadic numeral itself has a Godel number, obtained by replacing 1 by 12 and 2 by 112.
Theorem 8.1 (a) If 1,2 are symbols of K, then the relation g(x) = y is f.r. over K. (b) For K the alphabet {1, 2}, the function g is recursive.
Proof (a) The relation g(x) = y is represented by "P" in that E.F.S. over K whose axiom schemes are
Pk1i12
Chapter 6. Introduction to formal systems and recursion
130
Pk2,112
Pkn, Hn
Px, y -+ Pz, W - Pxz, yw (b) By (a) the relation g(x) = y (among dyadic numerals) is r.e., hence by Theorem 5.1, the function g is recursive. For use in a later chapter, we will need the following generalization of (a) of Theorem 8.1.
Theorem 8.2 Let K be an alphabet (al, ... , an) and let A be an admissible notation whose alphabet is a sub-alphabet of K. For any expression X in K, let X. be the .-numeral for the dyadic Godel number of X. Then the set R of all ordered pairs (X, Xo), where X is a string in K, is f.r. over K.
Proof Assume the given conditions. For each number n, let n be the A-numeral for n. By Theorem 4.3 the relation of dyadic concatentation is fir. in .-notation, and since the alphabet of A is a sub-alphabet of K, then by Theorem 3.4, there is an elementary formal system (E) over K in which some predicate C represents the set of all tuples (x, y, z) such that x * y = z. Now take a new predicate K and add the axioms
Raj,1 Ran, n Rx, y -i Rz, w -> Cy, w, v -* Rxz, v
Then "R" represents the desired relation. Remark S."
Our main application (in fact the only one) of Theorem 8.2 will be that in which A is the Peano notation. .re
Suppose now that K is an alphabet of two or more signs and W is a relation of words in K. We wish to show that W is f.r. over K if and only if the corresponding relation Wo of dyadic Godel numbers is r.e. Well, let
§8 Dyadic Godel numbering
131
K be an alphabet (al, a 2 ,-- . , a,n). The set of Godel numbers of all strings
in K is the set of all numbers that can be expressed as concatenations and so is recursively enumerable, since it is of the numbers represented by G in the dyadic arithmetic whose axioms are G12 G112
Gll Gx -+Gy -+ Gxy
Ian
Now let (E) be an elementary formal system over K in which P represents W. We construct a dyadic arithmetic (E)o in which P represents WO as follows. We first take a predicate G that doesn't occur in any of the axioms of (E) and we start (E)o with the above n+1 axioms. Thus G represents the set of Godel numbers of the strings in K. Next, let X1, ... , X, be the axioms of (E). For each axiom X of (E), let X' be the result of replacing each symbol ai of K by its dyadic Godel numeral IIi. [For example, .'."
if X is the string Qa2xaly, then X' is the string Q112xl2y.] Next,'let X* ... - > Gxn - + X', where x1, ... , xn are the variables be the string Gx1 that occur in X. [For example, if X is the string Qa2xaly, then X* is the string Gx -4 Gy -> Q112x12y.] Then we add to the n + 1 axioms already at hand, the axioms Xi , ... , X,*. In this dyadic arithmetic, P represents ...
(L)
`"'zoo
WO. Thus, if W is f.r. over K, then WO is r.e. Going in the other direction, suppose Wo is r.e. Without any real loss of generality, we can assume that 1,2 are symbols of K (if not, we use any two others in their place). Since WO is f.r. over the sub-alphabet D of K, it is f.r. over K (by Theorem 3.4). Also by (a) of Theorem 8.1, the relation
g(x) = y is f.r. over K, and since W = g(Wo), then W is f.r. over K by Theorem 2.4. We thus have
Theorem 8.3 W is f.r. over K if WO is r.e. Next we get
Theorem 8.4 Suppose that K has at least two symbols and that L is an extension of K. Then for any relation W of words in K, W is f.r. over L if W is f.r. over K.
Chapter 6. Introduction to formal systems and recursion
132
Proof We already know that if W is f.r. over K then it is f.r. over L (Theorem 3.4). Now suppose that W is f.r. over L. Let us order L so that the symbols of K occur at the beginning; let gl be the dyadic Godel numbering of the strings of K and let 92 be the dyadic Godel numbering of the strings of L. Thus 92 is an extension of the function gl, i.e., for any string X in K, 91(X) = 92 (X). Since W is f.r. over L, then g2(W) is r.e. (by Theorem 8.3), hence gi(W) is r.e. (since g1(W) = 92(W)), hence W is f.r. over K by Theorem 8.3.
§9 Lexicographical Godel numbering and n-adic notation We now wish to show that for each n > 2, a numerical relation is formally representable in n-adic notation if it is r.e. (formally representable 0..
in dyadic notation). For each n > 2, let Zn(x, y) be the relation "x is an n-adic numeral, y is a dyadic numeral, and the number designated by x in n-adic notation is the same as the number designated by y in dyadic notation."
Lemma 9.0 For each n > 2 the relation Zn (x, y) is formally representable over the alphabet Dn of n-adic digits.
Proof We are assuming that D2 is a sub-alphabet of Dn. By Theorem 4.1, the successor relation is f.r. in n-adic notation over Dn. Also by Theorem 6.1, the successor relation is f.r. in dyadic notation over D2, hence by Theorem 3.4, it is f.r. in dyadic notation over the larger alphabet D. Let (E) be an elementary formal system over Dn in which P represents the successor relation in dyadic notation and Q represents the successor relation in n-adic notation. [Thus, Px, y is provable in (E) if x and y are dyadic numerals and for some number k, x is the dyadic numeral for k and y is the dyadic numeral for k + 1, whereas Qx, y is provable in (E) if x and y are n-adic numerals for some k and k + 1 respectively.] Now take a new predicate Z and add to (E) the axioms
Z1,1 Zx, y -> Qx, X1 -> Py, yl -4 Zx1, yl
Then Z represents the relation Zn(x, z).
Theorem 9.1 For each n > 2, a numerical relation R(x1, ... , xt) is f.r. in n-adic notation if it is r.e.
§9 Lexicographical Godel numbering and n-adic notation
133
Proof Its
(a) Suppose R is representable in n-adic notation. Let (E) be an E.F.S. over Dn in which "R" represents R and "Z" represents the relation Zn(x, y) of the above lemma. Add a new predicate B and the axiom
Rxl,...,xt -4 Zxi,yi -+ ... -4 Zxt,yt -4 Byl,...,yt Then B represents the relation R in dyadic notation but over the alphabet Dn. However, R is then f.r. in dyadic notation over the smaller alphabet D2 by Theorem 8.4.
(b) Conversely, suppose R is f.r. in dyadic notation over D2. Then by Theorem 3.4, it is formally representable in dyadic notation over the
larger alphabet Dn. Let (E) be an E.F.S. over Dn in which R is represented in dyadic notation by a predicate "R" and in which "Z" represents the relation Zn(x, y). Then take a new predicate C and add the axiom
Ryi,...,yt -> Zxi,yi -+ ...Zxt,yt -4 Cxl,...,xt Then C represents R in n-adic notation (over the alphabet Dn).
Corollary 9.2 For each n > 2, a numerical relation is solvable in n-adic notation if and only if it recursive. n-adic concatenation For any n > 2, by x *n y we mean the number designated in n-adic notation
by the n-adic numeral for x followed by the n-adic numeral for y. The relation x *n y = z - which we call n-adic concatenation - is obviously f.r. in n-adic notation - it is represented by "C" in that E.F.S. over Dn whose sole axiom is Cx, y, xy. Hence by Theorem 9.1 we have
Corollary 9.3 For each n > 2, the relation of n-adic concatenation is r.e.
Lexicographical Godel numbering
Another method of Godel numbering that we have found useful is the following. For K an ordered alphabet (al, ... , an), arrange all strings in K in lexicographical order and then assign to each string X in K its position in that sequence. We call that number the lexicographical Godel number
Chapter 6. Introduction to formal. systems and recursion
134
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of X. Equivalently, if we look at the symbols a1,.. . , an as n-adic digits, then the number thus designated (in n-adic notation) is the lexicographical Godel number of X - in other words if X is a string a;,,.a;,,._1 ... ai0, then the lexicographical Godel number of X is io + iln + i2n2 + ... + irnr. We note that in this Godel numbering, all positive integers are Godel numbers. We let h(X) be the lexicographical Godel number of X. [We still sue.
write g(X) for the dyadic Godel number of X.] And for any relay tion W(xl,... , xt) of strings in K, we let Wh be the corresponding relation of lexicographical Godel numbers, i.e., the set of all t-tuples (h(X1),... , h(Xt)) such that W(X1,... , Xt). It is immediate that W is f.r. over K if Wh is a numerical relation that is f.r. in n-adic notation, ms's
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which is (by Theorem 9.1) in turn the case if Wh is r.e. From this it also follows that W is solvable over K if Wh is recursive. And so we have
Theorem 9.4 For h the lexicographical Godel numbering of all strings .'3
in an ordered alphabet K, a set or relation W of words in K is f.r. over K if Wh is r.e., and is solvable over K if Wh is recursive. On the whole, dyadic Godel numbering will play a much more important role in this volume than lexicographical Godel numbering.
Chapter 7
A universal system and its applications Part I A universal system We now wish to construct a "universal" system (U) in which we can, so to speak, express all propositions of the form that such and such a number is in such and such a recursively enumerable set, or more generally that
such and such an n-tuple of numbers is in such and such a recursively enumerable relation. A complete knowledge of which sentences in this C/)
system are true, and which are not, would imply a complete knowledge of all formalized mathematics. But we will see that the problem is undecidable - a result that Post[13] calls "Godel's Theorem in Miniature." Also, our universal system will be the basis for two fundamental results in recursion theory - the Enumeration Theorem and the Iteration Theorem, which are indispensable for many of the chapters that follow. (1)
§1 The universal system (U) Preparatory to the construction of (U), we need a device for "transcribing" all dyadic arithmetics (elementary formal systems over the alphabet {1, 2}) into one single finite alphabet. We now define a transcribed dyadic arith-
metic - abbreviated "T.A." - to be a system like a dyadic arithmetic mar,
except that instead of taking our variables and predicates to be individual symbols, we take three signs v, ',p and define a transcribed variable 'T3
- abbreviated "T.A. variable" - to be any of the strings v', v", v"', ..., and we define a transcribed predicate - abbreviated "T.A. predicate" to be a string of p's followed by a string of accents; the number of p's is to indicate the degree of the predicate. A transcribed dyadic arithmetic,
Chapter 7. A universal system and its applications
136
cad
then, is like a dyadic arithmetic except that we use transcribed variables instead of individual symbols for variables, and transcribed predicates instead of individual symbols for predicates. We thus have one single alphabet K7 (namely the seven symbols 1 2 v ' p , -+ ) from which all T.A.'s are constructed. It is obvious that representability in an arbitrary dyadic arithmetic is equivalent to representability in one in transcribed notation. We define "T.A. term," "atomic T.A. formula," "T.A. formula," as we do term, atomic formula, formula, sentence, using "transcribed variable", "transcribed predicate" instead of "variable," "predicate," respectively. We now construct our system (U) as follows. We first extend K7 to an alphabet K8 by adding the symbol *. To any T.A. whose axioms are X1,.. . , XK we wish to associate a string in K8; this string shall be X1*X2* ... *Xk (or X1 alone, if k = 1). Such a string B is called a base (after Post). The T.A. formulas X1,... , Xn are called the components of the base X1* ... *Xk. Next, we extend K8 to an alphabet K9 by adding the symbol I-, and we define a sentence (of (U)) to be an expression of the form B F- X, where B is a base and X is a T.A. sentence. We define the sentence B F- X to be true if X is provable in that T.A. whose axioms are the components of B. Thus X1* ... *Xk F- X is true if X is provable in that transcribed dyadic arithmetic whose axioms are X1i ... , Xk. By a predicate H of (U) (not to be confused with a T.A. predicate!) we shall mean an expression of the form B F- P, where B is a base and P is a transcribed predicate; the degree n of H is, by definition, that of P. A predicate H of (U) of degree n is said to represent in (U) the set of all n-tuples (al, ... , an) of numbers (dyadic numerals) such that the sentence Ha 1, . . . , an is true. Thus a predicate X1* ... *Xk F- P of degree n represents in (U) the set of all n-tuples (al, ... , an) such that Pal,... , an is provable in that T.A. whose axioms are X1, ... , Xk - in other words it represents the very same relation as P represents in that T.A. whose axioms are X1, ... , Xk. And thus a relation is representable in (U) if it is r.e. In this sense, (U) is called universal for the collection of r.e. relations. 'c7
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We let T be the set of true sentences and To be the set of Godel numbers of the true sentences (under the dyadic Godel numbering). We now wish to prove the basic result that the set T is formally representable over K9 (and ..j
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cm.
thus (U) is a formal system, though not an elementary formal system), and that the set To is recursively enumerable. [The entire development of recursive function theory from the viewpoint of elementary formal systems rests on this result, which is a variant of a result of Post[13].] We shall now construct an E.F.S. W over K9 in which the set T is represented.
§1 The universal system (U)
137
Construction of the system W
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We must first note that the implication sign of W is to be distinct from the implication sign of T.A.'s. We could continue to use "-p" for the implication sign of (U); we prefer however to use ....... for the implication sign of W, since it will occur so frequently, and we shall now take the implication sign of T.A.'s to be "imp." Similarly, we shall now use the ordinary comma for our punctuation sign of W, and "com" for the punctuation sign of T.A.'s. For variables of W (not to be confused with T.A. variables) we will use the letters x, y, z, w, with or without subscripts. Predicates of W (not to be confused either with predicates of (U) or T.A. predicates) will be introduced as needed. We now introduce the axioms of W in groups, first explaining what each newly introduced predicate of W is to represent. N represents the set of numbers (dyadic numerals). p.,
Ni N2
Nx-+ Ny-Nxy Acc represents the set of strings of accents. Acc' Acc x -> Accx'
V represents the set of T.A. variables. Accx -+ Vvx P represents the set of T.A. predicates.
Accx-Ppx Px -+ Ppx t represents the set of T.A. terms.
Nx-+tx tx-+ty-4 txy FO represents the set of atomic T.A. formulas.
Accx -+ty -+Fopxy Fox ->ty-4 Fopxcomy
Chapter 7. A universal system and its applications
138
F represents the set of T.A. formulas.
Fox->Fx "ti
Fox -+ Fy -4 Fx imp y So represents the set of atomic T.A. sentences.
Accx ->Ny ->Sopxy Sox-+ Ny ->Sopxcomy F-+
S represents the set of T.A. sentences.
Sox-+Sx Sox
->Sximpy
d represents the relation "x and y are distinct variables."
Vx-+Accy->dx,xy dx,y->dy,x Sub represents the relation "x is any string compounded from numerals, T.A. variables, T.A. predicates, com, imp; y is a variable; z is a numeral and w is the result of substituting z for all occurrences of y in x (that is, all occurrences which are not followed by more accents).
Nx->Vy-Nz-Subx,y,z,x Vx Sub x, x, z, z dx, y -+ Nz -+ Sub x, y, z, x
Subx,y,z,x Vy - Nz -+ Sub com, y, z, com Vy - Nz -4 Sub imp, y, z, imp Sub x, y, z, w -+ Sub xl, y, z, wl -> Sub xxl, y, z, wwl
pt represents the relation "x is a T.A. formula and y is the result of substituting numerals for some (but not necessarily all) variables of x (such a formula y is called a partial instance of x).
Fx -> Subx,y,z,w -+ ptx,w
ptx,y-+pty,z-4ptx,z in represents the relation "x is a T.A. formula and y is the result of substituting numerals for all variables of x (y is an instance of x).
ptx,y-+Sy-4 inx,y
§2 The recursive unsolvability of (U)
139
B represents the set of bases of (U).
Fx->Bx Bx -+Fy -+ Bx*y C represents the relation "x is a component of the base y."
Fx -+ Cx, x
Tr represents the set of true sentences of (U). Cx, y -4 in x, xl -+ Tr y F- x1
TryF- x -iTryF- ximpz ->Sox ->Tryl- z This concludes the construction of the system W in which T is represented. To represent To over {1, 2}, just take all the above axioms and replace each symbol of K9 by its dyadic Godel number. And so we have proved:
Theorem 1 The set T of true sentences of (U) is formally representable over K9 and its set To of Godel numbers is recursively enumerable.
§2 The recursive unsolvability of (U) One of the fundamental results of recursion theory is that there exists a recursively enumerable set that is not recursive. Well, we are about to see that To is such a set. The norm operation introduced in Chapter 1 will
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now be used for a good purpose. For any string X in K9, we define its norm to be the string XXo, i.e., X followed by its own Godel number (written in dyadic notation). We note that if H is a predicate of (U) of degree 1, then its norm HHo is a sentence
that is true if the Godel number of H lies in the set represented by H in (U). If n is a dyadic numeral, then n itself has a Godel number no, hence also
a norm nno. Since our Godel numbering is an isomorphism with respect to concatenation, it follows that if n is the Godel number of X, then its norm
nno is the Godel number of the norm Xn of X. Thus the Godel number of XXo is XoXoo We let r(x, y) be the number xyo. Thus the norm of x is r(x, x). For any set A of numbers we let A# be the set of all numbers x whose norm is ..-"r
in A.
Chapter 7. A universal system and its applications
140
Lemma 2.0 (a) The function r(x, y) is recursive,
(b) If A is r.e. so is A#.
Proof (a) By (b) of Proposition 8.1 of Chapter 6, the relation xo = y is r.e. cad
Then r(x, y) = z is the r.e. relation 3w (w = yo A xw = z), hence r is a recursive function.
(b) Suppose that A is r.e. Then x c A# if ly(r(x,x) = yAy c A). Thus A# is r.e. Godel sentences
Call a sentence X of (U) a Gddel sentence for a set A of numbers if either X is true and its Godel number is in A, or X is false and its Godel number
is not in A - in other words X is true
X0 E A
Theorem 2.1 For any r.e. set A there is a Godel sentence X for A. Proof Suppose A is r.e. Then so is A# by the above lemma. Then A# is represented in (U) by some predicate H. Then for every number n,
Hn is true H n E A#, hence Hn is true H nno E A Taking h for n, where h is the Godel number of H, we have: Hh is true
hhoEA. However, hho is the Godel number of Hh, and so Hh is a Godel sentence for A.
Remark
Having proved (b) of Lemma 2.1, Theorem 2.0 is actually a corollary of Theorem 1 of Chapter 1. Can the reader see why this is so?
Exercise 1 Why is this so? Also, how is Theorem 2.1 a special case of a result proved in Chapter 4? What representation system is appropriate for the system (U)?
§3 The enumeration theorem
141
The non-recursiveness of To
There obviously cannot be a Godel sentence for the complement To of To, CDn
for such a sentence would be true if and only if its Godel number was not the Godel number of a true sentence, which is impossible. And so by Theorem 2.1, it follows that To is not r.e. Thus we have
Theorem 2.2 The set To is recursively enumerable but not recursive.
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Exercise 2 A set A is said to be (many-one) reducible to a set B if there is a recursive function f (x) such that for all x, x E A f (x) E B (in other words, A = f-1(B)). A set B is called universal if every r.e. set is reducible to B. Show that the set To is universal,
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Exercise 3 Show that To has the stronger property that there is a recursive function f (x, y) such that for every r.e. set A, there is a number i such that for all x: x E A H f (i, x) E To.
Exercise 4 Show that every universal set B has this stronger property. [Hint: To is reducible to B.] C71
Exercise 5 Show that no universal set can be recursive. [This generalizes Theorem 2.2.]
Exercise 6 Show that if A is reducible to B and A is non-recursive, then B is non-recursive.
Part II Enumeration and iteration theorems We now turn to the study of two basic theorems of recursion theory - the enumeration theorem, due to Post and to Kleene, and the iteration theorem (a variant of the s-m-n theorem of Kleene, also discovered independently by Martin Davis). After having proved these two theorems, the universal system (U) will have served its purpose and will no longer be needed.
§3 The enumeration theorem We wish to arrange all r.e. sets in an infinite sequence w1, w2, ... , w,n, .. . (allowing repetitions) in such a way that the relation "wx contains y" is
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142
an r.e. relation. For this purpose we use the universal system (U). We are letting T be the set of true sentences of (U), and To the set of Godel cad
numbers of the true sentences of (U). We are letting r(x, y) be the number xyo, and we have already shown that the function r is recursive, and since To is an r.e. set, then the relation r(x, y) E To is r.e. We now define wn to
be the set of all x such that r(n, x) E To. Thus X E wn H r(n, x) E To. Since the relation r(x, y) E To is r.e. and the r.e. relations are closed under introduction of constants, then for each n, wn is an r.e. set. We call n an index of wn. Thus every set with an index is r.e. Conversely, every r.e. set
ran
has an index, because if A is r.e., then it is represented in (U) by some predicate H with Godel number h, and so for all x, x E A Hx E T <--+ r(h, x) E To H X E wh. Thus every r.e. set A has an index. We have thus proved.
Proposition 3.0 A set is r.e. if and only if it has an index.
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C'3
We have thus succeeded in enumerating all r.e. sets in an infinite sequence w1, L02.... , wn, ... in such a way that the relation "wx contains y" (which is the relation r(x, y) E To) is r.e. We next wish to show that for each n > 2, we can enumerate all r.e. relations of degree n in an infinite sequence Rl , R2 , ... , R,... in such a manner that the relation Rx (y1, ... , yn) is an r.e. relation among x, yi, ... , yn. yam,
To this end it will be particularly useful to use the indexing of r.e. sets that we already have and to use the recursive pairing function J(x, y) and the n-tupling functions Jn(x1i ... , xn) of Chapter 6. We simply define Ri (x1, ... , xn) if Jn(x1, ... , xn) E wi. Thus Ri is the set of all n-tuples (x1, ... , xn) such that Jn(x1, ... , xn) E wi. Since wi is r.e., so is R2 (by (a) of Theorem 7.3, Chapter 6). Also, given any r.e. relation R(xi..... .xn), the set A of all numbers Jn(x1i ... , xn) such R(x1, .... xn) - this set A is r.e. (by (b) of Theorem 7.3, Chapter 6), hence is wi for some i. Therefore R(xii... , xn) H Jn(x1, ... , x,) E wi H Ri (xl, ... , xn), and so R = Ri . cad
x0.
cad
Thus every r.e. relation has an index. We are regarding sets as special cases of relations, and so we let Ri be wi. We thus have:
Theorem 3.1 [The enumeration theorem] For each positive n, the enucar
cad
meration Ri , R2 , ... , Rn, ..., of all r.e. relations of degree n is such that the relation Rx (y1, ... , yn) (as a relation among x, y1, ... , yn) is r.e.
cam'
For each n we let Un+1 be the set of all (n + 1)-tuples (x, y1i ... , yn) such that Rx (yl,... , yn). Thus Un+1(x, yl, ... yn) H Rx (yl, .... yn). The relations U2, U3'. .. , Un,... are sometimes referred to as the universal re-
§3 The enumeration theorem
143
lations (U2 (X, y) is the universal relation for r.e. sets, U3 (X, yl, y2) is the universal relation for r.e. relations of degree 2, and so forth).
No ambiguity will result if we henceforth write Ri(x1i... , x,z) for Cad
U'.
R2 (xl, ... , x,,) and U(x, yl, ... , y,,,) for Un(x, yl, ... , y,, ), since the number of arguments makes the superscript clear. And we repeat that we are using "R" synonymously with "w2." Note C,3
It is also possible to give an enumeration theorem for r.e. relations without using a pairing function. We let c be the Godel number of the punctuation symbol com of (U) and for every n > 2 we define rn+1(x, y1, ... , yn) oho
to be the number xz1cz2c... czn, where for each i < n, zi is the Godel number of (the numeral) yi. Thus if x is the Godel number of some expression X of (U), then r(x, yi, ... , yn) is the Godel number of the expresCSI
sion X Y1 com yl corn ... com yn. We could then define Rz (x1, ... , xn) if r(i, x1i ... , xn) E To. Then every r.e. relation R(xl,... , xn) is represented in (U) by some predicate H with Godel number h, and so R = R. However, we prefer the indexing using the functions J,- (xl, ... , xn),
since it is a technical advantage that R(xl,... , xn) is equivalent to J(xl, ... , xn) E wi Exercise 7 We have proved the enumeration theorem for r.e. relations of positive integers. From this we can obtain an enumeration theorem for r.e. relations of natural numbers as follows. For any positive n and natural number i define Si as the set of all n-tuples (x1, ... , xn) of natural numbers such that R+1(x1 + 1, ... , xn + 1). [Thus S? (xl, ... , xn) R +1(x1 + 1, ... , xn + 1).] Show that for every r.e. relation S(xl,... , xn) of natural numbers, there is a natural number i such that S = Sin.
Exercise 8 Let us go back to the enumeration theorem for r.e. relations of positive integers. Given an r.e. set wi, the number i is not necessarily the Godel number of 4.,
some predicate which represents wi in (U). Prove that there is a recursive function q(x) such that for any number i, 0(i) is the Godel number of a predicate which represents wi in (U). [Informally, this means that given an r.e. set, in the sense of given an index of it, we can effectively find a predicate of (U) to represent it.] In fact there is such a recursive function O(x) with the additional property that x < O(x), for every x. Prove this. cad
Exercise 9 Prove that every r.e. set has infinitely many indices.
Chapter 7. A universal system and its applications
144
§4 Iteration theorem The second basic tool of recursive function theory is the iteration theorem, which we will shortly state and prove. To fully appreciate the power and significance of this theorem, the reader should first try solving the following exercises. 'C7
Exercise 10 We know that for any two r.e. sets A and B, their union 04'
AU B is r.e. Is there a recursive function O(x, y) such that for all numbers i and j, wb(ij) = wiUwj? [The question can be phrased informally: given two r.e. sets (in the sense that we are given indices of them), can we effectively find an index of their union?]
Exercise 11 Show that there is a recursive function O(x) such that for every number i, R0(i) (x, y) is the inverse of the relation Ri (x, y) (i.e., for all x and y: Rb(i) (x, y) <-+ Ri(y, x). [This can be paraphrased: Given an index of an r.e. relation of two arguments, we can effectively find an index of its inverse.]
Exercise 12 Show that for any recursive function f (x) there is a recursive function t(x) such that for every number is wt(i) = f -' (wi) (i.e., for all x, x c wt(i) <-+ f(x) E wi).
The reader who has solved the above problems has probably found himself repeatedly having to go back to the universal system (U). As we will see, the iteration theorems below will free us from this once and for all.
Theorem 4.0 [Simple iteration theorem] For every r.e. relation R(x, y) there is a recursive function t(y) such that for every x and is R(i, x) <--* x E wt(i).
'O""
Proof The r.e. relation R(x, y) is represented in (U) by some predicate H of degree 2 with Godel number h. We let c be the Godel number of the punctuation symbol com. And so R(i, x) <-a Hi coma E T F-* hiocxo E
To # x E Whi0c. And so we take 0(y) = hyoc and we have R(i, x) " x E Wt(i).
From the above we obtain:
§4 Iteration theorem
145
Theorem 4.1 [Iteration theorem] For any positive integer m there is a recursive function q5(xl,... , xm) such that for any positive n and any r.e. relation R(xi,... , xm, y1, , yn) the following holds for all 21i...,Zm,yl,...,yn: RO(i..... i.a) (yl, ... , yn) t-+ R(il, ... , im, yl, ... , Yn)
Proof Given an r.e. relation R(xl,... , xm, yl,... , yn), by Theorem 7.6, Chapter 6, there is an r.e. relation M(x, y) such that for all xl, ... , xm, iii, ... , yn: R(xl, ... , xm, Yl, ... , yn) holds if M(Jm(xl, ... )xm), Jn(y1, ... , yn)). Now, by Theorem 4.0 there is a recursive function t(y) such that for all x and /y, M(x,y) <--* y E wt(x), and therefore R(xj,...,xm, Yl, ... , yn) F--r M(Jm(xl, ... , xm), Jn(yl, ... , Yn)) t-a Jn(yl, ... , yn) E wt(J,n(xl,...,x,n)) f* Rt(J,.(x1,...,xr))(yl,... , yn). And so we take q(xl, ... , xm) = t(Jm(x1, ... , Xm)).
By applying Theorem 4.1 to the universal relation U(x, xi, ... , xm, yl,
.... yn) we have:
Theorem 4.2 [Uniform iteration theorem] For every positive m there is a recursive function sm(x, xl, ... , xm) such that for all i, il, ... , im, Y1,...,yn: Rsm (z,2i,...,im) (Y1, ... , yn) H Ri (21, ... , 2m, yl, ... ) yn)
Theorem 4.1 says informally that given an r.e. relation R(xl,... , xm, , yn), if w e plug in numbers i1 .... , im for the variables x1 7. .. , xm, Yi, we can find an index of the relation R(ii,... ) im, yl, ... , yn) (as a relation among yl, .... yn) as a recursive function of i1, ... , im. Moreover (in our version of the above theorem) the recursive function sm is independent of the number n. [The more usual version is that for every m and n, there is a recursive function sm (dependent on both m and n) such that the above equivalence holds (replacing "sm" by "sn ").] Let us now see how the iteration theorem gives easy solutions to the
three exercises above. .4r
W+0
For Exercise 10, let R(x, y, z) be the r.e. relation z E wx V z E wy. [It is r.e., since it is the union of the relation A xyzU2 (x, z) with the relation Axyz U2(y, z), both of which are explicitly definable from the relation U2(x, y).] Then by the iteration theorem there is a recursive function 0(yl, y2) such that for any numbers il, i2i x: x E wlk(il,i2) <-4 R(il, i2i x), 0'N
141
which means that x E wlk(il,i2) <-+ x E wil Uwi2 and thus wO(il,i2) = wil Uwi2 . 7
Chapter 7. A universal system and its applications
146
For Exercise 11, let R(x, y, z) be the r.e. relation Rx(z, y) (which is the relation U3(x, z, y), which is explicitly definable from U3(x, y, z)). Then by the iteration theorem there is a recursive function q(x) such that for all i, x, y: Rk(z) (x, y) H R(i, x, y). Hence Rqz) (x, y) <-+ Ri(y, x). Thus 170(j) = RS 1.
For Exercise 12, let R(x, y) be the r.e. relation f (y) E Wx (f is a given recursive function). [Why is this relation r.e.?] Then by Theorem 4.0 there is a recursive function O(x) such that for all i and x: x E W,(i) <-a R(i, x) <--* f (x) E Wi
X (Z-
f-1(Wi). Thus Wo(i) = f-1(Wi).
Other applications of the iteration theorem are given in the following
exercises.
Exercise 13 (a) Prove that there is a recursive function q(y) such that for all i, x: x c Wk (i) " Ri(x, x).
(b) Prove that there is a recursive function q(y) such that for all i,x: x c Wk(i) - 3yRi(x, y) (in other words, for all i, w,5(j) is the domain of R?).
Exercise 14 Here is a cute one. For any number n by {n} is meant the set whose only element is n. Prove that there is a recursive function O(x) such that for every n, Wo(n,) = {n}. More generally, prove that for any recursive function f (x) there is a recursive function q(x) such that for every n: Wq(n) = If (n)}.
Exercise 15 Prove that for any r.e. set A there is a recursive function q(x) such that for every number n in A, Wi(n) is the set of all positive integers, and for every n not in A, Wq(n) is the empty set!
Exercise 16 [Chapter 2 revisited] (a) Using the iteration theorem, show that there is a recursive function F(x, y) such that for all numbers a and b: WF(a,b) = {x : Ra(x, b)}.
(b) Let F(x, y) be such a function and let d(x) = F(x, x). Show that the hypotheses of Problems 3 and 4 of Chapter 2 are satisfied (where W1, W27 ... ,Wn, ... are the r.e. sets). Conclude therefore that (1) for
every recursive function f (x) there is a number n such that Wn = W f(n); (2) for every r.e. relation R(x, y) there is a number n such that Wn = Ix: R(x, n) }.
§4 Iteration theorem
147
[These are two (weak) forms of a result known as the recursion theorem. Other proofs of this are given in a later chapter.]
Exercise 17 We call a set A generative if there is a recursive function
:°s
S."
O(x) such that for every number i, 0(i) E A <-+ 0(i) E wi. Such a function O (x) is called a generative function for A, and we say that A is generative under O(x). The set C of all numbers x such that x c wx is generative - generative under the identity function I (x). The set C is also r.e., and so r.e. generative sets exist. We have defined a set A to be (many-one) reducible to B if A = f -1(B) for some recursive function f, and we have defined a set U to be universal if every r.e. set is reducible to U. Using the iteration theorem, prove that if A is reducible to B and A is generative, then B is generative. [Hint: suppose A = f-1(B), where f is recursive, and suppose A is generative under 0(x). By the iteration theorem there is a recursive function t(x) such that for all i, wt(i) = {x : f (x) E wi}. Show that B is generative under the function
f We have proved the enumeration and iteration theorems for r.e. relations of positive integers, and in Exercise 7 we showed how to obtain an enumeration theorem for r.e. relations of natural numbers. We can also obtain an iteration theorem for r.e. relations of natural numbers as indicated in the following exercise.
Exercise 18 Let Sz be the relation of natural numbers as defined in Exercise 10 (Si (x1i ... , xn) if R +1(x + 1, ... , xn + 1)). Now, for each positive n, define pn(x, x1, ... , xn) = sn(x + 1,x1 + 1,.. . , xn + 1) - 1 (x, x1, ... , xn are natural numbers). Prove that for all natural numbers 2,21i...,2m,X1,.... xn: Sp .(i,i1,...,i .) (x1i ... , xn)
H Si m(21, ... , 2m, xl) ... , xn)
II
Systems with Effective Properties
Chapter 8
Arithmetization of formal systems CDm
In Part I of this chapter we prove the fundamental result that if W is formally representable, then the corresponding relation Wo of dyadic Godel numbers is E1. This has several applications including a neat use of elementary formal systems in incompleteness proofs, which we explain in the next chapter. Next, it leads easily to the crucial fact that every r.e. relation is the existential quantification of a recursive relation - a fact needed for our "double enumeration" of Part II, which is basic to all of Chapters 10 and 11. In Part III we first prove the stronger fact that every r.e. relation is the existential quantification of a Eo-relation, and thus that the classes of r.e. relations and E1-relations are the same. Then we establish some closure properties of recursive relations that are of more specialized interest'. Part III is not necessary for any subsequent chapters; the results are of interest in their own right (particularly those of §4). e~+
Part I Arithmetization of elementary formal systems
°a."PC"
`.off
We now turn to the proof that if W is formally representable, then Wo is E1. Our proof will be wholly constructive in that given an elementary formal system in which W is represented, we can explicitly write down a E1-condition that describes Wo. To make this more precise, we must first define Eo-formulas and E1-formulas. We recall the language Lo briefly mentioned in Chapter 5, whose formulas are those of L in which the exponential symbol doesn't appear, which are called arithmetic formulas (note the small a!). We define x < y to be llz(x + z = y). For any variable 'It provides a neat substitute for that known as "course of values recursion."
Chapter 8. Arithmetization of formal systems
150
v, formula F, and variable or numeral a, we abbreviate Vv(v < a i F) by (Vv < a)F, and we abbreviate 3v(v < a A F) by (3v < a)F. By an atomic Eo-formula we mean an expression of one of the two forms a +,3 = ry, a /3 = ry, where each of a, /3, ry is either a variable or a numeral. Then the Eo-formulas are defined by the following inductive scheme.
(1) Every atomic Eo-formula is a Eo-formula.
(2) If F1 and F2 are Eo-formulas, so are -F, and F1 D F2 (and hence so are F1 V F2, F1 A F2, F1 - F2).
(3) If F is a Eo-formula then for any variable v and any a that is either a numeral or a variable distinct from v, the expressions (Vv < a)F and (2v < a)F are Eo-formulas. Then, of course, we define a E1-formula to be an expressions of the form
3vF, where v is a variable and F is a Eo-formula. As with all formulas, a E0 or E1-formula is called open if it has at least one free variable, otherwise it is said to be closed, or to be a sentence. Obviously every open Eo-formula expresses a Eo-relation and every open E1-formula expresses a El-relation.
What we meant, when we said that the proof we will give is wholly constructive, is that there is an effective procedure whereby from any elementary formal system in which W is represented we can construct a E1-formula that expresses Wo. For this proof, several preliminaries are necessary.
§1 Some facts about Eo-relations We let x * y be the number expressed in dyadic notation by following x by y (e.g., 212 * 1121 = 2121121). We are also writing xy for x * y (which we recall is not to be confused with x times y, which we write x y). Our first objective is to show that the relation x * y = z is E1.
Proposition 1.1 The relation x divy (x divides y) is E1.
Proof xdivyH Now we use a modification of a clever and most useful observation of John Myhill2. 2Myhill showed this to me sometime around 1953. I showed it to Quine, who was delighted. In later years, Myhill credited the idea to Quine, and Quite credited it to me! But I can assure you that the idea is unquestionably due to Myhill.
§1 Some facts about Eo-relations
151
Proposition 1.2 For any prime number p the set of all powers of p is E1.
Proof For any prime p, x is a power of p if and only if every proper divisor of x is divisible by p. We let powp(x) be the condition that x is a power of p. Then powp(x) F-r (Vy < x)((y divp A y
1) D p divy)
Note
If the reader is worried about the sudden appearance of the implication sign "D", it should be recalled that for any propositions p and q, p D q can be alternatively expressed as -p V q. For any number x, by L(x) we mean the length of x (in dyadic notation).
Proposition 1.3 The relation y = 2L(x) is Eo. Proof For any number r, the smallest number of length r is 11 ... _1, which is 2" - 1, and the largest number of length r is 2\2
, which is
2-(2' -1).
Thus a number x has length r if and only if the following condition holds: (*)
Next we show that 2L(x) = y if the following two conditions both hold. C1: y is a power of 2 C2:
y = 2L(x). Let r = L(x). Then of course y is a power of 2, and since x has length r, then 2r - 1 < x < 2 (2r - 1), and so
C2holds.
Conversely, suppose C1 and C2 both hold. Then by C1, y = 2r for some r. Then by C2, 2r - 1 < x < 2(2r - 1), and hence r = L(x), and so y = 2L(x) It remains only to show that condition C2 is Eo (condition C1 is Eo by Proposition 1.2). So we must show that each of the conditions y - 1 < x (a)
y-1<x-(E1z
Chapter 8. Arithmetization of formal systems
152 x < 2(y - 1)
(b)
F-r
x < (y - 1) + (y - 1) (2y1 < y)(EIy2 < Y) (X = y1 + y2)
Now we have
Theorem 1.4 The relation x * y = z is E0. And so x*y = z iff (Elz1 < z)(EIz2 < z)(2Lh> _ Proof x*y = z1 =z2Az2+y=z).
The relations xBy, xEy, xPy ..O
We define xBy (x begins y) to mean that x = y or xz = y for some z (which is necessarily < y). We define xEy (x ends y) to mean that x = y or zx = y
for some z. We define xPy (x is part of y) to mean that x begins some z that ends y (or equivalently, z ends some z that begins y, or equivalently that x either begins or ends y, or z1xz2 = y for some z1 and z2).
Proposition 1.5 The relations xBy, xEy, xPy are Eo.
Proof xBy xEy
F-r
((Elz < y) (xz = y) V x = y)
<-a
((2z < y)(zx = y) V x = y)
xPy H ((az < y)(xBz A zEy) Proposition 1.6 For each n > 2 the following relations are Eo.
(a) x1*...*xn=y (b) x1 * ... * xnPy
Proof (a) By Theorem 1.4, this holds for n = 2. For n > 3, x1 * ... * xn = y if the following condition holds: (2y2 < y)...(2yn-1 < y)
(yix2 = Y2 A y2x3 = y3 A
... A yn-1xn = y)
§1 Some facts about E0-relations
153
(b) xl * ... * xn,Py if (3z < y)(xl * ... * x,,, = z A zPy) We henceforth write xPy for -xPy. Dyadic Godel numbers cad
We recall that we are letting 11n be the number consisiting of a string of l's of length n followed by 2. We let Gn be the set of dyadic numerals compounded from 111, ... , 11n. [Thus for any alphabet K of n symbols, Gn is the set of dyadic Godel numbers of all strings in K.] We let Gm be the set of all dyadic numerals compounded from any members of the infinite CAD
sequence 111,112, .
, H ... .
Proposition 1.7 (a) The set Gm is Eo. (b) For each positive n, the set Gn is E0.
Proof (a) x E Gm <-a 1Bx A 2Ex A -22Px
(b) For each n, x c Gn * (x c Gm A -Hn+iPx) Substitution lemma
For our final preliminary, we establish a simple but basic lemma.
Consider an alphabet L of symbols ordered in some sequence (al,
.... am) and a string X compounded of the symbols of L and variables
`YT
,-r
xl, ... , Xn - as an example, suppose X = x2ala3xlx2al. We let g be the dyadic Godel numbering of the words in L. We let I (X) be the set of all instances of X (all strings resulting from substituting strings in L for the variables of X) and we let lo(X) be the set of dyadic Godel numbers of the strings in I(X). We wish to show that I0 (X) is E0. We first illustrate the proof with the example X = x2ala3xlx2al. Then x E Io(X) if
-o<
(2x1 <_ x)(EIx2 < x)(G3(XI) AG3(x2) Ax = x2121222x1x212).
numeral 11ii. We thus have
;-.
More generally, the condition x E 10(X) can be written (3xl < x) ... (2xn < x)(Gm(xl) A ... A G,n(xn) A x = X*), where X* is the string obtained from X by replacing each symbol ai of L by its dyadic Godel
154
Chapter 8. Arithmetization of formal systems
Lemma 1.8 [Substitution lemma] For any finite alphabet L, if X is any string composed of symbols of L and variables, the set of dyadic Godel numbers of all instances of X is E0.
§2 Arithmetization of elementary formal systems Now consider an E.F.S. (E) over K. Let L be the set of symbols of K together with the predicates of (E) and the comma and arrow. [Thus L is the alphabet out of which all the sentences of (E) are built.] We order L so that the symbols of K come at the beginning and we let g be the dyadic Godel numbering of all words in L. We let T be the set of provable sentences of (E). We recall that by a proof in (E) we mean a finite sequence X1, . . . , X,
of sentences of (E) such that each XZ is either an instance of an axiom of (E) or is derivable from earlier terms of the sequence by the rule of detachment. W e now need to assign to every finite sequence X1, ... , Xt of
D,0
expressions in L a number which we call the Godel number of the sequence,
which we do as follows. We let n be the number of symbols of L and we
let m = n + 1. Then to the sequence X1,... , Xt we assign the number IImx1II,n ... IImxtIIm, where xl,... , xt are the respective Godel numbers of X1i .... Xt. We let Ox be the condition that x is a sequence number of some sequence X1, ... , Xt. We let x c y be the condition that y is a sequence number of some sequence and that x is the Godel number of one of the terms of the sequence. We let x -- y be the relation "z is a sequence number of some sequence Z of which yzis the Godel number of some term Y, and x is the Godel number of some term X which occurs earlier than the first occurrence of Y.
Lemma 2.1 The conditions 9x, x E y, x z y are E0. Proof We recall (Proposition 1.7) that the sets Gn and Gm are E0. (a) 9x F-' (Gm(x) A IImBx A IImEx A -Hm = x A -IImHmPx) (b) x E y H (9y A IImxHmPy A Gn(x))
(c) x z y
(9z) A y E z A (3w < z) (9w A wBz A x E w A -y E w) )
We now define Pf (x) to mean that x is the sequence number of a proof CID
in (E) and we define y pfx to mean that y is the sequence number of a proof X1, ... , Xt such that x is the Godel number of one of its terms XZ (i < t).
§2 Arithmetization of elementary formal systems
155
Proposition 2.2 The relations Pf (x) and y pfx are E0. 1z7
Proof Let Y1,. .. , Y,. be the axioms of (E) and for each i < r, let Ai be the set of Godel numbers of all the instances of Yi. Each of the sets A,,-, A, is Eo by Lemma 1.8, hence their union Al U ... A, is E0. Thus
mob.
car
the set A of Godel numbers of all instances of the axioms of (E) is Eo. Next, we let II,,, be the Godel number of the implication sign and we let Der(x, y, z) be the Eo-relation "y = xlIaz A -IIa,Px." If x, y, z are respective Godel numbers of sentences X, Y, and Z, then Der(x, y, z) holds
if X is atomic and Y = X -+ Z, and thus Der(x, y, z) holds if Z is derivable from X and Y by the rule of detachment.
(a) Then Pf(x) is Eo, for it is equivalent to Ox A (Vy < x) (y E x D (A(y) V (Elz < y) (]w < y) (z x y A w x y A Der(z, w, y)))).
(b) ypfx<--*(Pf(y)AxEy) Now we have
Theorem 2.3 If
f.r. over K then WO is E1.
IAA
Proof Let H represent W over K and let h be its Godel number. If W is a set, then x c Wo - Ely (y pfh * x). If W is a relation of n arguments, where n > 2, then Wo(xi,... , x,,,) <--* 2y(Elz < y)(ypfz A z = hx1cx2C... Cxn), where c is the Godel number of the comma. As an obvious special case of Theorem 2.3 we have
Corollary 2.4 If A is r.e. then AO is E1.
ate'
0-0
In Part II of this chapter we prove that every recursively enumerable relation is E1, but for now, let us note that Corollary 2.4 at once yields the following weaker result (which is all that we will need for our intended applications). emu,
Theorem 2.5 If A is r.e. then A is the existential quantification of a recursive relation.
Proof We illustate the proof for A a set. Suppose A is r.e. Then Ao is E1i which means that it is the existential quantification of a Eo-relation
Chapter 8. Arithmetization of formal systems
156
R(x, y), and of course R(x, y) is recursive (Theorem 6.4, Chapter 6). Also the function g(x) is recursive (Theorem 8.1, Chapter 6). Then
xEA H g(x)EAo
2y(g(x) = y A y c Ao) 3y(g(x) = y A 3z(R(x, z))) 3y3z(g(x) = y A R(x, z))
lw(]y < w)(Ilz < w)(g(x) = y A R(x, z))
[The last equivalence follows from Proposition 6.6, Chapter 6.] Thus A is the existential quantification of the recursive relation
(2y < w) (3z < w)(g(x) = y A R(x, z))
Exercise 2.1 Prove the above theorem for A a relation of two or more arguments.
Part II Double enumeration §3 Some applications We now turn to some applications of Theorem 2.5. A separation principle I(1
We wish to show that for any r.e. sets A and B there are disjoint r.e. sets A', B' such that A - B C A' and B - A C B' (such a pair (A', B') is said to separate A - B from B - A). Moreover, we will see that given indices of A and B we can effectively find indices of A' and B. The relation "wx contains y" is r.e. by the enumeration theorem, and so by Theorem 2.5 it is the existential quantification of a recursive relation
tea.
P(x, y, z). Thus y c wx if ]zP(x, y, z). We read P(x, y, z) as "z puts y in wx", and so y is in wx if there is some z that puts y in wx. And now
ran
car
C."
we shall say that a number n is in wx before n is in wy - which we write B(n, x, y) - if some z puts n in wx and no number w < z puts n in wy. [If n is in wx but not in wy, it is automatic that n is in w1 before n is in wy.] Thus B(n, x, y) iff Elz(P(x, n, z) A (Vw < z) -P(y, n, w)). Since the relation P(x, y, z) is recursive, so is -P(x, y, z), from which it easily follows that the relation P(x, n, z) A (Vw < z)-P(y, n, w) is recursive, and hence the relation B(x, y, z) is r.e. Let us now note the following facts. First, since the relation B(x, y, z) is r.e., then for any numbers i and j, the set of all x such that B(x, i, j) is r. e., and by the iteration theorem, there is a recursive function Q(x, y) such that WO(ij) is this set. Thus
§3 Some applications
157
OWN
wa(i,j) _ {x : B(x, i, j)} - thus w,(i j) is the set of x's that are in wi before they are in wj. Next, it is obviously impossible for x to be in wi before it is in wj, and also to be in wj before it is in wi, and so WQ(i j) is disjoint from w,(j i). Finally, whether wi and wj are disjoint or not, wi - wj C wa(i,j)
(because if x c wi and not in wj, B(x, i, j) holds), and also, of course,
Wj - wi C wou i). And so we have:
Theorem 3.0 There is a recursive function o,(x, y) such that for all numbers i and j: (1) w,(i j) is disjoint from w,(j i),
(2) wi -wj C wo(i,j), (3) If wi is disjoint from wj then wi = w,(ij). From the above theorem we obtain:
Theorem 3.1 There is a recursive function o (x, y) such that for every n, i, and j, (1)
is disjoint from Rn
i)
(2) Ri - Rjn C Rtr(i i)' (3) If RZ is disjoint from Rjn then Ri = RQ(i j).
E^?
Proof By Theorem 3.0 and the fact that for any n, i, x1,.. . , xn,, Ri (xl,
...,xn.) - Jn(xl,...,xn.) E wi. Exercise 1 Complete the details of the above proof. A double enumeration
We now use the recursive pairing function J(x, y) of §7, Chapter 6, and its inverse functions K(x), L(x). For any number i, we define ai to be the set of all x such that x E WK(i) before x E WL(i), and /3i to be the set of all x such that x E WL(i) before
x E WK(i). Thus ai = wv(Ki,Li) and ,Qi = wo(Li,Ki) Of course if i is such that WKi is disjoint from wLi then ai = WKi and /3 = wLi. Also, for
any pair (A, B) of r.e. sets, there is a number i such that A = WKi and B= WLi, namely i = J(a, b) where a is any index of A and b is any index
Chapter 8. Arithmetization of formal systems
158
of B. Therefore, for every disjoint pair (A, B) of r.e. sets, there is some i such that (A, B) is the pair (ai, /3i). It is easily verified that the relations "x E ay" and "x c my" are r.e., and so we have enumerated all disjoint pairs of r.e. sets in a sequence ) .. , (an, Nn), ... such that 02) the relations x c ay, x E /3y are both r.e. 1,31
Doubly universal pairs
A disjoint pair (A1i B1) of sets of numbers is called doubly universal if for every disjoint pair (A, B) of r.e. sets there is a recursive function f (x) such
that for all x, x E A H f (x) E A1i and x E B <-a f (x) E A2 - in other words, A = f-1(A1) and B = f-1(A2). We now take U1 to be the set of all numbers J(x, y) such that y c ax, and U2 to be the set of all numbers J(x, y) such that y E /3x. The sets U1 and U2 are r.e. and disjoint. Also for every disjoint pair (ai, /3i) and every x, x E ai <-a J(i, x) E U1, and x E ,Qi <--+ J(i, y) E /32i and so if we take f (x) to be the recursive function
J(i,x), then ai = f-1(U1) and /3i = f 1(U2), which means that (U1, U2) is doubly universal. We thus have
Theorem 3.2 The pair (U1, U2) is a doubly universal pair of r.e. sets. We remark that we are now in a position to verify the claim made in Part II of Chapter 3 that the collection of r.e. sets satisfies conditions Q1 and Q2.
Exercise 2 Show that this claim holds (with J(x, y) as the function called :tom
.-+
"x * y" in Chapter 3). Conclude, therefore, that the, pair (U1, U2) is recursively inseparable (i.e., that there exists no recursive set A such that U1 C_ A and U2 C A).
Exercise 3 Show that for any numbers i and j (a) wi - wj C aJ(ij) (b) aJ(zj) = OJ(j,i)
Exercise 4 Define Ai(x, y) if J(x, y) E ai and /3i(x, y) if J(x, Y) E /3i. Show that for any disjoint r.e. relations R(x, y), S(x, y) there is a num-
ber i such that R = Ai and S = Bi (a "double enumeration" of all r.e. relations of two arguments).
§4 All r.e. relations are E1
159
Part III More on arithmetization §4 All r.e. relations are E, We have shown (Theorem 2.4) that if A is r.e. then AO is E1i but to get from the fact that AO is El to the fact that A is E1 is a bit of a problem.
What is needed is to show that the relation g(x) = y is E1i and this is trickier than meets the eye! [We recall that g(x) is the Godel number of the dyadic numeral for x.] We need the following lemma (which will have other applications as well).
Lemma F [Finite sequence lemma] There exists a E1-relation K(x, y, z) such that (1) For any finite set {(al, bi), ... , (an, b,,,)} of ordered pairs of numbers
(positive integers) there is a number z such that for all x and y, K(x, y, z) holds if (x, y) is one of the pairs (al, b1), ... , (an, bn),
(2) For all x,y,z,K(x,y,z)=(x
of 1's longer than any string of 1's that is part of any of the numbers a1, b1i ... , an, bn, and let f be the number 2t2 and let z be the number f f al f a2 f f ... f f an f bn f f. We shall call such a number z a code number of the sequence. From the number z, the sequence (al, b1), ... , (an, bn) can
be completely determined, and a pair (x, y) is one of the pairs (ai, bi) if f f x f y f f is part of z and t is not part of x or of y. We thus define K(x, y, z) to be the following E0-relation. (3t < z)(3f < z) (1,
(2Pt A f = 2t2 A tlPz A f f x f y f f Pz A tPx A tPy)
This relation K(x, y, z) will be constant until further notice, and we shall sometimes write (x, y) E z for K(x, y, z).
The relation g(x) = y
We let Rl (x, y) be the E0-relation "(x = 1 A y = 12) V (x = 2 A y = 112)." [This means that g(x) = y and x consists of a single digit.] We let R2(x, y, z, w) be the relation "(z = xl A w = y12) V (z = x2 A w = y112)."
We note that R2(x, y, z, w) implies x < z and y < w. Now, the point to
Chapter 8. Arithmetization of formal systems
160
observe is that g(x) = y if there is a finite set S of ordered pairs such that the following two conditions hold:
C1: (x,y)ES C2: For any pair (a, b) E S, either Rl (a, b) or there is a pair (c, d) in S such that R2 (c, d, a, b).
yaw
To see this, it follows from C2 that for every pair (a, b) in S, it must be that g(a) = b, as can be seen by complete induction on the length (number 'L7
of dyadic digits) of a. And so if S satisfies conditions C1 and C2, then g(x) = y. Conversely, suppose g(x) = y. Let x = dld2 ... d,, where each di is the dyadic digit 1 or 2. Then y = 9192 ... gn, where gi is the Godel number of the digit di (12, if di = 1, and 112, if di = 2). Then take S to be the set {(d1i 91), (dld2, 9192), ... , (d1d2 ... dn, 9192 ... gn)} and S satisfies cad
the above conditions. From this and Lemma F it follows that g(x) = y if 3zA(x, y, z), where
A(x, y, z) is the Eo-relation (x, y) E z A (Va < z) (Vb < z) ((a, b) E z = (RI (a, b) V (Ela' < a) (Elb' < b) ((a', b') E z A R2 (a', b', a, b))). cc--
(CD
To see this equivalence, suppose that g(x) = y. Then there is a finite set satisfying C1 and C2. By Lemma F there is a number z such that for all x and y, (x, y) E z iff (x, y) is a number of S. Then for such a number z, the relation A(x, y, z) must hold (verify!). Conversely, suppose there is some z such that A(x, y, z) holds. For such a number z, let S be the set of all ordered pairs (a, b) such that A(a, b, z) holds. Then it is obvious that S obeys conditions C1 and C2, and hence g(x) = y. Thus g(x) = y H E]zA(x, y, z). We have thus proved:
Proposition 4.0 The relation g(x) = y is E1. Now we can obtain
Theorem 4.1 If A is r.e. then A is E. Proof This is an obvious modification of the proof of Corollary 2.4 and we illustrate the proof for A a set. Suppose A is an r.e. set. Then AO is E1, and is thus the existential quantification of a Eo-relation R(x, y). Now we use Proposition 4.0, and so the relation g(x) = y is the existential quantification of a Eo-relation S(x, y, z), i.e., R(x, y) H ]zS(x, y, z). Then
xEA
<--*
g(x)EAo
§5 More on arithmetization (optional)
161
Ely(g(x)=yAyEAo) E1y(E1zS(x, y, z) A 3wR(y, w))
]y]z]w(S(x, y, z) A R(y, w)) w(Ely < v)(2z < v)(Elw < v)(S(x, y, z) A R(y, w)) Since the relation xy = z is r.e., we thus have:
Corollary 4.2 The exponential relation xy = z is E1. Note
The above proof of Corollary 4.2 utilized the fact that the exponential relation is r.e. where "r.e." was defined in terms of elementary formal systems. A direct proof of the above corollary, which does not depend on elementary formal systems, is this: xy = z if the following condition holds:
Elw((y,z)EwA(Va<w)(Vb<w)((a,b)EwD((a=1Ab=xV (2C<w)(Eld<w)((c,d) E
Exercise 5 Prove the above equivelence.
*§5 More on arithmetization (optional) Consider a set S fixed for the discussions and a subset A of S and a relation
R(xl,... , xn,, y) of elements of S. By an A-R sequence we shall mean a sequence (x1, ... , xt) of elements of S such that for each i < t, either x;, E A,
'.Y
con
'.3
or there are numbers jl, ... , jn (not necessarily distinct), each of which is < i, such that xj,, , x) holds. And we call such a sequence an A-R sequence for its last term xt. By the closure of A under R we mean the set of all x such that there exists an A-R sequence for x. Of course, given an A-R sequence, if repetitions are deleted, the resulting sequence is also an A-R sequence, and so x belongs to the closure of A under R if there is an A-R sequence for x without repetitions. The following is an alternative characterization of the closure of A under R. Call a subset B of S closed under R if for any elements xl, ... , x,,, of B, if R(xl,... , xn, y) holds, then y is also in B. Then the closure of A under R can be alternatively defined to be the intersection of all supersets of A which are closed under R. The closure of A under R is itself closed under R. Now let us consider the case where S is the set of positive integers. If A is an r.e. set and R(xl,... , xn) y) is an r.e. relation, then the closure of A under R is r.e., for given a dyadic arithmetic in which "A" represents
Chapter 8. Arithmetization of formal systems
162
A and "R" represents R, we can take a new predicate "C" and add the axioms
Ax - Cx Cxl -f ... - Cxn -+ Rxl,... , xn, y - Cy Then "C" represents the closure of A under R. Now, let us say that R is progressing if R(x1i... , x,, y) implies that each of x1 ,. .. , x, is less than y. Our objective is to show that if A and R are both recursive and if R is progressing, then the closure of A under R is recursive. And for this, we must establish more about sequences. We shall say that z is a sequence number of a finite sequence (al, ... , an) p-'
if z is a code number of the sequence (1, al), ... , (n, an) of ordered pairs. Thus if t is a string of l's longer than any string of l's that is part of any of the numbers 1,. .. , n, a1i ... , an and if f = 2t2, then 'L3
`ti
f f 1 f a1 f f 2 f a2 f f ... f f n f an f f is a sequence number of the sequence
(al,... , an). We now wish to exhibit a recursive function h(x, y) such that for any sequence (al, ... , an), if m is the maximum of the numbers two
al, ... , an, n, then there is a sequence number z of the sequence (al, ... , an) that is less than or equal to h(n, m). We now rely heavily on dyadic concatenation. We recall that x * 1 =
We letL(x)bethe
length of x, and so L(1) = 1; L(2) = 1; L(x * y) = L(x) + L(y).
*Lemma 5.0 (a) 2L(x) < x + 1 (b) x < 2L(x)+1 (in fact x < 2L(x)+1 - 2)
Proof As we noted in the proof of Proposition 1.3, x has length r if 2' - 1 < x < 2 (2T - 1). Therefore 2L(x) - 1 < x < 2 (2L(x) - 1). (a) Since 2L(x) - 1 < x, then 2L(x) < x + 1. (b) Since x < 2.(2L(x) -1), then x < 2.2L(x) - 2 = 2L(x)+1 -2 < 2L(x)+1
Now consider a sequence (al, ... , an). Let m be the maximum of the numbers n, a1,... , an. Then m is at least as long as any of the numbers 1.... , n, a1 ,.. . , an. Let t be a string of l's whose length is L(m) + 1
and let f = M. Then L(f) = L(m) + 3. Let z be the number f f l f al f f
... f f n f an f f. Since t is a string of l's longer than any of the
§5 More on arithmetization (optional)
163
numbers 1 , ... , n, al, ... , an, then z is a sequence number of the sequence 2(5n+2)-(sm+s) (a1i ... , an). We show that z < 2 For each i < n, let gi = f f i f ai. Then z = 9192 ... gn f f , and each gi has length <_ 5 L(f ), so L(z) < (5n + 2) L(f ). Then by (b) of the above 2. lemma, z < Now, L(f) = L(m) + 3, hence 2L(f) = 2L(m)+3 = 8.2L(m) < 8. (m + 1) 2.2(5n+2).(8m+8). And so we take (by (a) of the above lemma), and so z < h(x, y) to be 2. ,(5x+2)-(8y+8)
We now take 'y(x) to be the recursive function h(x, x) and we have
*Theorem 5.1 If (a1i ... , an) is a sequence without repetitions such that for each i < n, ai is < an, then it has a sequence number < -y(an). Now we can prove
*Theorem 5.2 If A and R are both recursive and if R is progressing, then the closure of A under R is recursive.
'L3
$w,
Proof We illustrate the proof for R a relation of two arguments. Thus x is in the closure of A under R(x, y) if and only if there is an A-R sequence for x without repetitions. Since R is progressing, then all terms of the sequence must be < x, and since there are no repetitions, the number of terms of the sequence is < x. Therefore, by Theorem 5.1, the sequence has a sequence number < y(x). And so x is in the closure of A under R if the IA
G'.
following recursive condition holds: (Elz < -/ (x)) (Ein < z) ((n, x) E z A (Vi < n) (Vy < z) ,z-
((i, y) E z = (y c A V (3j < i) (E]w < z) ((j, w) E z A R(w, y)))))
°'U
To see that this condition is recursive, it is obvious that the part following "(3z < -I(x)" is recursive, and since ry is a recursive function, then the condition is recursive by (b) of Theorem 6.3, Chapter 6. Note
It can also be shown that if A and R are primitive recursive, and if R is progressing, then the closure of A under R is primitive recursive (the key point is that the function y(x) is primitive recursive).
Consider now a set W of strings in an alphabet K and a relaC."
tion R(X1,... , Xn, Y) of strings in K. We shall call R progressing if
164
Chapter 8. Arithmetization of formal systems
R(X1, ... , X,n, Y) implies that each of X1,... , Xn, is part of Y. Now, suppose W is solvable over K and R is solvable over K and progressing. Let
V be the closure of W under R. We wish to show that V is solvable over K. Well, order the symbols of K in any way you like and let h be the corresponding lexicographical Godel numbering (see §9, Chapter 6). Let Wh, Rh, Vh be the corresponding set (relation) of lexicographical Godel numbers. Since h is 1-11 onto the set N of positive integers, then it is easily seen that Vh is the closure of Wh under Rh. Also, since P is a progressing relation of expressions, then Rh must be a progressing relation of numbers. And since W and R are solvable over K, then Wh and Rh are recursive (Theorem 9.4, Chapter 6). Hence by Theorem 5.2 above, Vh is recursive, hence V is solvable over K (again by Theorem 9.4, Chapter 6). And so we have
*Theorem 5.3 If W is a solvable set of expressions in K and R is a progressing relation of strings in K and also solvable over K, then the closure of W under R is solvable over K.
Chapter 9
Elementary formal systems and incompleteness proofs We wish to show here how elementary formal systems can facilitate incompleteness proofs. We avoid use of the usual "Beta-function," or any of the related devices of Part III of Chapter 8. We consider the language Lo briefly mentioned in Chapter 5 - the system of arithmetic based on just plus, times, and successor. Our immediate ...
objective is to show that truth in Lo is not formalizable, i.e., that there exists no elementary formal system in which the set of true sentences of Go is represented. This means that no formally representable set W of true sentences of Lo can contain all the true sentences. In fact, there is an effective procedure whereby given an E.F.S. in which a set W of true sentences is represented, we can find a true sentence of Lo that is not a f2,
coo
car
member of W. This means that given any mathematical axiom system .M in the language of Go, if M is correct for Lo and if the set of provable sentences of M is formally representable, then M is incomplete. Moreover, if M is correct for Lo, then given any elementary formal system (E) in which the provable sentences of M is represented, an undecidable sentence of M can be explicitly found. It is relatively easy to give an elementary formal system for the famous axiom system of Peano Arithmetic, which
we do in Part II of this chapter. From it we then see how to obtain an car
undecidable sentence of Peano Arithmetic (under the assumption that all provable sentences of Peano Arithmetic are true).
Chapter 9. Elementary formal systems and incompleteness proofs
166
Part I Truth in ,CO cannot be formalized §1 The non-formalizability of arithmetic truth We let K be the alphabet of the language Lo, which consists of the 12 symbols 0
1
-
(
)
`d
v
1
For our present purposes it will be more convenient to use dyadic Godel
°a..
numbering, instead of the Godel numbering of Chapter 5. The twelve symbols of K, in the above order, will have the respective Godel numbers 22, 23, ... , 212, which in dyadic notation are 12,112, ... ,1111111111112. To avoid confusion let us carefully note that for any number n, by n we do not mean the dyadic numeral for n; we mean the Peano numeral for n - the expression consisting of the symbol 0 followed by n accents (e.g., 5 is the expression "0""'," not "21"). For any expression X in K, by X we mean the Peano numeral that designates the dyadic Godel number of X. As in Chapter 5, by the Tarskification of X we mean the expression Vvi(vi = X i X). We let t(X) be the Tarskification of X.
Lemma 1.0 The Tarskification function t(X) is formally representable over K.
Proof Since the Peano notation is admissable (Theorem 4.2, Chapter 6), then by Theorem 8.2, Chapter 6, the set of all ordered pairs (X, X) is formally representable over K; let (E) be an E.F.S. over K in which P represents this relation. Then take a new predicate Q and add the axiom Px, z -> Qx, `dvl (Vi = z D X)
Then Q represents the relation t(X) = Y.
For any set W of expressions in K, we let Wt be the set t-1(W), i.e., the set of all expressions X in K whose Tarskification is in W. If W is f.r., then since the relation t(X) = Y is f.r. over K, the set t-1(W) is f.r. over K. More constructively, given an E.F.S. over K in which W is represented, we can take a common E.F.S. over K in which W is represented by some
predicate - say "W" - and the relation t(X) = Y is represented by some S".
predicate - say "Q". We then take a new predicate "P" and add the axiom Qx, y -4 Wy -4 Px. Then in this system, P represents Wt. We thus have
coo
§ I The non-formalizability of arithmetic truth
167 car
Proposition 1.1 Given an elementary formal system in which W is represented, an elementary formal system can be found in which is represented the set Wt of all expressions whose Tarskification is in W. Now, in Part I of Chapter 8, we proved that if W is formally rep-
chi
.'3
car
car
(
resentable, then WO is E1, and we saw that given an elementary formal system in which W is represented, we can effectively find a E1-formula that expresses WO. Now, suppose we are given an elementary formal system in which W is represented. Then by Proposition 1.1 we can construct an elementary formal system in which Wt is represented, and then we can ruin
construct a E1-formula that expresses (Wt)o, which is W*. And so we have
Proposition 1.2 For any set W of expressions of K, given an elementary CDR
formal system in which W is represented, a E1-formula can be found that expresses the set W* of all numbers n that are Godel numbers of expressions whose Tarskification is in W. 'CO
Next, suppose that W is a set of expressions of K that is formally representable. Then, by Proposition 1.2, given an E.F.S. in which W is represented, we can find a E1-formula F(vi) that expresses the set W*. Let H(vi) be the formula -F(vi) and let h be the Godel number of H(vi). (s+
'31
.s:
Then H(vi) expresses the complement of W* and so its Tarskification H[h]
.'3
is a sentence that is true if and only if it is not in W (since H[h] is true H h $ W* H H[h] ¢ W). If all sentences of W happen to be true, then H[h] is true but not in W. We thus have cue
Theorem 1.3 If W is a set of expressions of K such that all sentences (
cad
in W are true in Lo, then given any elementary formal system in which W is represented, a true sentence of Go not in W can be found.
Now suppose that we have a formal mathematical system M in the car
goo
language of Lo which is correct in that all provable sentences of .M are true in Go. As with first-order systems in general, a sentence is called refutable ..j
vii
if its negation is provable. Then for a correct system M, all refutable sentences must be false (not true), since for any sentence X, one and only
one of the sentences X, -X is true. We let P be the class of provable cad
formulas of M (in most formulations of systems in first-order logic, open formulas can also be provable) and let R be the class of formulas whose negations are provable. If M is a formal system - if the set P is formally representable - then given any elementary formal system in which P is represented, we now see how to find a true sentence of Lo that is not in P G-'
'.3
cad
168
Chapter 9. Elementary formal systems and incompleteness proofs
- a true sentence X of Go that is not provable in M. Its negation -X is not provable in M either, since it is not a true sentence. Thus X is undecidable in M. And so we have:
Theorem 1.4 For any formal system M that is correct for Go, given any elementary formal system in which the set P of provable formulas of M is represented, an undecidable sentence X of M can be found.
This shows how, in general, elementary formal systems can be used for finding undecidable sentences in formalized first-order systems of arithmetic. As a concrete application, the system of Peano Arithmetic is considered in the next section and completely formalized in §3.
Exercise 1 [Tarski's theorem for Go] (a) In Chapter 5 we defined the notion of a Godel numbering g being adequate for a first-order language G. Let us now say that g is weakly adequate for G if the following two conditions hold.
(1) There is a relation C(x, y, z) expressible in L such that for any expressions X, Y with respective Godel numbers x, y, the relation C(x, y, z) holds if z is the Godel number of XY. ,.,
(2) There is a relation N(x, y) such that for any Godel number x, the relation N(x, y) holds if y = x. First, show that any adequate Godel numbering is weakly adequate.
Next, show that Theorem T of Chapter 5, Part I, holds under the weaker assumption that g is weakly adequate for G.
(b) Using Theorem 8.2, Chapter 6 and Theorem 2.3, Chapter 8, show that the dyadic Godel numbering of the strings of the symbols of Go is weakly adequate for Go.
(c) Conclude that the set of dyadic Godel numbers of the set of true sentences of Go is not expressible in Lo. [Thus this set is not only not r.e.; it is not even arithmetic.] *(d) In Part III of Chapter 8 we showed that every r.e. relation is E1 (and hence arithmetic). Using this fact, give another proof of (c) above.
coy
§2 Peano arithmetic
169
Part II A concrete incompleteness proof §2 Peano arithmetic For a concrete illustration of how elementary formal systems can be helpful in incompleteness proofs, we shall consider the system of Peano Arithmetic. We will give but a brief sketch of this system. In Chapter 5 we defined the notions of numerals, variables, terms, formulas, and free and bound occurrences of variables in formulas. For any
variables x and y and any formula F, we say that x is bound by y in F if there is some formula G such that x has at least one free occurrence in G and VyG is part of F. For any term t, variable x and formula F, we say that t is free for x in F if x is not bound in F by any variable that occurs in t. The axioms of Peano Arithmetic are any expressions of one of the folQ..
lowing 14 forms. We introduce these axioms in three groups - the first Q04
are axioms of propositional logic, the second are additional axioms of firstorder logic, and the third are the arithmetical axioms. In displaying these forms, F, G, H are formulas, x is a variable, t is a term, A(x) is a formula
and for any term t, A(t) is the result of substituting t for all occurrences of x in A(x). Group I
(1) (FD(GDF)) (2) ((F D (G D H)) D ((F D G) D (G D H)))
(3) ((^'F D G) D ((-F D -G) D F)) Group II
(4) (VxA(x) i A(t)), if t is free for x in A(x) (5) (Vx(F D G) D (VxF D VxG)) 'L7
Group III
(6) (x=yD(y=zIx=z)) (7) (x=yDx'=y') (8) (x'=y'Dx=y) (9) mix' = 0
170
Chapter 9. Elementary formal systems and incompleteness proofs
(10) (x + 0) = x
(11) (x+y')=(x+y)' (12) (x 0) = 0 (13) (x y') = ((x y) + x) (¢3
(14) (A(0) D (Vx(A(x) D A(x')) D VxA(x)))' In addition to these axioms, there are also two rules of inference.
'L7
coo
1. We say that formula H is inferable from formulas F and F i G by modus ponens if H is the formula G - in other words, G is inferable by modus ponens from F and F D G. 2. We say that formula G is inferable from formula F by generalization if for some variable x, G is the formula VxF - in other words, VxF is inferable from F by generalization.
BCD
By a proof in Peano Arithmetic is meant a finite sequence of formulas such that each member of the sequence is either an axiom, or is inferable from two earlier members by modus ponens, or is inferable from one earlier member by generalization. Thus: (i) every axiom is provable; (ii) if F and F D G are provable, so is G; (iii) if F is provable, so is VxF; (iv) no formula is provable unless its being so is a consequence of (i), (ii), and (iii). A formula F is called refutable if its negation -F is provable. This concludes our definition of the provable and refutable formulas of Peano Arithmetic. We let P be the set of provable formulas and R be the set of refutable formulas.
§3 Peano arithmetic is a formal system We now show that Peano Arithmetic is a formal system. We will construct an elementary formal system Rover the alphabet K of Peano Arithmetic
in which we represent the set of provable formulas, the set of refutable formulas, and several other sets and relations along the way. We shall introduce the predicates and axioms of P in groups, first explaining what each newly introduced predicate is to represent. It is important not to confuse variables of the elementary formal system P with variables of Peano
Arithmetic - for the former, we use the letters x, y, z, t, f, g, sometimes with subscripts, whereas variables of P.A. (Peano Arithmetic) consist of the letter v followed by subscripts. Also, "Y is the implication symbol 'These are the axioms of induction.
§3 Peano arithmetic is a formal system
171
"-'
of P.A., whereas "-f" is the implication symbol of the elementary formal system P. s represents the set of strings of subscripts 81
Note
Axiom sl says that 1 is a string of subscripts. Axiom sx -i sxl says that if x is a string of subscripts, so is xl (i.e., so is x followed by another subscript). In place of the the second axiom, we could have alternatively taken: sx -4 sy -4 sxy. Next, V represents the set of variables. 8X --+ Vvx
Note
Alternatively, we could have taken the two axioms: (1) Vvi; (2) Vx -4 Vx1.
D represents the relation "x and y are distinct variables."
Vx -+sy -Dx,xy Dxy, x N represents the set of numerals. Vx
NO
Nx -+Nx' T represents the set of terms.
VxTx Nx->Tx Tx - Tx' Tx-+Ty->T(x+y) F represents the set of formulas.
Tx->Ty-#Fx=y Fx -> F-x Fx -Vy
-FVyx
172
Chapter 9. Elementary formal systems and incompleteness proofs
So represents the relation "t is a term, x is a variable, y is a term, and z is the result of substituting t for all occurrences of x in y."
Tt-+Vx-+Sot, x,x,t Sot,x,y,y
Tt-Dx,y->Sot, x,y,y Sot, x, y, z -+ Sot, x, y', z'
Sot, x, y, Z -4 Sot, x, y2, z2 - Sot, x, (y + y2), (z + Z2) Sot, x, y, z -+ Sot, x, y2, z2 - Sot, x, (y . y2), (z . z2)
S represents the relation "t is a term, x is a variable, f is a formula, and g is the result of substituting t for all free occurrences of x in f ." Sot, x, y, z - Sot, x, Y2, z2 -+ St, x, y = Y2, z = Z2
St,x,f,9
St,x,f2,92 -+ St,x,(fl D f2),(91 'J 92) St, x, f, g -> St, x, -f, -g St, x, f,g -+ Dx,y -4 St,x,`dyf,Vyg
Tt ->Ff ->Vx-+ St,x,bxf,bxf O represents the relation "x is a variable, f is a formula, and x does not occur free in f."
Dy, x - Sy, x, f, f -+ Ox, f B represents the relation "x and y are variables, f is a formula, and x is not bound by y in f ."
Tw-+Bx,y,z=w Bx, y, f -+ Bx, y, -f Bx, y, f - Bx, y, g -i Bx, y, (f
g)
Bx,y, f - Dy,z-Bx,y,Vzf Bx, y, f -> Ox, f -> Bx, y, Vyf
E represents the relation "t is a term, x is a variable, f is a formula, and t is free for x in f." Bx, y, f -+ Ey, x, f
Vx -+Ny -+Ff -Ey,x,F Et, x, f -+ Et', x, f
Etl,x, f -+ Et2,x,f - E(tl +t2),x,f Etl, x, f - Et2, x, f - E(tl . t2), x, f
§3 Peano arithmetic is a formal system
173
A represents the set of axioms.
Ff-Fg-'A(f(gDf)) Ff-#Fg-+Fh-#A((fD(gDh))D((fog)D(fDh))) Ff -+ Fg - A((-f D g)
((^'f D ^'g) D f))
St, x) f, g - Et, x, f - A(Vxf D g) Dg)D(Vxf DVxg))
Vx - A(x + 0) = x
Vx-4 Vy-A(x+y')=(x+y)' SO, x, f, g - Sx', x, f, h - A(g D (Vx(f D h) D Vx f )) P represents the set of provable formulas.
Af - Pf Pf-'P(fDg)-+1'g Pf -4 Vx-4 P`dxf R represents the set of refutable formulas.
P f -> R- f We have thus proved
Theorem 3.1 The sets P and R are formally representable over the alphabet K of Peano Arithmetic. Remark
In the elementary formal system that we constructed, all the items preceding P and R are not only formally representable over K, but solvable over K. This can be shown using Theorem 5.3, Chapter 8 (see appendix to Chapter 4, G.I.T. for a related treatment). From Theorem 3.1 and Proposition 1.2 we have
174
Chapter 9. Elementary formal systems and incompleteness proofs
Theorem 3.2 The sets P* and R* are E1. We remark that from the elementary formal system that we have constructed, we can explicitly construct a E1-formula that expresses P* and another that expresses R*. From Theorems 2.1 and 1.4 we have
Theorem 3.3 Peano Arithmetic is incomplete (assuming it is correct). Again, the finding of an undecidable sentence of Peano Arithmetic is wholly effective: once a E1-formula for P* has been found, the Tarskification of its negation is undecidable in Peano Arithmetic (assuming correctness). Alternatively, the Tarskification of any formula that expresses R* is undecidable in the system (why?). co.
§4 The Godel and Rosser proofs
car
Let us now briefly discuss the proofs of the incompleteness of Peano Arithmetic that do not employ the notion of truth. We will now take advantage of results proved in Chapter 4 by associating with Peano Arithmetic a representation system that we will call Z : the expressions, sentences, provable sentences, refutable sentences of Z are those Peano Arithmetic. For each n > 1, the predicates of degree n of Z shall be the formulas F(vl,... , vn) of Peano Arithmetic in which all the free variables are in the set {Vi, ... , vn}, '17
and to apply such a formula F to an n-tuple (al, ... , an) of numbers is to take the sentence `dv1 ...Vvn((vl = al A ... A vn = an) D F) - denoted , . ... , an]. Then the notions of representability, contrarepresentability, complete representability, definability, separability - all these notions of Chapter 4 are now applicable to Peano Arithmetic. We say that Peano Arithmetic is w-inconsistent if there is a formula F(vi) such that the sentence 11v1F(vl) is provable, yet all the infinitely many sentences F(0), F(1), .... F(n) are refutable. And Peano Arithmetic is called w-consistent if it is not w-inconsistent. [We informally defined [a-1
these notions in Chapter 4, before we introduced the definition of formulas.] Now, it is well known that under the assumption that Peano Arithmetic is w-consistent, all E1-sets are representable in Peano Arithmetic. [We refer
.'3
COD
the reader to G.I.T.; Chapter 5, for a careful proof, or to R.M.; Chapter 0, for a more sketchy proof.] From this and Theorem 3.2, the sets P* and R* are representable in P.A. (Peano Arithmetic), under the assumption that P.A. is w-consistent. Then, as we saw in Chapter 4, once we have a formula H(vi) representing P*, the sentence H[k] is undecidable, where k
§4 The Godel and Rosser proofs
175
ti,
is the Godel number of -H(vi) - or alternatively, if F(vi) represents R*, then F[f] is undecidable, where f is the Godel number of F(vi). And so in G-'
.'S
this manner, P.A. can be shown incomplete, replacing the assumption that P.A. is correct by the weaker assumption that P.A. is w-consistent. This incompleteness proof follows the lines of Godel. O..
We also showed in G.I.T.; Chapter 6 (and sketchily in Chapter 0 of R.M.) the well known fact that if P.A. is simply consistent (no sentence and its negation are both provable), then any two disjoint E1 sets are strongly separable in P.A., hence R* is separable from P*, and therefore an undecidable sentence of P.A. can be found by taking the Tarskification of a formula that separates R* from P*. In this manner we obtain Rosser's proof of the incompleteness of P.A. under the mere assumption of simple consistency. We shall not take the space to carry out the arguments here; a connected exposition of all this can be found in G.I.T. The key fact, however, in all of these proofs, is that the sets P* and R* are El, and the car
maw'
(On
ms's
--n
c..
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lot
main purpose of this chapter was to show how the use of elementary formal systems provides a smooth proof of this.
Chapter 10
Doubly indexed relational systems In this chapter we turn to some results that generalize results in recursion theory. These will be used in the next chapter to establish some generalizations of results in metamathematics. What we will now do falls under the category of generalized recursion theory, in the sense of Fitting[5].
Part I Indexed relational systems §0 Definitions We use the expression "indexed relational system" in a far more general setting than that of Fitting[5]. By an indexed relation system R we shall
mean simply an ordered triple (N, E, ') where N is a set, E is a set of functions with arguments and values in N, and
is a mapping that assigns
X09
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to every positive integer n and every element a E N a relation Rn of n arguments on N. [For n = 1, Rl is a subset of N, which we also write Wa.] Without ambiguity we can write Ra(x1, ... , xn) for Rn(xl, ... , xn). The relations Rn will be called the R-relations. The elements of E will be
coo
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ran
called the E-functions. [For recursion theory, the R-relations are of course the r.e. relations, and in our intended applications, the E-functions are the recursive functions. There are other useful applications, however, in which the class of E-functions is a subclass of the class of recursive functions, such as the class of 1-1 recursive functions, or the class of primitive recursive functions. Other applications are also possible. For arithmetic relations (as defined in Chapter 5), the R-relations are the arithmetic relations indexed by the Godel numbers of formulas that express them, and the E-functions are the recursive functions (or a subclass of them). The theory of indexed relational systems also generalizes the theory of hyperarithmetical relations,
§ 0 Definitions
177
as explained in Fitting[5]. Standard systems
We shall call an indexed relational system R a standard system if the following conditions hold:
(1) The class of R-relations is closed under unions, intersections, explicit transformations, introduction of constants, and contains the identity
relation and the set N and the empty set
.
(2) The class of E-functions is closed under composition, explicit transformations, introduction of constants, and contains the identity function. Cad
(3) For any R-relation R(xl,... , xk) and any E-functions fl (xl,... , xn), so.
.... fk(xl,...,xn), the relation R(fl(xl,...,xn),...,fk(xl,...,xn)) is an R-relation. Also, for any R-relation R(xl, ... , xn) and any Efunction f (xl, ... , xn), the set of all elements f (x1, ... , xn) such that
R(xl,... , xn) - this set is an R-set. (4) [Iteration properties] (a) [Uniform iteration property] For any positive integer m, there is a E-function s(x, xl, ... , xm) such that for all elements x, xl, .... xm, y1 i ... , yn: R.8(x,x,,...,xm)
(yl,... ,
yn) t-- Rx (x1.... , xm) yl, ... , yn)
(b) [Iteration property] For every m and every R-relation R(x1i ... , xm, yl, ... , yn) there is a E-function 0(x1 i .... Xm) such that for E^?
all xl.... ,xm,yl,...,yn: Rc1(xi,...,xm) (y1) ... , yn) t-a R(x1 i ... , xm, Y1, ... , yn) C'1
(c) [Extended iteration property] - For any E-functions g(x),
,y),. , fk (x, y1, , yn) there is a E-function 0(x, Y1.... , yn) such that for all x, yl, ... , yn: f (x, y1,
Ron(x) (yl, ... , yn) -
Rk(x) (f (x, yl, ... , yn), ... , fk (x, yl, ... , yn))
We remark that it was not necessary to list (b), since it follows from (a) (because any R-relation R of n + m arguments has some index a, and so we can take q5(x1,... , xm) to be s(a, xl, ... , xm). But we wish to list (b) separately in order to compare it with (a) and (c), which we will shortly do.
Chapter 10. Doubly indexed relational systems
178
Universal systems We call R universal if for each positive integer n, the relation R x (yl, ... , yn)
(as a relation among x, yl,... , yn) is an R-relation. What Fitting[5] calls an indexed relational system,is essentially what we are calling a universal standard system. Many of the fixed point properties of recursion theory which Fitting generalized to his indexed relational systems do not really require the hypothesis of universality. But more about this in a later chapter.
Let us note that for a universal system, the iteration properties (a)
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and (c) are both consequences of (b), because the relation Rx(xl, . . . , xrn, , yl, ... , yn), and yl, , yn) is itself an R-relation (among x, xl) ... , X the relation R9(x) (f 1(x, yl, . , y , ) ), . ... , fk (x, Y17 .... yn)) is an R-relation among x, Y 1 ,- .. , yn. And so for a universal system, the iteration properties (a) and (c) come automatically with (b). In particular, the indexed system of recursion theory has the iteration property (c). We shall assume throughout this chapter that R is a standard system,
but we do not yet assume that it is universal. We shall also make the `i1
inessential but highly convenient assumption that our indexing is like our indexing in recursion theory, in that for each n > 2 and every element a, Ra is the set of all n-tuples (x1, ... , xn) such that Jn(x1, ... , xn) E Wa. [Such an indexing we might call "nice."]
§1 Universal sets Given two subsets A and B of N, we say that a function f (x) reduces A to
S-'.
B if A = f -1(B) (i.e., if for all x in N, x E A H f (x) E B). We say that A is (many-one) reducible to B if there is a E-function f (x) that reduces A to B. We call B universal if every R-set is reducible to B. We call B uniformly universal if there is a E-function f (x, y) - under which B will be said to be uniformly universal - such that for every element a, the .s.
function f (a, x) (as a function of x) reduces wa to B. If f reduces A to B and g reduces B to C, then obviously the function f (g(x)) reduces A to C, from which follows:
Proposition 1.1 Suppose B is reducible to C. Then (a) If B is universal, so is C. (b) If B is uniformly universal, so is C. As an obvious corollary, we have
§2 Generative sets
179
Proposition 1.2 If there is a uniformly universal R-set, then every universal set is uniformly universal.
Post's complete set K Suppose now that R is a universal system and also that E contains a pairing function J(x, y). Let K be the set of all elements J(x, y) such that wx contains y. Under our assumption that R is universal, the relation "wx contains
y" is an R-relation, and so K is then an R-set. It is obviously uniformly universal under J(x, y), since for every i and x, x c wi <-+ J(i, x) E K. And so we have:
Proposition 1.3 If R is universal and E contains a pairing function J(x, y), then there is a uniformly universal R-set - to wit, the set K of all elements J(x, y) such that y c wx. Note
For recursion theory the set K is what Post[13] calls the complete set. We see now that for recursion theory, since a uniformly universal r.e. set exists, then every universal set is uniformly universal.
§2 Generative sets We shall call a set A of elements of N a generative set if there is a E-function
f (x) - called a generative function for A, or a function under which A is generative - such that for every element i, f (i) E wi F-r f (i) E A.
Proposition 2.1 If A is reducible to B and A is generative, so is B. Moreover if A is generative and A is reducible to B under f (x) then there is a E-function k(x) such that B is generative under the function f (k (x)).
Proof Suppose A is generative under h(x) and f (x) reduces A to B. By the iteration property (c) there is a E-function O(x) such that for all x and y, y E wk(x) F-r f (y) E wx. [To see this, we take the iteration property (c) with g(x) the identity function, and so there is a E-function O(x) such that R1,k(,;) (y) H R.1 (f (y)), which in w-notation is y E wb(x) H f (y) E wx.] We show that f (h(q(x))) is a generative function for B, and so we then take k(x) = h(q(x)). Well, for any i,
Chapter 10. Doubly indexed relational systems
180
f (h(q(i))) E B
<--*
Fr
h(q(i)) E A h(q(i)) E w,5(2) f (h(q(i))) E w2
(since f reduces A to B) (since h is generative for A)
(since for any y, y c
wo(2)
Fr
f (y) E w2, and so we take h(q(i)) for y) As an obvious corollary we have:
Corollary 2.2 If there exists a generative R-set, then every universal set is generative. Next we have:
Proposition 2.3 Suppose B is uniformly universal under f (x, y). Then (a) The set B of all x such that f (x, x) E B is generative. (b) The set B is generative.
Proof (a) The set d is obviously generative under the identity function, since i c wz +-+ f (i, i) E B +-+ i E B.
(b) It is understood that B being uniformly universal under f (x, y) implies automatically that f (x, y) is a E-function. Then by the iter-
ation property, there is a E-function t(x) such that for all x and y, y E Wt(x) H f (y, y) E wx. [Verify!] Then t(x) E wt(x) <--> f (t(x), t(x)) E wx. But also, t(x) E wt(x) H f (t(x), t(x)) E B. And so f (t(x), t(x)) E wx * f (t(x), t(x)) E B, which means that B is generative under the function f (t(x), t(x)).
Corollary 2.4 If there exists a uniformly universal R-set U, then every universal set is generative. Note
The above corollary can be proved in two different ways. Suppose U is a uniformly universal R-set. Then, on the one hand, every universal set is uniformly universal by Proposition 1.2, and hence generative by Proposition 2.3. On the other hand, U itself is generative, and since U is an
§2 Generative sets
181
R-set, it is reducible to any universal set B, hence B is again generative by Proposition 2.1. [We might remark that if one is interested in getting a generative function for a universal set B, the two different methods of proving B generative yield different generative functions for B.] '.3
For recursion theory, in which we have universality and a recursive pairing function, Post's complete set K is uniformly universal, and the K cm.
c0+
90+
yam
- the set of all numbers x such that x E wx - which is the set of all x such that J(x, x) E K - is generative. The standard symbol for this set is "C" and is known as Post's creative set. Let us say more about this. Post defines a set A to be productive if there is a recursive function O(x)
gyp
u,'
such that for every x such that wx is a subset of A, the number O(x) is in A, but not in wx. A set A is called co-productive if its complement is productive, which means that there is a recursive function O(x) such that
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for every x such that wx is disjoint from A, the number O(x) is outside both A and wx. Then Post calls a set creative if it is both co-productive and r.e. A set A is sometimes called completely creative if it is r.e. and there is a recursive function O(x) such that for every x, whether wx is disjoint from A or not, O(x) E A - O(x) E wx. [Of course, if wx happens to be disjoint from A, then q(x) is outside both A and wx, so every completely creative set is also creative.] Thus a set is completely creative if it is both r.e. and what we are calling generative. [We believe that generative sets, whether
41>
r.e. or not, are of sufficient importance to warrant a name.] Of course car
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Post's creative set C is immediately completely creative - a fact that Post should have capitalized on. Actually, creative sets and completely creative sets have turned out to be the same thing, but the proof of this involves a fixed point result known as the recursion theorem, which we study in Part III of this volume. In recursion theory, every universal r.e. set is generative, hence creative. A celebrated result of John Myhill is that every creative set is universal, and the proof of this appears to require the recursion theorem. As pointed out to me many years ago by Daniel Lacombe, no fixed point argument is required to show that every completely creative set is universal, and we shall study this later on. "'S
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coo
It-
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ten
cm.
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>>.
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All I have said generalizes to standard universal relational systems, and we shall carry the above terminology "creative," "productive," "coproductive," "completely creative" intact, replacing "r.e. set" by "R-set," and "recursive function" by "E-function."
Chapter 10. Doubly indexed relational systems
182
Part II Double indexing §3 Double enumeration and double universality Until further notice we shall assume that R is universal and that E contains a pairing function J(x, y) and inverse functions K(x), L(x) such that
KJ(x, y) = x; LJ(x, y) = y; J(K(x), L(x)) = x. We shall also assume
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that as in the case of recursion theory, E contains a "separation" function a(x, y) such that for all i and j in N: (i) w,(i j) is disjoint from w,(j i); (ii) wi - Wi C wQ(i j); (iii) if wi is disjoint from wj, then wi = wa(ij). Then, as in recursion theory (see §3, Chapter 8), we take ai to be WQ(K(i) L(i)) and /3i to be w.(L(i),K(i)). Then every pair (A, B) of disjoint R-sets is (ai, /3i) for some i (and of course for every i, ai is disjoint from /3i). Such a system R might be called doubly indexable. For the rest of this chapter, R will be assumed to be both doubly indexable and universal. From this it follows that the relations "ax contains y" and "/3w contains y" are R-relations. We will also assume that our indexing of R-relations is "nice", i.e., that Rn (xl, ... , xn) t-- Jn(x1i ... , xn) E wi. Hence it follows that for all n, and any elements i and j: (1) RQ(i ) is disjoint from Rn(i i),
(2) R2 - Rjn C Rn(i,i)' (3) If RZ is disjoint from Rjn, then RZ = Rn(i
i).
Next,
for every n > 2 we define An = Rn(Ki Li) and B? _
Ra(Li,xi)
Since our indexing is "nice," it follows that AZ (xl, ... , xn) <--*
01
Jn(x1,...,xn) E cai and BZ (xl,...,xn)
2
Jn(xl,:..,xn) E A. We take
A2 = ai and B2 = /3i. Since ai is disjoint from /3i, it follows that for every n, AZ is disjoint from B2 . Also, by (2) above, taking Ki for i and Li for j, we have RKi - RLi C An. . If instead we take Li for i and Ki for j, we have RLi - RKi C_ BZ . Next, by (3), if RKi is disjoint from RLi then RKi = An and RLi = B!'. Equivalently, if Rz is disjoint from Rin, then RZ = An(i and Rin = BAid) . We thus have:
Proposition 3.0 For every n and any elements i and j: (a) Ai is disjoint from B!'.
(b) RKi - RLi C An (hence also Rn - Rn C Anj(i j)); RLi - RKi C BZ (hence also Rin - R2 C
i)).
(c) If Rni is disjoint from Rjn then Rn = Anj(i j) and Rjn = Bn(i ).
§3 Double enumeration and double universality
183
Now we have:
Theorem 3.1
[Double enumeration theorem] For every positive number
n:
(a) For every element i, the relations An and BZ are disjoint.
(b) Given any disjoint relations Ra and Rb , there is an element i such that the pair (An, Bn) is the pair (Ra, R6 ).
Proof (a) is a repetition of (a) of Proposition 3.0. As for (b), take i = J(a, b), so that Rn =RKi and R6 = RLi. Since RKi is given to be disjoint from RLi, then by (c) of Proposition 3.0, RKi = AZ and RLi = BZ .
Exercise 1 Prove the following: (a)
Rq(j,i)'
J(i,j) = R_(i,j) and
An
O..
(b) For any R-relations Ml and M2 of n arguments there is an element
isuch that M1-M2CAn and M2-M1CBZ. Exercise 2 Show that for any number n and any E-functions fl(xl, .... xn) and f2 (X1, ... , xn) there is a E-function f (x1, ... , xn) such that for any m and any elements X1 ,- .. , xn such that R'L from Rf2(X1 x.), the following holds:
is disjoint
(1) Af(= R.f.(x.,...,xv.) 2)
Bf(x......
Rf2(xl,...,x, )
Next we have:
Proposition 3.2 (a) For any n there are E-functions O(x), )(x) such that for all x: An(x) _ Rn and BV) (x) = Rte
(b) For each n there is a E-function O(x) such that for all x: An = RK(k(x)' and Bx = RLq,(x)
Chapter 10. Doubly indexed relational systems
184
Proof (a) Let a be any index of the empty set of n-tuples, i.e., any element such that Ra is empty. Then obviously for every element x, Rn is disjoint from Rn, and so by (e) of Proposition 3.0, An = R j(x a) and BX = R j(a x). And so we take O(x) = J(x, a) and V(x) = J(a, x).
(b) Take O(x) = J(v(Kx, Lx), Q(Lx, Kx)), so that KO(x) = v(Kx, Lx) and La(x) = v(Lx, Kx). Then RKo(x) = Rn = A and RLc(x) - Rv(Lx,Kx) - Bx The following result is basic:
Theorem 3.3 [Uniform double iteration theorem] For every m and n there is a E-function q(x, xl, ... , xm,) such that for all elements x, xl, ... , Xm, Y1, ... , yn:
(1) Ac6(x,xl,...,xr) (yl,... , xn) H Ax(xl,... , Xm, Y1, ... , yn)
(2) Bm(x,x1,...,x",) (yl, ... , xn) - Bx(xl, ... , xm, yl, ... , yn)
Proof The relations Ax(xl,... , xm,, yl, ... , yn) and Bx(x1,... , xm, y1,
.... yn),
as relations among x,x1,... , xm, y1, ... , yn, are disjoint R.relations, so by the iteration property there are E-functions 01(x,x1i ...,xm), 02(x,xl,...,xm,) such that for all x, (a) Rc1 (x,xi,...,x,.,.) (y1, ... , yn) + Rn2(x,x1,...,x,,,)(Y1,...,Yn) ,
(b)
Ax(xl, ... , xm, y1, ... , yn) Bx(x1,...,xm,Y1,...,Yn)
We take lp(x, x1, ... , xm) = J(01(x, xl, ... , xm), 02 (x, x1, ... , xm)). The relation xl x,,.) is disjoint from Rc62(x x, x ), hence by (c) of Proposition 3.0 Rn = An(P(x,x1,.......) and Rnc12(x,x1,...,x,n) -
B(x xl
-
x
)
and so (1) and (2) of the conclusion follow from (a) and (b).
Given m and n, and taking O(x, X1,. .. , xm) as in the above theorem, if we replace the variable x by a constant a, the function q(a, x1i . . . , xm) (as a function of x1i ... , xm) is a E-function, and so as a corollary we have:
§3 Double enumeration and double universality
185
Theorem 3.4 [Double iteration theorem] For any n and m and any disjoint R-relations Mi(xi, , xm, Y1, , Yn), M2(xl, , xm, Y1, .... Yn) there is a E-function ' ( x 1 , . ... , xm) such that for all x1, ... , xm, y1, ... , Yn:
)(y1,...,ym) H Ml(xl,...,xm,y1,...,yn)
(1) A,i(xi,...,x
(2) BV,(x1....,x,n) (yl,... , ym) <--> M2 (xl, ... , xm, y1, ... , yn)
The following special case will be frequently used.
Proposition 3.5 For any disjoint R-relations Mi(x, y), M2(x, y) there ;..
is a E-function O(x) such that for all i and x:
(1) x E ak(i) H MI(i,x) (2) x E I,5(i) H M2(i, x) As a particularly useful corollary we have
Corollary 3.6 For any E-function f (x) there is a E-function q(x) such that for all is
(1) ao(i) = f-1(ai) (2)
3,k(j) = f-1(0i)
Proof By applying Proposition 3.5 to the disjoint R-relations "f (y) E aaxn
"f (y) E
Qx."
From the double iteration theorem (Theorem 3.4) we have:
Proposition 3.7 For any two R-relations M l (xl, ... , xn, Y 1 ,- .. , Ym), M2(x1, ... , xn, yl, ... , ym) there is a recursive function V) (x1, ... , xn) such that for all xi, , xn, Yi, ... , Ym:
(1) [Ml (xl, ... , xn, Yl, ... , ym) A ^'M2 (xl, ... , xn, Yl, ... , Ym)] (y1, ... , Ym) (2)
[M2(xl,...,xn,Yl,...,ym) A' Ml(xl,...,xn,Yl,...,Ym)]
(yl,...
,
ym)
186
Chapter 10. Doubly indexed relational systems
Proof Take disjoint R-relations Mi, M2 of n + m arguments such that Ml - M2 C Mi and M2 - M1 C_ M2 and then apply the double iteration theorem to Mil and M2. Involution USA
For any element x we define x = J(Lx, Kx). [We might call x the involute of x.]
Proposition 3.8 (a) J(x, y) = J(y, x)
(b) Kx = Lx and Lx = Kx
Proof Obvious.
Proposition 3.9 For any positive n and x, An = Bx, and Bn = A. Proof An= Ra(Kx,Lx) = Ra(Lx,Kx) = Bx B. The proof that Bx = An is symmetric.
D. U. pairs of sets
ti,
We shall say that a function f (x) reduces a pair (A1, A2) of sets of elements of N to a pair (B1, B2) if f reduces Al to B1 and A2 to B2. Thus f reduces
(A1,A2) to (B1,B2) if Al = f-1(B1) and A2 = f-1(B2), or equivalently, for all x, x E Al H f (x) E B1, and x E A2 <-a f (x) E B2. We say that
,:C
(A1, A2) is reducible to (B1, B2) if there is a E-function f (x) that reduces (A1, A2) to (B1, B2). As in recursion theory, we call a disjoint pair (A, B) doubly universal abbreviated D.U. - if every disjoint pair of R-sets is reducible to (A, B). We call the pair uniformly D.U. under a function f (x, y) if f (x, y) is a Efunction and for every element i, f (i, x), as a function of x, reduces (a;,, /32) to (A, B). And we say that (A, B) is uniformly D.U. if it is uniformly D.U. under some E-function f (x, y). As in recursion theory, we take U1 to be the set of all elements J(x, y) G".
..O
such that y E ax, and U2 to be the set of all elements J(x, y) such that y E /3x. Thus the pair (U1, U2) is uniformly D.U. under the function J(x, y).
Also, since the relations y E ax and y E 3x are both R-relations, the sets
§4 Doubly generative pairs
187
U1 and U2 are both R-sets. And so there exists a uniformly D.U. pair of R-sets. Suppose now that (A1i A2) is D.U. Then (U1, U2) is reducible to (A1, A2)
under some E-function f (x). Then obviously (A1i A2) is uniformly D.U. under the function f (J(x, y)). And so we have:
Proposition 3.9 Every D.U. pair is uniformly D.U.
§4 Doubly generative pairs We call a disjoint pair (A, B) doubly generative - abbreviated D.G. - if there is a E-function g(x) - called a D. G. function for (A, B) - such that for all elements x:
(1) g(x) E ax H g(x) E A
(2) g(x) E /3 - g(x) E B If g(x) is a D.G. function for (A, B), we also say that (A, B) is D.G. under g(x). Note
In R.M. we defined a disjoint pair (A, B) to be D.G. if there is a E-function h(x, y) such that for any elements i, j such that Wi is disjoint from w3: m
(1) g(i,j) E Wi H g(i,j) E A (2) g(i,j) E Wj `-' g(i,j) E B Actually, the two definitions are equivalent (though we find the present definition much neater to work with).
Exercise 3 Prove that the two definitions are equivalent. Next we note
Proposition 4.0 If (A, B) is D.G. then A and B are both generative. Proof Suppose (A, B) is D.G. under g(x). By (a) of Proposition 3.2 there is a E-function O(x) such that for all x, ao(x) = w x. Then, since g is a D.G. function for (A, B), g(q5(x)) E A g(q(x)) E agx) # g(O(x)) E Wx. Thus A is generative under the function g(q(x)). The proof that B is generative is similar.
Chapter 10. Doubly indexed relational systems
188
Exercise 4 Suppose that (A, B) is D.G. under g(x). Show that (B, A) is D.G. under the function g(x). [We recall x = J(Lx, Kx).] Now we shall "double up" on some results of §2.
Proposition 4.1 If (A1i A2) is D.G. and (A1, A2) is reducible to (B1, B2), where B1 is disjoint from B2, then (B1, B2) is D.G.
Proof Let g(x) be a D.G. function for (A1i A2) and let f (x) be a Efunction that reduces (A1, A2) to (B1, B2). By Corollary 3.6 there is a
:-'
E-function q(x) such that for all x, ao(x) = f-1(ax) and 00(x) = f-1(Ox). We will show that (B1, B2) is D.G. under the function f (g(q(x))).
f(gMx))) E ax
t-' g(O(x)) E f-'(ax) F-'
s-'
g(o(x)) E ao(x) g(q5(x)) E Al
H f (g(o(x))) E B1 The proof that f (g(q(x))) E Nx H f(g(o(x))) E B2 is similar. car
did
Proposition 4.2 Suppose (B1, B2) is uniformly D.U. under f (x, y). Let
B1 be the set of all x such that f (x, x) E B1 and let B2 be the set of all x such that f (x, x) E B2. Then
(CD
(a) The pair (B1, B2) is D.G. (b) The pair (B1, B2) is D.G.
Proof .mss
bbl
(a) The pair (Bl, B2) is D.G. under the identity function, since x E ax f (x, x) E B1 <-a x E Bl. Similarly x E Nx +-+ x E B2. G"+
(b) The function f (x, x) is a E-function, and so by Corollary 3.6 there is a E-function O(x) such that for all x and y, y E ak(x) <-a f(y, y) E ax and y E 00(x) `-' f (y, y) E 3x. Then O(x) E ao(x) +-a f (O(x), O(x)) E ax. But also O(x) E acb(x) f (O(x), O(x)) E B1. And so f (fi(x), O(x)) E ax <-a f (q(x), O(x)) E B1. Similarly it is proved that f (q(x), O(x)) E Nx H f(q(x), fi(x)) E B2. And so (B1, B2) is D.G. under g(O(x), O(x)).
§4 Doubly generative pairs
189
The Kleene pair (K1, K2)
The pair (U1, U2) is uniformly D.U. under J(x, y). We let K1 be the set of all x such that J(x, x) E U1 and K2 be the set of all x such that J(x, x) E U2. By Proposition 4.2 we have:
Proposition 4.3 (a) The pair (K1, K2) is D.G. (b) The pair (U1, U2) is D.G.
Proposition 4.4 Every D.U. pair is D.G.
Proof 1 Every D.U. pair is uniformly D.U., hence also D.G. by (b) of Proposition 4.2.
Proof 2 By Proposition 4.3 there exists a D.G. pair of R-sets. It is reducible to any D.U. pair, hence any D.U. pair is D.G. by Proposition 4.1. Exercise 5
(a) Suppose (A, B) is D.G. under g(x). Show that for any E-functions h(x) and f (x) there is a E-function O(x) such that for every element x:
(1) g(O(x)) E A `-' f(O(x)) E ah(x) 0.1
(2) g(O(x)) E B - f(O(x)) E)3h(x)
(b) Suppose (A1, A2) is D.G. and (A1, A2) is reducible to (B1, B2). Using (a), show that for any E-function h(x) there is a E-function q5(x) such
that for all x: /(1) f
/(O(x)) E B1 H f (O(x)) E ah(x) '0-
(2) f (fi(x)) E B2 t-a f lcb(x)) E Oh(x)
(c) How does (b) generalize Proposition 4.1?
Exercise 6 Show that a disjoint pair (A, B) is D.G. if and only if there G.1
is a E-function h(x) such that for every i for which WKi is disjoint from WLi, h(i) E A +-4h(i) E WKi, and h(i) E B H h(i) E WLi.
Chapter 10. Doubly indexed relational systems
190
§5 Semi-D.G. pairs We shall say that a disjoint pair (A, B) is semi-D. G. if there is a E-function
h(x) - under which (A, B) will be said to be semi-D.G. - such that for all x:
(1) h(x) E ay = h(x) E A (2) h(x) E,3, w
h(x) E B
arc
Thus a semi-D.G. function for (A, B) is like a D.G. function, except that the equivalences are weakened to implications. We now wish to show that for any semi-D.G. pair (A, B): (i) if A is an R-set, then A is generative; (ii) if A and B are both R-sets, then (A, B) is D.G. Part (ii) is particularly important for metamathematical applications. For the first part we need the following lemma:
Lemma A If A is an R-set then there is a E-function O(x) such that for every element i,
(1) wi - A C ao(i) (2) A - wi C am(i)
Proof Let a be an index of A (A = Wa) and take O(x) = J(x, a). The result then follows from (b) of Proposition 3.0 (taking n = 1).
Theorem 5.1 If (A, B) is semi-D.G. and A is an R-set, then A is generative. Moreover if (A, B) is semi-D.G. under h(x) and A is an R-set, then there is a E-function O(x) such that A is generative under h(O(x)).
Proof Suppose that (A, B) is semi-D.G. under h(x) and that A is an Rset. Let O(x) be as in the above lemma. Since (A, B) is semi-D.G. under h(x), then for any i, h(q(i)) E ao(i) = h(q(i)) E A, and h(q(i)) E ,30(i) = h(0(i)) E B = h(0(i)) 0 A. And so: (1) h(cb(i))
wi - A = h(q(i)) E am(i) = h(q(i)) E A
(2) h(q(i))
A - wi = h(cb(i)) E 00(i) = h(cb(i)) 0 A
Thus, if we let k be the element h(O(i)) we have:
(1)' (kewiAkOA)=*keA (2)' (keAAkOwi)=kOA
§ 5 Semi-D. G. pairs
From (1)' it follows that k c wi
191
k E A, and from (2)' it follows that
kEwi. Thus So A is generative under k(O(x)).
Our next theorem and its proof are more interesting. We first prove the following lemma:
Lemma B For any R-sets A and B, there is a E-function O(x) such that for all is (1) (ai U B) - (/3i U A) C a,5(i) (2) (/3i U A) - (ai U B) C /30(i)
Proof Since B is an R-set, the relation y c
a relation between x and y) is an R-relation, hence by iteration property (b) there is a E-function tl(x) such that for all is wt, (j) = ai U B. Similarly there is a E-function t2(x) such that for all i, Wt2(i) = Ni U A. Also wtl(i) - Wt2(i) C aa(tl(i),t2(i)) and Wt2 (i) - Wtl(i) C 01(tl (i),t2(i)), and so (ai U B) - (/i U A) C aa(tl(i),t2(i)), and (/3i U A) - (ai U B) C (3Q(tl(i) t2(i)). We therefore take O(x) = cr(tl(x),t2(x))
-
Theorem 5.2 If (A, B) is semi-D.G. and A and B are R-sets, then (A, B) is D.G. Moreover, if (A, B) is semi-D.G. under h(x) and A, B are R-sets, then there is a E-function O(x) such that (A, B) is D.G. under the function h(O(x)).
[We remark that the above "Moreover" will be quite significant in applications to metamathematics.]
C)/.
Proof Suppose (A, B) is semi-D.G. under h(x) and A and B are both R-sets. Then take O(x) as in the above lemma. We show that (A, B) is D.G. under the function h(O(x)). Take any element i and let k = h(o(i)). Since (A, B) is semi-D.G. under h(x), then k c ao(i) = k E A and k c /30(i) = k e B. And so we have:
(1) kE((aiUB)-(,3iLA))=kEao(i)=kEA (2) Similarly, kE((/3iUA)-(aiuB))=: k E
We are to show that kEaiHkEAand that kE0iHkEB (i.e., h(O(i)) E ai <--> h(O(i)) E A and h(O(i)) E 13i <-4 h(o(i)) E B). But this
Chapter 10. Doubly indexed relational systems
192
follows by propositional logic from (1), (2), and the facts that ai is disjoint from ,3i, and A is disjoint from B. To see this, suppose k c ai but k V A.
Then k c (ai U B) but since k 0 /3i and k V A, k V (Qi U A). Then k c (ai U B) - (i3 U A), hence by (1), k E A, contrary to the assumption
that k 0 A. Therefore k c ai = k c A. Conversely suppose k c A. If, also, k 0 ai, then k c Qi U A and k V ai U B (since k V ai and k 0 B, since B is disjoint from A), hence k c (Qi U A) - (ai U B), hence k E B, which is impossible, since k E A and B is disjoint from A. Thus k c A and k 0 ai
is not possible, which means that k c A = k c ai. Thus k c ai H k E A. The proof that k G ,3i +-* k E B is symmetric. Note 0-4
The propositional argument used in the above proof has a generalization
that is worth noting in its own right, and will be used again in a later
a.8
chapter in which we discuss some results of John Shepherdson. The argument in its general form is as follows. Suppose we have four propositions P1, p2, q1, q2 such that pl is incompatible with p2 (i.e., pl A P2 is not true), ql is incompatible with q2, and the following two facts hold: (1) ((pi V q2) A ^'(p2 V qi)) = ql
(2) ((p2 V qi) A -(pl V q2))
Then pl H ql and p2
q2
q2.
Exercise 7 (a) Prove the above. (b) Show how this is relevant to the proof of Theorem 5.1. In that proof, what are the propositions p1, P2, qi, q2? Discussion The notions of double generativity and semi-double generativity were introduced in R.M. and there we proved Proposition 5.2 by a more roundabout method involving the notion of complete effective inseparability, which we will now briefly discuss. In recursion theory, a disjoint pair (A, B) is called recursively inseparable
if there exists no recursive superset of A that is disjoint from B. [We call a set C a superset of A if A is a subset of C.] Equivalently, (A, B) is recursively inseparable if for any disjoint r.e. superset wi, wj of A, B respectively, there is a number outside both wi and wj. The pair (A, B) is
§5 Semi-D.G. pairs
193
called effectively inseparable - abbreviated E.I. - if there is a recursive function f (x, y) such that for any i and j such that wi and wj are disjoint and respective supersets of A and B, the number f (i, j) is outside of both wi and wj. This notion has played an important role in recursion theory and metamathematics. Now, in R.M. we defined a disjoint pair (A, B) to be completely E.I. (completely effectively inseparable) if there is a recursive
function 6(x, y) such that for any i and j, if A C wi and B C wj, then, regardless of whether or not wi is disjoint from wj, 6(i, j) E wi H 6(i, j) E wj. One of the first pairs to be recognized as recursively inseparable was Kleene's pair (K1, K2), which was soon after recognized to be effectively inseparable. Actually, it is just as easy to see that (K1, K2) is completely E.I. (which we pointed out in R.M.). As a matter of fact, every E.I. pair of r.e. sets is completely E.I., but the proof of this involves a clever fixed point argument - the "double recursion theorem" that we will study later on. Now, in R.M. we showed that every semi-D.G. pair is completely E.I. (this is quite easy) and that every completely E.I. pair of r.e. sets is D.G., from which it follows that every semi-D.G. pair of sets is D.G. This is how we proved Proposition 5.2 (for recursion theory) in R.M. But our present proof is more direct. Nevertheless, since the notions of effective inseparability and complete effective inseparability have played important roles historically, then the reader is encouraged to look at the following exercises.
Exercise 8 In the general context of index relational systems, we define a disjoint pair (A, B) to be completely E.I. if there is a E-function 6(x, y) such
that for any supersets wi, wj of A and B respectively, S(i, j) E wi A E wi.
* 6(i,
(a) Show that a disjoint pair (A, B) is completely E.I. iff there is a Efunction h(x) such that for any i for which A C- wKi and B C wLi, h(i) E WKi H h(i) E wLi.
(b) Show that every semi-D.G. pair is completely E.I. Conclude that Kleene's pair (K1, K2) is completely E.I., and so is (U1, U2) (and in fact so is every D.U. pair).
(c) Show that every completely E.I. pair of R-sets is D.G. [The proof is a modification of that of Proposition 5.2.1
Semi-reducibility
We shall say that a disjoint pair (A1, A2) is semi-reducible to a disjoint pair (B1, B2) if there is a E-function f (x) - under which (A1, A2) is said to be
Chapter 10. Doubly indexed relational systems
194
semi-reducible to (B1, B2) - such that for all x, x c Al = x e B1, and x E A2 = x E B2 - in other words, Al C f-1(B1) and A2 C f-1(B2) (instead of Al = f -1(B1) and A2 = f -1(B2), as in the case of reducibility). Thus, unlike the case of reducibility, if x is outside Al U A2, f (x) is not necessarily outside B1 U B2.
Proposition 5.3 If (A1i A2) is semi-D.G. and is semi-reducible to (B1, B2), where B1 is disjoint from B2, then (B1, B2) is semi-D.C. Moreover, if in addition f is a E-function under which (A1, A2) is semi-reducible to (B1, B2), then there is a E-function O(x) such that (B1, B2) is semi-D.G. under the function f (O(x)).
Proof The proof is very much like that of Proposition 4.1. Suppose h(x) is a semi-D.G. function for (A1, A2) and (A1, A2) is semi-reducible to
(B1, B2) under f (x) and B1 is disjoint from B2. By Corollary 3.6 there
is a E-function t(x) such that for all x, at(x) = f-l(ax) and ,Qt(x) =
f Then (B1, B2) is semi-D.G. under the function f (h(t(x))), because f (h(t(x))) E ax = h(t(x)) E f -l(ax) +-* h(t(x)) E at(x) = h(t(x)) E Al = f (h(t(x))) E B1. Similarly, f (h(t(x))) E 13 = f (h(t(x))) E B2. And so (B1, B2) is semi-D.G. under f (O(x)), where O(x) = h(t(x)).
Theorem 5.4 Suppose (A1, A2) is semi-D.G. and is semi-reducible to (B1, B2) under f (x). Then: (i) if B1 is an R-set, then for some E-function h(x), B1 is generative under f (k(x)); (ii) if B1 and B2 are both R-sets, then for some E-function k(x), (B1, B2) is D.G. under f (k(x)).
Proof Suppose (A1, A2) is semi-D.G. and is semi-reducible to (B1, B2)
C/]
under f(x). Then by Proposition 5.3 there is a E-function 01(x) such that (B1, B2) is semi-D.G. under f (01(x)). If B1 is an R-set, then by Theorem 5.1 there is a E-function 02(x) such that B1 is generative under f(01(02(x))), and we then take k(x) = 01(02(x)). If B2 is also an R-set, then by Theorem 5.2 there is a E-function 03(x) such that (131, 132) is D. G.
under f (01(03(x))), and we then take k(x) = 01(03(X))Semi-D. U. pairs mss'
We call a disjoint pair (A, B) semi-D. U. if every disjoint pair of R-sets is semi-reducible to (A, B). If (A, B) is D.U., then certainly some pair of semi-D.G. sets is semi-reducible to (A, B) (since there exists a semi-D.G. pair of R-sets, e.g., (U1, U2)), and so Theorem 5.4 yields: ue.
§6 Closing the circle
195
Theorem 5.5 (a) Every semi-D.U. pair of R-sets is D.G.
(b) If (A, B) is semi-D.U. and if A is an R-set, then A is generative. In particular, if (A, A) is semi-D.U. and A is an R-set then A is generative.
We shall say that a disjoint pair (A, B) is uniformly semi-D. U. if there
is a recursive function f (x, y) - under which (A, B) will be said to be uniformly semi-D.U. - such that for every i, is semi-reducible to (A, B) under f (i, x) (as a function of x).
Proposition 5.6 Every semi-D.U. pair is uniformly semi-D.U. Proof Suppose (A, B) is semi-D.U. Then the D.U. pair (U1, U2) is semireducible to (A, B) under some recursive function g(x). Then (A, B) is uniformly semi-D.U. under the function g(J(x, y)) (for x c ai <-4 J(i, x) E U1 = g(J(i, x)) E A, and similarly x E Qi = g(J(i, x)) E B).
§6 Closing the circle We have proved that every universal set is generative and that every D.U. pair is D.G. We will now prove that, conversely, every generative set is universal and that every D.G. pair is D.U.
Lemma 6.1 (a) For every R-set A there is a E-function t(x) such that for all is
iEA=wt(i)=N i 0AWt(i) = 0 ..y
(b) For every disjoint pair (A1, A2) of R-sets there is a E-function t(x) such that for all is i
Al
i 0A,
at(i) = N at(i) = 0
iEA2,(3t(i)=N
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Proof (a) Suppose A is an R-set. Define M(x, y) iff x c A. Since M is explicitly
definable from A, then M is an R-relation. Then by the iteration property there is a E-function t(x) such that for all i and x, x c
wt(i) H M(i, x). Thus x e wt(i) H i E A. And so if i c A, then x E wt(i) holds for every x, hence in this case wt(i) = N. If i 0 A, then x E wt(i) holds for no x, so in this case wt(i) = 0.
(b) Suppose A1i A2 are disjoint R-sets. By (a) there are E-functions tl (x), t2(x) such that for all is i Al i Al i c A2 i 0 A2
= wtl(i) = N wtl(i) = Wt2(i) =
0
N
Wt2(i) = 0
0
Now, i cannot be in both Al and A2, so at least one of the two sets wtl(i), Wt2(i) must be empty, and so these two sets are disjoint. Then by (c) of Proposition 3.0, wtl(i) = aJ(tl(i),t2(i)) and Wt2(i) = QJ(tl(i),t2(i)). We let t(x) be the E-function J(tl(x),t2(x)). Then for all i, at(i) = wtl(i) and Qt(i) = wt2(i), and the conclusion obviously follows.
Lemma 6.2 (a) Suppose B is generative under g(x). Then for any i,
(1) wi=N=g(i)EB (2) wi = 0 = g(i) V B (b) Suppose (B1i B2) is D.G. under g(x). Then for any i,
(1) ai = N = g(i) E B1
(2) ai=0=g(i)OBl (3) 13i = N = g(i) E B2
(4) 0i=0=g(i)OB2 Proof Obvious.
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197
Theorem 6.3 (a) Every generative set B is universal. Moreover if g(x) is a generative function for B then for every R-set A there is a E-function t(x) such that g(t(x)) reduces A to B. (b) Every D.G. pair (B1i B2) is D.U. Moreover if g(x) is a D.G. function for (B1, B2) then for every disjoint pair (A1i A2) of R-sets there is a E-function t(x) such that g(t(x)) reduces (Al, A2) to (Bl, B2).
Proof yam'
(a) Suppose that B is generative under g(x) and that A is any R-set. Take a E-function t(x) related to A as in Lemma 6.1. Then for any is
(1) i (2) i
A = wt(i) = N = g(t(i)) E B (by Lemma 6.2) A = wt(i) = 0 g(t(i)) 0 B (by Lemma 6.2)
Therefore g(t(x)) reduces A to B.
`.m
(b) Suppose (B,, B2) is D.G. under g(x). Let (A,, A2) be any disjoint .'3
pair of R-sets. Take a E-function t(x) related to (Al, A2) as in Lemma 6.1. Then by Lemma 6.1 and Lemma 6.2, for every is
(1) iEAl=at(i)=NZg(t(i))EBl (2) iEA2=Qt(i)=N=g(t(i))GB2 (3) i V (Al U A2)
(at(i) = 0 A,3t(i) _ 0) g(t(i)) V Bl A g(t(i)) V B2
Thus g(t(x)) reduces (A1i A2) to (B1i B2).
Some Consequences
Since every semi-D.U. pair of 1Z-sets is D.G. (Theorem 5.5), then by Theorem 6.3 we have:
Theorem 6.4 Every semi-D.U. pair of R-sets is D.U. Theorem 6.4A If (A, A) is semi-D.U. and A is an R-set, then A is universal.
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Remarks
Theorem 6.4 makes no mention of the notion of double generativity, although this notion was used in the proof. Another proof will emerge in the next chapter which makes no appeal to the notion of double generativity (the proof is a modification of a clever fixed point argument of John Shepherdson [26]) .
Obviously, if Al C_ B1 and A2 C B2 and B1 is disjoint from B2 then (A1, A2) is semi-reducible to (B1, B2) (under the identity function), and so as a corollary of Theorem 6.4 we have:
Theorem 6.5 If (A1i A2) is D.U. (or even semi-D.U.) and B1i B2 are C/]
disjoint 7Z-sets such that Al C B1 and A2 C B2, then (B1, B2) is D.U. Next, from Theorems 5.4 and 6.3(b) we have:
Theorem 6.6 Suppose that B1, B2 are disjoint R-sets. Then if some D.G. (or even semi-D.G) pair is semi-reducible to (B1, B2) then (B1, B2) is D.U. - moreover, if some semi-D.U. pair is semi-reducible to (B1, B2) under f (x), then for any disjoint pair (A1, A2) of 7Z-sets, there is a recursive function 0(x) such that (A1i A2) is reducible to (B1i B2) under f (O(x)).
(=U
Proof Assume the hypothesis. Then by Theorem 5.4 there is a E-function k(x) such that (B1, B2) is D.G. under f (k(x)). Then by (b) of Theorem 6.3 there is a E-function t(x) such that (A1, A2) is reducible to (B1, B2) under f (k(t(x))). And so we take O(x) = k(t(x)).
§7 Summary of main results Let us now summarize the main results of this chapter - particularly those that will have applications to the metamathematical results of the next chapter. (1) A set is generative if and only if it is universal. (2) A pair of sets is double generative if and only if it is doubly universal.
(3) Every semi-D.U. pair of R-sets is D.U. (4) If (A1, A2) is D.U. (or even semi-D.U.), and Al C B1, and A2 C B2, and B1, B2 are disjoint 1Z-sets, then (B1, B2) is D.U.
§7 Summary of main results
199
(5) If some semi-D.G. pair is reducible to (B1, B2) under f (x), and B1 and B2 are disjoint 1Z-sets, then there is a E-function k(x) such that (B1i B2) is D.G. under f (k(x)). (6) Under the hypothesis of (5), for every disjoint pair (A,, A2) Of R-sets there is a E-function O(x) such that (A1, A2) is reducible to (B1, B2) under f (O(x)).
Chapter 11
Effective representation systems Part I Effective representation 'nom'
!=h
(¢p
In this chapter we apply the results of the last chapter to the representation systems of Chapter 4. For any positive n we define rn+i (x, yl, , yn) to be the Godel number of Ex(y1, ... , yn). We call Z an effective representation system if: (1) for each positive n, the function rn+1(x) yl, ... , yn) is recursive; (2) the set of Godel numbers of the sentences is recursive, and for each n, the set of Godel numbers of predicates of degree n is recursive. Throughout this chapter Z will be assumed to be an effective representation system. We let r(x, y) be r2 (X, y) - thus r(x, y) is the Godel number of Ex(y) and the diagonal function d(x) is r(x,x). We shall also make an assumption (which in fact automatically holds for all representation systems associated with first-order systems) that for any positive n and any predicate K of n + 1 arguments and any number i, there is a predicate L of n arguments such that for all elements x1i ... , xn, the sentence L(xl,... , xn) is equivalent to K(xl,... , xn, i) and furthermore that for each such K, there is a recursive function O(x) such that for each number i, 0(i) is the Godel number of such a predicate L (thus EE(i) is a predicate of n arguments and for all x1, ... , xn, is EE(2) (x1, ... , xn) is provable, refutable in Z if K(xl, ... , xn, i) is respectively provable, refutable in Z). s^.
Proposition 0 (a) The diagonal function d(x) is recursive.
(b) If A is r.e. (recursive), so is A# (A# = d-1 (A)). (c) If Wo is r.e. (recursive), so is W*.
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201
(d) If the sets Po and Ro are r.e. (recursive), then all sets representable or contrarepresentable in Z are r.e. (respectively recursive).
Exercise 0 Prove Proposition 0.
§1 Undecidability We shall call Z an r. e. system, or say that Z is axiomatizable, if the sets Po and Ro are r.e. We call Z decidable if the sets PO and Ro are recursive; otherwise Z is called undecidable. It is perhaps unfortunate that the word undecidable has two quite different meanings - one as applied to a sentence being undecidable in a given
system, and the other as applied to the system as a whole. The second meaning is tied up with recursion theory; the first is not.
Theorem 1.1 If every recursive set is representable in Z, then Z is undecidable (more specifically, PO is not recursive).
Proof Suppose all recursive sets are representable in Z. If PO were recursive, then P* would be recursive (by (c) of Proposition 0), hence P* would be recursive, hence representable in Z, contrary to Proposition 1.2 of Chapter 4 (taking P for K). Hence PO is not recursive, so Z is undecidable.
Theorem 1.2 If some non-recursive set is representable in Z, then Z is undecidable.
Proof This is equivalent to the statement that if Z is decidable then This latter is true by (d) of
every set representable in Z is recursive. Proposition 0.
The following is obviously a corollary of Theorem 1.1, and also of The-
orem 1.2, together with the fact that there exists an r.e. set that is not recursive.
Corollary 1.3 If all r.e. sets are representable in Z, then Z is undecidable.
Exercise 1 (a) Show that if every r.e. set is representable in Z then Po is not r.e.
Chapter 11. Effective representation systems
202
(b) Show that if the complement of every r.e. set is representable in Z, then P0 is not r.e. Normality We recall that we are defining Z to be normal if whenever A is representable
in Z, so is A#. By (b) of Proposition 0 we have:
Proposition 1.4 If the sets representable in Z are precisely those that are r.e., then Z is normal. From the above and (d) of Proposition 0 we have:
Proposition 1.5 If Z is an r.e. system in which all r.e. sets are representable, then Z is normal. Undecidability and incompleteness
Theorem 1.6 If Z is axiomatizable but undecidable, then Z is inconsistent or incomplete.
Proof We will prove the equivalent proposition that if Z is consistent, complete, and axiomatizable, then Z is decidable. And so suppose that Z is consistent and complete and that P0 and Ro are r.e. sets. Thus So is the disjoint union of P0 and_Ro. We are assumin throughout this chapter that So is recursive, and so_So is r.e. Now, Po = So U Ro, hence Po is r.e. Thus P0 is recursive. Also Ro = So U Po, so Ro is r.e. and hence Ro is recursive. Thus Z is decidable.
Proposition 1.7 If Z is axiomatizable, then there is a recursive function O(x) such that for any predicate H of degree 1 with Godel number h, WO(h)
is the set represented by H in Z. Proof We recall that we are letting r(x, y) be the Godel number of E,,(y) and we are assuming that r(x, y) is a recursive function. Let R(x, y) be the relation "r(x, y) E Po." Since r is a recursive function and P0 is r.e. (by hypothesis), then R(x, y) is an r.e. relation, and so by the iteration theorem there is a recursive function O(x) such that for any numbers i and x, x E wo(2) H R(i, x). Then for any unary predicate H with Godel number h, x E WO(h) +-* R(h, x) +-* r(h, x) E Po H H(x) is provable. And so WO(h) is the set represented by H.
§1 Undecidability
203
Universal systems
We call Z universal if all r.e. sets are representable in it. From Proposition 0, (d), Corollary 1.3, Proposition 1.5, and Theorem 1.6 we have:
Theorem 1.8 Suppose that Z is universal and axiomatizable. Then (1) The representable sets of Z are precisely those that are recursively enumerable.
(2) Z is normal. (3) Z is undecidable. ;-s
(4) The set Po is r.e. but not recursive. car
(5) Z, if consistent, is incomplete.
Exercise 2 Suppose that Z is universal and axiomatizable. Show that for every r.e. set A there is a Godel sentence for A, i.e., a sentence X such that X E P H X0 E A (X0 is the Godel number of X). Effectively universal systems
We shall say that Z is effectively universal if there is a recursive function O(x) such that for every number i, 0(i) is the Godel number of a predicate that represents wi in Z. [Informally, this means that not only is it the case that every r.e. set is representable in Z, but given any r.e. set A, in the sense of given an index of it, we can effectively find a predicate that represents A.]'
Exercise 3 Suppose Z is axiomatizable and effectively universal. Show that there is a recursive function O(x) such that given any predicate H (of degree 1) with Godel number h, the number 0(h) is the Godel number of a Godel sentence for the set represented in Z by H.
Theorem 1.9 If the relation x E wy is representable in Z then Z is effectively universal. 'Such systems were called effectively Godelian in R.M., but the present terminology seems more appropriate.
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204
Proof Suppose K is a binary predicate that represents the, relation x E WY. Let O(x) be a recursive function such that for every number i, Eo(2) is a unary predicate such that for all x, Eo(2) (x) is equivalent to K(x, i). Then x E Wi 4-4 K(x, i) is provable in Z H Eo(2) (x) is provable in Z. And so Eo(i) represents wi in Z.
§2 Generative systems For the remainder of this chapter we will be using definitions and theorems
of the last chapter in the special case in which R is the collection of r.e. relations and E is the class of recursive functions. We call Z generative if the set Po is generative (which for recursion theory means that there is a recursive function O(x) - a generative function
for P0 - such that for all n, 0(n) E Po H 0(n) E Wn. We shall call a CAD
(unary) predicate H a generative predicate if for every n, H(n) is provable if its Godel number is in W. Of course if H is generative, then the system Z is generative (under the function r(h, y), where h is the Godel number of H). We shall say that H is a near generative predicate if there is a recursive function k(x) such that for every number n, H(k(n)) is provable in Z if its Godel number is in wn. If H is even a near generative predicate for Z, then
Z must again be generative - generative under the function r(h, k(x)), where h is the Godel number of H.
Theorem 2.1 If H represents some generative set then H is a near generative predicate.
acs
Proof Let A be a generative set represented by H. For any x, let f (x) be the Godel number of H(x). Then f is a recursive function, and since H represents A, then f clearly reduces A to the set P0. Then by Proposition 2.1 of Chapter 10, there is a recursive function k(x) such that P0 is generative under the function f (k(x)), and so for every n, H(k(n)) is provable .s2
H f (k(n)) E Po H f (k(n)) E wn, but f (k(n)) is the Godel number of H(k(n)). Thus H is a near generative predicate. Since there does exist an r.e. generative set (e.g., the set C of all x such that x E wx) we have as corollary:
Theorem 2.2 If Z is universal then Z has a near generative predicate.
§3 Doubly generative systems
205
Remark We will see later on that if Z if effectively universal, then Z has a generative predicate. [The proof involves a combination of two fixed point arguments - one of them a variation of an argument of Shepherdson[26].]
We will say that a predicate H almost represents a set A if there is a recursive function O(x) such that for all n, n E A +-* H(O(n)) is provable in Z.
Theorem 2.3 If H is a near generative predicate, then for every r.e. set A, H almost represents A.
Proof Suppose k(x) is a recursive function such that for all n, H(k(n)) is provable if its Godel number is in Wn. Let g(n) be the Godel number of H(k(n)). Then g(x) is a recursive function and for all n, g(n) E Po <--> g(n) E wn, so Po is generative under g(x). Then by (a) of Theorem 6.3 of the last chapter, for every r.e. set A, there is a recursive function t(x) such that g(t(x)) reduces A to Po. Since g(t(x)) is the Godel number of H(k(t(x))), then for all n, H(k(t(n))) is provable if g(t(n)) E Po if n E A.
We thus take O(x) = k(t(x)), and so for all n, H(q(n)) E P <--> n E A, which means that H almost represents A. Generative sets and incompleteness
We recall from the last chapter that Post's generative set C (the set of all n such that n E wn) is generative under the identity function.
Theorem 2.4 Suppose C is contrarepresentable in Z and Z is axiomatizable and consistent. Then Z is incomplete. More specifically, suppose Z is axiomatizable and consistent and that ,i.
H contrarepresents C in Z. Then if w2 is the set represented by H, the sentence H(i) is undecidable in Z.
Exercise 4 Prove Theorem 2.4.
§3 Doubly generative systems We call a consistent system Z a D.G. system if the pair (Po, Ro) is D.G., and we call Z semi-D. G. if the pair (Po, Ro) is semi-D.G. We shall now call a (unary) predicate H a D.C. predicate if (P0,R0) is D.G. under the functions r(h, x), where h is the Godel number of H. In other words, H
Chapter 11. Effective representation systems
206
(CD
fir"
is a D.G. predicate if for every number n, H(n) is provable in Z if its Godel number is in an, and is refutable in Z if its Godel number is in On. We shall call H a near D.C. predicate if there is a recursive function k(x) such that for all n, H(k(n)) is provable if its Godel number is in an, and refutable if its Godel number is in On (in other words, (Po, Ro) is D.G. under the function r(h, (k(x))).
Theorem 3.1 Suppose that H is a near D.G. predicate for Z. Then for every disjoint pair (A1i A2) of r.e. sets, there is a recursive function O(x) such that for all n: H(O(n)) is provable if n E A1, and refutable if n E A2. '32
Proof Suppose that H is a near D.G. predicate for Z. Let k(x) be a recursive function such that for all n, H(k(n)) is provable (refutable) in
Z if n E an (n Eon). Let g(n) be the Godel number of H(k(n)). Then the function g(x) is recursive and (Po, Ro) is D.G. under g(x). Then by Theorem 6.3 (b) of Chapter 10, for any disjoint pair (A,, A2) of r.e. sets, there is a recursive function t(x) such that (A,A2) is reducible to (Po, Ro) under g(t(x)). Since g(t(x)) is the Godel number of H(k(t(x))), then for all n, H(k(t(n))) is provable in Z if n E A1, and refutable in Z if n E 01. And so we take O(x) = k(t(x)), and then H(cb(n)) is provable if n E A1, and refutable if n E A2. 'z3
Semi-D. G. predicates coo
H(i) We shall call H a semi-D. G. predicate if for all i, gn(H(i)) E ai is provable, and gn(H(i)) E 13i = H(i) is refutable. [By gn(H(i)) we mean the Godel number of H(i).] We shall, say that H is a near semi-D. G. predicate, or that H is almost a semi-D.G. predicate, if there
is a recursive function k(x) (which we will call an associate of H) such that for all i, gn(H(k(i))) E ai = H(k(i)) is provable, and gn(H(k(i))) E Qi = H(k(i)) is refutable. [These near semi-D.G. predicates will turn out to be more useful than their definitions might suggest!]
Theorem 3.2 If Z is consistent and axiomatizable, then every near semiD.G. predicate is a near D.G. predicate.
Proof Suppose that Z is consistent and axiomatizable and that H is a near semi-D.G. predicate with associate k(x). Thus for all n, gn(H(k(n))) E an = H(k(n)) E P, and gn(H(k(n))) E On = H(k(n)) E R. Let f (n) be the Godel number of H(k(n)). Then the pair (Po, Ro) is semi-D.G. under f (x). Also by hypothesis, PO and Ro are r.e. and
§4 Rosser systems
207
disjoint, Then by Proposition 5.2 of the last chapter, there is a recursive function O(x) such that (Po, Ro) is D.G. under f (q(x)). Thus for an <--> f (O(n)) E Po H H(k(O(n))) is provable, and all n, f (O(n)) f(O(n)) E On <-> f(O(n)) E Ro f-* H(k(O(n))) is refutable. We thus let
(l+
%(x) = k(O(x)), and so H(z/&(n)) is provable if its Godel number is in an,
and is refutable if its Godel number is in On. Thus H is a near D.G. predicate.
Theorem 3.3 Suppose that some semi-D.G. pair (A, B) is strongly separated by H in Z and that Z is consistent. Then (1) H is a near semi-D.G. predicate.
(2) If, also, Z is axiomatizable, then H is a near D.G. predicate.
Proof Suppose H strongly separates A from B in Z, where (A, B) is semi-D.G. Let h(x) be the Godel number of H(x). Then h(x) is a recursive function, and (A, B) is semi-reducible to (Po, Ro) under h(x). Then by Proposition 5.3 of the last chapter, there is a recursive function O(x) such that (Po,Ro) is semi-D.G. under the function h(0(x)), which means that H is a near semi-D.G. predicate with O(x) as an associate. If, also, Z is axiomatizable, then H is a near D.G. predicate by Theorem 3.2.
Theorem 3.4 If (A, B) is semi-D.G. and H weakly separates A from B in Z and if Z is axiomatizable, then H is a near generative predicate.
Proof Assume hypothesis. Let h(x) be the Godel number of H(x). Since H weakly separates A from B, then (A, B) is semi-reducible to AI Po) under h(x). Since PO is r.e., then by Theorem 5.4, Chapter 10, there is a recursive function O(x) such that Po is generative under f (O(x)). Hence H is a near generative predicate.
§4 Rosser systems The definition of two relations being separable in a representation system Z is quite analogous to that for sets. Given a predicate K of n arguments, we let Kp be the relation represented by K, and KR the relation contrarepre(CD
sented by K (the set of all n-tuples (al, ... , an) such that K(al,... , an) is refutable in Z). Then for any disjoint relations R1, R2 of n arguments, we say that K strongly separates R1 from R2 if R1 C Kp and R2 C_ KR. [For n = 1, this coincides with our definition of a unary predicate strongly sep-
Chapter 11. Effective representation systems
208
arating a pair of sets.] From now on the word separable will mean strongly separable, unless specified to the contrary. We now define Z to be a Rosser system for sets if every disjoint pair of r.e. sets is (strongly) separable in Z. More generally, for any positive n, we define Z to be a Rosser system for relations of degree n if every disjoint pair of r.e. relations of n arguments is separable in Z. We define Z to be a Rosser system if for every n, Z is a Rosser system for relations of degree n (this includes the case n = 1). We shall say that Z is effectively a Rosser system for sets - or that Z is an effective Rosser systems for sets - if there is a recursive function II(x) ,s"
- which we call a Rosser function for Z - such that for every number i, the number 11(i) is the Godel number of a (unary) predicate that separates ai from Qi in Z. Note
In R.M. we defined Z to be effectively a Rosser system for sets if there is a recursive function 11(x, y) of two arguments such that for all numbers i and j, 11(i, j) is the Godel number of a predicate that separates wi - wj from wj - wi in Z. The two definitions are actually equivalent (but we have found the present definition more convenient). car
Exercise 5 Prove that the two definitions are equivalent. For later use we will need:
Theorem 4.1 If Z is a Rosser system for binary relations, then Z is effectively a Rosser systems for sets. In fact if the relation x E ay is (strongly) separable from the relation x E Qy, then Z is effectively a Rosser system for sets.
Proof Suppose that K (strongly) separates the relation x E ay from x E Qy. Let 11(x) be a recursive function such that for all i, En(i) is a unary predicate such that for all x, En(i)(x) is equivalent to K(x, i).
Then for any x: x E ai = K(x, i) E P = En(i) (x) E P. x E Qi
Similarly,
En(i) (x) E R. Then 11 (x) is a Rosser function for Z. t"'
Since there exists a D.G. pair of r.e. sets, the following result is a consequence of Theorem 3.3:
Theorem 4.2
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209
(a) If Z is a consistent Rosser system for sets, then there is a near semiD.G. predicate for Z.
(b) If, also, Z is axiomatizable, then there is a near D.G. predicate for Z.
Remarks
We will see later that if Z is a consistent axiomatizable effective Rosser system for sets, then Z contains a D.G. predicate. Whether this holds if we drop "effective," we do not know. Semi-Rosser systems
We shall say that Z is a semi-Rosser system for n-ary relations if for any disjoint r.e. relations R1, R2 of n arguments, R1 is weakly separable from R2
in Z (i.e., there is a predicate K of degree n such that R1 C Kp and Kp is disjoint from R2 - in other words, for all x1, ... , xn, if Rl (xi, ... xn), then K(xl,... , xn) is provable in Z, and if R2(x1i ... , xn), then K(xl,... , xn) is not provable in Z). We shall now be concerned with semi-Rosser systems )
for sets. Since there is a semi-D.G. pair of r.e. sets, Theorem 3.4 yields:
Theorem 4.3 If Z is an axiomatizable semi-Rosser system for sets, then Z contains a near generative predicate. Stem functions
By a stem function we shall mean a function o(x) such that for some number h, or(x) = J(x, h) for all x. Thus the stem functions are J(x, 1), J(x, 2), ... , J(x, n).... (as functions of x). [Stem functions are not Fir
only primitive recursive, but even constructive arithmetic.]
Theorem 4.4 If Z is a Rosser system for sets, then for any recursive function f (x) there is a predicate H and a stem function or(x) such that for all x, H(o-(x)) is provable if its Godel number is in a f(x), and refutable if its Godel number is in Of (.) .
Proof Given a recursive function f (x), let Al be the set of all numbers and let A2 be the set of all J(x, y) such that gn(E,, (J(x, y))) E a numbers J(x, y) such that gn(E,, (J(x, y))) E 3,,. The sets Al and A2 are r.e. and disjoint, and so by hypothesis there is a unary predicate Eh that strongly separates Al from A2. Therefore, for any x:
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(1) gn(Eh(J(h, x))) E a f(x) = J(h, x) E Al = Eh(J(h, x)) E P (2) gn(Eh(J(h, x))) E Of(.)
J(h, x) E A2 = Eh(J(h, x)) E R
And so we take o(x) = J(h, x). Taking f to be the identity function, we have the following strengthening of (a) of Theorem 4.2.
Corollary 4.5 If Z is a consistent Rosser system for sets then Z contains a near semi-D.G. predicate with a stem function for an associate.
§5 Extensions We shall now see a significant property of near semi-D.G. predicates.
We shall say that a representation system Z' is an extension of a representation system Z - or that Z' extends Z - if the expressions of Z and Z' are the same and all sentences, provable sentences, and refutable sentences of Z are, respectively, sentences, provable sentences, and refutable sentences of T. The following lemma is obvious.
Lemma 5.1 Suppose that Z' is a consistent extension of Z and that X is a sentence of Z such that :
(1) If X is provable in Z', then X is refutable in Z. (2) If X is refutable in Z', then X is provable in Z. Then X is undecidable in Z'. We now let Z be a denumerable sequence Z1, Z2, ... , Zn, ... of representation systems. We shall call Z an effective sequence, or an r.e. sequence, if the relations "E., is provable in Z," and "Ex is refutable in Z," are r.e. relations. We shall now define a predicate H of Z to be a master predicate if for every r.e. sequence Z1, Z2, ... , Zn, ... of extensions of Z, there is a recursive function O(x) such that for every number i for which Zi is consistent, the sentence H(O(i)) is undecidable in Zi. We wish to prove that every near semi-D.G. predicate is a master predicate. We first prove:
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211
Lemma 5.2 If Z1, Z2, ... , Zn, ... is an r.e. sequence, then there is a recursive function q(x) such that for any i for which ZZ is consistent, aq(i) is the set of Godel numbers of the refutable sentences of ZZ and Oq(i) is the set of Godel numbers of the provable sentences of Zi.
Proof Since the relations "E,, is provable in Zy," "Ex is refutable in Zy" are r.e., then by the iteration theorem there are recursive functions pl(x),p2(x) such that for all i, wpj(i),wp2(i) are, respectively, the set of Godel numbers of the provable, refutable sentences of Zi. Let q(x) _ J(p2(x), p1(x)). Then if ZZ is consistent, the set wp2(i) is disjoint from wpl(i), hence by (c) of Proposition 3.0, Chapter 10, aq(i) = Wp2(i) and Qq(i) = wpl(j), and so aq(i) is the set of Godel numbers of the refutable sentences of Zi, and 13q(i) is the set of Godel numbers of the provable sentence of Zi. Now we can prove:
Theorem 5.3 Every near semi-D.G. predicate is a master predicate. Proof Given an r.e. sequence Zl, Z2, ... , Zn, ... of extensions of Z, let q(x) be as in Lemma 5.2. Now, suppose H is a near semi-D.G. predicate for Z. Let k(x) be an associate of H. Then for all i, H(k(q(i))) is provable in Z if its Godel number is in aq(i), and refutable in Z if its Godel number is in,3q(i). This means that if ZZ is consistent, then H(k(q(i))) is provable in Z if it is refutable in Zi, and refutable in Z if it is provable in Zi, and hence by Lemma 5.1, is undecidable in ZZ (if Zi is consistent). We thus take O(x) = k(q(x)), and so for every i for which ZZ is consistent, H(O(i)) is undecidable in Z. From Theorems 5.3 and (a) of Theorem 5.2 we have:
Theorem 5.4 If Z is a consistent Rosser system for sets then Z contains a master predicate.
Theorem 5.4 tells us that if Z is a consistent Rosser system for sets, then for any r.e. sequence Z1, Z2, ... , Zn, ... of extensions of Z there is a predicate H and a recursive function O(x) such that for every i for which ZZ is consistent, H(O(i)) is undecidable in Zi. But we have the following strengthening of this:
212
Chapter 11. Effective representation systems
Theorem 5.5 If Z is a consistent Rosser system for sets, then for any r.e. sequence Z1, Z2,. . ., Zn, ... of extensions of Z, there is a predicate H and a stem function Q(x) such that for every i for which ZZ is consistent, H(Q(i)) is an undecidable sentence of ZZ
Proof Suppose that Z is a consistent Rosser system. Given an r.e. sequence Z1, Z2, ... , Zn, ... of extensions of Z, let q(x) be as in Lemma 5.2. Then by Theorem 4.4, there is a predicate H and a stem function o,(x) such that for every i, H(o(i)) is provable in Z if its Godel number is in aq(2), and refutable it its Godel number is in Nq(Z). Then, by a similar argument to the latter portion of the proof of Theorem 5.3, if ZZ is consistent, then H(Q(i)) is undecidable in Z.
Part II Exact and effective Rosser systems §6 Exact Rosser systems We shall call Z an exact Rosser system for sets if every disjoint pair (A, B) of r.e. sets is exactly separable in Z (which we recall means that there is a predicate H1 that represents A and contrarepresents B). We also recall that a function f (x) is called admissible in Z if for every (unary) predicate H there is a predicate H1, such that for all n, H1(n)
H(f (n)) (i.e., Hl (n) E P <-4H (f (n)) E P, and Hl (n) E R
H(f (n)) E
R).
Theorem 6.1 If there is a near D.G. predicate for Z and if all recursive functions of one argument are admissible in Z, then Z is an exact Rosser system for sets.
Proof Suppose that H is a near D.G. predicate for Z and that all recursive functions of one argument are admissible in Z. Let (A1, A2) be any disjoint pair of r.e. sets. By Theorem 3.1 there is a recursive function O(x) such that for all n, H(O(n)) is provable (refutable) in Z if n E Al (n E A2). By hypothesis O(x) is admissible in Z, and so there is a (unary)
predicate H1 such that for all n, H1(n) - H(q(n)). Then H1 exactly separates (A1i A2) in Z. From Theorem 4.2(b) and Theorem 6.1 we have the following Theorem of Putnam and Smullyan [15]:
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213
Theorem 6.2 If Z is a consistent, axiomatizable Rosser system for sets in which all recursive functions of one argument are admissible, then Z is an exact Rosser system for sets. The above Theorem can also be obtained as a corollary of the following two propositions.
Proposition 6.3 If Z is a consistent axiomatizable Rosser system for sets then some D.U. pair is exactly separable in Z.
Proposition 6.4 If some D.U. pair is exactly separable in Z and if all recursive functions of one argument are admissible in Z, then Z is an exact Rosser system for sets.
To prove Proposition 6.3, assume the hypothesis. Then some D.U. pair (B1, B2) of r.e. sets is separable in Z - let H be a predicate that effects the separation. Then B1 C_ Hp and B2 C HR, and since Z is axiomatizable,
the sets Hp and HR are r.e. sets. Also, Hp is disjoint from HR by the
s-'
a^'
assumption of consistency. Then by Theorem 6.5 of the last chapter, the pair (Hp, HR) is D.U. And of course H exactly separates Hp from HR. As for Proposition 6.4, suppose that H exactly separates (B1, B2) in Z, where (B1i B2) is D.U. Now let (A1i A2) be any disjoint pair of r.e. sets. Then there is a recursive function f (x) that reduces (A1i A2) to (B1i B2), and hence for any n, H(f (n)) is provable (refutable) iff f (n) E Bl (f (n) E B2) if n E Al (n E A2). And so, as in the proof of Theorem 6.1, if the function f (x) is admissible, then there is a predicate H1 such that for all x, Hi(x) - H(f (x)), and H1 then exactly separates Al from A2 in Z. Since every exact Rosser system for sets is obviously universal, then 111
Theorem 6.2 yields the following theorem of Ehrenfeucht and Feferman [4]:
Theorem 6.5 If Z is a consistent axiomatizable Rosser system for sets in which all recursive functions of one argument are admissible, then Z is universal.
Actually, Theorem 6.5 can be strengthened to (b) of the theorem below:
Theorem 6.6 Suppose all recursive functions are admissible in Z. Then (a) If Z contains a near generative predicate, then Z is universal. (b) If Z is an axiomatizable semi-Rosser system for sets, then Z is universal.
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Proof Suppose all recursive functions are admissible in Z. cad
(a) Suppose H is a near generative predicate. Then by Theorem 2.3, for any r.e. set A, H almost represents A, which means that there is a recursive function O(x) such that for all n, H(e(n)) E P H n E A. By hypothesis, q(x) is admissible, hence there is a predicate Hl such
that for all n, Hl(n) - H(q(n)). Then Hl represents A. (b) This is a consequence of (a) and Theorem 4.3. Historical remarks
It was known prior to 1960 that the system of Peano Arithmetic is +a-
universal provided it is w-consistent. The first proof of the universality of Peano arithmetic that did not need to assume w-consistency was that of Ehrenfeucht and Feferman[4]. Then came the Putnam-Smullyan result that strengthened the conclusion from "universal" to "exact Rosser system for sets." The original proofs of both the Ehrenfeucht and Feferman theorem and the Putnam-Smullyan theorem used fixed point arguments (the recursion theorem for the first and the double recursion theorem for the second). It was in R.M. that we showed that the original fixed point arguments were not necessary.
Theorem 6.7 If Z is a consistent axiomatizable exact Rosser system for sets, then all recursive functions of one argument are admissible in Z.
Proof Assume the hypothesis. Let f (x) be any recursive function and H any unary predicate. Let A be the set of all n such that H(f (n)) is provable in Z, and B the set of all n such that H(f (n)) is refutable in Z. Since Z is axiomatizable, the sets A and B are r.e. Since Z is consistent, the sets A and B are disjoint. Therefore some predicate Hl exactly separates (A, B) in Z, and so for all x, Hl (x) E P +-4x E A H H(f (x)) E P, and likewise Hl (x) E R - H(f (x)) E R. And so f (x) is admissible in Z.
§7 Variations on a theme by Shepherdson John Shepherdson[26] proved by an entirely different method, which used virtually no recursion theory, that for systems couched in first-order logic, if the system is axiomatizable, consistent, and a Rosser system for binary relations, then it is an exact Rosser system for sets. His proof goes through intact for representation systems. The hypothesis appears to be incomparable in strength to that of the Putnam-Smullyan theorem. On the one hand,
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215
the hypothesis that Z is a Rosser system for binary relations is stronger than that Z is a Rosser system for sets. On the other hand, Shepherdson's conditions do not involve the necessity that all recursive functions of one argument be admissible in Z. And so the two theorems are apparently of incomparable strength.
In R.M. we showed how a modification of Shepherdson's argument yielded an alternative proof that every semi-D.U. pair of r.e. sets is D.U. This argument avoids all appeal to double generativity - and indeed, to indexing. We also stated that the two results could be obtained from a common theorem, which we promised to give in this volume, and which we will shortly do.
Then we gave a third argument - another variation of Shepherdson's argument - which strengthened the Putnam-Smullyan theorem by replacing the hypothesis that all recursive functions of one argument are admissible in Z by the much weaker hypothesis that all stem functions are admissible in Z. It turns out pleasantly that this third result, together with the other two, can be obtained from one common theorem. We now give this theorem and show how all three of the above results can be obtained as corollaries. The theorem that follows is a pure theorem of recursion theory, which could have been stated and proved in Chapter 7.
"r30'
Theorem 7.1 Suppose that C is a set and that Si(x, y, z), S2(x, y, z) are disjoint r.e. relations such that for any two disjoint r.e. relations Mi(x, y), M2 (x, y) there is a number k in C such that for all x and y, Mi (x, y) = S1 (k, x, y), and M2 (x, y) = S2 (k, x, y). Then for any two disjoint r.e. sets A1, A2 there is a number k in C such
that for all x:
(a) x c Al H Sl (k, x, k)
(b) xEA2HS2(k,x,k) Proof Assume the hypothesis. Let A1, A2 be disjoint r.e. sets. Let Rl(x, y) be the r.e. relation x c Al V S2(y, x, y), and let R2(x, y) be the r.e. relation x E A2VS1(y, x, y). Let Mi(x, y), M2(x, y) be disjoint r.e. relations such that R1-R2 C_ M1 and R2-R1 C M2. Now take a number k in C such
that for all x and y, Mi(x, y) = Si(k, x, y), and M2 (x, y) = S2 (k, x, y). Then for all x and y: (1) [(x E Al V S2(y, x, y)) A -(x E A2 V S1(y, x, y))] = S1(k, x, y) (2) [(x e A2 V S1(y, x, y)) A -(x E Al V S2 (y, x, y))] = S2 (k, x, y)
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We now take k for y and we have: (1)' [(x E Al V S2(k, x, k)) A -(x E A2 V S1(k, x, k))]
S1(k, x, k)
(2)' [(x E A2 V Si(k, x, k)) A -(x E Al v S2(k, x, k))]
S2(k, x, k)
Since Al is disjoint from A2 and the relations S1, S2 are disjoint, it follows from (1)' and (2)' by propositional logic that x c Al S1(k, x, k), and x e A2 H S2(k, x, k) (see Exercise 7 of the last chapter). Note
We proved the above theorem for R the collection of r.e. relations. Of course
the argument goes through for any standard indexed relations system R (replacing "r.e. relation" by "R-relation," and "r.e. set" by "R-set." For our first application of Theorem 7.1 we have:
Theorem 7.2 [Shepherdson] If Z is a consistent axiomatizable Rosser system for binary relations, then Z is an exact Rosser system for sets.
Proof Assume the hypothesis. In Theorem 7.1 take S1 (x, y, z) to be the r.e. relation "Ex (y, z) is provable in Z" (it is the relation 'Y2 (x, y, z) E Po"), and P0 is r.e. by hypothesis), and let S2(x, y, z) be the r.e. relation "Ex (y, z)
SAC
is refutable in Z." These two relations are disjoint by the assumption of consistency. We let C be the set of Godel numbers of the predicates of degree 2. Then the hypothesis of Theorem 7.1 holds (since Z is a Rosser system for binary relations). Therefore by Theorem 7.1, for any disjoint r.e. sets Al and A2, there is a number k in C - a Godel number of some predicate K of degree 2 - such that for all x, (a) and (b) of the conclusion hold, which means that x E A, if K(x, k) is provable, and x E A2 if K(x, k) is refutable. As stated in §0, we are assuming that there is a predicate H of degree 1 such that for all x, H(x) - K(x, k), and H then exactly separates (Al, A2) in Z. As another consequence of Theorem 7.1 we have:
Corollary 7.3 Suppose that B1, B2 are disjoint r.e. sets and that C is some set and f (x, y) is a recursive function such that for any two disjoint r.e. sets Al and B1, there is a number bin C such that (A,, A2) is semireducible to (B1, B2) under f (b, x) (as a function of x). Then for any disjoint r.e. sets Al and A2 there is a number h in C such that (A1, A2) is reducible to (B1, B2) under the function f (h, J(h, x)).
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217
s.4
Proof Assume the hypothesis. Let Sl (x, y, z) be the relation f (x, J(z,y)) E B1 and S2(x, y, z) be the relation f (x, J(z, y)) E B2. Then cad
(Z?
the relations S1 and S2 are r.e. and disjoint (since B1 and B2 are r.e. and disjoint). Given disjoint r.e. relations Rl (x, y), R2 (x, y), let Al be the set of all numbers J(x, y) such that R1(y, x), and let A2 be the set of all J(x, y) such that R2(y, x). Then by hypothesis there is a number bin C such that for all x and y, Ri(x, y) = J(y, x) E Al = f (b, J(y, x)) E B1 = S1(b, x, y), and similarly R2(x, Y) = S2(b, x, y). Thus the hypothesis of Theorem 7.1 is fulfilled, and so for any disjoint r.e. sets Al and A2, there is a number c7'
h in C such that for all x, x c Al H Sl (h, x, h) - f (h, J(h, x)) E B1, and similarly, x e A2 +-4f (h, J(h, x)) E B2, which means that (A1, A2) is reducible to (B1, B2) under f (h, J(h, x)). Of course the conclusion of Corollary 7.3 implies that (B1, B2) is D.U., and so we have the following alternative proof that every semi-D.U. pair of r.e. sets is D.U. Suppose (B1, B2) is a semi-D.U. pair of r.e. sets. Then by Proposition 5.6, Chapter 10, the pair (B1, B2) is uniformly semi-D.U. under some recursive function f (x, y). Then the hypothesis of Corollary 7.3 is obviously fulfilled, and so by Corollary 7.3, (B1, B2) is D.U. And now Corollary 7.3 yields the following strengthening of Theorem 6.2 (which was stated and proved in R.M.).
Theorem 7.4 [A strengthening of the Putnam-Smullyan theorem] If Z is a consistent axiomatizable Rosser system for sets, and if every stem function is admissible in Z, then Z is an exact Rosser system for sets. °"'
Proof Assume the hypothesis. Then the hypothesis of Corollary 7.3 holds, taking f (x, y) = r(x, y) (the Godel number of Ex (y)), B1 = Po, B2 = R0, and C the set of Godel numbers of unary predicates. Then by Corollary 7.3, for any disjoint pair (A1, A2) of r.e. sets there is a number h that is the Godel number of some predicate H such that for all x:
x e A,
H(J(h, x)) E P
xEA2HH(J(h,x))ER By hypothesis the function J(h, x) (as a function of x) is admissible, and so there is a predicate H1 such that for all x, Hl (x) is equivalent to H(J(h, x)). Then H1 exactly separates the pair (A1, A2) in Z.
Exercise 6 Suppose that the hypothesis of Theorem 7.1 holds. Derive the stronger conclusion that for any disjoint r.e. relations Ri(x, y), R2(x, y) there is a number k in C such that for all x:
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(1) Ri(x, k) <-4S, (k, x, k) (2) R2 (x, k) +-4S2 (k, x, k)
Why is this a stronger conclusion?
Exercise 7 Suppose that Z is a consistent axiomatizable Rosser system for binary relations. Let Rl (x, y), R2(x, y) be disjoint r.e. relations
(a) Show that there is a predicate K of degree 2 with Godel number k such that for all x: (1) K(x, k) is provable +-4R, (x, k)
(2) K(x, k) is refutable
R2(x, k)
(b) Show that there is a unary predicate H with Godel number h such that for all x: (1) H(x) is provable +-4R, (x, h) (2) H(x) is refutable R2(x, h)
Pry
(c) Show that there is a unary predicate H with Godel number h such that for all x: (1) H(x) is provable <-4R, (x,gn(H(x))) (2) H(x) is refutable R2(x, gn(H(x))) [By gn(H(x)) we mean the Godel number of H(x).]
Exercise 8 We have already proved that if Z is a consistent Rosser system for sets, then Z contains a near-D.G. predicate. Prove that if Z is a consistent axiomatizable Rosser system for binary relations, then Z contains a D.G. predicate.
Exercise 9 Let us say that S is a semi-Rosser system for binary relations if for any disjoint r.e. relations R1(x, y), R2(x, y) there is a predicate K of degree 2 that weakly separates Rl from R2 (i.e Rl C Kp and Kp is disjoint from R2 - or what is the same thing, for all x and y: Rl (x, y) K(x, y) is provable, and R2(x, Y) = K(x, y) is not provable). Show that if Z is consistent, axiomatizable, and is a semi-Rosser system for binary relations, then Z is universal (all r.e. sets are representable in Z).
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Exercise 10 Suppose that S(x, y, z) is an r.e. relation and C is a set such that for any disjoint r.e. relations Mi(x, y), M2 (x, y) there is a number k in C such that for all x and y: (1) Mi(x,y) z S(k,x,y) (2) M2(x,y) = - S(k,x,y)
Prove that for any r.e. set A there is a number k E C such that for all x:
xEAHS(k,y,k) Exercise 11 Suppose that B is an r.e. set and that g(x, y, z) is a recursive .S2
function such that for any disjoint r.e. relations Mi(x, y), M2(x, y) there is a number k in C such that for all x and y:
(1) Ml (x, y) = g(k, x, y) E B (2) M2 (x, y) z g(k, x, y) O B
Prove that B is a universal set.
§8 Effective Rosser systems We recall that we are defining Z to be effectively a Rosser system for sets, or to be an effective Rosser system for sets, if there is a recursive function
1I(x) - called a Rosser function for Z - such that for every number i, 1I(i) is the Godel number of a unary predicate H that separates ai from Oi in Z. We shall say that Z is effectively an exact Rosser system for sets if there is a recursive function O(x) - called an exact Rosser function for Z - such that for every number i, O(i) is the Godel number of a predicate that exactly separates in Z. Effective Rosser systems for sets have some very interesting properties. For one thing, as we showed in R.M., any consistent axiomatizable effective
Rosser system for sets is an exact Rosser system for sets - in fact is effectively an exact Rosser system for sets. We will now establish this as Q..
a corollary of a pure result of recursion theory - namely, that if f is a recursive function such that for all i, ai C_ af(i) and 3i C 3f(i), then there is a recursive function O(x) such that for all i, ai = a f(O(i)) and A = 13g(cb(i)).
We first need two lemmas, the first of which is actually an easy corollary of one of the "double recursion" theorems that we prove in Part III of this study, but rather than borrow the result, we will now give a direct proof
Chapter 11. Effective representation systems
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- which incidentally is somewhat simpler than the proof via the double (DD
recursion theorem, considering the antecedent machinery involved. We now use our double enumeration of binary r.e. relations, as well as sets.
Lemma A For any disjoint r.e. relations R, (z, x, y), R2 (z, x, y) there is a recursive function 0(y) such that for all i and x: (a) x c ao(i) E-a Rl (2, x, 0(2))
(b) x E 0o(i) - R2(i, X, O(i))
Proof The relations Ay(x, y) and By(x, y) are r.e. and disjoint, so by the double iteration theorem (Theorem 3.4, Chapter 10) there is a recursive function d(x) such that for all x and y: (1) X E ad(y) H Ay (x, y), and x E Qd(y) H By (x, y).
Now, suppose the relations Ri(z, x, y) and R2(z, x, y) are r.e. and disjoint. Then the relations Ri(z, x, d(y)), R2 (Z, x, d(y)) are r.e. and disjoint, so again using the double iteration theorem, there is a recursive function t(x) such that for all i, x, and y: (2) At(i) (x, y) E-' Rl (i, x, d(y)), and Bt(i) (x, y) 4-' R2(i, x, d(y)).
In (1) and (2) we take t(i) for y, and we have: X E ad(t(i)) <--> At(i) (x, t(i)) F--> Rl (i, x, d(t(i)))
x E Qd(t(i)) <--> Bt(i) (x, t(i)) H R2(i, x, d(t(i)))
And so we take q(x) = d(t(x)). As a corollary of Lemma A we have:
Lemma B For any r.e. relations Ml (z, x, y), M2 (z, x, y) (not necessarily disjoint) there is a recursive function O(x) such that for all i and x: (1) [M1(i, x, 0(i)) A ,,,M2(i, x, O(i))] = x E ao(i)
(2) [M2(i, x, 0(i)) A-M, (i, x, O(i))] = x E Op(i)
Proof Given M1 and M2, we take disjoint r.e. relations R1, R2 (of three arguments) such that M1- M2 C R1 and M2 - M1 C R2 and apply Lemma A to R1 and R2. Now we come to the heart of the matter.
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Theorem 8.1 Suppose f (x) is a recursive function such that for all is a2 C a f(i) and ,Oi C Of (i). Then there is a recursive function O(x) such that for all is ai = a f(q5(i)) and 13i = Of(O(i)) ,-1
Proof The hypothesis and Lemma B imply that for any r.e. relations Mi (z, x, y), M2 (z, x, y) there is a recursive function O(x) such that for all i and x: ..,
(1) [Mi(i, x, 0(i)) A -M2(i, x, 0(2))] = x E af(O(i))
(2) [M2(i, x, O(i)) A - Mi (i, x, O(i))] = x E ,Qf (q5(i))
We now take Ml (z, x, y) to be the r.e. relation x E Oz U Of(y), and M2 (z, x, y) the r.e. relation x E /3z U a f(y). Then there is a recursive function O(z) such that for all is -0-
(1)' (ai U Of(cb(i))) - (Qi U a.f(o(i))) C afwi)) -0-
(2)' (Qi U af(O(i))) - (ai U,Qf(O(i))) C Of(O(i))
From (1)', (2)' it follows by propositional logic that ai = a f(q5(i)), and ,Oi = Of (q5(i)) (see Exercise 7, Chapter 10).
And now, from Theorem 8.1, we have a new proof of the following result (stated and proved in R.M.)
Theorem 8.2 If Z is a consistent axiomatizable effective Rosser system -e-
CAD
for sets, then Z is effectively an exact Rosser system for sets. More specifically, if Z is consistent and axiomatizable and if II(x) is a Rosser function for Z, then there is a recursive function O(x) such that II(0(x)) is an exact Rosser function for Z.
Proof Suppose Z is consistent and axiomatizable and that II(x) is a Rosser function for Z. Then the relations "En(x) (y) E P," and "En(x) (y) E R" are disjoint and r.e., hence there is a recursive function f (x) such that for any i, a f(i) is the set represented by En(i), and Of (i) is the set contrarepresented by En(i). Since 11 is a Rosser function for Z, then ai C a f(i) and 13i C Of(i), and so by Theorem 8.1, there is a recursive function 0(x) such that for all i, ai = a f(O(i)) and /3i = Of (O(i)), and so En(o(i)) represents ai and contrarepresents ,Oi. Thus 11(0(x)) is an exact Rosser function for Z.
Theorem 8.1 affords yet another proof that every semi-D.U. pair of r.e. sets is D.U. and also gives the following additional information:
Chapter 11. Effective representation systems
222
Theorem 8.3 If (B1, B2) is uniformly semi-D.U. under g(x, y) and B1, B2 are r.e., then there is a recursive function O(x) such that (B1, B2) is uniformly D.U. under &(x), y). Briefly, Theorem 8.3 is proved by taking a recursive function f (x) such
that a f(2) is the set of x such that g(i, x) E B1 and Of (i) is the set of x such that g(i, x) E B2. Then the hypothesis implies that ai C a f(i) and ,Oi C Of (i), so by Theorem 8.1, there is a recursive function O(x) such that cai = c f(q5(i)) and ,Oi = Of(ci)), which means that (B1,B2) is uniformly
D.U. under &(x), y). Remarks
Theorem 8.3 appears to be weaker than Theorem 8.1, but it is strong enough to give Theorem 8.2, because if Z is a Rosser function for Z, then (Po, R0) is semi-D.U. under the function r(II(x), y) (we recall that r(x, y) is the Godel number of Ex(y)). Theorem 8.1 is really of the same strength as the following: =.y
Theorem 8.1' Suppose that S1(x, y), S2(x, y) are disjoint r.e. relations such that for all i and x, x c ai = Si(i, x), and x c ,Oi = S2(i, x). Then there is a recursive function O(x) such that for all i and x, x c ai H 'L2
Si(O(i), x), and x E ,Oi H S2(O(i), x).
Exercise 12 (a) Derive Theorem 8.1' from Theorem 8.1. (b) How can Theorem 8.1 be derived as a consequence of Theorem 8.1'?
Exercise 13 Suppose that f satisfies the hypothesis of Theorem 8.1. Prove the following:
(a) For any recursive function g(x, y) there is a recursive function O(x) Cam
such that for all i, a9(i,q5(i)) = of(O(i)) and Qg(i,q5(i)) =,Of(O(i)). [This
is a strengthening of Theorem 8.1. Why?] (b) For any recursive function h(x) there is a recursive function O(x) such that for all i, ah(i) = of (O(i)) and Oh(i) = ,O f(o(i)). (c) For any recursive function h(x) there is a recursive function O(x) such that for all i, ah(o(i)) = a f(O(i)i and Qh(q(i)) _ ,O f(o(i)).
§9 Some simpler related results
223
Exercise 14 Suppose that Z is a consistent exact Rosser system for sets (not necessarily axiomatizable). Show that for any disjoint r.e. relations Rl (z, x, y), R2 (z, x, y) there is a recursive function O(x) such that for every i, EO(i) is a unary predicate that exactly separates x: R1(i, x,.O(i)) from x:
R2(i, x, 0(i))
*§9 Some simpler related results Theorem 8.1 and 8.2 are respective "double analogues" of the following two theorems:
Theorem 9.1 Suppose f is a recursive function such that for all i, ai g w f(i) and Qi C w f(i). Then there is a recursive function O(x) such that for all i, wi = wf(O(i)).
Theorem 9.2 If Z is a consistent, axiomatizable, and effectively a semiRosser system for sets, then Z is effectively universal. More specifically, if Z is consistent, axiomatizable, and if 1I(x) is an effective semi-Rosser function for x (i.e., if for all i, En(i) is a unary predicate that weakly separates ai from /3i), then there is a recursive function O(x) such that for every i, 1I(O(i)) is the Godel number of a unary predicate that represents wi in Z. To prove Theorem 9.1, we use Lemma B of §8 taking Ml (z, x, y) to be
the relation "x E w,z" and M2(z, x, y) to be "x e w f(y)." Then there is a recursive function O(x) such that for all i, wi - w f(O(i)) C uf (,p(i)), and w f(q5(i)) - wi C w f(q5(i)). Then by Propositional logic, wi = w f(ci)).
Theorem 9.2 is derived from Theorem 9.1 by taking w f(i) to be the set represented by En(i) in Z.
§10 Unconquerable systems and omniscient functions NON
car
Let us say that a predicate H covers a sequence Z1, Z2, ... , Zn, ... of extensions of Z if for every n for which Zn is consistent, H(n) is undecidable in Z. We shall call Z unconquerable if for every r.e. sequence of extensions of Z, there is a predicate H that covers it. At the end of §5, we stated in so many words that we will show that every consistent effective Rosser system for sets is unconquerable. Actually, we will show something stronger: Given an r.e. sequence Z1, Z2, ... , Zn, ... of representation systems, by an index of the sequence we shall mean a number J(a, b) such that Ra(x, y) is the relation: "Ex is provable in Zr" and Rb(X, y) is the relation "Ex is refutable in Z,." Thus if i is an index of the sequence, then for all x and
Chapter 11. Effective representation systems
224
y: RK(2) (x, y) if E. is provable in Z,,, and RL(2) (x, y) if Ex is refutable in Z,. And now we shall define Z to be effectively unconquerable if there is a recursive function M(x) - which we call an omniscient function for Z - such that for every r.e. sequence Z1, Z2,. .., Zn, ... of extensions of
Z, if i is any index of the sequence, then M(i) is the Godel number of a predicate that covers the sequence. [Informally, this means that given any r.e. sequence of extensions of Z, in the sense of given an index, we can effectively find a predicate that covers the sequence.] What we will 0
show is that every consistent effective Rosser system for sets is effectively unconquerable. Discussion
inn
(D¢
.'3
;_4
For a first-order system S, even unconquerability (let alone effective unconquerability) is apparently a strengthening of uniform incompletability, as we defined it in R.M., as follows. Given a consistent first-order system S, we let Sn be that system whose axioms are those of S together with all formulas whose Godel number is in Wn. The sequence S1, S2, ... , Sn, ... is CAD
an r.e. sequence of extensions of S, and we defined S to be uniformly incom-
car
iii
pletable if there is a predicate H that covers that particular r.e. sequence. Of course if S - or rather the representation systems associated with S is unconquerable, then S is uniform incompletable, and thus for first-order systems, unconquerability is a strengthening of uniform incompletability. It was shown by Marion Pour-El[14] that Peano Arithmetic is uniformly incompletable - indeed, she showed that the following conditions are jointly sufficient that an axiomatizable consistent system be uniformly incomail
;21
pletable: (i) all formulas of the form v1 < n H (vi = 0 V ... V = n) are provable in S; (ii) all formulas of the form vl < nV n < vi are provable
in S; (iii) all primitive recursive functions of one argument are definable
in S. By a very different method, we showed in R.M. that (iii) can be car
ran
ear
replaced by the weaker condition that all stem functions are definable in S. We also showed that if S is a consistent axiomatizable Rosser system for sets in which all stem functions are definable, then S is uniformly incompletable. [This result, by the way, is an easy corollary of Theorem 5.2 of this chapter.] And then by a completely different argument (a fixed point argument using an application of the double recursion theorem similar to Lemma A of this chapter) we showed the apparently incomparable result that every consistent axiomatizable effective Rosser system for sets is uniformly incompletable (the definability of stem functions is not required). It is this result that we will now extend to representation systems, and in a strengthened form. We first obtain the following corollaries of Lemma A of §8: ...
Q.,
ail
cam
§10 Unconquerable systems and omniscient functions
225
Corollary Al If Z is effectively a Rosser system for sets, then for any disjoint r.e. relations Si(x, y, z), S2(x, y, z) there is a recursive function %(x) such that for every number i, the expression EG(i) is a unary predicate such
that for all n: (1) Sl (i, n, 0(i)) implies that EO(i) (n) is provable in Z,
(2) S2(i,n,0(i)) implies that Ep(i)(n) is refutable in Z.
Proof Let II(x) be a Rosser function for Z. Given disjoint relations Sl (z, x, y) and S2 (z, x, y), let Rl (z, x, y) be the relation Sl (z, x, II (y)) and
R2(z, x, y) the relation S2(z, x, II(y)). Then the relations R1 and R2 are ((DD
r.e. and disjoint, hence by Lemma A there is a recursive function O(x) such
that for every i and n: Si (i, n, II(q(i)) H R1(i, n, O(i)) H n E ao(i) En(,k(i))(n) is provable in Z (since H is a Rosser function for Z. Similarly, S2(i, n, II(O(i))) implies that En(,i(i)) (n) is refutable in Z. And so we take O(x) = II(O(x)).
Corollary A2 If Z is effectively a Rosser system for sets, then for any disjoint r.e. relations Mi (z, x, y) and M2(z, x, y) there is a recursive function
,O(x) such that for every number i, the number 'O(i) is the Godel number of a unary predicate H such that for all n:
(1) M1(i, n, gn(H(n))) implies that H(n) is provable in Z, (2) M2(i, n, gn(H(n))) implies that H(n) is refutable in Z [We recall that gn(H(n)) means the Godel number of H(n).]
Proof Suppose Z is effectively a Rosser system for sets and that Ml (z, x, y), M2 (z, x, y) are disjoint r.e. relations. We let Sl (z, x, y) be the r.e. relation M, (z, x, gn(Ey(x))) and S2 (z, x, y) be the r.e. relation M2(z,
x,gn(Ey(x))). Then the relations S1 and S2 are r.e. and disjoint, and so by Corollary A1i there is a recursive function %(x) such that for every i, the expression El(i) is a unary predicate and for every n, if Ml (i, n, gn(E p(i) (n))) then S1(i, n, iP(i)), hence EO(i) (n) is provable in Z. Similarly, M2(i, n, gn(Ep(i) (n))) implies that EO(i) (n) is refutable in Z. Remarks
If Z happens to be effectively an exact Rosser system for sets, then Corollaries Al and A2 both hold good replacing "implies that" by "if and only if." But these stronger properties are not required for that which follows.
Chapter 11. Effective representation systems
226
Theorem 10.1 If Z is a consistent effective Rosser system for sets then Z is effectively unconquerable. -4,
Proof Suppose Z is a consistent effective Rosser system for sets. In Corollary A2, we take Mi(z, x, y) to be the relation Bz(y, x), and M2(z, x, y) the relation Az(y, x), and so there is a recursive function M(x) such that for every number i, M(i) is the Godel number of a predicate H such that for all n: s0.
.°i
C."
(1) Bi (gn(H(n)), n) implies that H(n) is provable in Z,
(2) Ai(gn(H(n)),n) implies that H(n) is refutable in Z.
U,,
We will show that the function M(x) is an omniscient function for Z. Consider any r.e. sequence Z1, Z2, ... , Zn, ... of extensions of Z and
let i be an index of the sequence. Then for any numbers x and n, we have RK(i) (x, n) if E., is provable in Zn, and RL(i) (x, n) iff Ex is refutable cad
in Zn. Therefore if Ex is provable in Zn and not refutable in Zn, then Ai(x, n) holds, and if is refutable in Zn and not provable in Zn, then Bi(x,n) holds. Now suppose that Zn is consistent. Then E., is provable in Zn z Ai(x, n), and Ex is refutable in Zn Bi(x, n). We now take the Godel number of H(n) for x, so that Ex = H(n), and we have: F.,"
°'W
(1) H(n) is provable in Zn = Ai(gn(H(n), n))
H(n) is refutable in Z,
(2) H(n) is refutable in Zn = Bi(gn(H(n), n)) = H(n) is provable in Z. Then by Lemma 5.1, H(n) is undecidable in Zn. There are many more fixed point properties for effective representation systems that we could prove now, but we prefer to derive them as corollaries of some of the more abstract fixed point theorems of the next several chapters.
III
Fixed Point Theorems in a General Setting
Chapter 12
Sequential systems It-
We now come to our main topic. Our purpose in this and the remaining chapters is to unify various fixed point properties of a more complex sort than those of Chapter 2 that arise in indexed relational systems, such as that of recursion theory, applicative systems, such as combinatory logic, and representation systems, with their applications to mathematical logic. To this end we introduce the notion of sequential systems, which will be the basis for the next several chapters.
Part I Definitions and purpose By a sequential system S we shall mean a triple (N, E, -*), where N is a set, E is a collection of functions of various numbers of arguments, all arguments and values being in N, and -> is a transitive binary relation on the set of all finite non-empty sequences of elements of N. We write
xl,...,xn -4 y1,...,yk (n > 1,k > 1) to mean (xl,...,xn) -4
(y1,...,yk)
We use letters a, b, c, x, y, z, w as variables ranging over N. Elements of E will be called E-functions. Elements of N will simply be called elements. There are various "fixed point" properties that S may or may not have. Here are some of them:
(1) [Weak fixed point property] For every element a (in N) there is an element b such that b -* a, b. (2) [Myhill Property] There is a E-function O(x) such that for all x E N : O(x) -> x, q(x). [This is a "uniform" version of the weak fixed point property and trivially implies it: given a, take b = 0(a) and we have
b-- a,b.]
Chapter 12. Sequential systems
229
(3) [Recursion Property] For every element a there is a E-function O(x) such that for every element x : O(x) -> a, x, O(x). (4) [Weak Kleene Property] For every E-function f, there is some element
a such that a -> f (a). ((DD
Let us see what these fixed point properties will mean in some of our intended applications. A Indexed relational systems
In Chapter 10 we defined an indexed relational system to be an ordered triple (N, E, (D) where N is a set, E is a set of functions with arguments and values in N, and 1 is a mapping that assigns to every positive integer n and
every element a e N a relation Rn of n arguments on N. [For n = 1, Rn
(CD
Q.1
is a subset of N, which we also write wa.] Without ambiguity we can write Ra(xl, ... xn) for Ra(xl, ... , xn). The relations Rn are called 1Z-relations. If 1Z is an indexed relational system (N, E, 4D), we define S(R) to be the sequential system (N, E) ->), where a, a1,. .. , an -> b, b1,. .. , bk is defined to mean that for all x in N, Ra(a,, . . . , an, x) if Rb(bl, . . . , bk, x). [For
n=k=0, we definea ->btomeanthatforallx,xEwaiffxEwb - in other words, Wa = Wb. For n = 0, k $ 0, we define a - > b, bl, ... , bk to mean that for all x E N, x c wa iff Rb(bl, ... ) bk, x) and we define b, bl, ... , bk -> a
car
'Z-
to mean the same thing.] This relation -> is of course transitive (indeed, it is an equivalence relation). To say that S(R) has the weak fixed point property is to say that for any a there is some b such that b -> a, b, which means that for all x E N, X E Wb if Ra(b, x) - in other words wb = {x : Ra(b, x)}. To say that S(R) has the Myhill property is to say that there is a E-function 0(y) such that for all y : wo(y) = x : Ry(O(y), x)). [For recursion theory, where the )Z-relations are the recursively enumerable relations, this is Myhill's fixed point theorem [10].] car
To say that S(R) has the recursion property is to say that for every R -relation Ra(x, y, z) there is a E-function 0(y) such that for all y E N : wo(y) = {x : Ra(y, 0(y), x)}. [For recursion theory, this is Kleene's recursion theorem in the form given in [28].]
Her
To say that S(R) has the weak Kleene property is to say that for any E-function f (x) there is an element a such that Wa = w f(a). [For recursion
theory, taking E to be the set of all general recursive functions, this is another form of the recursion theorem.]
What we have labeled S(R) let us now label S1(R). More generally, for any positive integer n, we define Sn (R) to be the sequential system (N, E, , ), where (a, a1, ... , ak) (b, b1, ... , br) is defined to mean that for all elements x1, ... , xn: Ra(al,... , ak, x1, ... , xn) if
Chapter 12. Sequential systems
230
Rb(bl, ... ) br, _ x1 ,. .. , x,,,). Then 5,,,(R) has the weak fixed point property
if for every R-relation Ra(x, yl, ... , yn) there is some b such that for all , yn) iff Ra(b, yl, ... , yn). The Myhill property is that , yn: Rb(y1, yl, such an element b can be expressed as a E-function of a. [Actually, all four of our listed fixed point properties hold for every positive n if the R-relations are the r.e. relations and the E-functions are either the general recursive or primitive recursive functions - or even the 1-1 primitive recursive functions.]
ors
We remark that sequential systems are also directly applicable to indexed systems of partial functions. We let Q be a triple (N, E, 0), where N, E are as before, and 0 is now a mapping that assigns to every positive integer n and every element a of N, a partial function ga(x1, ... , xn) on N of n arguments. We write ga(xl,... , xn) _- gb(yl, ... , yn) to
mean that both sides are defined and equal, or both sides are undeBy Sn(Q) we mean the sequential system (N, E, , ), where we/ define a, al, ... , an ; b, b1, .... br to mean that for all x1i ... , xn ga(al,... , ak, x1, ... , xn) -_ gb(bl,... br, xl, ... , xn). To say that Sn(Q) fined.
:
,
has the weak fixed point property is then to say that for every a there is car
some b such that for all x1, ... , xn: gb(x1, ... , xn) ^ _ qa(b, x1, ... , xn). [For
the theory of partial recursive functions, this is Kleene's original form of his second recursion theorem.] B Applicative systems
As explained in Chapter 2, an applicative system A is a set N together with an operation that maps every pair (x, y) of elements of N to an element x y - also written xy. Following the usual convention of combinatory logic, we write xyz for (xy)z; xyzq for ((xy)z)w, etc. Thus when parentheses are deleted, it is understood that association is to the left. For any element a E N and every positive integer n we let [a]n be that function from Nn into N defined by: [a]n(xl, ... , xn) = ax1 ... xn. [Thus,
e.g., [a]3(x, y, z) = ((ax)y)z.] We let E be the collection of all functions [a]n and we define S(A) to be the sequential system (N, E, -*), where
a,xl,...,xk -> b,y1i...,y, is defined to mean axl,...,xk = car
C7,
[For k = r = 0, a -> b means a = b.] Obviously the relation -* is transitive (in fact an equivalence relation). To say that S(A) has the weak fixed point property is to say that for every element a there is an element b such that b = ab - in other words, every element a has a fixed point b. To say that S(A) has the Myhill property is to say that there is a E-function [c]1 such that for every element x, [c] 1(x) = x [c] 1(x) - in other words, for all x, cx = x(cx). [Such an element c is called a fixed point combinator.] The recursion property for
Chapter 12. Sequential systems
231
S(A) is that for every element a there is some element c such that for all x: cx = ax(cx). The weak Kleene property for S(A) is the same thing as the weak fixed point property for S(A). C Sentential systems
;-4
By a sentential system (T) we shall mean a triple (Z, -, E), where Z is a representation system, - is an equivalence relation on the sentences of Z, and E is a collection of functions with arguments and values on the set N of numbers'.
'-'
'DD
car
Given a representation system Z, there are several equivalence relations on sentences that are of interest. First, we might define two sentences X and Y to be weakly equivalent if they are equi-provable, i.e., if either one is provable, so is the other. We might call two sentences strongly equivalent if they are both equi-provable and equi-refutable. A still stronger equivalence relation arises in representation systems which incorporate propositional logic, in which case there is a logical connective of equivalence, and we can then define X to be very strongly equivalent to Y if the very sentence X Y is provable in Z. Since all three of these equivalence relations are of interest, we shall leave the choice of equivalence relations open. As to the class E of functions, in most of the applications we have in mind, E can be the class of recursive functions, or perhaps the class of
0
F"-.
primitive recursive functions, or perhaps the narrower class of 1-1 primitive recursive functions. We will also leave open which class E is chosen. For technical reasons, it will be convenient to have separate indexings for the sentences, the unary predicates, the binary predicates, and so forth. We do this as follows. For each natural number n, we arrange all predicates of degree n in the sequence H1 , H2, ... , Hk , ... according to the magnitude of their Godel numbers. We regard sentences as predicates of degree 0, and .-"r
we write Sk for Hk, and so, in particular, the sentences are arranged in the sequence S1, S2,. . ., Sk,... according to the magnitude of their Godel ''M
numbers. We call k the index of Sk, and more generally, for each n, we call k the index of HH . We sometimes delete superscripts when no ambiguity can result, e.g., we write Hk(al,... , an) for HH (al, ... , an). We call a sentence So, a fixed point of a unary predicate H if So, is equivalent to H(a). s0.
And now, if (T) is a sentential system (Z, -, E), we define the sequential system So(T) to be (N, E, o ), where N is the set of numbers, b, b1, ... , bt we shall is defined as follows. By a, a1, ... , an and mean that Ha (a1, ... , an) is equivalent to 14 (b,, ... , bt). [For n = 0, III
..p
b,b1,...,bt means Sa - Hb(bl,...,bt); for t = 0, a, a,,..., an --,4 b III
a
'Again, the word "number" can mean either positive integer or natural number, and all results are correct under either interpretation.
Chapter 12. Sequential systems
232
CAD
III
III
means Ha(a1 i ... , an) - Sb, and for n = t = 0, a b means that So, = Sb.] For positive n, by Sn(T) we mean (N, E, n ), where we define a, 01 b, B2 (01 i 02 are finite sequences, each or both of which may be empty) to mean that for all yl, ... , yn: Ha(01, yi, , yn) ° Hb(02, y1, , yn) We say that (T) has a given fixed point property if for every natural number n, the associated sequential system Sn(T) have the property. Let us see what our four fixed point properties so far considered look like for a sentential system (T): III
;],
T*
III
(1) To say that So(T) has the weak fixed point property is to say that for every predicate H of degree 1, there is a number b such that Sb - H(b)
c-r
- in other words, that every unary predicate H has a fixed point. To say that Sn(T) has the weak fixed point property (n > 1) is to say that for every predicate H of degree n + 1 there is a number b such that for all x1,. .. , xn: Hb(xl,... , xn) - H(b, xl,... , xn). And thus to say that (T) has the weak fixed point property is to say that for every natural number n, and every predicate H of degree n + 1, there is a number b such that for every n-tuple 0, Hb(0) - H(b, 0). III
s0.
[If n = 0, 0 is empty and Hb(B) is then the sentence Sb.]
car
(2) To say that (T) has the Myhill property is to say that for every n > 0 there is a E-function O(x) such that for every n-tuple 0, Hk(x)(9) HH(O(x), 0) (and for n = 0, this says that for every x, Si(x) is a fixed point of H, i.e., So(x) HH(O(x))).
(3) To say that (T) has the recursion property is to say that for every III
predicate H of n+2 arguments there is a E-function O(x) such that for all x, and every n-tuple 0: He(x) (0) - H(x, O(x), 0). [In particular, for n = 0, So(x) - H(x, fi(x)).]
s5.
(4) To say that (T) has the weak Kleene property is to say that for every n > 0 and every E-function f (x) there is a number b such that for every n-tuple 0, Hb(6) - HH(b)(B).
c4'
cc-
We now see how sequential systems unify various areas of metamathematics. Any fixed point theorem we prove about sequential systems in general will be simultaneously applicable to indexed relational systems (and in particular to recursion theory), to applicative systems (in particular to combinatory logic), and to sentential systems, which in turn have further applications to first-order arithmetical systems such as Peano Arithmetic. We have considered only a small portion of the interesting fixed point +', mho
properties that a sequential system might have. Other more elaborate c-t
properties will be considered in later chapters.
§1 Types 0,1,2
233
Part II Preliminary definitions Before turning to a systematic development of sequential systems, several preliminary definitions are necessary. cap
§1 Types 0,1,2
We shall say that E is of type 0 if E contains the identity function and is closed under explicit definability, i.e., for every E-function g(x1i ... , xk), '.O
all functions explicitly definable from g are also in E. We shall say that E is of type 1 if E is of type 0 and is closed under composition, i.e., for any E-functions g(xl, ... , xk), f1(x, ... , xn), ... , fk(xl, , ... , xn), the function g(f1(x1i ... , x ) , . .. , fk(xl, ... , xn)) is also in E. We shall say that E C,"
is of type 2 if E is of type 0 and is closed under introduction of constants for variables, i.e., if for any E-function g(x, yl, ... , yn) and any element c c N, the function g(c, yl, . . . , yn) (as a function of yl, ... , yn) is in E. We shall say that E is of type 1 and 2 if it is both of type 1 and type 2. By the type of a sequential system (N, E, ->), we shall mean the type of E. In all that follows, our sequential systems will be assumed to be at least of type 0. [In most of our applications, the sequential systems involved will be of types 1 and 2.]
Let us note that if E is of type 0, then any E-function g(xl, ... , xn) can be expressed as a E-function of x1, ... , xn, yl, ... , y,,,,, for the function f (x1, ... , xn, y1, ... , y,,,,) = g(x1, ... , xn) is explicitly definable from g. Also, for i < n, the function f (X 1, ... , xn) = xi is explicitly definable from the identity function I (f (x1, . . . , xn) = I (xi)) and so if E is of type 0, then each of X1 ,-- . , xn can be expressed as a E-function of x1, ... , xn. Connectivity
Now we come to a key notion: we shall say that S is weakly connected if for every element a and any E-functions fi(yi,... , yn), , fk(y1, , yn) (n > 1) there is an element b such that for all yl, ... , yn:
b, yl,..., yn -4 a, fl(yl)...)yn))...) fk(yl].... yn) In all that follows, S will be assumed to be at least weakly connected (and at least of type 0, as already noted). Illustrations
(A) Consider an indexed relational system R. We shall say that R is weakly connected if for every positive m, the sequential
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system 5,,,, (R) is weakly connected, which is to say that for every Rrelation R(xl,... , xk, yl, .... ym) and any E-functions fl(x1, ... , xn),
..., fk(xl,//..) xn), the relation
R(fl(xl,...,xn),..., fk(xl,...) xn),Y1,...,ym) is again an R-relation. Of course this condition holds for the indexed relational system of recursion theory, i.e., when the R-relations are the r.e. relations and E is the class of recursive functions. [Indeed, any indexed relational system that is standard, as defined in Chapter 10, is weakly connected.] (B) Weak connectivity for applicative systems (i.e., for the sequential sys-
tems associated with applicative systems) boils down to this: For any positive n and any elements a, cl, ... , ck there is an element b
such that for all elements xl,... , xn: bxl ... xn = a(clxl ... xn) ...
... xn). This condition is a special case of what is called combinatory completeness, which is roughly this: given any well formed expression W(xl, ... , xn) in variables x1, ... , xn (and possibly also (Ckxl
names of elements of N), there is an element b in N such that for all elements a1, . . . , an the equation bal ... an = W(al,... , an) holds. As is well known, combinatory completeness is equivalent to the ex-
istence of elements S and K such that for all x, y, z: Kxy = x and Sxyz = xz(yz). [We give the proof of this in a later chapter.] (C) For a sentential system (T), to say that SO(T) is weakly connected is to say that for any positive integers n and k and any element a and any E-functions fl(xl,... , xn), ... , fk(xl, ... , xn) there is a number b such that for any numbers al, ... , an, the sentence Hb(al,... , an)
is equivalent to Ha(fl(al,...,an),...,fk(al,...,an)). [This condition holds for the sentential system associated with Peano Arithmetic, with E the class of recursive functions. It also holds for the complete theory (N), with E the class of arithmetic functions.
Indeed, in both cases, Sr(T) is weakly connected for any natural number r.]
Exact functions
For any n > 2, we shall call a function E(xl,... , xn) an exact function (for S) if for all elements x1 ,. .. , xn:
E(xl,... xn) -> xl, ,
... , xn
Let us see what exact functions are like in some of our intended applications.
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235
(A) Given an indexed relational system R, for any positive k and any n > 2, to say that E(xl,... , xn) is an exact function for Sk(R) is to say that for all elements a 1, ... , an, x1, ... , xk: RE(ai,...,a,.) (x1, ... , xk) -4 Rai (a2, ... , an, x1, ... , xk)
For recursion theory, the existence of exact recursive functions for all
n > 2 and k > 0 is given by the iteration theorem (or Sn theorem, for partial recursive functions instead of r.e. relations). Indeed, if R is any standard system (as defined in Chapter 10), then for each k > 0, n > 2, there is an exact function E(xi, ... , xn) for Sk(R).
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(B) For applicative systems, if N contains an element I (an identity combinator) such that Ix = x for all x, then for any positive n, [I]n(xl, ... , xn) = x1 ... xn, hence [I]n is an exact function. And >0e
so if N contains an identity combinator, then it is trivial that exact functions of n arguments exist in E for all n > 2.
(C) For a sentential system (T), an exact function for So(T) is a function E(xl,... , xn) such that for all numbers a,,..., an: SE(al,...,an) is equivalent to Hal (a2, ... , an). Well, we simply take E(al, a2, ... , an) to be the index of the sentence Hal (a2, ... , an), and this SE(aI....,a, ) is the sentence Hal ( a 2 ,- .. , an). In standard applications, the Godel numbering is such that for each n > 2, the function E(al,... , an) is a recursive function (in fact a primitive recursive function) and so if E contains all recursive functions (or even all primitive recursive functions) then E contains all exact- functions of two or more arguments for So(T).
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More generally, for any natural number k, to say that E(xl,... , xn) is an exact function for Sk(T) is to say that for all elements x1, ... , xn, and every k-tuple 0 of elements of N, HE(al,...,a,,.)(0) is equivalent to Hal (a2, ... , an, 0). Again, in standard applications, with standard Godel numberings, there is for each n > 2, and each k > 0, a (primitive) recursive function E(xl,... , xn) such that a,,.) (0) is the sentence Ha, (a2, ... , an, 0). cad
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HE(a......
Some notations
4-a
We are using the symbol 0 to denote any finite possibly empty sequence of elements of N. As an example of our use, to say that S is weakly connected is to say that for any E-functions f l (x, 0), ... , fk (x, 0) of n + 1 arguments (n > 0)
Chapter 12. Sequential systems
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and any element a there is an element b such that for every x and every n-tuple 0:
b,x,0-4a, fl(x,B),..., fk(x,0) To say that a function E(x, y, 0) of n + 2 arguments (n > 0) is an exact function is to say that for all elements x, y of N and for any n-tuple 0:
E(x,y,0)->x,y,0. [For n = 0, this reduces to E(x, y) -> x, y.]
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Exercise 1 Suppose that we are not given that S is weakly connected, but that we are given that -> is transitive and that for every element a and every E-function f (xl, ... , xn) and every m > 0, the following two conditions hold:
(1) There is an element b such that for all x1i ... , xn and every m-tuple 0:
b,x1i...,xn,9 - a,f(xl,...,xn),0 (2) There is an element c such that for all x1, ... , xn and every m-tuple 0:
c,x1i...,xn,0 - a,xl,...,xn,f(xle...,xn),0 Prove that S is weakly connected.
Part III Some fixed point properties §2 The weak fixed point property Now things will get more interesting. Diagonalizers
We shall define a function D(x, 0) of n + 1 arguments (n > 0) to be a diagonal function, or more briefly, a diagonalizer if for every element x and in'
every n-tuple 0: D(x, 0) -> x, x, 0. [For n = 0, this of course reduces to D(x) -> x, x.] It is obvious that if E(x, y, 0) is an exact function, then the function E(x, x, 0) is a diagonal function. [We shall sometimes refer to the function E(x, x, 0) as the diagonalization of the function E(x, y, B).] Let us note that if E(x, y, 0) is a E-function, so is its diagonalization E(x, x, 0)
§2 The weak fixed point property
237
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(which is explicitly definable from E(x, y, 0)), and so, since E is of type 0, we see that for any natural number n, if E contains an exact function of n + 2 arguments, then E contains a diagonalizer of n + 1 arguments. We shall define a function Q(x, 0) of n + 1 arguments (n > 0) to be a quasi-diagonalizer if for every element a there is an element a' such that for every n-tuple 0: Q(a', 0) -> a, a', 0. [Obviously any diagonalizer is also a quasi-diagonalizer - just take a' = a.] We might call a function F(x, y, 0) quasi-exact if for every a there is some a' such that for all y, 0: F(a', y, 0) -* a, y, 0. An exact function is also quasi-exact (take a' = a) and its diagonalization is a quasi-diagonalizer. We have observed (somewhat to our surprise) that many of the things normally accomplished using diagonalizers can be accomplished with quasidiagonalizers [A little foretaste of this was given in Part III of Chapter 2. Another example is Theorem 1.1 below.] We recall that we are defining S to have the weak fixed point property if for every a there is some b such that b -4 a, b.
Theorem 1.1 If E contains a quasi-diagonalizer Q(x) of one argument, then S has the weak fixed point property. Proof Assume hypothesis. Since S is weakly connected, then given any element a there is some element c such that for every element x: (1) c, x --4a, Q(x) S-+
Also, since Q(x) is a quasi-diagonalizer, there is some element c' such that Q (c') -> c, c'. Now, by (1), taking c' for x, we have c, c' -4 a, Q(c'). Thus Q(c') , c, c' --4a, Q(c'), and since --f is transitive, we have Q(c') -> a, Q(c'). Thus b -- a, b, where b = Q(c').
Corollary 1.2 If S contains a diagonal function of one argument, then S has the fixed point property.
Corollary 1.3 If E contains an exact - or even quasi-exact - function E(x, y) of two arguments, then S has the weak fixed point property. The following theorem can be obtained as a corollary of Theorem 1.1.
Theorem 1.4 For n > 1, suppose E contains a function Q(x) such that for any element a there is an element a' such that for all x1 , .. , xn:
Q(ar ), xl,... , xn -> a, ar , x1i ... xn
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Then for every a there is some b such that for all x1, ... , xn: b, x1, ... , xn ->
a,b,xli...,xn. Proof One can, of course, prove this from scratch by only a light modification of the proof of Theorem 1.1, but the following proof (using Theorem 1.1, which has already been proved) is more interesting. Given n, we define a new relation T on non-empty finite sequences as follows. W e define a1 ,.. . , ak -; b 1 , . .. , br if for all elements x1, ... , xn:
a1,...,ak,x1,...,xn --> b1,.... br,x1,.... Xn We then let Sn be the sequential system (N, E, ). It is a routine matter is) and that Sn is weakly connected to verify that n is transitive (since (since S is). The hypothesis of the theorem is then that Q(x) is a quasidiagonalizer for the system Sn, and so by Theorem 1.1 applied to Sn, for any a there is some b such that b
ab, which means that for all x1, ... , xn:
b,x1,...,xn - a,b,xl,...,xn Exercise 2 Verify that in the above proof, n is transitive and that Sn is weakly connected. Next we have:
Theorem 1.5 The weak fixed point property is equivalent to the following condition: for any element a and any E-functions f, (x), ... , fk (x) there is an element b such that b --+ a, f1(b),..., fk(b).
Proof Obviously the weak fixed point property is a special case of this
condition - the case k = 1 and fi (x) = x (f, is the identity function). But the weak fixed point property implies this more general condition as follows:
Suppose S has the weak fixed point property. Then given a and Efunction f, (x), .
. .
,
fk (x) there is some element c such that for all x: c, x --f
a, fi(x),..., fk(x). But by the weak fixed point property there is some b such that b -p c, b. Also c, b -4 a, f, (b),
.
.
.
,
fk(b), so, since -> is transitive,
b -> a, fl(b),..., fk(b).
Exercise 3 Show that the hypothesis of Theorem 1.4 implies that for any E-functions f, (x, y1, ... , yn), ... , fk (x, y1, ... , yn), and any element a, there is some b such that for all xi,... , xn:
b,xl,...,xn - a,fl(b,xl,...,xn),..., fk(b,xl,...,xn)
§3 Universal systems and the weak Kleene property
239
§3 Universal systems and the weak Kleene property For any positive n, we shall say that S is n-universal if there is an element
e - which we call an n-universal element - such that for all x1, ... , xn: e, x1i ... , xn -> x1, ... , xn. We call S universal if S is n-universal for every positive n. Examples
(A) Consider an indexed relational system R. To say that S(R) is n-universal is to say that there is an element e such that for all x1,.. . , xn: e, xl, ... , xn -> x1,... , xn, which is to say that for all x, x1i ... , xn: Re(x, xl, ... , xn) iff Ry(x1i X2.... , xn) - in other
...
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words, the relation R.(yl, ... , yn) (as a relation among x, yl, . . . , yn) is itself an R-relation. We call R universal if for each n, Sn (R) is universal. The indexed system of r.e. relations is universal, as we have seen. [An important example of an indexed relational system that is not universal is the system of arithmetic relations, as indexed by the Godel numbers of formulas that express them.] z.,
(B) For an applicative system A, if N contains an identity combinator I, then S(A) is obviously universal (for each n, just take e = I).
(C) For a sentential system (T), to say that Sk(T) is n-universal (k >
0, n > 1) is to say that there is a number c such that for every x1i ... , xn and every k-tuple 9, the sentence He(xl,... , xn, 9) is equivalent to ( X 2 ,- ... , xn, 9) - in other words, for n > 1, He(xl, ... , xn, 9) is equivalent to HH1(x2, ... , xn, 9), but for n = 1, He(x1, 9) is equivalent to H.,1(9). For n = 1 and k = 0, this reduces
to He(x) - Sx, for all x. We shall say that (T) is k-n universal if Sk(T) is n-universal, and we shall say that (T) is universal if (T) is k-n universal for every k > 0, n > 1. Note
The definition of unversality for a sentential system (T) is independent of the class of E-functions; it depends merely on the representation system Z and the equivalence relation -. And so for a representation system Z and an equivalence relation - on the sentences of Z, we can speak of the pair (Z, -) as being universal or not. If it is, then we shall say that Z is universal with respect to the equivalence relation -. Similar remarks apply to "k-n universal" in place of "universal." Consider now the representation system Zp of Peano Arithmetic. It is indeed universal with respect to weak equivalence (assuming consistancy),
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but it is definitely not universal - not even 0-1 universal -,with respect to very strong equivalence, i.e., there is no predicate H such that for all n, the sentence H(n) <* SS,, is provable in Peano Arithmetic.
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Exercise 4 Assuming Peano Arithmetic is consistent, why is it that there can be no predicate H such that for all n, the sentence H(n) Sn is provable? [Hint: take an n such that the sentence Sn H 'H(n) is provable.]
Exercise 5 Why is Peano Arithmetic universal, with respect to weak equivalence?
Now let us go back to sequential systems in general.
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Lemma 3.1 If S is universal then for any E-functions f1(xl,... , xn), .... fk(x1i ... , xn) there is an element a such that for all x1, ... , xn:
a,xl,...,xn
fl(xl,...,xn),...,.fk(x1,...,xn)
Proof Suppose S is universal. Let e be an n-universal element. Then there is some a such that for all X1 ,-- . , xn:
a,xl,...,xn
--+
e,f1(x1i...)xn),...,fk(x1,...,xn) f1(x1,...,xn),... ,fk(x1i...,xn)
Then, by transitivity, a, x1,... , xn -> f 1(x1, ... , xn), ... , fk (x1, ... , xn).
Theorem 3.2 If S is universal and has the weak fixed point property, then S has the weak Kleene property. Proof Suppose that S is universal. Given a E-function f (x), by the above lemma there is an element a such that for all x: a, x --+ f (x). If now S has the weak fixed point property, then there is some b such that b --4a, b, but a, b -4 f (b) and so b --4f (b).
Corollary 3.3 If S is universal and E contains a diagonalizer - or even a quasi-diagonalizer - of one argument then S has the weak Kleene property. Exercise 6 Give a direct proof of the above corollary (one that does not r-1
use Theorem 1.1).
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241
Exercise 7 Let Kl be the weak Kleene property. Let K2 be the property that for any E-function F(x, y) and any element a there is an element b such that b -> F(a, b). Show that K2 implies K1.
Theorem 3.4 If S is of type 2 and has the weak Kleene property, then S has property K2 (for any E-function F(x, y) and any element a there is an element b such that b --+ F(a, b)).
Proof Assume the given conditions. Given a E-function F(x, y) and given an element a, let f (x) = F(a, x). Since S is of type 2, then f (x) is a E-function, hence by hypothesis there is some b such that b - f (b), which means that b --+ F(a, b).
§4 Integrated systems We shall define S to be integrated if for any element a and any natural number n there is a E-function t(x) such that for all x, yl, ... , yn:
44Y1,...,yn -4 a, x,yl,...,yn It is obvious that if S is integrated, then for every positive m, the system S,,,, is integrated. Integrated systems will play a key role in several of the
chapters that follow. Meanwhile, let us take a look at what integration means in our intended applications.
(A) For an indexed relational system 1Z, we say that 1Z is integrated if S(R) is integrated (which automatically implies that for every pos-
itive k, the system Sk(R) is integrated). And to say that S(R) is integrated is to say that for every element a and every number n, there is a E-function t(x) such that for all x, yl, ... , yn:
Rt(x)(yl,...,yn) H Ra(x,y1,...,yn) For recursion theory this condition is another instance of the iteration theorem. Indeed, any standard indexed relational system is integrated.
(B) For applicative systems, integration is automatic: given a, take t(x) = [a]1(x) - thus t(x) = ax. Then for any n, t(x), y1, ... , yn -> a, x, Y 1 ,-- . , yn simply means that axyl ... yn = axyl ... yn, which is obviously true.
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0'4
(C) For sentential systems, integration means that for every predicate Ha of degree n+1 (n > 1) there is a E-function t(x) such that for all numbers x, yi, ... , yn: Ht(x) (yl, , yn) is equivalent to Ha (x, yl, , yn) In all standard applications, for each a and n, there is a (primitive)
recursive function t(x) such that for all x, yi, ... , yn, the sentence Ht(x) (yl, ... , yn) is the sentence Ha(x, y1, ... , yn).
Exercise 8 Prove that if E is of type 2, then a sufficient condition for S to be integrated is that for each positive n, there is an exact function E(x, y) for S. Does this hold replacing "exact" by "quasi-exact?"
Theorem 4.1 If S is integrated and if S has the weak Kleene property, then S has the weak fixed point property.
Proof Suppose S is integrated. Then, given a, there is a E-function t(x) such that for all x: t(x) -> a, x. Now suppose S also has the weak Kleene property. Then there is an element b such that b -> t(b). But t(b) -> a, b, SUB
hence b -> a, b.
From Theorems 3.2 and 4.1 we have:
Theorem 4.2 If S is universal and integrated then the weak fixed point property for S is equivalent to the weak Kleene property for S.
Part IV Summary and applications
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Before turning to the more interesting properties of sequential systems, which we do in the next four chapters, let us pause and summarize some of the main things we have done so far and consider their applications. We shall temporarily say that S is minimally adequate if E contains an exact function E(x, y) of two arguments (and also, of course, that E is of type 0 and S is weakly connected).
Theorem M Suppose S is minimally adequate. Then
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(a) For any E-functions f, (x), . . . , fk(x) and element a there is an element b such that b -> a, fi (b), ... , fk (b). In particular (taking k = 1 and fi(x) the identity function) for any a there is some b such that b -> a, b. [This comes from Theorems 1.1 and 1.5.]
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243
(b) If S is also universal, then for any E-function f (x) there is an element b such that b -> f (b). If, also, S is of type 2, then for any E-function F(x, y) and any element a there is an element b such that b -> F(a, b). [From Corollary 1.3, Theorems 3.2 and 3.4.1 Applications
We shall say that an indexed relational system R is minimally adequate if for each n > 1, the sequential system Sn(R) is minimally adequate. We shall say that a sentential system (T) is minimally adequate if for every n > 0, the sequential system Sn(T) is minimally adequate. And we shall say that an applicative system A is minimally adequate if the sequential system S(A) is minimally adequate. Of course the indexed relational system of recursion theory is minimally adequate. An example of a minimally adequate sentential system is the sentential system associated with Peano Arithmetic. Of course combinatory logic (which is a complete applicative system) is minimally adequate (and much more). As applications of Theorem M we have: Theorem I
Suppose R is minimally adequate. Then
(a) For any R-relation R(x, yl,... , yn) there is an element b such that for
,yn: Rb(y1, ... , yn) E-a R(b, yi,... , yn)
In particular, for any R-relation R(x, y) there is an element b such that for all y: y E Wb H R(b, y)
(b) If R is universal, then for any n > 1 and any E-function f (x) there is an element b such that R6 = Rf(b). Theorem II
For any minimally adequate sentential system (T):
(a) For any predicate H of k arguments and any E-functions fi(x), ... fk(x) there is a number b such that Sb - H(f1(b),... , fk(b)). More generally, for any predicate H of k+n arguments (k > 1, n > 0)
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244
and any E-functions f, (x), ... , fk (x) there is a number b such that for every n-tuple 0: Hb(0) - H(fl(b), ... , fk(b), 0). In particular, for every unary predicate H there is a number b such that Sb - H(b) (Sb is a fixed point of H). m""
(b) If (T) is universal, then for any n > 0 and any E-function f (x) there is a number b such that for every n-tuple 0, Hb (B) is equivalent to Hf b) (0). In particular, for any E-function f (x) there is a number b such that Sb Sf(b). v.,
Theorem III
Suppose A is a minimally adequate applicative system. Then for any elements a, al, ... , ak there is an element b such that b = aalb ... bakb. Also for any element a there is an element b such that b = ab (b is a fixed point of a). CAA
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The applications of this chapter are quite meager as compared to the applications of the next four chapters.
Chapter 13
Strong fixed point properties In this chapter we deal with the recursion property, the Myhill property, and some related "strong" fixed point properties.
Part I The recursion and extended recursion properties §1 The recursion property C."
We recall that we are defining S to have the recursion property if for every element a there is a E-function O(x) such that for all x: O(x) --+ a, x, 5(x). We also recall that S is called integrated if for every element a and every positive number n there is a E-function t(x) such that for every n-tuple 0: t(x),0--f a, x, 0. We recall that S is said to be of type 1 if S is of type 0 and E is closed w''
under composition. We will shortly prove that if S is integrated and of type 1, and if E contains a diagonalizer D(x) of one argument, then S has the recursion property. But we will prove something a bit more general, which we will need in a later chapter. Normalizers H-,
We shall call a function N(x, 0) of n + 1 arguments (n > 0) a normalizes if for every element a there is a E-function t(x) such that for every x and every n-tuple 0: N(t(x), 0) --f a, x, t(x), 0
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246
Proposition 1.1 If S is integrated then every diagonalizer is a normalizer.
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Proof Suppose S is integrated and that D(x, B) is a diagonalizer of n+1 arguments (n > 0). Given an element a, there is a E-function t(x) such that for every x and y and every n-tuple 0, t(x), y, 0 --f a, x, y, 0. Hence t(x), t(x), 0 -4 a, x, t(x), 0. But also D(t(x), 0) -- t(x), t(x), 0. Therefore D(t(x), 0) --f a, x, t(x), 0. Thus D(t(x), 0) is a normalizer. Next we have the following theorem that will have applications in later chapters as well.
Theorem 1.2 If E is of type 1 and contains a normalizer N(x) of one argument then S has the recursion property.
Proof Assume hypothesis. Given an element a, let a' be such that for all x, y: a', x, y -4 a, x, N(y). Then let t(x) be a E-function such that for all x: N(t(x)) -> a', x, t(x). But a', x, t(x) -4 a, x, N(t(x)), and so N(t(x)) --+ a, x, N(t(x)). Since E is of type 1, the function N(t(x)) is a E-function, and so we take O(x) = N(t(x)) and we have O(x) - a, x, 0(x). From Proposition 1.1 and Theorem 1.2 we have
Theorem 1.3 If S is integrated and of type 1 and if E has a diagonalizer D(x) of one argument then S has the recursion property.
Theorem 1.4 If S is integrated of type 1 and if E contains an exact .'3
function E(x, y) of two arguments, then S has the recursion property. Remarks
(1) It may be helpful to look at a direct proof of Theorem 1.4. Given an element a, by integration there is a E-function t(x) such that for
all x and y: t(x),y --f a, x, E(y, y). Then E(t(x),t(x)) -> t(x),t(x) -> Q)-
a, x, E(t(x), t(x)), and so we take O(x) = E(t(x), t(x)). (2) Comparing this theorem with Corollary 1.3 of Chapter 12, we see that if E contains an exact function E(x, y), then S automatically has the weak fixed point property, and if also S is integrated and of type 1, then S has the recursion property. Actually, Theorems 1.3 and 1.4 of this chapter go through under a slightly weaker hypothesis - see Exercise 1 below. cad
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247
Exercise 1 Call a function D(x, 0) a uniform quasi-diagonalizer - or, '-r
more briefly, a near diagonalizer - if there is a E-function h(x) such that for all x, 0: D(h(x), 0) -> x, h(x), 0. [Obviously every diagonalizer is also a near-diagonalizer - take h(x) the identity function. Also every near diagonalizer is also a quasi-diagonalizer - given a, take a' = h(a) and we have D(a', 0) -> a, a', 0.]
(a) Prove that if S is of type 1, then every near-diagonalizer is a normalizer.
(b) Conclude, therefore, that Theorem 1.3 holds replacing "diagonalizer" by "near diagonalizer." .20
(c) Call a function E(x, y, 0) nearly exact (or uniformly quasi-exact) if there is a E-function t(x) such that E(t(x), y, 0) - x, y, 0. Show that 1.4 holds, replacing "exact" by "nearly exact." Normalization lemma car.
Theorem 1.2 is actually but a special case of the following lemma that will have other applications in the next chapter.
amp
Lemma N [Normalization lemma] If N(x, 0) is a normalizer of n+1 argucar
ments, then for any element a and any E-functions f, (x, y, 0), ... , fk (x, y, 0)
of n + 2 arguments, there is a E-function t(x) such that for every x and every n-tuple 0: ?f'
N(t(x), 0) - a, f, (x, t(x), 0), ... , fk (x, t(x), 0)
Proof Take a' such that a', x, y, 0 --f a, f, (x, y, 0), ... , fk (x, y, 0). Then there is a E-function t(x) such that N(t(x), 0) -4 a', x, t(x), 0 --> a, fl (x, t(x), B), ... , fk (x, t(x), 0)
Now, Theorem 1.2 follows from the above lemma, taking n = 0, k = 2, f, (x, y) = x, and f2 (x, y) = N(y). [Then again, of course, we take 0(x) _ N(t(x)).] S-.
More generally, from Lemma N, it easily follows that if N(x) is a normal-
izer, then for any element a and any E-function f (x), there is a E-function t(x) such that N(t(x)) --+ a, x, f (t(x)). Theorem 1.2 is essentially that special case where f (x) is N(x).
Chapter 13. Strong fixed point properties
248
cad
Suppose now that S is of type 1 and that E contains a normalizer N(x) of one argument. It then easily follows from Lemma N that for any element a and any E-functions hl (x, y), ... , hk (x, y) there is a E-function O(x) such that for all x: O(x) --f hl (x, O(x)), ... , hk (x, O(x)). However, we have the following apparently stronger fact: (fl
Theorem 1.5 If S has the recursion property then for any element a and any E-functions hl (x, y), . . . , hk(x, y) there is a E-function 0(x) such that for all x: O(x) - > a, hl (x, O(x)), ... , hk (x, O(x)).
Proof Assume hypothesis. Take an element a' such that for all x and y: a', x, y -* a, hl(x, y), ... , hk(x, y). Then take a E-function 0(x) such that O(x) - a', x, O(x), but a', x, O(x) - a, hl (x, O(x)),..., hk (x, O(x)), and so 0(x) - a, hl(x, 0(x)), .... , hk(x, 0(x))
Corollary 1.6 If S has the recursion property then for any element a and any E-functions g1(x), ... , 9k (X) there is a E-function O(x) such that for all x: O(x) -
a,x,g1(O(x)),...,gk(O(x))
Exercise 2 Prove Corollary 1.6. Exercise 3 Show that the recursion property implies the weak fixed point property. The following is a generalization of Theorem 1.2.
Theorem 1.7 If S is integrated and of type 1 and if E contains a normalizer N(x1, ... , xn,) of n arguments, then for any element a there is a E-function O(x1,.. . , x,,,) such that for all xl,... , x,,:
O(xl,...,xn)
a,xl,...,xn,O(xl,...,xn)
Proof Suppose E is of type 1 and contains a normalizer N(xl,... , xn). By Lemma N, for any element a and any E-functions fl,..., fn+l of n + 1 arguments there is a E-function t(x) such that for all X 1 ,- .. , xn:
N(t(xl), x2, ... , xn) -4a, fl (x1, t(xl), x2, ... , xn), ... , fn+1(x1, t(xl), x2, ... , xn)
§2 A type 2 approach
249
For each i < n, take f2(xl, y, x2i ... , xn) = x2 and take fn+l (x, y, x21 ... , xn) = N(y, x2, ... , xn). Then
N(t(x1),x2i...,xn) -f
a,x1,...,xn,N(t(xl),X2,...,xn)
And so we take 0(x1, ... , xn) = N(t(xl), X2.... , xn).
§2 A type 2 approach FMS
We now consider a completely different method of achieving the recursion property. The hypothesis of the following theorem (which does not presuppose either integration or being of type 1) is incomparable in strength with that of Theorem 1.3.
Theorem 2.1 If E is of type 2 and contains a quasi-diagonalizer Q(x, y) of two arguments, then S has the recursion property.
Proof Assume hypothesis. Given a, let a' be such that for all x and y: a', x, y -* a, y, Q(x, y)
(1)
Ltd
[Such an element a' exists by weak connectivity, since y can be expressed as a E-function of x and y.] Since Q(x, y) is a quasi-diagonalizer, there is an element b such that for all y: Q(b, y) -* a', b, y. But by (1) (taking b for x) a', b, y -* a, y, Q(b, y). Therefore Q(b, y) -> a, y, Q(b, y). We thus take 0(y) = Q(b, y), and Q(y) is a E-function of y (since E is of type 2). As a corollary we have:
Theorem 2.2 If E is of type 2 and contains an exact (or even quasiexact) function E(x, y, z) of three arguments, then S has the recursion property.
Fixed point functions
Both Theorem 2.1 above and Theorem 1.1 of Chapter 12 are really corollaries of the lemma below - which will have many more applications later on.
We shall call a function g(x, B) of n + 1 arguments a (weak) fixed point function if for every element a and any E-functions fl (X, 9), ... , fk(x, B) of n + 1 arguments, there is an element b such that for every n-tuple B: g(b, 9) -* a, f1(b, B), ... , .fk(b, 9)
Chapter 13. Strong fixed point properties
250
Lemma Q [Quasi-diagonal lemma] Every quasi-diagonal function is a fixed point function. `-'
Proof Suppose Q(x, 0) is a quasi-diagonalizer. Given a and E-functions f, (x, B), ... , fk (x, 0), let a' be such that a', x, 0 -* a, fl (x, B), ... , fk (x, 0). Then let b be such that Q(b, 0) -* a', b, 0. Also a', b, 0 -> a, fl (b, 0), .... fk (b, 0), and so by transitivity, Q(b, 0) -* a, fl (b, 0), ... , fk (b, 0). Thus Q(x, 0) is a fixed point function. Lemma Q has the following consequence:
Corollary Q1 If Q(x, 0) is a quasi-diagonalizer of n + 1 arguments, then for any E-functions f, (x, 0), ... , fk(x, 0) of n+1 arguments and any element a there is an element b such that for every n-tuple 0: Q(b, 0) -* a, 0, .fl (b, 0), ... , .fk (b, 0)
Proof For n = 0, this reduces to Q(b) -> a, f, (b), ... , fk (b), which holds by Lemma Q. Now suppose n > 0. For each i < n, let gi(x, y1i ... , yn) = yi. Then by Lemma Q there is some b such that
Q(b,0) -' a,gl(b,0),...,gn(b,0),.fl(b,0,...,.fk(b,0)) But the sequence (g1(b, B), ... , gn (b, B)) is simply 0, and so Q(b,0) -> a,0,.fl(b,0),...,.fk(b,0) Corollary Q1 in turn has the following special case, which simultaneously yields Theorem 2.1 of this chapter and Theorem 1.1 of Chapter 12: Q(b, 0) -> a, 0, Q(b, 0)
This, of course, is the special case in which k = 1 and f, (x, 0) = Q(x, 0).
Now, for n = 0, we have Q(b) -* a, Q(b), which is Theorem 1.1 of the last chapter. For n = 1, we have Q(b,y) -> a, y, Q(b, y), which gives Theorem 2.1 of this chapter. For later use, we will need the following consequence of Lemma Q.
Corollary Q2 If S is universal and if Q(x, 0) is a quasi-diagonalizer of n + 1 arguments, then for any E-functions f, (x, 0), . . . , fk(x, 0) of n + 1 arguments there is an element b such that for every x and n-tuple 0: Q(b, 0) -> fl(b, 0), ... , fk(b, 0)
§3 The extended recursion property
251
Proof This follows from Lemma Q, taking a such that a, yl,... , yk
yl,...Yk, for all yl,...,yk
§3 The extended recursion property We shall say that S has the extended recursion property' if for any element a and any E-functions g1 (x), ... , gk (x) there is a E-function O(x) such that for all x: O(x) -> a, x, 0(9,(x)), ... , O(9k(x))
fir I+,
This is not to be confused with the existence of a E-function O(x) such that for all x: O(x) -> a, x, g,(O(x)), ... , gk(o(x))! This latter is a consequence '-t
of the recursion property (Corollary 1.6), but we see no reason why the recursion property implies the extended recursion property. The extended recursion property will play an interesting role in a later chapter in connection with multiple fixed point properties. Of course the extended recursion property implies the recursion property (which is the 'C7
ay'
coo
special case where k = 1 and gl (x) is the identity function), but it is
p.:,;
.'3
coo
doubtful that the two properties are equivalent. Indeed, even if S obeys the hypothesis of Theorem 1.2 - or even the stronger hypothesis of Theorem 1.3 - it is doubtful that it necessarily follows that S must have the extended recursion property. But a simple modification of the proof of Theorem 2.1 yields: 'dam
Theorem 3.1 If E is of type 1 and type 2 and if E contains a quasidiagonalizer Q(x, y) of two arguments, then S has the extended recursion property.
Proof Assume hypothesis. Given an element a and E-functions gl(x),
...,9k(x), let fi(x,y) =Q(x,91(y)),...,fk(x,y) =Q(x,9k(y)). Since E is of type 1 then f, (x, y), . . . , fk (x, y) are all E-functions. By Corollary Q1i there is an element b such that for all x:
Q(b,x) -* a,x,fl(b,x),...,fk(b,x) Thus Q(b, x) -> a, x, Q(b, g, (x)), ... , Q(b, gk(x)). We let 0(x) = Q(b, x). Since E is of type 2, O(x) is a E-function. Also, for each i < k, Q(b, gi(x)) _ 0(gi(x)). And so we have 0(x) - > a, x, 0(g1(x)), ... , 0(gk(x)). Later we will have need of the following: 'For recursion theory, this property was introduced in R.M.
Chapter 13. Strong fixed point properties
252
Theorem 3.2 If S has the extended recursion property, then for any element a and any E-functions Fl (x, y),.. . , Fk (x, y), g1(x), ... , 9k (X) there is a E-function O(x) such that for all x:
Q(x) -*a,Fl(x,0(gl(x))),...,Fk(x,O(gk(x))) Proof Suppose S has the extended recursion property. Given a and the functions Fl, ... , Fk, gl, ... , gk, let a' be such that for all x and yl, ... , yk: a', x, yl, ... , yk -> a, F, (x, yl), ... , Fk(x, yk). Then let O(x) be a E-function such that/ for all x, O(x)) -> a', x, 0(g1(x)), ... , O(gk(x)). Then O(x) -> a, Fl(x, Y'(gl(x))), ... , Fk(x, c(gk(x))) As a corollary, we have .OS
Theorem 3.3 If S has the extended recursion property then for any element a and any E-functions fl (x), ... , fr (x), gl (x), ... , gk (x) there is a E-function O(x) such that for all x: O(x) -> a, fi(x), ... , .fr(x), 0(81(x)), ... , O(gk(x))
Proof Exercise. Exercise 4 Prove Theorem 3.3.
Part II The Myhill property and related properties We are saying that S has the Myhill property if E contains a function O(x)
such that for all x: O(x) -> x, O(x). [We recall that the Myhill property implies the weak fixed point property, for given a, we have b -> a, b, where b = 5(a). The Myhill property is a uniform version of the weak fixed point property.]
If S is universal and has the recursion property, then it also has the Myhill property (for then there is an element e2 such that for all x and y: e2, x, y -* x, y, and there is a E-function O(x) such that O(x) -> e2, x, O(x), hence O(x) -* x, O(x)). However, in application to indexed relational sys-
tems and sentential systems, we are interested in systems that are not universal but that nevertheless have the Myhill property, In such cases we cannot obtain the Myhill property via the recursion property, and so we
§4 Strongly connected systems
253
must turn to a different set of sufficient conditions that a system have the Myhill property.
§4 Strongly connected systems ..fir
We will say that S is strongly connected (or uniformly connected) if for any E-functions fi(yi,... , yn), ... , fk(yi, ... , yn) there is a E-function t(x) such that for all x, Y 1 ,- .. , yn:
t(x),y1,...,yn -' x,fl(yl,...,yn),...,fk(y1,...,yn)
SO.
It is obvious that strong connectivity implies weak connectivity, for given a, take b = t(a) and then b, yl, ... , yn -4 a, f1(y1, , yn), , fk(y1, .... yn)For an indexed relational system R, to say that S,n(R) is strongly connected is to say that for any E-functions f, (yl, ... , yn), , fk (yl, , yn) there is a E-function t(x) such that for all x, yl, ... , Yn, z1, , zm: 00.
Rt(x) (y1, ... , yn, z1, ... , zm)
4-' Rx(f1(y1,...,yn),..., fk(y1,...,yn),z1,...,.Zm) For recursion theorem this condition is a consequence of the extended iteration property (see §0, Chapter 10). For a sentential system (T), to say that Sn(T) is strongly connected is to say that for any E-functions fl (yl, ... , yn), ... , .fk(y1,... , yn) there is a E-function t(x) such that for all x, yl, ... , yn, and any k-tuple 0: Ht(.) (yl, ... , yn, 0) = H. (fl (yl, ... , Yn), ... , fk (Y1, ... , yn, B))
[This condition holds for Peano Arithmetic where E is the class of recursive functions.]
For an applicative system A, to say that S(A) is strongly connected is to say that for any elements al, ... , ak there is an element b such that for all x, yl, ... , yn: bxyl ... yn = xal (yl ... yn) ... ak (yl ... yn). This, of course, is guaranteed by combinatory completeness.
We wish to show that if S is strongly connected and of type 1 and if E contains a diagonalizer D(x) of one argument then S has the Myhill property. However, we will need to prove something apparently stronger. Strong fixed point functions
We shall define a function F(x, 0) of n + 1 arguments to be a strong fixed point function if for any E-functions fl (x, B), ... , fk (x, 0) of n+1 arguments there is a E-function t(x) such that for every x and every n-tuple 0: F(t(x), 0) - ' x, .f1(t(x), 0), ... , fk(t(x), 0)
CV+
Chapter 13. Strong fixed point properties
254
Of course every such function is also a weak fixed point function (given a,
take b = t(a), and then F(b, 0) -> a, fl(b, B), ... , fk(b, B)). Now, Lemma Q tells us that every diagonal function (in fact every quasi-diagonal function) is a weak fixed point function. [The proof, of course, used the weak connectivity of S.] The following lemma is a uniform version of this fact. FAY
Lemma F If S is strongly connected then every diagonalizer is a strong fixed point function.
Proof Suppose S is strongly connected and D(x, 0) is a diagonalizer of n + 1 arguments. Let fl(x, 0), ... , fk(x, 0) be any E-functions of n + 1 arguments. Then there is a E-function t(x) such that for every x, y and n-tuple 0: t(x), y, 0 -> x, fl (Y, 0), ... , fk(Y, 0)
Hence t(x), t(x), B -> x, fl (t(x), B), ... , fk (t(x), B). Also D(t(x), B) -> t(x), t(x), B. Hence D(t(x), B) -> x, fl(t(x), B), ... , fk(t(x), B). And so D(x, 0) is a strong fixed point function. Next we have:
Theorem 4.1 If E is of type 1 and contains a strong fixed point function F(x) of one argument then S has the Myhill property.
Proof Assume hypothesis. Since F(x) is a strong fixed point function, then (taking n = 0, k = 1) for any E-function f (x) there is a E-function t(x) such that for all x: F(t(x)) -> x, f (t(x)). Hence there is a E-function t(x)
such that for all x: F(t(x)) -> x, F(t(x)). And so we take O(x) = F(t(x)). By Lemma F and Theorem 4.1 we get:
Theorem 4.2 If S is strongly connected and of type 1 and if E contains a diagonalizer D(x) of one argument then S has the Myhill property. (DD
Exercise 8 Show that if S is of type 1 and strongly connected, then every near diagonalizer is a strong fixed point function. [It then follows that Theorem 4.2 holds replacing "diagonalizer" by "near diagonalizer."] It of course follows from Theorom 4.2 that if S is strongly connected of type 1 and if E contains an exact (or even a nearly exact) function E(x, y), then S has the Myhill property.
§5 The Myhill and recursion properties compared
255
From Theorem 1.2 and Theorem 4.2 we have: Summary
Suppose S is of type 1 and contains a diagonalizer (or even a near diagonalizer) D(x) of one argument. Then
(a) If S is integrated, S has the recursion property. (b) If S is strongly connected, S has the Myhill property.
Let us also note that from Theorem 2.1 we have:
Theorem 4.3 If S is universal of type 2 and if E contains a quasidiagonal function q(x, y) of two arguments, then S has the Myhill property. Later we will need the following:
'.,
Theorem 4.4 Suppose S is of type 1. Then the Myhill property is equivalent to the following condition: for any E-function g(x) there is a E-function 0(x) such that for all x: 0 (x) -* g(x), 0(x).
-0-
Proof The Myhill property is the special case of the above condition in which g(x) is the identity function. Now, suppose S is of type 1 and that there is a E-function O(x) such that O(x) -* x, 5(x). Then 0(g(x)) -> g(x), O(g(x)), and so we take O(x) = 0(g(x)) and we have O(x) -* g(x), 'b(x). Since E is of type 1, then %(x) is a E-function.
Theorem 4.5 If E is of type 1 and contains a fixed point function F(x1i , xn) of n arguments, then E contains a function 0(x1, ... , xn) such that for all x1, ... , xn: O(x1, .. , xn) I x1, ... , xn, O(x1, , xn).
. .
Exercise 9 Prove Theorem 4.5
§5 The Myhill and recursion properties compared Theorem 5.1 If S is integrated and of type 1 then the Myhill property for S implies the recursion property for S.
Chapter 13. Strong fixed point properties
256
Proof Suppose S is integrated and of type 1 and that O(x) is a Efunction such that for all x: O(x) -* x, 5(x). Since S is integrated, then given any element a, there is a E-function t(x) such that for all x and y: t(x),y -> a, x, y. Then 0(t(x)) -> t(x), O(t(x)) -> a, x, 0(t(x)). Hence O(x) -> a,x,%(x), where '(x) = 0(t(x)). Since E is of type 1, then 'O(x) is a E-function, and so S has the recursion property.
car
It of course follows from Theorem 5.1 that if S is universal, integrated, and of type 1, then the recursion property for S and the Myhill property for S are equivalent. But we will shortly prove something much better, namely, that if S is universal, integrated, and of type 1, then S has both the Myhill and the recursion properties (without any other hypothesis).
§6 Very strong connectivity We shall say that S is very strongly connected, if for any E-function g(x) and any E-functions f, (x, 9), ... , f k (x, 9) of n + 1 arguments there is a E-function t(x) such that for every x and every n-tuple 9: t(x), 0 -4 9(x), fl (x, 9), ... , fk(x, 9)
It is obvious that very strong connectivity implies strong connectivity (since we can take the identity function for g and for n > 1, any E-function Of Y 1 ,- .. , yn is also a E-function of x, yl, .. . , yn)
Exercise 10 Suppose that S is strongly connected and has the following additional property: for any E-function g(x) and every number n, there is a E-function h(x) such that for every x and every n-tuple 9: h(x), 9 -> g(x), x, 9
Show that S is very strongly connected. The following theorem may be a bit surprising:
Theorem 6.1 If S is very strongly connected and of type 1, then S has the Myhill property.
Proof Suppose S is very strongly connected. Then, taking n = 0, k = 1, g(x) = x, f, (x) = x, there is a E-function t(x) such that for all x: t(x) -> x, x. Thus t(x) is a diagonal function! The result then follows by Theorem 4.2.
§7 Universal integrated systems
257
Next we wish to prove that if S is very strongly connected and of type 2, then S is integrated. We first need a lemma that will have other uses as well.
Lemma 6.2 If S is of type 2, then for any element a and any n > 1, there is a E-function g(x1, . . . , xn) such that for all x1, . . . , xn: g(x1, ... , x,,) = a. PAW
Proof Suppose E is of type 2. Then E is also of type 0, so there is a (4D
E-function f (x, x1i ... , xn) such that f (x, x1, ... , xn) = x (because such a function f is explicitly definable from the identity function I(x) = x). Then, given a, take g(x1i... , xn) = f (a, xl,... , xn). Since E is of type 2, g is a E-function, and 9 ( X 1 ,- .. , xn) = f (a, xl,... , xn) = a.
Theorem 6.3 If S is very strongly connected and of type 2, then S is integrated.
Proof Suppose S is very strongly connected and of type 2. Given an element a, let g(x) be the constant function g(x) = a. Then g is a Efunction by the above lemma (taking n = 1). Let I(x) be the identity function. Since S is very strongly connected, then for every natural number
n there is a E-function t(x) such that for every x and every n-tuple 0: t(x), 0 -> g(x), I (x), 0, which means t(x), 0 -> a, x, 0. Thus S is integrated. By Theorems 6.1, 6.3, and 5.1 we have:
Theorem 6.4 If S is very strongly connected and of type 1 and 2, then S has the Myhill and the recursion properties.
§7 Universal integrated systems We will shortly see that if S is both universal and integrated, then S is not only very strongly connected, but obeys the following still stronger condition: for any E-functions f1(x, 0), ... , fk(x, 0) of n1 arguments, there is a E-function t(x) such that for every x and every n-tuple 0: t(x), 0 -> fl (x, s), ... , fk (x, s)
A system S satisfying this strong condition will be called hyperconnected.
Proposition 7.1 If S is universal and integrated then S is hyperconnected.
Chapter 13. Strong fixed point properties
258
-l.
Proof Suppose that S is universal and integrated. Consider any Efunctions f, (x, 6), ... , fk (x, 6) of n + 1 arguments. Since S is universal, then by Lemma 3.1 of the last chapter, there is an element a such that for all x and every n-tuple B: a, x, 8 -> f1(x, 8), . . . , fk(x, B). And since S is integrated, there is a E-function t(x) such that t(x), 6 -* a, x, B. And so t(x), 8 -> fl (x, B), ...
, .fk (x, B).
(DD
Q..
It of course follows from the above proposition that if S is universal and integrated then S is very strongly connected. So then by Theorem 6.1 and Theorem 5.1 we have: ?'+
Theorem 7.2 If S is universal, integrated, and of type 1, then S has the Myhill property and the recursion property. coo
More on hyperconnectivity
.fix
Hyperconnectivity is really a very powerful property, as the following proposition shows. .'3
...
Proposition 7.3 If S is hyperconnected then S is universal. .'3
Gam
Proof Suppose S is hyperconnected. Then surely for any positive n
:'D
there is a E-function t(x) such that for all x, yi, ... , yn: t(x), y1, , yn -> Y1, . . , yn (since for each i < n, y1 can be expressed as a E-function of x, y1, ... , yn, that is, we can take f2(x, y1, ... , yn) = y1). Then take any x and take c = t(x) and we have rye
From Proposition 7.3, Theorem 6.3, and Theorem 7.1 we have:
off
Theorem 7.4 For S of type 2, S is hyperconnected if and only if it is both universal and integrated.
§8 The strong Kleene property In the last chapter we defined S to have property K2 if for every E-function
F(x, y) and every element a, there is an element b such that b -> F(a, b). We now turn to a uniform version of this property. We will say that S has the strong Kleene property if for any E-function F(x, y) there is a E-function O(x) such that for all x: O(x) -> F(x, O(x))
§8 The strong Kleene property
259
-.0
Obviously the strong Kleene property implies property K2 (given F(x, y) and a, take b = 0(a)), which in turn implies the weak Kleene property.
Theorem 8.1 If S is universal and has the recursion property, then S '0'
has the strong Kleene property.
cps
Proof Suppose S is universal. Given a E-function g(x, y), by Lemma 3.1 of the last chapter there is an element a such that a, x, y --> g(x, y), for all x and y. If now S has the recursion property, then there is a E-function O(x) such that for all x: O(x) --> a, x, O(x), but a, x, q(x) --4 g(x, O(x)), and
so O(x) - g(x, O(x)) From the above theorem and Theorem 7.2 we have: p..
Theorem 8.2 If S is universal, integrated and of type 1, then S has the strong Kleene property.
`1l
Part III Summary and applications We now summarize the main results of this chapter and consider some of their applications to indexed relational systems, sentential systems and applicative systems.
CAD
Let us call S an exact system if for every n > 2, E contains an exact function E(xl, ... , xn) of n arguments. We shall say that S is of type A if it is exact, integrated, and of type 1. We shall say that S is of type B if it is exact and of type 2. We shall say that S is of type C if it is exact, strongly connected and of type 1. We shall call S unified if it is of all three types A, B, and C. To summarize the main points of this chapter: NEB
'0"
(A) Suppose S is unified - or even of type A or of type B. Then S has the recursion property - and more generally, for any positive integer n and any element a, there is a E-function 0(xl,... , xn) such that for all X 1 ,---, xn: ( x 1 , . . . , xn) - > a, xl, ... , xn, Oxl, ... , xn). Also, for any element a and any E-functions h1(x, y),. .. , hk (x, y) there is a E-function O(x) such that for all x: O(x) - a, hi(x, O(x)), ... hk(x, O(x)) ,
Chapter 13. Strong fixed point properties
260
(B) If S is unified - or even of type B - then S has the extended recursion property: for any element a and any E-functions 91(x), ... , gk (x) there is a E-function O(x) such that for all x:
0(x) - a, x, 0(91(x)), ... 0(9k(x)) ,
(C) If S is unified - or even of type C - then S has the Myhill property - and, more generally, for any n > 1 there is a E-function such that for all x1,..., x,,,: xl, .... x,n, O(xl, ... , x,a)
O(xl, ...
Also, under the same hypothesis, for any E-function g(x) there is a E-function '(x) such that for all x: '(x) -4 g(x), z/ (x). (D) If S is unified and universal - or even if S is universal and of either
type A or type B - then S has the strong Kleene property: for any E-function F(x, y) there is a E-function '(x) such that for all x: O(x) - F(x, O(x)).
Applications
ate-'
We shall call an indexed relational system 1Z unified if for every n > 1, the sequential system SOR) is unified. An example is the indexed relational system of recursion theory - or, for that matter, any system that is standard in the sense of Chapter 10. We shall say that a sentential system (T) is unified if for every n > 0 the sequential system S,n(T) is unified. An example is Peano Arithmetic (with E the class of recursive functions). We shall call an applicative system A unified if the sequential system S(A) is unified. Any complete applicative system (such as combinatory .V4"
p.,
logic) is unified. Here are some applications:
a+"
I Indexed relational systems For any unified indexed relation system R: (a) [Recursion theorem] For any R-relation R(x, y, z1i ... , zn,) there is a E-function 0(x) such that for all x, yl, ... , yn: RO(x) (yl, ... , y-n) F--* R(x, 0(x), y1, ... , yn)
II Sentential systems
261
More generally, for any R-relation R(xl,... , xn, y, z1i ... , zyn) there is a E-function q(x1 i ... , xn) such that for all x1, ... , xn, y1, ... , Ym: RO(xi,...,x,) (y1, .... ym) <---> R(xl, ... , xn, O(xl,... , xn), y17 .... ym) IVY
r-1
gyp
(b) [Extended recursion theorem] For any R-relation R of k + n + 1 arguments (k > 1, n > 1) and any E-functions gl (x), ... , gk (x) there is a E-function 0(x) such that for all x, Y1, , yn: RO(x) (yl, ... , yn) - R(x, 0(g1(x)), ... , 0(gk (x)), y1, ... , yn)
(c) [Myhill properties] For any n > 1 there is a E-function q(x) such that
for all x,yl,...,yn: RO(x) (y,.... , yn) F--* Rx (0(x), yl, ... , yn) '0-
More generally, for any E-function g(x) there is a E-function O(x) such that for all x, yl, , yn: RO(x) (y1, ... , Yn) H R,(x) (O(x), y1, ... , yn)
(d) If R is universal, then for every n > 1 and any E-function F(x, y) there is a E-function O(x) such that for all x, Rn(x) = RF(x fi(x)).
II Sentential systems For any unified sentential system (T): G.,
(a) For any n > 0 and any predicate H of n + 2 arguments there is a E-function O(x) such that for all x and every n-tuple 0, Hk(x)(8) is equivalent to H(x, O(x), 0). In particular (taking n = 0), this reduces to SO(x) - H(x, O(x)).
(b) For any predicate H of k + n + 1 arguments (k > 1, n > 0) and any E-functions 91 W, ... , gk(x) there is a E-function O(x) such that for
all x,y1,...,yn: He(x) (yl, ... , yn) = H(x, 0(gl(x)), ... , 0(gk(x)), yl, ... , yn)
(c) For every n > 0 there is a E-function q(x) such that for every x and every n-tuple 0: HO(x) (0) - Hx (0(x), 0)
[For n = 0 this reduces to Sq5(x) - Hx(O(x)).]
Chapter 13. Strong fixed point properties
262
(d) If (T) is also universal then for any E-function F(x, y) there is a (CD
E-function O(x) such that for every number x, SO(x) - SF(x,c(x)) More generally, if (T) is unified and universal, then for any n > 0 and any E-function F(x, y) there is a E-function O(x) such that for all x and every n-tuple 6: HO(x)(0) - HF(x,O(x))(0). III
a-=
III Applicative systems For any complete applicative system A there is an element 6 (called a "fixed point combinator") such that for every element x, bx = x(bx) (for every x, the element bx is a fixed point of x). [This is the Myhill property.]
Also, for any element a there is an element b such that for all x, bx = ax(bx). [This is the recursion property.]
Chapter 14
Multiple fixed point properties In this chapter we unify various "double" fixed point properties, and other multiple fixed point properties that arise in recursion theory, combinatory logic, and metamathematics.
Part I Double fixed points §1. The weak double fixed point property We shall say that S has the weak double fixed point property if for any elements al, a2 there are elements bl, b2 such that: (1) b1 -* al, bl, b2 (2) b2 -> a2i bl, b2
Let us see what the weak double fixed point property is in applications.
(A) For an indexed relational system R, to say that Sl (R) has the F-3
cad
weak double fixed point property is to say that for any R-relations Rai (x, y, z), Rae (x, y, z) there are elements bl and b2 such that wbi = {x : Ral (b1, b2, x)} and wb2 = {x : Rae (b1, b2, x)}. More
generally, for any positive n, to say that Sn (R) has the weak double fixed point property is to say that for any R-relations Rai (xl, x2, Y1.... , Yn), Ra2 (x1, x21 y1, .... yn,), there are elements bl, b2 such that for all yl,... , yn: Rbi(yl, ... , Yn) Ral (b1, b2, y1, .... yn), and Rb2 (y1, ... , yn) + Ra2 (b1, b2, y1, ... , yn). As we will see, the indexed relational system of recursion theory has "
this property (this is known as the weak double recursion theorem),
Chapter 14. Multiple fixed point properties
264
and so does the indexed system of arithmetical relations - and indeed, so does any standard system.
Q.,
car
CDs
car
(B) For an applicative system, A, to say that S(A) has the weak double fixed point property is to say that for any elements al, a2 there are elements bl, b2 such that alblb2 = bl and a2blb2 = b2. As we will see, a complete applicative system (such as that of combinatory logic) has this property (this result is known as the double fixed point theorem for combinatory logic).
(C) For a sentential system (T), to say that So(T) has the weak double fixed point property is to say that for any two-place predicates Hal, Hat, there are numbers b1, b2 such that Sbl - Hal (bl, b2) car
Imo
and Sb2 - Ha2(bl,b2). More generally, to say that Sn(T) has the double weak fixed point property is to say that for any predicates Hal, Ha2 of n + 2 arguments, there are numbers bl, b2 such that for every n-tuple B, Hbl (B) - Hal (bl, b2, B) and Hb2 (B) - Ha2 (b1, b2, B). [This condition for Peano Arithmetic is a special case of the generalized diagonal lemma of Boolos[2].]
Now let us get back to sequential systems in general.
Theorem 1.1 If E contains a quasi-diagonalizer Q(x, y) of two arguments, then S has the weak double fixed point property.
Proof Suppose Q(x, y) is a E-function and a quasi-diagonalizer. By Lemma Q of the last chapter, Q(x, y) is a fixed point function, hence for any al and a2 there are elements Cl, c2 such that for every x: (1) Q(cl, x) -. a,, Q(Cl, x), Q(x, C2) (2) Q(C2, x) -* a2, Q(x, C2), Q(Cl, X)
In (1) we take c2 for x and in (2) we take cl for x and we have: (1)' Q(C1, c2) -* a1, Q(C1, C2), Q(C2, Cl) (2)' Q(C2, CO -+ a2, Q(Cl, C2), Q (C2, CO a-'
And so we take b1 = Q(cl, c2); b2 = Q(C2, c1).
Corollary 1.2 If E contains an exact - or even quasi-exact - function E(x, y, z) of three arguments, then S has the weak double fixed point property.
§1 The weak double fixed point property
265
Discussion
I would like to point out another proof of the above corollary modeled after Boolos' proof of his generalized diagonal lemma[2]. Given elements al, a2, take elements cl, c2 such that for all x and y:
(1) E(cl, x, y) - a,, E(x, x, y), E(y, x, y) (2) E(c2, x, y) - a2, E(x, x, y), E(y, x, y) Then take b1 = E(cl, cl, c2) and b2 = E(c2, ci, c2).
Exercise 1 Show that the weak double fixed point property is equivalent to the following property: for any elements al, a2 there are elements bl, b2 such that: (1) b1 -* al, b1, b2
(2) b2 -> a2, b2, b1
Exercise 2 Suppose E contains a quasi-diagonalizer Q(x, y, z) of three arguments. Prove that S has the weak "triple" fixed point property, i.e., for any a1i a2, a3 there are elements b1, b2, b3 such that: (1) b1 -> a1, b1, b2, b3
(2) b2 -* a2, b1, b2, b3
(3) b3 - a3, b1, b2, b3
Solution (sketch) Take elements c1, c2, c3 such that: (1) Q(cl, x, y) -. a1, Q(cl, x, y), Q(x, y, cl), Q(y, cl, x) (2) Q(C2, x, y) -* a2, Q(y, C2, x), Q(C2, x, y), Q(x, Y, C2) (3) Q(C3, x, y) -* a3, Q(x, Y, C3), Q(y, C3, X), Q(C3, x, y)
Then take b1 = Q(cl, c2, c3), b2 = Q(c2, c3, cl), b3 = Q(c3, Cl, c2). '"'
Exercise 3 Suppose E contains a quasi-diagonalizer Q(xli... , xn) of n C!]
arguments. Show that S then has the weak "n-fold" fixed point property, i.e., for any elements a1i . . . , an there are elements b1, ... , bn such that for each i G n: bi -> ai, b1,.. . , bn. ,.p
Chapter 14. Multiple fixed point properties
266
.-.
Exercise 4 An easier way of achieving the n-fold fixed point property is possible if E contains a quasi-exact function E(x, yl, ... , yn) of n + 1 arguments. For then, given elements al, ... , an, for each i < n, let ci be such that E(ci, yl, .. , yn) ai, E(yl, yl, , yn), E(y2, yl, , yn), E(yn, Y1.... , yn). Then take bi = E(ci, c1, ... , cn) and show that for each i < n, bi -> ai, b1,... , bn.
Exercise 5 Suppose E is of type 1 and contains a normalizer N(x) of one argument. Does S necessarily have the weak double fixed point property? [The solution is very tricky and will emerge from a result of a later chapter.] Another weak double fixed point property
Theorem 1.3 If E contains a quasi-diagonalizer Q(x, y, z, w) of four arguments then for any elements a1i a2 there are elements b1, b2 such that (1) b1
al,a2,bl,b2
..o
(2) b2 -* a2, a1, b1, b2
Proof By the quasi-diagonal lemma, given a1 and a2 there are elements c1, c2 such that for all x, y, z: (1) Q(cl, x, y, z) -+ a1, z, Q(cl, x, y, z), Q(x, ci, y, z) (2) Q(c2, x, y, z) -> a2, y, Q(x, c2, y, z), Q(c2, x, y, z)
In (1) we substitute c2 for x, a1 for y and a2 for z. In (2) we substitute c1 for x, a1 for y and a2 for z, and we have: (1)' Q(cl, c2, al, a2) -. al, a2, Q(cl, c2, al, a2), Q(c2, cl, al, a2) (2)' Q(c2, c1, a1, a2) --a a2, a1, Q(c1, c2, a,, a2), Q(c2, cl, a,, a2)
And so we take b1 = Q(c1i c2, a1, a2) and b2 = Q(c2i c1, a1, a2).
The weak double Kleene property
We will say that S has the weak double Kleene property if for any Efunctions F, (x, y), F2(x, y) there are elements b1 and b2 such that b1 -> F1 (b1, b2) and b2 -. F2 (b1, b2).
Theorem 1.4 If S is universal and has the weak double fixed point property, then S has the weak double Kleene property.
§2 Double recursion
267
Proof Exercise. Exercise 6 Prove Theorem 1.4 and give the applications.
Part II Double recursion properties §2 Double recursion There are three properties that we call "double recursion properties" that a system S may or may not have, which will interest us: DRo: For any elements a1 and a2 there are E-functions 01(x), 02 (x) such that for/ all x: r-1
/
F'S
(1) O1(x) -* a1, x, 01(x), 02(x) (2) 02 (x) - a2, x, 01(x), 02 (x)
DR1: For any elements a1 and a2 there are E-functions 01(x, y), 02(x, y) such that for all x and y: (1) 01(x, y) - a1, x, 01(x, y), 02(x, y) (2) 02 (x, y) - a2, y, 01(x, y), 02 (x, y)
DR2: For any elements al and a2 there are E-functions 01(x, y), 02(x, y) such that for all x and y: (1) 01(x, y) - a1, x, y, 01(x, y), 02(x, y) (2) 02 (x, y) - a2, x, y, 01(x, y), 02 (x, y) CAD
These three properties are of increasing strength: DR1 obviously implies
DRo (just identify the variables x and y) and DR2 implies DR1, because given a1 and a2 we can take a', a2 such that for all x, y, z, w: ai, x, y, z, w -. a, x, z w and a2 i x, y, z, w -* a, y, z, w, and so applying DR2 to a', a2 we have DR1 for a1 and a2. We will now prove the following "double" analogues of Theorem 1.2 of the last chapter.
Theorem 2.1 Suppose S is of type 1. Then (a) If E contains a normalizer N(x, y) of two arguments then S has property DR1 (and hence also DRo).
(b) If E contains a normalizer N(x, y, z) of three arguments then S has property DR2 (and hence also DR1 and DRo).
Chapter 14. Multiple fixed point properties
268
Proof Suppose S is of type 1. (a) Suppose E contains a normalizer N(x, y). Then N(x, y), N(y, x) are both E-functions of x and y, hence for any elements a1 and a2 there are elements a', a2 such that for all x, y and z: a'1, x, y, z -* al, x, N(y, z), N(z, y) and a', x, y, z -. a2i y, N(z, y), N(y, z). Now take E-functions tl (x), t2 (x) such that N(tl (x), y) -> a', x, .,.,
t1(x), y and N(t2 (x), y) -> a2, x, t2 (x), y. Then
(1) N(tl(x),y) -* a1, x,N(ti(x),y),1V(y,tl(x)) (2) N(t2(x), y) -+ a2, x, N(y, t2(x)), N(t2(x), Y) In (2) we interchange x and y and get:
(2)' N(t2(y), x) - a2, y, N(x, t2(y)), N(t2(y), X) In (1) we take t2(y) for y and in (2)' we take tl(x) for x and we have:
(1)" N(ti(x),t2(y)) -> a1, x,N(ti(x),t2(y)),N(t2(y),ti(x)) (2)" N(t2(y),ti(x)) -> a2, y, N(ti(x),t2(y)), N(t2(y),ti(x))
We thus take 01(x,y) = N(tl(x),t2(y)) and 02(x,y) = N(t2(y), ti(x)). (b) Suppose E contains a normalizer N(x, y, z). Given al, a2, by Lemma N, §1 of the last chapter, there are E-functions t1(x), t2(x) such that for all x, y, z:
(1) N(tl (x), z, y) -' al, x, y, N(tl (x), z, y), N(z, ti (x), y) a2, x, y, N(z, t2(x), y), N(t2(x), z, y) (2) N(t2(x), z, y)
We then take 01(x, y) = N(tl(x), t2(x), y) and q52(x, y) = N(t2(x),
tl(x),y) From Theorem 2.1 and from Proposition 1.1 of the last chapter we have:
Theorem 2.2 Suppose S is integrated and of type 1. Then (a) If E contains a diagonalizer (or even a near diagonalizer) D(x, y), then S has properties DR1 and DRo. (b) If E contains a diagonalizer (or even a near diagonalizer) D(x, y, z), then S also has property DR2.
§3 A type 2 approach
269
Corollary 2.3 Suppose S is integrated and of type 1. Then (a) If E contains an exact function E(x, y, z) of three arguments then S has properties DR1 and DRo.
(b) If E contains an exact function E(x, y, z, w) of four arguments then S has property DR2. Question
Suppose S is integrated and of type 1 and that E contains a diagonal Q)-
function D(x, y) of two arguments. Does S necessarily have property DR2? We do not know (but see exercise 7 below).
CD.
fir"
Exercise 7 Call S n-integrated (n > 1) if for every natural number m and every element a there is a E-function t(xl, ... , xn) such that for all x1, ... , xn and every m-tuple 9: t(xl,... , xn), 0 -. a, x1, ...,x,9. [Thus 1-integrated is what we have been calling "integrated."] Suppose S is 2-integrated and of type 1 and that E contains a diagonalizer D(x, y) of two arguments. Prove that S has property DR2. coy
§3 A type 2 approach
As with recursion properties, we have a type 2 approach to double recursion
properties as well as a type 1 approach. The type 2 approach to double recursion is particularly neat.
Theorem 3.1 Suppose E is of type 2. Then (a) If E contains a quasi-diagonalizer Q(x, y, z) of three arguments, then S has property DRo. (b) If E contains a quasi-diagonalizer Q(x, y, z, w) of four arguments, then S has property DR2 (and hence also DR1 and DRo).
We first prove
Lemma QQ [Double quasi-diagonal lemma] Suppose Q(x, y, 9) is a quasi-diagonalizer of n+2 arguments (n > 0). Then for any elements a1 and a2 and any E-functions fl (x, y, 9), ... , f k (x, y, 9), 91(x, y, 9), ... , 9, (x, y, 9) of n + 2 arguments there are elements b1, b2 such that for every n-tuple 9:
Chapter 14. Multiple fixed point properties
270
(1) Q(bi, b2, 0) -> ai, fl (bl, b2, 0), ... , fk (bl, b2, 0) (2) Q(b2, b1, 8) -+ a2, 9i (bi, b2, 0), ... , 9r (bi, b2, 0)
Proof By Lemma Q, Chapter 13, there are elements b1, b2 such that for every 'y and n-tuple 8: (1) Q(bi, y, 8) -> ai, fi (bi, y, 0), ... , fk (b1, y, 8) (2) Q(b2, y, 8) -+ a2,91(y,b2,0),...,9r(y,b2,8)
Then take b2 for y in (1) and b1 for y in (2) and the conclusion follows. Proof of Theorem 3.1
Suppose E is of type 2. cad
(a) The functions z, Q(x, y, z), Q(y, x, z) are all E-functions of x, y, z, so by the above lemma there are elements b1, b2 such that for all x: ..p
(1) Q(b1, b2, x) -' ai, x, Q(b1, b2, x), Q(b2, bi, x) (2) Q(b2, b1, x)
a2, x, Q(b1, b2, x), Q(b2, b1, x)
And so we take 01(x) = Q(bi, b2i x) and 02(x) = Q(b2, bi, x). (b) Again by Lemma QQ there are elements b1, b2 such that for all x and
(1) Q(bi, b2, x, y) - a1, x, y, Q(bi, b2, x, y), Q(b2, bi, x, y) (2) Q(b2, bi, x, y) -> a2, x, y, Q(b1, b2, x, y), Q(b2, bi, x, y)
And so we take 01(x, y) = Q(b1i b2, x, y) and 02(x, y) = Q(b2, b1, x, y).
Corollary 3.2 Suppose E is of type 2. Then (a) If E contains a quasi-exact function of four arguments, then S has property DRo.
(b) If E contains a quasi-exact function of five arguments then S has property DR2.
§3 A type 2 approach
271
Discussion a--
I would like to point out another proof of the above corollary. Suppose E is of type 2 and contains a quasi-exact function E(x, y, z, w) of four arguments. Given al, a2, take cl so that for all y1 i Y2, x: E(cl, y1, y2, x) -* al, x, E(yl, y1, y2, x), E(y2, y1, y2, x)
and take c2 such that E(c2i y1, y2, x) -> a2, x, E(y1, y1, y2, x), E(y2, y1, y2, x)
Then take 01(x) = E(cl, cl, c2i x) and 02(X) = E(c2i c1, c2, x). We then get (a). We get (b) the same way, everywhere replacing "x" by "x, y."
Exercise 8 Prove that S has property DR, if and only if for every al, a2 there are E-functions 01(x, y), 02(x, y) such that for all x and y: (1) 01(x, y) - al, x, 01(x, y), 02 (x, y)
(2) 02(x, y) -a2, y, 02(x, y), 01(x, y)
We will later need
Theorem 3.3 Suppose S has property DRo. Then for any elements a1 and a2 and any E-functions gl (x, y), g2 (x, y) there are E-functions 01(x), 02(x) such that for all x: (1) 01(x) -> a1,x,g1(01(x),02(x)) (2) 02(x) _+ a2,x,92(01(x),02(x))
Proof Take ai, a2 such that ai, x, y, z - a1, x, g1(y, z) and a2/,x , y, z -> a2i x, g2(y, z). Then take E-functions 01(X), 02(X) such that 01(x) -* al, x, 01(x), 02(x) and 02(X) - a2, x, 01(x), 02(x)
Exercise 9 Suppose S has property DR2.
Show that for any ele-
ments al, a2 and any E-functions gl (x, y), g2(x, y) there are E-functions 01(x, y), 02(x, y) such that:
(1) o1(x,y) - a1,x,y,gl(o1(x,y),02(x,y)),92(01(x,y),02(x,y)) (2) 02 (x, y) -* a2, x, y, 9101(x, y), 02(x, y)), 9201(x, y), 02 (x, y))
Chapter 14. Multiple fixed point properties
272
Unified systems and applications
We see from Corollaries 2.3 and 3.2 that if S is either of type 1, integrated and has an exact function of four arguments, or of type 2 and has an exact
function of five arguments, then S has the double recursion properties DR2, DR1, and DRo. This gives two very different ways of proving that any unified sequential system has the double recursion properties, which in turn has the following applications: (A) For any R-relations M1(x, y, z1, z2, w1, ... , wn) and M2 (x, y, z1, z2, car
w1,. .. , wn) of a unified indexed relational system R, there are Efunctions 01(x, y), 02(x, y) such that for all x, y, wl, ... , wn:
(1) Rol
(x,y)(wl,...,wn) H Ml(x,y,01(x,y),02(x,y),w1i...,wn)
(2) R02(x,y)(wl, .... wn) (B)
M2(x, y, 01(x, y), 02(x, y), wl, ... , wn)
For a unified sentential system (T), for any predicates K1i K2 of k + 4 CAD
arguments (k > 0) there are E-functions 01(x, y) and 02(x, y) such that for every x, y and k-tuple 0: (1) Hm1(x,y) (0) = K1(x, y, 01(x, y), 02 (x, y), 0)
(2) H02(-,Y)(0) = K2(x, y, 01(x, y), 02(x, y), O)
p..
(C) The double property DR2 for complete applicative systems is of less interest than the double Myhill property, which we consider in Part III, but for the sake of the records, for any elements a1, a2 there are elements b1, b2 such that for all x and y: bixy = alxy(bixy)(b2xy), and b2xy = a2xy(blxy)(b2xy). Strong double Kleene property .tee
ry.
its
We will say that S has the strong double Kleene property if for any Efunction Fl (x, y, z, w) and F2 (x, y, z, w) there are E-functions 01(x, y), 02(x, y) such that for all x and y: (1) q51(x, y) -* F1(x, y, 01(x, y), 02 (x, y))
(2) 02(x, y) -> F2(x, y, 01(x, y), 02(x, y))
Theorem 3.4 If S is universal and has property DR2, then it has the strong double Kleene property.
§4 The double Myhill property
273
Exercise 10 Prove Theorem 3.4. Conclude that every unified universal sequential system has the strong double Kleene property. What does this tell us about unified indexed relational systems and unified sentential systems?
Exercise 11 Suppose S is universal and has property DRa. Show that for any E-functions Fi(x, y, z) and F2(x, y, z) there are E-functions 01(x) and 02 (x) such that for all x: 01(x) -. Fl (x, 01(x), 02 (x) ), and 02 (x) -* F2(x, 01(x), 02(x))
Part III Double Myhill properties and related results §4 The double Myhill property We shall say S has property DM (the double Myhill property) if there are E-functions 01(x, y), 02(x, y) such that for all x and y: (1) 01(x, y) -* x, 01(x, y), 02 (x, y)
(2) 02(X, y) - y, 01(x, y), 02 (x, y)
It is obvious that property DM implies the weak double fixed point coo
property (it is the uniform version of it - given a1, a2, take b1 = 01(a1, a2) and b2 = 02(a,, a2)). It is also obvious that if S universal and has property DR1i then it has property DM(just take al to be a three-universal element and a2 = a1). We are now interested in seeing how DM can be obtained without universality. ...
c0.
coo
As the reader might expect, we will prove that if E is of type 1 and contains a strong fixed point function F(x, y) of two arguments, then S has the double Myhill property. As we are at it, we will prove something more general.
For any k > 0 we will say that S has property DMk if there are Ecad
functions 01(x, y, 0), 02 (x, y, 0) of k2 arguments such that for every x, y and k-tuple 0:
(1) q1(x,y,B)-*x,0,01(x,y,B),02(x,y,e) (2) 02(x, y, 0) - y, 0, 01(x, y, 0), 02(x, y, 0)
Thus DMo is the double Myhill property DM. Property DM1 is a uniform version of the double recursion property DRo, and implies it if E is of type 2. DM2 is a uniform version of DR2, and implies it, if E is of type 2. Thus
274
Chapter 14. Multiple fixed point properties
for a system S of type 2, property DM2 is even stronger than DR2, and DM1 is even stronger than DRo.
Theorem 4.1 If E is of type 1, then for any natural number k, if E contains a strong fixed point function F(x, y, 0) of k + 2 arguments, then S has property DMk.
Proof [Very much like that of (a) of Theorem 2.1] - Take E-functions t1(x), t2(x) such that for every x, y, and k-tuple 0: (1) F(ti(x), y, 0) -* x, 0, F(ti(x), y, 0), F(y, ti(x), 0) (2) F(t2(y), x, 0) -* y, 0, F(x, t2(y), 9), F(t2(y), x, 0)
Then take q1(x, y, B) = F(tl (x), t2 (y), B) and 02 (x, y, B) = F(t2 (y), ti (x), 0)
By Theorem 4.1 and by Lemma F of the last chapter we have:
Theorem 4.2 Suppose S is strongly connected and of type 1. Then for any natural number k: (a) If E contains a diagonalizer D(x, y, 0) of k + 2 arguments, then S has property DMk.
(b) If E contains an exact function E(x, y, z, 0) of k + 3 arguments, then S has property DMk. Remark
We will subsequently prove stronger versions of (a) and (b) above - that is, the hypotheses lead to stronger conclusions than those stated. The special case (k = 0) of Theorem 4.2 is:
Theorem 4.3 Suppose S is strongly connected and of type 1. Then (a) If E contains a diagonal function D(x, y) of two arguments, then S has the double Myhill property. (b) If E contains an exact function E(x, y, z) of three arguments, then S has the double Myhill property.
§4 The double Myhill property
275
It of course follows that any unified sequential system has the double Myhill property. This has the following applications.
(A) For a unified indexed relational system R, for any n > 1 there are E-functions 01(x, y), 02(x, y) such that for all x, y, wl,. , w,,,: (1) R01(x,y)
(wl,... ,
wn) E---* Rx(c1(x, y), 02(x, y), w1i
... wn)
(2) R02(x,y) (wl, .... wn) - Ry(01(x, y), 02(x, y), w1, ... , wn) Al
(B) For a unified sentential system (T), for any k > 0 there are Efunctions ct1(x,y), g52(x,y) such that for every x, y, and k-tuple 0: (1)
Hx(01(x, y), 02(x, y), 0)
(2)
H02(x,y) (0) = Hy(01(x, y), 02(x, y), 0)
(C) For a complete applicative system A, there are elements 7l,'Y2 such that for all elements x and y: (1) 71xy = x('Y1xy)('Y2xy)
(2) 72xy = y('Y1xy)(l2xy)
[This is a uniform version of the double fixed point theorem of combinatory logic.] The property DM
Later we shall need a variant of the property DM. We shall say that S has property DM' if for any E-functions g1(x), g2(x) there are E-functions 01(x), 02(x) such that for all x: r-1
(1) 01(x) - 91(x), 01(x), 02(x) (2) 02(x) -* 92(x), 01(x), 02(x)
Theorem 4.4 Suppose S is of type 1 and has the double Myhill property. Then G".
(a) For any E-functions 91(X),92(X) there are E-functions V)1(x, y), b2(x, y) such that for all x and y: (1) 01(x, y) - 91(x), 4'1(x, y), 02(x, y)
(2) 02(x, y) - 92(y), 01(x, y), 02(x, y)
(b) S has property DM'.
Chapter 14. Multiple fixed point properties
276
Proof (a) Given E-functions 01(x, y), 02(x, y) that witness the double Myhill property, take 01(x,y) = 01(91(x),92(y)), and 02(x,y) = 02(91(x), 92(y))
(b) This follows from (a), by taking 01(x) = 01(x,x) and 072(x) _ 02(x, x).
Remark
We will see in the next chapter that if E is of type 1 and contains a pairing function J(x, y) with inverses K(x), L(x), then the properties DM and DM' are equivalent. Later we will need: C-'
Theorem 4.5 If S has property DM' and is strongly connected and of F'S
type 1, then for any E-functions g1(x), 92(x), hl(x, y), h2(x, y) there are E-functions 01(x) and 02(x) such that for all x: (1) 01(x) -. 91(x), h1(01(x), 02(x)) -0-
(2) 02(x) -+ 92(x), h2(01(x), 02(x))
Proof (sketch)
Assume the hypothesis. Since S is strongly connected then there are Efunctions t1(x), t2(x) such that for all x, y, z: (1) tl (x), y, z - 91(x), hl (y, z) (2) t2(x), y, z -. 92(x), h2(Y, z)
Then there are E-functions 01(x), 02(x) such that for all x: (1)' 0/,1(x) -4 t1(x), 01(x), 02(x) - 91(x), h1(01(x), 02(x)) (2)' 02(x) -* t2(x), 4'1(x), 02(x) - 92(x), h2(01(x), 02(x))
§5
The properties DMk
277
coy
§5 The properties DMk For any natural number k, we will say that S has property DMk if there
are E-functions 01(x, y, B), 02(x, y, B) of k + 2 arguments such that for all x, y, and every k-tuple B: (1) 01 (X, y, e) -. x, 01(x, y, e), 02 (x, y, e)
(2) 02(x, y, B) -* y, 02(x, y, B), 01(x, y, e)
[The significance of these properties will be seen in Part IV.] An obvious modification of the proof of Theorem 4.1 yields:
Theorem 4-i If E is of type 1 and contains a strong fixed point function F(x, y, B) of k + 2 arguments then S has the property DMk. We remark that we will see in Part IV that the above hypothesis yields an even stronger conclusion.
It of course follows that if E is strongly connected and of type 1, if E contains a diagonal function D(x, y, B) of k + 2 arguments, then S has the property DMk. But this also follows from (a) of Theorem 4.2 and the following result:
Theorem 5.1 If S is strongly connected and of type 1, then the properties DMk and DMk for S are equivalent.
Proof Suppose S is strongly connected and of type 1. We will show that DMk implies DMk (the proof that DMk implies DMk is quite similar). And so suppose that S has property DMk. Since S is strongly connected, there is a E-function g(y) such that for any y, z, w, and any k-tuple B: g(y), B, z, w -+ y, B, w, z. Now take E-functions 01(x, y, B), 02(x, y, B) such
that (1) 01(x, y, e)
x, B, 01(x, y, 0),'2(x, y, e)
(2) 02 (X, y, B) -' y, e, 02(x, y, B), 061(x, y, B)
Then (1)' 01(x, 9(y), B) -* x, e,01(x, 9(y), e),'b2(x, 9(y), e) (2)'
02(x, 9(y), B) -* 9(y), B,02(x, 9(y), e),'b1(x, 9(y), e) - y, e, 01(x, 9(y), e), 02(x, 9(y), B)
And so we take 1(x, y, B) = 01(x, g(y), B) and 02 (x, y, B) = 02 (x, g(y), B) and so we have property DMk.
Chapter 14. Multiple fixed point properties
278
Exercise 12 Without assuming strong connectivity, or that E is of type 1, prove:
(a) If S has property DM then S has the weak double fixed point property.
(b) If S is of type 2, then: (1) property DM1 implies DRa; (2) property DM implies DR2.
§6 Very Strongly Connected Systems .-y
We know that if E contains an exact function of n + 2 arguments, then E contains a diagonalizer of n + 1 arguments. We now show: .'3
CAD
Proposition 6.0 If S is very strongly connected of type 1 and E contains an exact function E(x, y, B) of n + 2 arguments, then E contains a diagonalizer D(x, y, B) of n + 2 arguments.
Proof Assume the hypothesis. Then there is a E-function t(x) such that for every x, y, and n-tuple B: t(x), y, 8 -* x, x, y, B, hence E(t(x), y, B) -> x, x, y, B, and so E(t(x), y, B) is a diagonalizer.
CDR
By (b) of Theorem 4.2, if S is strongly connected of type 1 and if E contains an exact function E(x, y, z, B) of k + 3 arguments, then S has the property DMk. But now, by the above proposition and (a) of Theorem 4.2 we have: CAD
Q..
.'3
Theorem 6.1 If S is very strongly connected of type 1 and if E contains an exact function E(x, y, B) of k + 2 arguments, then S has the property DMk.
§7 Integrated systems of type 1 The following theorem contains "double analogues" of Theorem 5.1, Chapter 13.
Theorem 7.1 Suppose S is integrated and of type 1. Then (a) If S has property DM', it has property DRo.
(b) If S has property DM, it has property DR1.
§8 Symmetric functions
279
Proof Suppose S is integrated and of type 1. (a) Suppose S has property DM'. Then given elements a1, a2 there are E-functions tl (x), t2 (x) such that for all x and y: tl (x), y -. al, x, y and t2(x), y -> a2i x, y. Since S has property DM' then there are E-functions 01(x), 02(x) such that: (1) 01(x) - t1 (x), 01(x), 02(x) - > a1, x, 01(x), 02(x)
(2) 02(x) -* t2 (X), 01(x), 02(x) - a2, x, 01(x), 02(x)
(b) Suppose S has property DM. Since S is integrated, there are Efunctions tl (x), t2 (x) such that for all x, y, and z: tl (x), y, z -* a1i x, y, z and t2(x), y, z - > a2i x, y, z. By Theorem 4.4 there are Efunctions 01(x, y), 02(x, y) such that for all x and y: (1) 01(x, y) -* t1(x), 01(x, y), 02(x, y) . a1, x, 01(x, y), 02(x, y) (2) 02(X, Y) - > t2(y), 01(x, y), 02(x, y) -* a2, y, 01(x, y), 02(x, y)
Thus S has the property DR1. Remark
The hypothesis that S is of type 1 was necessary for (b), but not for (a).
Part IV Symmetric functions and nice functions am.
We now consider two other approaches to double fixed point properties which in fact give strengthenings of earlier results.
§8 Symmetric functions ..r
By a symmetric function of k + 2 arguments we mean a function S(x, y, 9) satisfying the following condition: S(x, y, 8) -. x, 8, S(x, y, 8), S(y, x, 8) r0,
Of course this condition also implies: S(y, x, 8) -. y, 8, S(y, x, 8), S(x, y, 9) '-'
And so if we take 01(x, y, 9) = S(x, y, 8) and 02(x, y, 9) = S(y, x, 9) we have:
Chapter 14. Multiple fixed point properties
280
(1) 01(x,y,0) - x, 0,01(x,y,0),02(x,y,0) (2) ct2 (x, y, 8) - y, 0, g52 (x, y, e), 01(x, y, B) (CD
Therefore, if S(x, y, 0) is a E-function, then 01(x, y, 0), 02 (x, y, 0) are both E-functions, hence S then has the property DMk, and then by Theorem 5.1 we have:
Proposition 8.0 If S is strongly connected of type 1 and E contains a 2S'
symmetric function S(x, y, 8) of k+2 arguments then S has property DMk. Now comes the crucial fact: ',3
.'3
F-+
Theorem 8.1 If E is of type 1 and contains a fixed point function car
F(x, y, 0) of k+2 arguments then E contains a symmetric function S(x, y, 0) of k + 2 arguments.
Proof Assume hypothesis. Then there is a E-function t(x) such that for every x, y, and k-tuple 0: F(t(x), y, 0) -> x, 0, F(t(x), y, 0), F(y, t(x), 8) '.O
Hence F(t(x), t(y), 0) -> x, 8, F(t(x), t(y), 0), F(t(y), t(x), 8). And so we take S(x, y, 0) = F(t(x), t(y), 0).
pp,
The above theorem, together with Lemma F of the last chapter, gives the following strengthening of (a) of Theorem 4.2.
Theorem 8.2 Suppose S is strongly connected of type 1 and E contains a diagonal function D(x, y, 0) of k + 2 arguments. Then (1) E contains a symmetric function S(x, y, 0) of k + 2 arguments.
(2) S has the property DMk (and also DMk).
BCD
From (1) of the above theorem it of course follows that for a unified system S, for every k > 0, E contains a symmetric function S(x, y, 8) of k + 2 arguments. This has the following applications: (A) For any unified indexed relational system R, for every k > 0, n > 1, there is a E-function S(x, y, 0) of k + 2 arguments such that for every x, y, z 1 , . . . , zn and k-tuple 0: Rs(x,y,e) (zl, ... , zn)
Rx(B,S(x,y,B),S(y,x,e),z1,...,zn)1
§9 Nice functions
281
(B) For a unified sentential system (T), for any k > 0, n > 0, there is a E-function S(x, y, 0) of k + 2 arguments such that for every x, y, k-tuple 0, and n-tuple 01: Hs(x,y,e) (01) = HH(9, S(x, y, 9), S(y, x, 0), 01)
In particular (for k = 0, n = 0), there is a E-function s(x, y) such that for all x and y: III
S8(X,y) = H.(s(x, y), s(y, x)) f')
(C) For any complete applicative system A, there is an element II such that for all x and y: IIxy = x(IIxy)(IIyx) car
CD'
Also, for any n > 1 there is an element II such that for all x,y,z1,...,zn: Ilxyz1 ... zn = x(Ilxyzl ... zn)(Ilyxz1 ... zn)
§9 Nice functions We now come to a particularly attractive approach to double fixed point properties. By a nice function of k + 3 arguments (k > 0) we mean a function H(x, y, z, 9) such that for all x, y, z, and every k-tuple 8: (*) H(z, x, y, 9) -+ z, H(x, x, y, 9), H(y, x, y, 8)
If we substitute x for z, we get
(1) H(x,x,y,9) -* x,H(x,x,y,0),H(y,x,y,8) If in (*) we instead substitute y for z, we get (2) H(y, x, y, 0) -> y, H(x, x, y, 8), H(y, x, y, 9) E-+
Thus, if we let 01(x, y, 9) = H(x, x, y, 0), and 02(x, y, 0) = H(y, x, y, 0), we get: (1)' 01(x, y, 0) - x, 01(x, y, 0), 02(x, y, 0) (2)1 02(x, y, 0) -. y, 01(x, y, 0), 02 (x, y, 9)
Thus, if H is a E-function, so are 01 and 02, and we then have property DMk! And so we have: 1For recursion theory, this was stated and proved in R.M.
Chapter 14. Multiple fixed point properties
282
Theorem 9.1 If E contains a nice function H(x, y, z, 8) of k + 3 arguments, then S has property DMk.
Corollary 9.2 If E contains a nice function H(x, y, z) of three arguments, then S has the double Myhill property. We showed (Theorem 4.2, (a)) that if S is strongly connected of type 1 and E contains an exact function E(x, y, z, 8) of k + 3 arguments, then S has property DMk. But now we have the following better result.
Theorem 9.3 If S is strongly connected of type 1 and E contains an exact function E(z, x, y, 8) of k + 3 arguments, then E contains a nice function H(z, x, y, 0) of k + 3 arguments. .4"
Proof Assume hypothesis. Then there is a E-function t(x), such that for all z, x, y, and every n-tuple 8:
t(z),x,y,8 -+ z,8,E(x,x,y,0),E(y,x,y,8) Then E(t(z), x, y, 0) -> z, 0, E(x, x, y, 0), E(y, x, y, 0). Hence
E(t(z), t(x), t(y), 8) -* z, 8, E(t(x), t(x), t(y), 8), E(t(y), t(x), t(y), 8) (CD
And so we take H(z, x, y, 8) to be E(t(z), t(x), t(y), 0), and H is a nice function To summarize, suppose S is strongly connected and of type 1. Then for
anyk>0: (a) If E contains a diagonal function of k + 2 arguments, then E contains a symmetric function of k + 2 arguments and S has property DMk (and also DMk).
(b) If E contains an exact function of k + 3 arguments, then the conclusions of (a) follow, and also E contains a nice function of k + 3 arguments.
From (b) above, it of course follows that if S is unified, then for every k > 0, E contains a nice function H(x, y, z, 8) of k + 3 arguments. This has the following applications:
(A) For a unified indexed relational system R, for any k > 0, n > 1 there is a E-function H(z, x, y, 0) of k + 3 arguments such that for all
§10 Universal systems of type 2
283
X, y, z, w1, ... , w,,, and every k-tuple 0: RH(z,x,y,e) (w1, ... , wn) Rz(8,H(x,x,y,0),H(y,x,y,0),wl,...,wn)2
(B) For a unified sentential system, for any k > 0, r > 0 there is a E-
III
function O(z, x, y, 0) of k + 3 arguments such that for every z, x, y, k-tuple 0, and r-tuple 01: Hq,(z,x,y,e) (0l) = Hz Mx, x, y, 0), O(y, x, y, 0), 01) CAD
(C) For a complete applicative system A, there is an element N (a "nice" combinator) such that for all z, x, y:
Nzxy = z(Nxxy)(Nyxy)3
§10 Universal systems of type 2 For universal systems of type 2, there is the following neat way of obtaining nice functions and symmetric functions: F".
,-a
Theorem 10.1 If S is universal of type 2 and if E contains a quasidiagonalizer Q(w, z, x, y, 9) of k + 4 arguments, then E contains a nice function H(z, x, y, 0) of k + 3 arguments.
Proof By Corollary Q2 of the last chapter there is some element b such that
Q(b,z,x,y,0) --> z,Q(b,x,x,y,0),Q(b,y,x,y,0) And so we take H(z, x, y, 0) = Q(b, z, x, y, 0).
It is of course a corollary of Theorem 10.1 that if S is universal of type 2 and if E contains an exact function of k + 5 arguments, then E contains a nice function of k + 3 arguments - a result of apparently incomparable strength with Theorem 9.3. 2For recursion theory, this was stated in [30] and proved in R.M. 3This was stated and proved in [29] and [30].
Chapter 14. Multiple fixed point properties
284
Theorem 10.2 If S is universal of type 2 and if E contains a quasiQ21
diagonal function Q(w, x, y, 8) of k + 3 arguments, then E contains a symmetric function S(x, y, 8) of k + 2 arguments.
Proof Exercise.
§11 Multiple fixed point properties 'JO
Virtually all the preceding results of this chapter can be generalized from double fixed point properties to n-fold fixed point properties, n > 2. We shall give the appropriate definitions and theorems here; proofs of these theorems will be left as exercises.
For n > 2, k > 0, by a grand function of type n-k we shall mean a .
.
, xn, 0) of n + k + 1 arguments such that for every z, 2r'
function G(z, x1,
xl, ... xn, and k-tuple 0: G(z, X1.... , xn, 9)
- z,8,G(xl,xl,...,xn,B),G(x2ix1,...,xn,B),..., G(xn,x1,...,xn,B) [Note that a nice function of k + 3 arguments is thus a grand function of type '2-k.]
By a multiple symmetric function of type n-k (n > 2, k > 0) we shall mean a function S(x1 i ... , xn, 0) of n + k arguments such that
S(x1i...,xn,8) xlA SY(xl, ... , xn, 8) i S(x27 ... , x,n,i xl7 O)7 ... , S(xn, xl, ... xn-l, 0)
We shall say that S has the n-k Myhill property (n > 1, k > 0) if there are E-functions 01(x1, ... , xn, 8), ... , On(xl,.. , xn, 8) of n + k arguments such that for each i < n and every X1.... , xn and k-tuple 8:
Oi(x1,...,x-n,) - xi,8,Ol(xli...,xn,8),...,On(x1,...,xn,0) [Note: the 2-k Myhill property is DMk.]
In'
We shall say that S has the n-m recursion property (n > 1, m > 1) if for any elements al,... , an there are E-functions 01(x1, . , xm), ... , On(xl, ... , xm) such that for each i < m and all x1, ... , xm,: Oi(x1, ... , xm,) -4 ai,x1,...,Xm701(x1,...,xm),... On,(x1,...,xm)
Note: The 1-1 recursion property is simply the recursion property. The 2-2 recursion property is DR2. Now, here are the facts:
X11 Multiple fixed point properties
285
Theorem 11.1 Suppose S is strongly connected and of type 1. Then: (a) If E contains an exact function E(z, x1 i ... , xn, 8) of n + k + 1 arguments then E contains a grand function G(z, x1i ... , xn, B) of type n-k. urn
(b) If E contains a grand function of type n-k then S has the n-k Myhill property.
Theorem 11.2 .ti
(a) If E is of type 1 and contains a fixed point function F(xl,... , xn, B) of n + k arguments (n > 2) then E contains a multiple symmetric function of type n-k. (b) If S is strongly connected of type 1 and E contains an exact function of n + k + 1 arguments then E contains a multiple symmetric function of type n-k.
(c) If S is strongly connected of type 1 and E contains a multiple symmetric function of type n-k then S has the n-k Myhill property.
Theorem 11.3 If S is of type 2, then for any positive n and m, if S has the n-(n + m) property, then S has the n-m recursion property.
Theorem 11.4 Suppose that S is universal of type 2. Then: F".
(a) If E contains a quasi-diagonal function Q(w, X1.... , xn, 8) of n+k+2
arguments then E contains a grand function G(z, xl, ... , xn, B) of type n-k. h-4
(b) If E contains a quasi-diagonal function Q(w, x1, ... , xn, B) of n+k+l arguments, then E contains a symmetric function S(x1i... , xn, B) of
type n-k.
Exercise 13 Prove Theorems 11.1 through 11.4 Exercise 14 State and prove appropriate generalizations of Theorems 3.4 and 4.4.
Chapter 15
Synchronization and pairing functions
.ii
In this chapter we consider a totally different approach to double and multiple fixed point properties. This approach is an outgrowth of the original form of the double recursion theorem[28] which involves a pairing function J(x, y) and its inverse functions K(x), L(x) for its very statement. We will deal with this in Part III of this chapter.
Part I Synchronized fixed point properties §0 Synchronized sequences of functions Let II be a finite sequence (Ill, ... , II,,,) of E-functions of one argument. We will say that II is synchronized if the following two conditions hold:
(1) For any n-tuple. (a1, .. , an) of elements of N, there is an element a of N such that for each i < n, IIi(a) = ai (thus (a1, ... , an) is the sequence (II, (a), ... , Ile(a)).
(2) For any n-tuple (fl, ... , fn) of E-functions of the same number of arguments - say, k arguments - there is a E-function f of k arguments such that for each i < n, fi(x1i ... , xk) = IIi(f (x1, ... , xk)). Synchronization arises with systems in which E contains a pairing function J(x, y) and inverse functions K(x), L(x), which we will study in Part III of this chapter. If II is a synchronized sequence II, (x), ... , IIe (x), we call II a synchronizer of order n (for S) or an n-synchronizer. We let ord(II) be the order of II. We shall say that S is n-synchronized if there is an n-synchronizer II
287
§0 Synchronized sequences of functions
for S. As we will see later, if S is 2-synchronized, then for every positive n, S is n-synchronized. Synchronized fixed point properties
.-.
We let S be a sequential system (N, E, -->) and II a synchronizer for S - let m be its order. We now define the sequential system Sr, to be the ordered triple (N, E, n ), where n is defined as follows. For any sequences 01, B2, we say that a, 81 n b, B2 if for every i < m, Hi (a), B1 ----> Hi (b), 02- [We
will shortly see that if II is synchronized, then n is transitive and Sr, is
'1"
weakly connected (assuming of course that -> is transitive and S is weakly connected), and so Sr, then satisfies our minimal assumptions concerning sequential systems.] We will now say that S has property SW - the synchronized weak fixed point property - if for every synchronized II, the system Sr, has the weak .7'
7-a
fixed point property - which means that for every element a there is an element b such that for all i < ord(II): Hi (b) ---> Hi(a), b
We will say that S has property SR - the synchronized recursion property - if for every synchronized IT, the system Sr, has the recursion property, which means that for every element a there is a E-function O(x) such that for every i < ord(II) and element x: Hi(0(x)) --i Hi(a), x, 0(x)
We say that S has the property SM - the synchronized Myhill property - if for every synchronizer II the system Sr, has the Myhill property, which means that there is a E-function O(x) such that for every i < ord(H) and every element x:
Ili(0(x)) -> IIi(x), 0(x)
More generally, we might say that for any property P of sequential systems, a system S has property SP if for every synchronizer 11, the system
Sr, has property P. Let us note the obvious fact that if S has property SP, then it certainly also has property P (because if II is the trivial synchronizer
of order 1, Sr, is the system S). What we shall now do is to show that in several of our earlier theorems in which a certain hypothesis implies a certain fixed point property P, the same hypothesis implies the stronger property SP. One reason for our interest in synchronized fixed point properties is their relation to double fixed point properties. Later we will see that if S is of type 1 and II is a synchronizer for S of order 2, then Sr, has the recursion
Chapter 15. Synchronization and pairing functions
288
property if and only if S has the double recursion properties DR2, DR,, DRo, and Sr, has the Myhill property if S has the double Myhill properties DM and DM' (and also Sr, has the weak fixed point property.if and only if S has the weak double fixed point property).
§1 Synchronized fixed point theorems
+.n
At this point it will be helpful to summarize some of the results proved (or all but proved) in Chapters 12 and 13. Summary A
(1) If E contains a quasi-exact function F(x, y) of two arguments then S has the weak fixed point property (Corollary 1.3, Chapter 12).
(2) If S is integrated of type 1 and E contains an exact function E(x, y) of two arguments then S has the recursion property (Theorem 1.4, Chapter 13). a-=
(3) If E is of type 2 and contains a quasi-exact function F(x, y, z) of three arguments, then S has the recursion property (Theorem 2.2, Chapter 13).
(4) If S is strongly connected of type 1 and E contains an exact function E(x, y) of two arguments then S has the Myhill property. [This is immediate from Theorem 4.2, Chapter 13.] (5)
If S is very strongly connected of type 1 then S has the Myhill property (Theorem 6.1, Chapter 13).
We now wish to prove the following "self-strengthening" of Summary A.
Theorem A* For S of type 1, all five parts of Summary A hold good, replacing "weak fixed point property" by "synchronized weak fixed point property", "recursion property" by "synchronized recursion property", and "Myhill property" by "synchronized Myhill property." Although the five parts of Summary A are but special cases of the corresponding parts of Theorem A*, we will derive Theorem A* from Summary A by showing that for S of type 1, in each of the five parts (1)-(5) of Summary A, if the hypothesis holds for S, then it also holds for Sn, where 11 is any synchronizer, and hence the conclusion also holds for Srj. However, we must first show two lemmas.
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289
Lemma 1.0 If H is a synchronizer for a sequential system S, then Sr, is a sequential system, i.e., n is transitive and Sr, is weakly connected.
c(D
Proof Transitivity is obvious. Next, since S is weakly connected, then for any element a and E-function fl (xl, ... , xn), ... , fk(x1, .... xn), for
each i < ord(II) there is an element bi such that for all X 1 ,- .. , xn: bi, xl, ... , xn --> IIi (a), fl (xl, ... , xn), ... , fk (x1, ... , xn). Then, since II car
is synchronized, there is an element b such that for each i < ord(n), IIi(b) = bi, and so IIi(b), xl, ... , xn -- IIi(a), fl(xl, ... , xn), fk(x1, ... , xn), which means that b, xl, .... xn II a fl (xl,... xn) ... fk (xl . ,
.... xn). Thus Sr, is weakly connected. Now we have
Lemma 1.1 [main lemma] Suppose that S is of type 1 and that II is synchronized for S. Then
(al) If E contains an exact function of n + 2 arguments for S (n > 0) then E contains an exact function of n + 2 arguments for Sll. (a2) Same with "quasi-exact" in place of "exact."
(b) If S is integrated, so is Sri (c) If S is strongly connected, so is Sll. (d) If S is very strongly connected, so is Sll.
Proof We let m be the order of H.
VIA
(al) and (a2) Given a E-function E(x, y, 8) of n + 2 arguments, for each i <_ m the function E(IIi(x), y, 8) is a E-function (since E is of type 1) and since II is synchronized, there is a E-function F(x, y, 8) such that for each i < m, IIi(F(x, y, B)) = E(IIi(x), y, 0).
Then for each i <
17m
(i) Suppose E(x, y, 8) is exact (for S).
m, IIi(F(x, y, B)) = E(IIi(x), y, 9)
-->
IIi(x), y, 8.
Hence
F(x, y, 9) n x, y, 8, so F(x, y, 9) is an exact function for S. (ii) Suppose E(x, y, 9) is only quasi-exact. Let a be any element.
Then for each i < in, there is some element bi such that C'1
E(bi, y, 8) - IIi(a), y, 8; hence by synchronization there is some
b such that for all i < ord(II): E(Hi(b), y, 9)
-->
IIi(a), y, 8.
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290
Hence for all i < m, IIi(F(b, y, B)) = E(IIi(b), y, 0) --> IIi(a), y, 0, and so F(b, y, 0) n a, y, 0. Thus F(x, y, 0) is a quasi-exact func-
tion for Sn.
(b) Suppose S is integrated. Let a be any element and n be any natural
number. Then for every i < m, there is a E-function ti(x) such that for all x and every n-tuple 0: ti (x), 0 --> IIi(a), x, 0. Then by synchronicity there is a single E-function t(x) such that for all i < m, Ili(t(x)) = ti (x), and hence Ili(t(x)), 0 ---> IIi(a), x, 0, and therefore
t(x), 0 n a, x, 0. Thus Sr, is integrated. [Note: the hypothesis that S is of type 1 is not necessary for this part.] (c) Suppose that S is strongly connected. Given E-functions f1(yl, ... , , yn), there is a E-function h(x) such that for all , ,fk(y1, yn),
x,y1,...,yn: h(x),yl,...,yn --i x,fl(y1.... ,yn,),...,fk(y1,.... Yn) and so for each i < m, h(IIi(x), yi, .... yn) -4 IIi(x), fl(y1.... , Then by synchronization there is a E, fk (yl, ... , yn). yn),
function t(x) such that for each i < m, IIi(t(x)) = h(Hi(x)), hence Ili (t(x)), y1, ... , yn - IIi (x), f1(y1, ... , yn), ... , fk (y1, ... , yn). Thus
t(x),yl,...,yn hI x,fl(yl,...,yn),...,fk(yl,...,yn), and so Sr, is strongly connected.
(d) Suppose S is very strongly connected. Then given E-functions fl(x, yl, ... , yn),... , fk(x, yl, ... , yn) and a E-function g(x), for each i < ord(II) there is a E-function ti (x) such that for all x, yl, ... , yn: ti(x), yl, ... , yn - IIi(9(x)), fl(x, y1, ... , yn), ... , fk(x, y1, ... , yn), hence
there is a E-function t(x) such that for all i < in, Hi(t(x)), yl, ...,yn -p IIi(9(x)),fl(x,yl.... ,yn).... ,fk(x,y1,.... yn), and so t(x), yl, ... , yn n 9(x), f1(x, y1, ... , yn), ... , fk (x, y1, ... , yn). Sr, is very strongly connected.
Thus
E-+
This concludes the proof of Lemma 1.1, and so the proof of Theorem A* is now complete.
§2 Relation to multiple fixed point properties Next we have:
Theorem B Suppose that 11 is a 2-synchronizer for S. Then (1) If Sr, has the weak fixed point property then S has the double weak fixed point property.
(2) If Sr, has the recursion property and S is of type 1, then S has the property DRo.
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291
(3) If Sr, has the Myhill property and S is strongly connected of type 1 then S has property DM'. As we are at it, we can just as easily prove the following generalization:
Theorem B* Suppose that II is an n-synchronizer for S (n > 1). Then
(1) If Sr, has the weak fixed point property then for any elements al, ... , an there are elements bl, ... , bn such that for each i < n:
bi-->ai,bl,...,bn (2) If Sr, has the recursion property and S is of type 1, then for any elements al, ... , an there are E-functions 01(x), ... , On (x) such that for each i < n and all x: Oi(x) - ai, x, 01(x), ... On(x) ,
(3) If Sri has the Myhill property and S is strongly connected of type 1 then for any E-functions gl (x), ... , gn(x) there are E-functions q1(x), .... 02(x) such that for all i < n and all x: Oi(x) - gi(x), 01(x), ... On(x) ,
Proof Suppose that II(x) is an n-synchronizer for E. (1) Suppose that Sr, has the weak fixed point property. Let al, ... , an be any elements of N. For each i < n let ai be such that for all x: a', x --> ai, IIi(x), ... , IIn(x). Then let a be such that for all i < n: IIi(a) = a? (such an a exists since II is n-synchronized). By hypothesis there is some b such that b n a, b, and so for each i < n: IIi (b) ---> IIi (a), b. But Hi(a) = a', hence Hi(b) -- a', b --> ai, II, (b) ... , IIn(b). We thus take bi = IIi(b) and we have bi ---> ai, b1, ... , bn.
(2) Suppose Sr, has the recursion property and S is of type 1. Given elements a1,... , an we can now take a such that for each i < n: IIi (a), x, y --> ai, x, IIi (y), ... , IIn (y). [We first take a? such that az, x, y -> ai, x, IIi (y), .... IIn (y)7 then take a such that for each i < n: IIi(a) = a'.] Then there is a E-function q(x) such that for all x, O(x) n a, x, O(x), which means that for each i < n, IIi(O(x)) ---> IIi(a), x, q(x), hence IIi(q(x)) -> a'i, x, O(x) -+ ai, x, II1(q(x)), ... , IIi(O(x)). We thus take Oi(x) = IIi(O(x)) and we have Oi (x) - ai, x, q1(x), .... On (x).
Chapter 15. Synchronization and pairing functions
292 (3)
Suppose S is strongly connected of type 1 and that Sr, has the Myhill property. Given E-functions gl (x), ... , gn(x) let g(x) be a E-function such that for all i < n: IIi(g(x)) = gi(x). Since S is strongly connected, so is Sr, (by (c) of Lemma 1.1), hence there is a E-function g(x) such that for all x, y: h(x), y n x, II'(y), ... , IIn(y). Then h(g(x)),y n g(x), Ei(y), ... , IIn(y). We let t(x) = h(g(x)) and since S is of type 1, t(x) is a E-function . Thus we have a E-function t(x) such that t(x), y n g(x), IIW(y), ... , IIn(y).
'0-
Since Sr, has the Myhill property, then by Theorem 4.4, Chapter 13 applied to Srj, there is a E-function O(x) such that for all x: O(x) n t(x), O(x). Thus O(x) n g(x), IIn(O(x)) (since t(x)) O(x) -> g(x), IIl(O(x)), ... IIn(O(x))). Thus for each i < n: IIi(O(x)) --i IIi(9(x)),IIl(O(x)),...,IIn(O(x)). Thus IIi(O(x)) 9i (X), 1110(x)), ... , IIn(O(x)). We thus take Oi(x) = Hi (O(x)) and we have Oi(x) ---p 9i (X), 01(x), ... , On(x).
Al
Exercise 1 Suppose that S is of type 1 and that there is a 2-synchronizer for S. Prove that for all n > 2 there is an n-synchronizer for S. [Use mathematical induction.]
Exercise 2 Suppose S is of type 1 and Sr, has the recursion property, where 11 is a 2-synchronizer for S. Then by Theorem B, S has the double recursion property DRo. Does S necessarily have the property DR2?
Exercise 3 Suppose S is strongly connected of type 1 and II is a 2synchronizer for S and Sr, has the Myhill property. Does S necessarily have the double Myhill property DM?
Part II Pairing functions We shall say that S is of type 1* (type 2*) if it is of type 1 (type 2) and E contains a 1-1 function J(x, y) and functions K(x), L(x) such that for all x and y:
(1) K(J(x, y)) = x (2) L(J(x, y)) = y We do not require that J be onto N, nor that J(K(x), L(x)) = x. [These requirements can be met in recursion theory but not in combinatory logic.]
§3 Exact and quasi-exact functions
293
n-tupling functions VV]
We assume S to be of type 1*. As in recursion theory, for every n > 2 we define the functions ( X1 ,- ... , xn) by the following inductive scheme:
J2(x1,x2) = J(x1,x2) J3 (x1, ... , x3) = J(J2 (x1, x2), x3)
Jn+l(xl,...,xn) = J(Jn(x1i...,xn),xn+1) As in recursion theory, for each n > 2 we define the inverse functions
Kl , ... , Kn by the following inductive scheme: (1) K12 =Kand K2 = L (2) For i < n, K? +1(x) = K? (K(x)). But Kn+l (x) = L(x) An obvious induction argument shows that for all n > 2:
Ki (Jn(xl, ... , xn)) = x1
Kn(Jn(x1i...,xn) = xn
m--
i.e., for each i < n: K? (Jn(x1, ... , xn)) = x2. It is obvious that for every n > 2 the sequence Ki (x), ... , Kn (x) of Efunctions is synchronized, for given any E-functions gl (B), ... , gn (B) we can take g(9) = Jn(gl(B),... , gn(9)), and so for each i < n: K2 (g(9)) = g2(8).
Thus if S is of type 1*, then for every n > 1 the system S is niii
synchronized. Thus all results of Part I of this chapter have applications to systems of type 1*, which we will consider in Part III of this chapter.
00.
We might note that any 2-synchronized system of type 1 is of type 1*, for suppose S is of type 1 and II is a 2-synchronizer for S. Let fl (x, y) = x and f2(x, y) = y. The functions fl and f2 are both E-functions (since they are explicitly definable from the identity function) and so there is a E-function f (x, y) such that II1(f (x, y)) = fl (x, y) and n2 (f (x, y)) = f2 (x, y); hence IIl (f (x, y)) = x and T12 (f (x, y)) = y. We can thus take J(x, y) = f (x, y),
K(x) = IIi(x),L(x) = 112(x), and so S is of type 1*. It then further follows that if S is of type 1 and is 2-synchronized, then for every n > 2, S is n-synchronized - a fact mentioned earlier in this chapter.
§3 Exact and quasi-exact functions Theorem 3.1 Suppose that S is of type 1*. Then
Chapter 15. Synchronization and pairing functions
294
(1) If E contains a quasi-exact function of two arguments, then for every n > 2, E contains a quasi-exact function of n arguments. Imo
(2) If E contains an exact function of two arguments and S is strongly connected, then for every n > 2, E contains an exact function of n arguments.
Proof Suppose S is of type 1*. car
(1) Suppose E(x, y) is quasi-exact. Let n > 2 and let F(x, x1, ... , xn) _ E(x, Jn(xl, ... , xn)). We assert that F is also quasi-exact.
To prove this, given any element a there is an element a' such that for any element y: a', y -> a, Ki (y), . . . , Kn (y). Then there is some b such that for all y: E(b, y) - > a', y, hence E(b, Jn(xl, ... , xn)) --> a', Jn(xl,... , xn) --> a, x1i ... , xn. Thus E(x, J(xl, ... , xn)) is quasiexact.
(2) Suppose E(x, y) is exact and S is strongly connected. Then there is a E-function t(x) such that t(x), y -> x, Ki (y), ... , Kn(y), hence E(t(x), y) -> x, Ki (y), ... , Kn(y). Then for all x, x1, ... , xn: E(t(x), Jn(x1, ... , xn)) --> x, x1, ... , xn, and so the function E(t(x), Jn(x1, .... xn)) is exact.
Corollary 3.2 Suppose S is of type 1*. Then (a) If E contains a quasi-exact function of two arguments, then for every positive n, E contains a quasi-diagonalizer of n arguments.
(b) If S is strongly connected and E contains an exact function of two arguments, then for every positive n, E contains a diagonalizer of n arguments.
We next wish to prove that if S is integrated and of type 1* and if E contains an exact function E(x, y) of two arguments, then for every positive
n, E contains a normalizer of n arguments. To this end we introduce a definition: .F,
Let us call a function M(x, 9) of n + 1 arguments a master function if for every element a there is a E-function t(x) such that for every x and every n-tuple B: M(t(x), B) --> a, x, 9.
§3 Exact and quasi-exact functions
295
Lemma 3.3 (a) If S is integrated, then every exact function is a master function. (b) If M(x, y, 0) is a master function then its diagonalization M(x, x, 0) is a normalizer.
Proof Obvious. Lemma 3.4 Suppose that S is of type 1* and that E contains a master function of two arguments. Then for every n > 2, E-contains a master function of n arguments.
E3y
Proof This proof is similar to that of (2) of Theorem 3.1. Suppose that S is of type 1* and that E contains a master function M(x, y). We assert that for every n > 2: M(x, J,,,(xl,... , x.,,,)) is a master function. For given a, let a' be such that a', x, y > a, x, K1 (y), .... Kn (y). Then let t(x) be a E-function such that M(t(x), y) --> a', x, y. Then M(t(x), y) --> a, x, Ki (y), ... , Kn (y) Then M(t(x), Jn(x1, ... , xn)) -' a, x x1 ... , xn. And so M(x, Jn (x1, ... , xn)) is a master function. Note
If E(x, y) is an exact function and S is strongly connected of type 1*, then there is a E-function t(x) such that E(t(x), Jn(x1i... , xn)) is a master function, whereas if M(x, y) is a master function, then it is M(x, Jn(x1i... , xn)) itself that is a master function. From Lemmas 3.3 and 3.4 we have:
Theorem 3.5 Suppose that S is of type 1* and integrated and that E contains an exact function of two arguments. Then for every positive n, E contains a master function of n + 1 arguments and a normalizer of n arguments.
We next have the following rather curious result:
Theorem 3.6 Suppose that S is of types 1* and 2* and that E contains a quasi-exact function of two arguments. Then for every positive n, E contains a normalizer of n arguments (in fact for any n > 0, E contains a master function of n + 1 arguments).
Chapter 15. Synchronization and pairing functions
296
We first prove:
Lemma 3.7 Suppose S is of type 2*. Then if F(x, y, B) is quasi-exact, then the function F(K(x), L(x), 9) is a master function. Proof Suppose S is of type 2* and that F(x, y, 9) is a quasi-exact function for S. Let M(x, B) = F(K(x), L(x), B). Since F(x, y, 9) is quasi-exact, then given any element a there is an element b such that for all x, 9: F(b, x, 6) --> a, x, 9. Also M(J(b, x), B) = F(b, x, 9) (because M(J(b, x), 9) = F(K(J(b, x)), L(J(b, x)), 9) = F(b, x, B)).
Hence M(J(b, x), 9) -> a, x, 6. We let t(x) = J(b, x). Since S is of type 2 then t(x) is a E-function and M(t(x), B) --> a, x, 9. Thus M(x, B) is a master function.
Proof of Theorem 3.6 Assume hypothesis. Since S is of type 1* and E contains a quasi-exact function of two arguments, then by (1) of Theorem 3.1, for any n, E contains a quasi-exact function F(x, y, B) of n + 2 arguments. Let M(x, 9) = F(K(x), L(x), B). Since S is of type 2 then M(x, 6) is a master function, and it is a E-function, since S is of type 1.
§4 Some consequences We proved in Chapter 14 that if S is integrated of type 1 and E contains a diagonalizer of three arguments, then S has the double recursion properties DR2, DR,, and DRo. Therefore if S is integrated of type 1 and if E contains
an exact function of four arguments then S has these double recursion -.z
properties.
CAD
Now, suppose that S is integrated of type 1* and that E contains an exact function of two arguments. Then by Theorem 3.6 E contains a normalizer of three arguments, and hence by (b) of Theorem 2.1, Chapter 14, S has the double recursion properties. And so we have: G".
Ham;
Theorem 4.1 If S is integrated of type 1* and if E contains an exact gym,
function E(x, y) of two arguments, then S has the double recursion properties DR2, DR,, DRa. Remarks
As we will see, an entirely different proof of Theorem 4.1, not using any material of Chapter 14, can be obtained from some of the synchronized results of Part I and in the next chapter, using yet another method, we
§5 Strongly connected systems of type 1*
297
will prove the following strengthening of Theorem 4.1: if S is integrated of type 1* and if E contains a diagonalizer D(x) of one argument, then S has the properties DR2, DR1, and DRo.
We also proved in Chapter 14 (Corollary 3.2) that if E if of type 2 and contains a quasi-exact function of five arguments, then S has the properties DR2, DR1, DRo. And now we have:
Theorem 4.2 If E is of types 1* and 2* and contains a quasi-exact function of two arguments, then S has properties DR2iDR1, and DRo.
Proof 1 Assume hypothesis. Then by (a) of Corollary 3.2,E contains a CAD
p_,
quasi-diagonalizer of four argument. The conclusion then follows by (b) of Theorem 3.1, Chapter 14.
pro
Proof 2 Under the same hypothesis, by Theorem 3.5, E contains a norr-1
malizer of three arguments. The conclusion then follows by (b) of Theorem 2.1, Chapter 14. Next we have: so.
Theorem 4.3 If S is of type 1*, then the properties DR2, DR1, DRo for S are all equivalent. G".
110
?+o
.s:
Proof Assume hypothesis. We already know from Chapter 14 (without the assumption that S is of type 1*) that DR2 DR1 DRo. S-'
But now that we are given that S is of type 1*, we can show that So suppose S has property DRo. Given elements al, a2 there are elements ai, a2 such that for all x, y, z: ai, x, y, z --> DRo implies DR2.
1-S -s0
al, K(x), L(x), y, z and a2, x, y, z --> a2i K(x), L(x), y, z. Then by DRo there are E-functions V)1(x), ' b2 (x) such that b1(x) - > a', x, 01(x), 02 (x) and b2(x) --> a2,x,V)1(x),02(x). Then take 01(x,y) = V)1(J(x,y))
and 02(x,y) = V)2(J(x,y)) and we have 01(x,y) = V)1(J(x,y)) all, J(x, y),')1(J(x, y)),')2(J(x, y)) - al, x, y, q5l(x, y), 02(x, y)
Similarly, 02 (x, y) ---> a2, x, y, 01(x, y), 02 (x, y), and so we have property DR2. coy
§5 Strongly connected systems of type 1* Theorem 5.1 Suppose S is strongly connected of type 1* and that E 'ca
contains an exact function of two arguments. Then S has the double Myhill
Chapter 15. Synchronization and pairing functions
298
properties DM and DM'.
Proof Assume hypothesis. Then by (b) of Corollary 3.2,E contains a diagonalizer of two arguments. Then by (a) of Theorem 4.3, Chapter 14, S has property DM. Then by (b) of Theorem 4.4, Chapter 14, S also has property DM'.
Theorem 5.2 If S is of type 1* then the properties DM and DM' for S are equivalent.
Proof We already know that DM implies DM'. Now suppose S is of type 1* and has property DM'. Then, taking gl(x) = K(x) and g2(x) = L(x), there are E-functions 01(x),02(x) such that for all x: 01(x) -> K(x),01(x),02(x) and 02(x) --> L(x),b1(x),02(x). Then for all x and y: V)1(J(x, y)) --> x, 01(J(x, y)), 02(J(x, y)) and V)2(J(x, y)) -->
y, /1(J(x, y)), 02(J(x, y)), and so we take 01(x, y) = 01(J(x, y)) and 02(x, y) _ 0(J(x, y)) and we have property DM. By (2) of Theorem 3.1 and by the results of Chapter 14 we have:
Theorem 5.3 Suppose S is strongly connected of type 1* and that E contains an exact function of two arguments. Then for every n, E contains
a nice function of n + 3 arguments and a symmetric function of n + 2 arguments (and hence S has properties DMn and DMn).
Exercise 4 Suppose S is strongly connected and of type 1*. (a) Show that if S has property DM3 then for every n > 3, S has property DMn.
-I,
(b) Show that if E contains a symmetric function of three arguments, then for every n > 3, E contains a symmetric function of n arguments. C!1
(c) Show that if E contains a nice function of four arguments, then for every n > 4, E contains a nice function of n arguments.
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299
Part III Synchronized recursion resumed §6 The property DR* In what follows, we shall let v be the 2-synchronizer (K, L), i.e., the domain of a is {1, 2} and vl(x) = K(x), 0'2 (X) = L(x).
We shall say that S has property DR* if for any elements al, a2 there is a E-function O(x) such that for all x: (1) K(O(x)) -' al, x, O(x) (2) L(O(x)) -' a2, x, O(x)
This can be equivalently stated that for any element a there is a E-function O(x) such that for all x:
(1) K(O(x)) - K(a), x, O(x) (2) L(0(x)) -> L(a), x, 0(x) Or, more concisely, for any element a there is a E-function O(x) such that for all x: (x) o a, x, 0(x)
Thus S has property DR* if and only if So has the recursion property. The original form of the double recursion theorem[28] was that for R the indexed relational system of recursion theory, for any n > 0, the system
Sn(R) has property DR* - which means that for any r.e. relations M1(x, y, z1, ... , zn) and M2(x, y, zl,... , zn) there is a recursive function
O(x) such that/ for all x, z1, ... , zn: (1)
(2) RL(O(x))(z1, ... , zn) H M2(x, c(x), z1i ... , zn)
We have the following generalization for sequential systems:
Theorem 6.1 If S is integrated of type 1* and if E contains an exact function E(x, y), then S has property DR*.
Proof By Theorem A*. Theorem 6.1 is also a consequence of the following apparently stronger result, which we will prove from scratch.
Chapter 15. Synchronization and pairing functions
300
Theorem 6.1# Suppose S is integrated of type 1* and that E contains functions di(x) and d2(x) such that for all x:
(1) di(x) -> K(x), y (2) d2(x) --> L(x), y
Then S has property DR*.
Proof Assume the given conditions. Let d(x) = J(dix, d2x). Then d(x) is a E-function and Kdx = dlx and Ldx = d2x. Thus K(d(x)) --> K(x), x and L(d(x)) --> L(x), x.
Next, given elements al and a2, since S is integrated, there are Efunctions t1(x),t2(x) such that for all x and y: (1) t1(x), y --> al, x, d(y)
(2) t2 (X), y - a2i x, d(y)
Next, we let t(x) = J(tl(x),t2(x)), and so Ktx = tlx and Ltx = t2x. Then for any x:
(1)' Kdtx - Ktx, tx = tlx, tx --> al, x, dtx (2)' Ldtx --> Ltx, tx = t2x, tx --> a2i x, dtx
And so we take O(x) = d(t(x)), and we have KOx
al, x, Ox and
Lax -> a2i x, Ox.1
Of course, the hypothesis of Theorem 6.1 implies the hypothesis of Theorem 6.1# (just take di(x) = E(K(x),x) and d2(x) = E(L(x),x)), and so Theorem 6.1 is indeed a consequence of Theorem 6.1#. Theorem 6.1 is also a consequence of Theorem 4.1 and the following basic result:
Theorem 6.2 If S is of type 1*, the properties DR2,DR1,DR0, and DR* for S are all equivalent.
Proof Suppose S is of type 1*. By Theorem 4.3 the properties DR2, G0.
DR1iDR0 for S are equivalent. We now show that DRo for S is equivalent to DR* for S. That DR* for S implies DRo for S follows from (2) of Theorem B (taking c for II). For the converse, suppose S has property DRo. Then 1For recursion theory, this is essentially the original proof of the double recursion theorem.
§7 The property DM*
301
by Theorem 3.3, Chapter 14, taking gl (x, y) = 92 (x, y) = J(x, y), for any elements al, a2 there are E-functions 01(X), 02(X) such that for all x: 01(x) -' a,, x, J(01 (X), 02(x)) and 02(x) - a2, x, J(01 (X), 02(x)). We let O(x) = J(01(x), 02(x)), and so K(O(x)) - a1, x, q(x) and L(q(x)) --> a2, x, 0(x), and so S has property DR*. Discussion
We now see that Theorem 6.1 can be alternatively derived as a consequence of Theorem 4.1, by virtue of Theorem 6.2. And so a synchronized approach (¢D
is not necessary for a proof of Theorem 6.1. But it is just as true that Theorem 4.1 can be derived as a corollary of Theorem 6.1 - again by virtue of Theorem 6.2. Thus Theorem 4.1 can be proved without appeal to any results of Chapter 14, but instead, by a synchronous fixed point argument.
Next, suppose S is of types 1* and 2* and that E contains a quasi-exact
function of two arguments. Then by (1) of Theorem 3.1, S contains a quasi-exact function of three arguments, and since S is of type 2, it follows from (3) of Theorem A* that S has property DR*. And so we have:
Theorem 6.3 If S is of type 1* and 2* and if E contains a quasi-exact function of two arguments, then S has property DR*.
cad
car
Alternatively, Theorem 6.3 can be proved as follows. By Theorem 4.2 the hypothesis implies that S has the double recursion properties DR2iDR1, and DRo, and so S then has property DR* by Theorem 6.2. Let us note that, alternatively, Theorems 6.3 and 6.2 yield another proof of Theorem 4.2. (_y
Exercise 5 Let DW be the weak double fixed point property and DW* the property that SQ has the weak fixed point property (S is assumed to be of type 1*). Prove that DW and DW* are equivalent. coy
§7 The property DM* ((DD
We shall say that S has property DM* if SQ has the Myhill property, i.e., if
there is a E-function q(x) such that for all x: K(q(x)) -4 K(x), O(x) and L(a(x)) -> L(x), O(x). By Theorem A* we have:
Theorem 7.1 For S of type 1*, if either S is very strongly connected, or if S is strongly connected and E contains an exact function of two arguments,
Chapter 15. Synchronization and pairing functions
302
then S has property DM*. We next wish to prove:
Theorem 7.2 If S is strongly connected of type 1*, then the properties DM*, DM, and DM' for S are all equivalent.
Proof Suppose S is of type 1* and strongly connected. Then by Theorem 5.2, the properties DM and DM' for S are equivalent. Also DM* implies DM' by (3) of Theorem B. It remains to show that DM' implies DM*. And so suppose that S has property DM'. Then by Theorem 4.5, Chap-
ter 14 (taking gl(x) = K(x),g2(x) = L(x), hl(x, y) = J(x, y), h2(x, y) _ J(x, y)) there are E-functions 01(x), 02 (x) such that for all x: 01(x) K(x), J(01(x), 02(x)) and 02(x) --' L(x), J(01(x), 02(x)). We take q(x) _ coo
J(01 (X), 02(x)) and we then have K(O(x)) --> K(x), O(x) and L(O(x)) --> L(x), q(x), and so S has property DM*.
§8 Integrated systems of type I* Lemma 8.0 Suppose S has property DM* (S, of course, is of type 1*). Then for any E-functions gl(x),92(x) there is a E-function 0(x) such that for all x: (1) K(O(x)) -' 91(x),O(x)
(2) L(0(x)) -' 92(x), 0(x)
cc-
a_0
Proof Assume hypothesis. Given E-functions gl(x),g2(x) we let g(x) = J(g1(x),g2(x)), and so K(g(x)) = gi(x) and L(g(x)) = g2(x). By property DM* there is a E-function O(x) such that K(O(x)) ---> K(x), fi(x) and L(&(x)) -- L(x), O(x). Then K(O(g(x))) -> K(g(x), O(g(x))) = 91(x),&(9(x)) and L(V)(9(x))) - L(9(x),V)(9(x))) = 92(x),"(9(x)). And (DD
so we take O(x) = V)(g(x)).
Remark
Of course property DM* is a special case of the conclusion of the lemma (the case gl(x) = K(x) and g2(x) = L(x)). And so DM* is equivalent to this seemingly stronger property. AO-.
Theorem 8.1 If S is integrated of type 1* and if S has property DM* then S has the double recursion properties DR2iDR1,DR0, and DR*.
§8 Integrated systems of type I*
303
Proof Suppose S is integrated of type 1* and has property DM*. Since S is integrated, then for any elements a1 and a2 there are E-functions gl (x), 92 (x) such that for all x and y: gl (x), y ---> al, x, y and 92 (x), y --> a2, x, y. Since S has property DM* then by Lemma 8.0 there is a E-function O(x) such that K(O(x)) -- g1 (x), q(x) and L(O(x)) --> 92 (X), O(x), and so
K(O(x)) - > a1, x, 0(x) and L(0(x)) -- a2i x, 0(x), and so S has property DR*. The rest follows by Theorem 6.2 From Theorems 7.1 and 8.1 we have:
Theorem 8.2 If S is integrated and very strongly connected and of type 1*, then S has the double Myhill properties DM, DM', DM* and the double recursion properties DR2iDR1,DRo, and DR*.
By Proposition 7.1, Chapter 13, if S is universal and integrated it is also very strongly connected (in fact hyperconnected), hence by Theorem 8.2 we have:
Theorem 8.3 If S is universal, integrated, and of type 1*, then S has properties DM, DM', DM*, and DR2iDR1,DRO, and DR*.
Chapter 16
Some further relations between fixed point properties Part I Single and double fixed point properties compared We shall now establish some interconnections between single and double fixed point properties. First we will show that if S has the recursion
property, then S also has the weak double fixed point property. [This generalizes results that are known for recursion theory and for combinatory logic.] Suppose S has the recursion property; does it necessarily have the double recursion properties DR2, DR1, and DRo? This appears
doubtful, but we will show that if S is of type 1*, then the answer is yes. Our proof will be divided into two parts as follows. For m > 1, we let R,1,L be the 1-m recursion property, i.e., the property that for every element a there is a E-function 0(x1,. .. , x,,,,) such that 0(xl,... , x,,,,) --> a, X1, ... , x,,,,, 0(1i ... , x,,,,) (m > 1). [Thus R1 is the recursion property.] Well, we will show that for S of type 1 (not necessarily of type 1*), R2 implies the double recursion property DR1 (it can also be shown that R3 implies DR2), and then (the easy part) we will show that for S of type 1*, R1 implies RZ (in fact R1 then implies Rn for any n > 1). We will also discuss some related results on Myhill and double Myhill properties.
§1 Recursion and double fixed points We first need two lemmas.
§1 Recursion and double fixed points
305
Lemma 1.0 If S has the recursion property, it also has the weak fixed point property.
'.O
Proof Suppose S has the recursion property. Given a, take a' such that a', x, y -> a, y. Then there is a E-function 0(x) such that for all x: 0(x) --> a', x, 0(x), hence 0(x) -> a, 0(x). Pick any particular x (say a) and let b=O(x). Then b ----> a, b.
Lemma 1.1 If S has the weak fixed point property then for any element a and any E-function g(x), there is an element b such that b --> a, b, g(b).
Proof Let a' be such that for all x: a', x --> a, x, g(x). Then let b be such that b --> a', b. Then b -- a, b, g(b). Remark
Actually this lemma is immediate from Theorem 1.5, Chapter 12, taking k = 2, fl the identity function, and f2 = g. Now we can prove:
Theorem 1.2 If S has the recursion property then S has the weak double fixed point property.
Proof Suppose S has the recursion preoprty. Take any elements al and a2. Since S has the recursion property then there is a E-function O(x) such that for all x: (1) O(x) -+ a2, x, q(x)
Next, since S has the recursion property, it has the weak fixed point property (by Lemma 1.0) and then by Lemma 1.1 there is an element c such that: (2) c -> al, c, O(c)
Then by (1), taking c for x, we have:
(1)' 0(c) - a2, c, 0(c)
We thus take bl = c and b2 = ct(c), and by (2) and (1)' we have bl --> al, bl, b2 and b2 -> a2, b1, b2, and so S has the weak double fixed point property.
Chapter 16. Some further relations between fixed point properties
306
Corollary 1.3 If S is of type 1 and E contains a normalizer N(x) of one argument, then S has the weak double fixed point property.
Proof This follows from the above theorem and Theorem 1.2 of Chapter 13.
Remark
This solves Exercise 5, Chapter 14. cap
§2 Recursion and double recursion '.Y
We now wish to prove that for S of type 1, R2 DR1. The following lemma plays the analogous role for this proof that Lemma 1.1 played in the proof of Theorem 1.2.
Lemma 2.1 Suppose that S has property R. Then for any element a and any E-function b(x, y) there is a E-function t(x, y) such that for all x,
t(x, y) - - > a, x, t(x, y), 0(y, t(x, y))
Proof Suppose that S has property R. Then given an element a and a E-function 0(x, y) there is an element a' such that for all x, y, and z: a', x, y, z -> a, x, z,b(y,z) By R2 there is a E-function t(x, y) such that for all x, y. t(x, y)
->
a', x, y, t(x, y)
->
a, x, t(x, y), V) (y, t(x, y))
oho
s0.
Theorem 2.2 Suppose S is of type 1 and has property R. Then S has `rd
property DR1.
Proof Assume hypothesis. Take any elements al and a2. By R2 there is a E-function b(x, y) such that for all y, z: (1) 0(y, z) --> a2, y, z,O(y, z)
By Lemma 2.1 there is a E-function 01(x, y) such that for all x, y
(2) 01(x,y) - a,,x,01(x,y),'b(y,01(x,y))
§3 Systems of type 1*
307
In (1) we replace z by 01(x,y) and we have (1) 1 0(y, 01(x, y)) - a2, y, 01(x, y), 0(y, 01(x, y))
We thus take 02(x, y) = L(y, 01(x, y)) and we have
01(x,y) - a1,x,01(x,y),02(x,y) 02 (x, y)
a2, y, 01(x, y), 02 (x, y)
Thus S has the double recursion property DR1.
Exercise 1 Prove that if S is of type 1 and has the property R3, then S has property DR2.
§3 Systems of type 1* Now we need:
Lemma 3.1 If S is of type 1* and has the recursion property R1, then for every n > 1, S has property R.
cad
Proof Assume hypothesis. We are given R1. Now let n be any number > 2 and let a be any element. Let a' be such that for all x and y: a', x, y -+ a, Ki (x), ... , Kn (x), y. Then take a E-function O(x) such that for all x, 0(x) --> a', x, 0(x) and take 0(x1, ... , xn) = 0 (Jn(x1i ... , xn)), and then O(xl,...,x,n,)
--i
a,Jn(xl,...,Xn,),O(x1,...,xn) a,x1i...,xn,Y'(x1,...,xn)
ace
Suppose now that S is of type 1* and has the recursion property. Then by the above lemma it also has property R2, hence by Theorem 2.2 it has property DR1. From this and Theorem 6.2, Chapter 15 we have:
Theorem 3.1 If S is of type 1* and has the recursion property, then S has the double recursion properties DR2, DR1, DRo, DR*.
By the above theorem and Theorem 1.2, Chapter 13 we have:
Theorem 3.2 If S is of type 1* and if E contains a normalizer N(x) of one argument then S has the double recursion properties DR2iDR1,DRo,DR*.
308
Chapter 16. Some further relations between fixed point properties
mom.
mo=d
We proved (Theorem 4.1, Chapter 15) that if S is integrated of type 1* and E contains an exact function E(x, y) then S has the double recursion properties. But now from Theorem 3.2 above and Proposition 1.1 of Chapter 13, we have the following stronger result (which we announced in Chapter 15 right after the proof of Theorem 4.1):
Theorem 3.3 If S is integrated of type 1* and E contains a diagonal function D(x) of one argument, then S has the double recursion properties.
mom"
§4 Myhill and double Myhill properties We shall let Mk' (k > 0) be the 1-k Myhill property. Thus Mo is the Myhill
property (there is a E-function q(x) such that for all x: O(x) --> x, 0(x)) and M1 is the property that there is a E-function O(x, y) such that for all x and y: O(x, y) - x, y, O(x, y). We will now show that if S is strongly connected of type 1, then if S has property M1, it also has the double Myhill property DM. We will also show that for a very strongly connected system of type 1*, the Myhill property implies property M11. But we showed in
Chapter 13 (Theorem 6.1) that a very strongly connected system of type 1 has the Myhill property. Putting all this together, it will follow that a very strongly connected system of type 1* has the double Myhill property! We first need the following analogue of Lemma 2.1:
Lemma 4.1 Suppose S is strongly connected of type 1 and has the property M11. Then for any E-function '(x, y) there is a E-function t(x, y) such that for all x and y: t(x, y) - x, t(x, y), 0(y, t(x, y))
Proof Assume hypothesis. By strong connectivity, given a E-function O(x, y), there is a E-function g(x) such that for all x, y, z: g(x), y, z --> x, z, 0(y, z). Then by M1 there is a E-function O(x, y) such that O(x, y) -> x, y, O(x, y), therefore O(g(x), y) --a g(x), y, q5(g(x), y) --> x, O(g(x), y),
0(y, q(g(x), y)). We thus take t(x, y) = q(g(x), y) and we have t(x, y) --> x, t(x, y), 0(y, t(x, y))
Theorem 4.2 If S is strongly connected of type 1 and has property Mil , then S also has the double Myhill property DM. Proof Like that of Theorem 2.2, deleteing "al" and "a2" and using Lemma 4.1 in place of Lemma 2.1.
§5 Bar fixed points
309
Exercise 2 Fill in the details of the above proof.
Theorem 4.3 If S is very strongly connected of type 1* then S has property Mil .
Proof Assume the hypothesis. By very strong connectivity, there is a E-function t(x) such that for all x and y: (1) t(x), y --> K(x), L(x), y
Also, by Theorem 6.1, Chapter 13, S has the Myhill property, hence
by Theorem 4.4, Chapter 13, there is a E-function b(x) such that for all x, O(x) --> t(x), &(x), but also t(x), O(x) --> K(x), L(x), /(x) and so O(x) --> K(x), L(x), b(x). Taking J(x, y) for x, we have '/ (J(x, y))
->
KJ(x, y), LJ(x, y), V (J(x, y)
=
x,y,i(J(x,y))
We thus take O(x, y) = V(J(x, y)), and we have O(x, y) -> x, y, O(x, Y). By Theorems 4.2, 4.3 above we have:
Theorem 4.5 If S is very strongly connected of type 1* then S has the double Myhill property.
Part II More on systems of type 1* §5 Bar fixed points In what follows, S will be assumed to be of type 1* and, furthermore, for III
VIII
all x and y, J(K(x), L(x)) = x. As in Chapter 10, for any x we define x = J(L(x), K(x)). Thus J(x, y) = J(y, x); K(x) = L(x); L(x) = K(x) and x = x. [We have called x the involute of x.] For any natural number n we will say that S has property B,,, if there is a E-function O(x, 0) of n + 1 arguments such that for every x and every n-tuple 9: O(x, 0) --> K(x), 0, O(x, 9), O(x, 0)
Exercise 3 Suppose S is of type 1*. (a) Show that if S has property Bo then S has the weak double fixed point property.
310
Chapter 16. Some further relations between fixed point properties inn
(b) Show that if S has property B1 and S is also of type 2, then S has the double recursion properties.
(c) Show that if S has property B1 then for every n > 1, S has property
B.
(d) Show that if E contains an exact function of two arguments then for
every n, S has property B. The following theorem (together with earlier theorems) yield solutions of (a) and (b) of Exercise 3.
Theorem 5.1 Suppose S is of type 1*. Then for any natural number n the following two conditions are equivalent: r-4
(1) S has property B. (2) S contains a symmetric function S(x, y, 6) of n + 2 arguments.
Proof Suppose S is of type 1*. (1) Suppose there is a E-function O(x, 0) of n + 1 arguments such that O(x, 0) --> K(x), 0, O(x, 0), O(x, 0). Let S(x, y, 0) = O(J(x, y), 6). Then S(x, y, 0) = O(J(x, y), 0) -> K(J(x, y)), 0, O(J(x, y), 0), q(J(x, y), 0) = x, 0, cb(J(x, y), 0), O(J(x, y), 0) = x, 0, ct(J(x, y), 0), O(J(y, x), 0) = x, 0, S(x, y, 0), S(y, x, 0). Thus S(x, y, 0) is a symmetric function.
(2) Conversely, suppose E contains a symmetric function S(x, y, 0) of n + 2 arguments. Let O(x, 6) = S(K(x), L(x), 6). Then O(x, 0) = S(K(x), L(x), 6) --> K(x), 6, S(K(x), L(x), 0), S(L(x), K(x), 0) = K(x), 0, S(K(x), L(x), 6), S(K(x), L(x), 6) = K(x), 0, O(x, 0), O(x, 6).
Thus S has property B.
Corollary 5.2 Suppose S is strongly connected of type 1*. (a) If S has property Bo then S has the double Myhill properties DM, DM', DM*.
(b) If S has property B2 then S has property DM1 (and hence also the double recursion properties, if S is also of type 2).
(c) If E contains an exact function of two arguments, then for every positive n, S has property B.
311
§ 6 Further topics
Exercise 4 Suppose S is of type 1*. Show that if E contains a symmetric function of three arguments then for every n > 3, E contains a symmetric function of n arguments. How does this solve (c) of Exercise 3? a-?
'.O
Exercise 5 Suppose that S is universal of type 1* and that for every element a, E contains a function O(x) such that for all x: O(x) --> a, x, O(x), O(T). Show that E contains a symmetric function S(x, y) of two arguments.
§6 Further topics 1 The extended recursion theorem revisited
We will say that S has property EM - the extended Myhill property - if for any E-functions fl (x), ... , fr(x), 91(4 ... , gk (x) there is a E-function O(x) such that for all x:
Y'(x) - fi(x),.. . , fr(x), 0(91(x)), ... , Y'(9k(x)) We showed in Chapter 13 (Theorem 3.3) that if S has the extended recursion property then for any a and any E-functions f, (x), ... , fr(x), gi (x), ... , gk (x) there is a E-function O(x) such that for all x: O(x) ---> a, fi (x), . . . , f, (x), 0(gi (x)), ... , 0(gk(x)), from which it obviously follows that if S has the extended recursion property and if S is universal, then (by suitable choice of a) S has the extended Myhill property. The extended Myhill property is of interest in that if S is of type 1* and has the extended Myhill property, then it not only has the double Myhill
properties - and in fact properties DMn for all n - but also E then contains nice functions and symmetric functions of any desired number of arguments (and all this without assuming strong connectivity). More generally (and we will soon see why this is more general) we have the following useful result.
Theorem EM Suppose S is of type 1* and has the extended Myhill property. Then for any n > 2 and any E-functions fi (x1, ... , x ) , .. , xn) there is a E-function O(x) fr(xl, , xn), 91(xl, , xn), .. , 9k (x1, .
,
such that for all x1 ,- .. , xn: O(Jn(x1, ... , xn)) -4
fi(xi,.... xn),...,fr(xl,...,xn),0(91(xl,...,xn))...... O(9k(x1, ... , xn))
312
Chapter 16. Some further relations between fixed point properties Before proving Theorem EM, we note the following._Let S be of type
1*.
For any function f (xl, ... , xn), n > 2, we let f (x) = f (Ki (x),
.... Kn (x)). [The function f is sometimes called the collapse' of f.] Then f (J(xl, ... , xn)) = f (xl, ... , xn). Of course for any E-function f of two or more arguments, its collapse f is also a E-function. Proof of Theorem EM `ti
Assume hypothesis. Then given the functions fi, ... , fr, 91, ... ,9k, by property EM there is a E-function O(x) such that for all x:
0(x)-
fl (X), ... , f, (4 0(91(x)), ... , 0(9k(x))
Replacing x by J(xl,... , xn) we get our desired conclusion.
..d
Now, let EM* be the property given by the conclusion of Theorem EM. Then the existence of nice functions and symmetric functions are not only consequences of property EM*, but are special cases of it, as we can now see:
(a) A special case of EM* is that for every n > 0 there is a E-function O(x) such that for all x and y and every n-tuple 0: c;,
0(Jn+2(x, y, e))
x, 9, 0(Jn+2(x, y, 0)), 0(Jn+2(y, x, e))
Then 0(Jn+2(x, y, B)) is a symmetric function. (b) Another special case of EM* is that for any n > 0 there is a E-function O(x) such that for every x, y, z, and every n-tuple 0: 0(Jn+3(z, x, y, B))
-4 z,0,0(Jn+3(x,x,y,e)),O(Jn+3(y,x,y,e)) Then cb(Jn+3(z, x, y, B)) is a nice function. coo
Still more generally, property EM* implies that E contains grand functions and multiple symmetric functions of all types, and hence also that for any n > 1, k > 0, S has the n-k Myhill property.
Exercise 6 Prove the above assertion.
§ 6 Further topics
313
2 Ideal functions Call a function b(x, 6) of k+1 arguments ideal (k > 0) if for any E-functions f, (x, 0), ... , fn (x, 6) there is an element b such that for every k-tuple 0:
0(b, 0) - fl(b, B), ... , fk(b, 0)
Call the system S ideal if for every k > 0, E contains an ideal function of k + 1 arguments.
Exercise 7 (a) Show that if S is universal then every quasi-diagonalizer is an ideal function.
(b) Show that if S is of type 2, then every ideal function is a quasidiagonalizer.
(c) Suppose that S is ideal. Show that E contains grand functions and multiple symmetric functions of any numbers of arguments and that S has the extended Myhill property. 3 Another type of synchronization
..S
H.0
There is another type of synchronization that is applicable to doubly indexed relational systems in the sense of Chapter 10. Define S,(R) (n > 1) to be the triple (N, E, ), where x, 6 n y, I' is defined to mean that for all z1, ... , xn, A.(0, zl, ... , zn) - Ay (I', zi, ... , zn) and also that Bx(0, z1, ... , zn) By J', zl, ... , zn). It is easily verified that for any n > 0, S2(R) is a unified sequential system (assuming that the
n
double indexing satisfies all the given conditions of Chapter 10) and hence
0-r,
that S2(R) has the recursion property. In particular, S2(R) then has the recursion property, which means that for any element a, there is a E-function O(x) such that for all x and is x E ao(i) E-4 Aa(x, i, 0(i)) and x E 00(i) - Ba(x, i, 0(i)). The special case of this for recursion theory is Lemma A of Chapter 11. We remark that this lemma is also an easy consequence of the fact that S1(R) has property DR*.
IV
Combinators and Sequential Systems
Chapter 17
Some special fixed point properties of combinatory logic Although many fixed point theorems of combinatory logic are special cases of fixed point theorems of sequential systems, others are not - in fact they are not even statable in terms of sequential systems. Applicative systems have very special features which we will look at in this and the next two chapters. In our final chapter, we will consider some other fixed point theorems for combinatory logic that do have generalizations to sequential systems.
In Part I of this chapter we give an introduction to combinatory logic by means of a series of problems concerning various relations between com-
binators - how some can be derived from others. Solutions are given at the end of the chapter. In Parts II and III we turn to various fixed point theorems that are special to applicative systems.
Part I Interdefinability of some 11.2
combinators Preliminaries
We recall that an applicative system A consists of a set N together with an operation called application that assigns to each ordered pair (x, y) of elements of N an element of N denoted xy. We also recall that parentheses are to be associated to the left, e.g., xyz is (xy)z; xyzw is ((xy)z)w, and so forth.
Given a set a1,... , a,,, of elements of N, an element b is said to be derivable from the set - or from the elements of the set - if it belongs
Chapter 17. Fixed point properties of combinatory logic
316
to every class C of elements such that C contains each of the elements It-
a1, ... , an, and such that C is closed under application, i.e., for any elements x, y in C, the element xy is in C. Thus the following facts hold:
(1) Each of the elements a1,... , a,,, is derivable from the set {al,... , a,,}.
(2) If x and y are derivable from the set {al, ... , ate,}, so is the element xy. (3)
No element is derivable from the set {a1, ... , an } unless its being so is a consequence of conditions (1) and (2). An element b is said to be derivable from an element a if b is derivable
from the unit set {a}.
§1 The combinator B We now consider an applicative system in which the set N of elements contains an element B satisfying the following condition:
Bxyz = x(yz)
..o
Of course the presence of B guarantees that the set E1 of all E-functions of one argument is closed under composition, for given any two such functions [a]1, [b]1, if we take c = Bab, then for all x, ex = Babx = a(bx), and so [c]1(x) = [a]1([b]1(x))
Problem 1 From B, derive elements D, B2, E, B3, D1, D2, D3, f satisfying the conditions below (for all elements x, y, z, w, v, etc.).
(a) Dxyzw = xy(zw) (b) B2xyzw = x(yzw) (c) Exyzwv = xy(zwv) (d) B3xyzwv = x(yzwv)
(e) Dlxyzwv = xyz(wv) (f) Dxxyzw = x(y(zw)) (g) D3xyzwv = x(yz)(wv) (h) Exyly2y3zlz2z3 = x(yly2y3)(zlz2z3)
§2 We add T
317
Problem 1.1 Using mathematical induction, prove the following generalization of (b) and (d) above. For every n > 1 there is a combinator Bn derivable from B satisfying the following condition (for all elements
X,y1,...,yn,z): Bnxy1 ... ynz = x(y1 ... ynz)
[For example, one can derive an element B4 satisfying the condition: B4xyly2y3y4z =
§2 We add T We continue to assume that N contains B. Now we assume further that N contains an element T satisfying the following condition: Txy = yx
The element T (unlike B) has a permuting effect. From B and T we can derive a host of combinators that parenthesize and permute.
Problem 2 (a) Show that from B and T we can derive an element R satisfying the condition:
Rxyz = yzx
(b) Show that from R (and hence from B and T) one can derive an element C satisfying
Cxyz = xzy (c) If one expresses C in terms of B and T via R, one obtains an expression of nine letters. It can be shortened by one letter. How?
(d) We have derived C from R. Also, if we start with C we can obtain R (that is, an element R satisfying Rxyz = yzx). How? Note
For any element x; Cx = RxR and Cx = B(Tx)R. car
Problem 2.1 Show that from B and C we can derive an element Q satisfying the relation
Qxyz = y(xz)
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318
Note
Q is more usually written B' and is a standard combinator. We will have a special use for it in fixed point problems.
Problem 2.2 It then follows that Q is derivable from B and T (since C is). Reducing C to B and T as in the solution to Problem 2(c), and expressing Q in terms of B and C, as in the solution to the last problem, we obtain an expression of nine letters. But this can be simplified to an expression of only six letters. How?
Problem 2.3 From Q alone, derive an element Qo satisfying the condition Qoxyzw = xz(yw)
Problem 2.4 It then follows that Qo is derivable from B and C, but it can be done with an expression of only four letters. How? Problem 2.5 From B and C - also from Q and C - derive elements Q1, Q2, Q3, Q4, Q5 satisfying the following conditions:
(a) Qlxyz = x(zy) (b) Q2xyz = y(zx)
(c) Q3xyz = z(xy) (d) Q4xyz = z(yx)
(e) Q5xyzw = z(xyw) (f) Q6xyzw = w(xyz) [We will be using Q3i Q5i and Q6.]
Problem 2.6 (a) We know that Q is derivable from B and C. Show that B is derivable from Q and C.
(b) It is also true that B can be derived from Q and T - moreover by an expression of only four letters. How?
(c) Since B is derivable from Q and T, and C is derivable from B and T, it follows that C is derivable from Q and T, but in fact it can be done with an expression of only four letters. How?
§2 WeaddT
319
Some terminology and discussion
An element A is called a proper combinator if there is some word
:t.
p..
W (X1, ... , xn) in variables xl,... , x,,, (but with no constants) such that the equation Ax1x2 ... xn = W(xl, ... , xn) holds. Any element derivable from proper combinators is called a combinator. All the combinators we have derived so far are proper. An example of a combinator that is not proper is TT, for TTx1 ... xn is x1Tx2 ... xn, which cannot be reduced further (and likewise TTx1 = x1T, which cannot be reduced further). A proper combinator A is called regular if it satisfies an equation of the form Axyl ... yn = xW1W2 ... Wk, where W1,. .. , Wk are words in the variables yl, ... , yn only. For example B and C are regular; T, R and Q are not. Order and bases
By the order of a proper combinator A is meant the smallest n such that there is a word W(xl,... , xn) in n variables such that the equation Ax1 ... xn = W(x1i ... , xn) is satisfied. As examples, T is of order 2; B and C are of order 3. For any combinators Xl,... , Xn, by {X1,. .. , Xn}+ we shall mean the class of all combinators derivable from X 1 ,.. . , Xn (they are all combinators, of course). We now see that the classes {B,T}+ and {Q,T}+ are the same, and also {B, C}+ = {Q, C}+. Also {B, C}+ C {B, T}+ (since C is derivable from B and T), but {B, C}+ # {B, T}+ - {B, C}+ is a strictly smaller class than {B,T}+ since it is known that T is not derivable from B and C. An important point to note about B and Q is that for any elements x and y, Bxy = Qyx. The reason is that we are taking CB as our definition of Q and so Qyx = CByx = Bxy. We shall have several occasions to make use of this fact. The combinator G
We shall have special use for a combinator G satisfying the following condition:
Gxyzw = xw(yz)
This combinator is derivable from B and T - in fact from B and C.
Problem 2.7 Derive G from B and C.
Chapter 17. Fixed point properties of combinatory logic
320
The combinator V
This combinator is used to construct pairing functions, and will play a key role in the next several chapters. It is defined by the condition:
Vxyz = zxy
Problem 2.8 Derive V from B and C. [It is easier to derive it from B, C, and T.]
Problem 2.9 Show that for each n > 2 there is a combinator Vn derivable from B and C satisfying the condition: Vnxy1 ... YnZ = XZY1 ... yn
Problem 2.10 Show that for each n > 2 there is a combinator Cn satisfying the condition: Cnxzy1 ... Y. = xy1 ... ynz
Problem 2.11 Show that from B and C one can derive a combinator F satisfying the following condition:
Fxyz = zyx
§3 We add M We now assume that in addition to B and T, we have a combinator M of order 1 satisfying the following condition:
Mx = xx This element M is sometimes called a duplicator and plays a basic role in fixed point constructions.
A basic standard combinator is W, defined by the condition: Wxy = xyy.
W can be derived from B, T, and M - in fact from B, C, and M. It is easier to first derive its converse W', defined by the condition:
W'xy = yxx
§3 We add M
321
Problem 3 (a) Derive W' from B, M, and R. Then express W' in terms of B, M, and C. Then express W' in terms of B, M, and T - which can be done in two ways, each involving only five letters.
(b) Then express W in terms of W' and C. (c) Now express W in terms of B, T, and M. This can be done with an expression of only ten letters, and there are two such expressions. Remarks (CD
Alonzo Church[3] was the first to show that W is derivable from B, T, M, and the identity combinator I. [Ix = x.] His method was both bizarre and ingenious - his expression involved twenty-four letters and thirteen
pairs of parentheses. Shortly after, J. Barkley Rosser found a ten letter expression, which moreover did not use the identity combinator I.
Problem 3.1 We have seen that W is derivable from B, T, and M. If we started instead with B, T, and W, could we derive M?
Problem 3.2 (a) Show that from B, T, and M, one can derive an element L satisfying the following condition:
Lxy = x(yy)
In fact L can be derived from B, C, and M, or alternatively from B, R, and M.
(b) Show that it is also true that L is derivable from B and W. Then express it in terms of Q, C, and W. (c) Also, L is derivable from M and Q. How? [This is a particularly neat derivation of L.] Remark
The combinator L (introduced in T.M.M.) plays a particularly significant role in the study of fixed points, as we will see.
Chapter 17. Fixed point properties of combinatory logic
322
Problem 3.3 (a) Show that from B and M one can derive a combinator M2 satisfying the following condition: M2xy = xy(xy)
(b) One can also derive such a combinator M2 from Q and W, and this fact will prove to be quite useful! What is the derivation? [It is tricky!]
Problem 3.4 Show that for any positive integer n, one can derive from B and W a combinator W,,, satisfying the condition: Wnxy1 ... 2Jnz = x7J1 ... Yn ZZ
Problem 3.5 We shall have good use for combinators P, P1, P2, and P3 satisfying the conditions:
(a) Pxyz = z(xyz) (b) Pixyz = y(xxz)
(c) P2xyz = x(yyz) (d) P3xyz = y(xzy)
They can be derived from B, T, and M - in fact from B, Q, and W (and hence from B, C, and W). How? The combinator S
0-0
rte
The classic combinator S is defined by the condition Sxyz = xz(yz). [We have already remarked that an applicative system is complete if and only if it contains S and the combinator K defined by Kxy = x.] There are several ways of deriving S from B, C, and W (and hence from B, T, and M). We shall consider several such derivations.
Problem 3.6 We will first derive the standard expression for S in terms of B, C, and W. We will do this in stages.
(a) From B, C, and W derive an element So satisfying the condition Soxyz = xzyz.
§3 We add M
323
(b) From So and B derive an element S' satisfying the condition S'xyzw = xywzw.
(c) Then construct S from B and S'. (d) Then reduce to B, C, and W.
Problem 3.7 The standard expression thus obtained has seven letters. In T.M.M. we found a more economical expression for S which has only six letters. It will be helpful to use the combinator G of Problem 2.7 (Gxyzw = xw(yz)). S can be easily derived from B, G, and W. How?
Problem 3.8 We know that S is derivable from B, C, and W. It is also true that W is derivable from B, C, and S - in fact from just C and S. How? [It helps to derive W' from S and R first.] Remark
We now see that the classes {B, C, W J+ and {B, C, S}+ are the same.
Problem 3.9 Since W is derivable from S and C, and C is derivable from B and T, then W is derivable from S, B, and T. Actually, W is derivable from just S and T - and quite economically. Also M is derivable from S
and T - in fact from W and T (a) Derive W from S and T. (b) Derive M from W and T. (c) Now write an expression for M in terms of S and T. Remark
We now see that the classes {B, T, W J+ and {B, T, S}+ are the same.
Some derivatives of B and S
From B and S can be derived the standard combinator (D satisfying the condition:
(Dxyzw = x(yw)(zw)
Then for each n > 2 one can derive from B and 4) a combinator (Dn satisfying the condition: (Dnxyzw1 ... wn = x(ywl ... wn)(zwl ... ton)
324
Chapter 17. Fixed point properties of combinatory logic
Problem 3.10 Derive (D, (D2i ... , (Dn,... (that is, show how for each n, (bn can be derived).
Problem 3.11 From B and S (and any other combinators already derived from them) derive combinators S1, S2, and S3 satisfying the following conditions:
(a) Sixyzw = xyw(zw) (b) S2xyzw = xzw(yzw) (c) S3xyzwv = xy(zv)(wv)
§4 Two special combinators We now turn to two combinators that will play a special role in the study of fixed point combinators. The first is the combinator 0 defined by the condition: Oxy = y(xy).
Problem 4 Derive 0 from Q and W. Problem 4.1 Show that P of Problem 3.5 can be derived from 0 and B.
Turing combinators
By a Turing combinator we shall mean a combinator U satisfying the condition
Uxy = y(xxy) Such a combinator U (discovered by Alan Turing[36]) allows for a fantastically simple construction of a fixed point combinator, as we will see.
It can be derived from B, Q, and W - in fact from L, Q, and W - in many different ways. It can also be derived from M, Q, and W, also from Q, M, and 0, and also from B, M, and 0, as we shall now see.
Problem 4.2 (a) Derive a Turing combinator from W1 and P1.
(b) Derive one from L and O.
(c) Derive one from P and W.
(d) Derive one from P and M.
§5 The combinator I
325
Problem 4.3 (a) Using (a) of the last problem, derive a Turing combinator from Q, L, and W, and reduce the expression to one in Q, B, and W. -4-=
(b) Using (b) of the last problem, derive a Turing combinator from L and 0 and then reduce to L, Q, and W (getting a different expression from that of (a)).
(c) Derive a Turing combinator from W, B, and 0 and then reduce to B, Q, and W (thus getting different expressions from those of (a)). (d) Derive a Turing combinator from Q, M, and 0, and then reduce to Q, M, and W.
§5 The combinator I This combinator (the "identity" combinator) is defined by the condition:
Let us now assume that I is present.
Problem 5 (a) Derive T from C and I. (b) Derive M from W and I. (c) Derive M from L and I.
(d) Derive M from S and I.
(e) Derive 0 from S and I.
(f) Derive M from 0 and I. (g) Derive 0 from P and I. (h) Derive M from P and I.
Chapter 17. Fixed point properties of combinatory logic
326
Bases
Let us say that a set {X1,.. X,,,} of combinators is a basis for the .
class {X1,..
.
,
,
X,,,}+ of all combinators derivable from X1,. .. , X,,,. Let
us call two bases equivalent if they are bases of the same class in other words, any combinator derivable from elements of the one is derivable from elements of the others. We now know that the bases
+'d
{B, T, M, I}, {B, C, W, I}, {B, C, S, I} are all equivalent (also the bases {B, T, I} and {B, C, I} are equivalent, since we have just seen that T is derivable from C and I). The combinators derivable from any of these three equivalent bases are known as combinators of the Al-calculus we might call them AI-combinators. Alonzo Church took {B, T, M, I} as the basis. Curry preferred the basis {B, C, W, I}. The class of AIcombinators is not combinatorially complete - it doesn't contain a combinator K satisfying the condition Kxy = x. The essential property of the AI-combinators is that for any word W (xl, ... , x7,,) which contains all the variables x1,. .. , x.,,,, there is a )I-combinator A such that the equation Ax, ... x,,, = W (xl, ... , x,,,) holds (e.g., there is a )I-combinator A such
that Axyz = (z(xxzy))(zx), but there is no )J-combinator A satisfying V.)
the condition Axyz = xz(zxx)z, since the variable y is missing in the expression on the right hand side of the equation). A proof of this partial completeness can be found in many books on combinatorial logic, or in
yeti
T.M.M. Another basis for the class of AI-combinators due to Rosser[21] is {I, J},
where J satisfies the condition Jxyzw = xy(xwz). For we can take T
to be JII, C to be JT(JT)(JT), B to be C(JIC)(JI), and M to be C(C(C(BJT)T)T)T.
§6 A basis equivalent to {B, T, I} The sub-class of the )J-combinators consisting of those derivable from just B, T, and I has been studied by Rosser and others. We showed in T.M.M. that these combinators B, T, and I could be replaced by the two combinators G and I. [We recall that Gxyzw = xw(yz).] Thus the bases {B, T, I} and {G, I} are equivalent.
Problem 6 Show that B and T are derivable from G and I. [Hint: first derive Q3 of Problem 2.5 from G and I, then C, then Q, then B, then T.]
§7 Some curiosities Problem 7 We say that x commutes with y if xy = yx.
§8 We now consider K
327
Find an element A derivable from B, T, and M that commutes with every element.
Problem 7.1 Is there an element X derivable from B, T, and M such that XXX = X (XX )?
Problem 7.2 Find an element A derivable from B, T, and M such that for all x, Ax = A(xx).
§8 We now consider K We recall that K is defined by the condition Kxy = x.
Problem 8 Show that B, T, M, and I are all derivable from S and K. Remark
We have stated that the class of combinators derivable from S and K is complete. In the next chapter we will give the standard proof of this.
Problem 8.1 (a) Prove the following cancellation law for K: if Kx = Ky then x = y. (b) Show that unless K is the only element of N, there is no element x such that Kx = K. (c) Assuming that K is not the only element of N, prove that there are infinitely many elements derivable from K. Some inequalities
Assuming that there is more than one element present, many inequalities can be derived - for example the following: K=A I
SCI S#K
CC =A C
VV #V W1 #1
T#K SSrS
RRrR QQrQ O0,
We leave the proofs of these inequalities to the reader'. Actually it 'They can be found in T.M.M., Chapter 22.
Chapter 17. Fixed point properties of combinatory logic
328
can be shown that no two of the elements B, C, W, S, R, T can be identical (assuming there is more than one element).
A pairing function and its inverses
For any elements x and y, we take the "representative" of the ordered pair (x, y) to be Vxy (V is a combinator satisfying the condition Vxyz = zxy. We are taking BCT for V). The element Vxy is sometimes written: "[x
.7]')
Problem 8.2 (a) Prove that the function [V]2 is 1-1, i.e., if Vxy = Vzw then x = z and y = w. sm.
(b) Find combinators K1 and K2 such that for all x and y, K1 (Vxy) = x and K2(Vxy) = y. Note
It is not the case that for all z there are elements x and y such that Vxy = z, and so we don't have the law: V(Klx)(K2x) = x. Klx is sometimes written (x)o and K2x, (x)1. Propositional combinators
We let t = K and f = KI. We might call t and f truth-combinators.
Problem 8.3 Find combinators neg (negation), c (conjunction), d (disjunction), i (material implication), and e (equivalence), all derivable from B, T, K, and I, such that for any elements p and q, each of which is either t or f, the following conditions hold: (a) negt = f and neg f =t (b) cpq = t iff p = t and q = t; otherwise cpq = f (c) dpq = t iff p = t or q = t; otherwise dpq = f
(d) ipq = t iff p = for q = t; otherwise ipq = f (e) epq = t if p = q; otherwise epq = f
§9 Idempotent and voracious elements
329
Problem 8.4 We call an applicative system extensional if for all elements a and b, if ax = bx for all x, then a = b.
Show that if A is extensional then neg can be taken to be a proper combinator. C09
§9 Idempotent and voracious elements We call an element X idempotent if XX = X. Of course I, is idempotent.
Problem 9 (a) Find an idempotent element that is derivable from B and M. (b) Find one derivable from L and M. (c) There is even one derivable from only L! Can you find one? We call an element X voracious if X y = X for every y.
Problem 9.1 171
(a) Find a voracious element that is derivable from B, K, and M.
(b) Find one that is derivable from K and L.
Part II Fixed points #A. The weak fixed point property §10 Some sufficient conditions We recall that in an applicative system, b is called a fixed point of a if ab = b.
Problem 10 (a) Show that if B and M are present, then every element has a fixed point.
(b) Show that if just L is present, then every element has a fixed point.
Chapter 17. Fixed point properties of combinatory logic
330
§11 Fixed points in relation to idempotent and voracious elements We shall now see how the solutions to Problems 9 and 9.1 were found.
Problem 11 (a) Show that any fixed point of M must be idempotent. This, together with the solution of Problem 10 yields solutions of (a) and (b) of Problem 9. How?
(b) Show that if x is a fixed point of LL, then xx must be idempotent. How is this related to (c) of Problem 9?
Problem 11.1 Prove that any fixed point of K must be voracious. This, together with the solutions of Problem 10, yield solutions to (a) and (b) of Problem 9.1. How?
§12 A fixed point principle We are considering an applicative system that is not necessarily combinatorially complete. We shall call a function f (xl,... , x,,,) from, N'z into N realizable if there is an element a such that axi ... x,,, = f (xi, ... , xn,) for all xi, ... , xn in N. We will say that such an element a realizes the function f. [What we call "realizable" is more commonly called "representable," but we prefer to use the latter term in a different context, which will be discussed in a later chapter.] A result that is commonly called the "fixed point theorem of combinatory logic," or sometimes the first fixed point theorem of combinatory logic," is the following: Tai
Theorem F If A is a complete applicative system, then for any expression W (x, yl,
. ,
yn,) in the variables x, yi, .
.
.
, yn
(and possibly constants
as well) there is an element a such that for all elements yi, ... , yn the cad
equation ayi ... yn = W (a, yl, ... , yn,) holds.
As an illustration of the use of this theorem, if we take the expression xlx(x3x2x)x, there is a combinator a such that for all elements xi, x2i x3: axix2x3 = xia(x3x2a)a. This basic result is a corollary of either of the following two theorems (which hold for a system not necessarily combinatorially complete).
§13 Some fixed point combinators
Theorem F1
331
If every element has a fixed point, then for every realizable
function f (x, y1, ... , yn) there is an element a such that for all Y1.... , yn,
ayl,...,yn = f(a,yl,...,yn). Theorem F2 For any function f (x, y1, ... , yn), if the function f (xx, Y1, , yn) is realizable, then there is an element a such that for all
y1,...,yn: ayl,...,yn=f(a,yl,...,yn). Theorems F1 and F2 provide different proofs of the fixed point theorem of combinatory logic.
Problem 12 Prove Theorems F1 and F2 and show how each one provides a proof of the fixed point theorem of combinatory logic.
#B. Some fixed point combinators We recall that we are defining a fixed point combinator as an element B such
that Ox = x(Ox), for all x. A fixed point combinator will also be called a sage2.
Here are some of the combinators of Part I (all derivable from B, M, and T - and all but M derivable from B, C, and W) with their defining characteristics, that will be useful in constructing sages.
Bxyz = x(yz) Rxyz = yzx Cxyz = xzy Qxyz = y(xz) Qoxyzw = xz(yw) Gxyzw = xw(yz) Lxy = x(yy)
Wxy = xyy M2xy = xy(xy) Sxyz = xw(yz) Oxy = y(xy) Pxyz = z(xyz) P1xyz = y(xxz) Uxy = y(xxy)
§13 Some fixed point combinators Problem 13 (a) Find a sage derivable from M, B, and R. (b) Now express one in terms of B, C, and M. (c) Find one derivable from M, B, and L. 21 introduced this term in T.M.M., but it has subsequently been used by others.
Chapter 17. Fixed point properties of combinatory logic
332
(d) Find one derivable from M, B, and W.
Problem 13.1 One of our favorite constructions of a sage is in terms of Q and M. How can this be done?
Let us define an assistant sage as a combinator A such that for every x, Axx is a fixed point of x.
Problem 13.2 (a) Show that if A is an assistant sage, then a sage is derivable from A and W. -ti
(b) Find an assistant sage that is derivable from Q and L. This with (a) yields a sage derivable from Q, L, and W. What sage is that? (c) One can not only derive an assistant sage from Q and L, but one can even derive one from Qo and L. How?
(d) There is also a combinator X derivable from Q and W such that XL
is a sage. What X is that? [This yields another proof that a sage can be derived from Q, L, and W.]
Problem 13.3 (a) Derive an assistant sage from G and L.
(b) Derive an assistant sage from B, C, and L.
Problem 13.4 Derive a sage from B, G, and W. Problem 13.5 Derive a sage from B, C, and W. Problem 13.6 Derive a sage from S and L. ova
Problem 13.7 Derive a sage from S, B, and W. [This is Curry's sage.]
§14 Turing sages
333
Problem 13.8 (a) Derive a sage from 0, B, and M. (b) Derive one from L and 0.
(c) Derive one from M and P. [Recall Pxyz = z(xyz).] (d) Derive one from P and W. (e) Derive one from P and I.
Problem 13.9 (a) There is a combinator X that we have derived from B, T, and M such that for any Y, if XYY = YY, then YY is a sage. What X is that? (b) Show that for any X with the above property, WX(WX) must be a sage.
§14 Turing sages We let U be a Turing combinator.
Problem 14 The marvelous thing about a Turing combinator U is that a sage can be derived from just U alone! How?
Problem 14.1 Prove that for any X, if UX = XX then XX must be a sage.
§15 Some special properties of 0 Problem 15 As a preliminary problem, show that if B is a sage and A is an element such that Ax = Ox for all x, then A must also be a sage. Remarks
ti'
L".
If course for an extensional system, the above problem is pointless, since the hypothesis would imply that A and B are the same element. But it is of interest that the solution of the above problem is valid without the hypothesis of extensionality.
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Problem 15.1 Prove that if 0 is a sage, so is 00. Problem 15.2 0 has the marvelous property that every fixed point of 0 is a sage! Why is this? .'3
Problem 15.3 If the system is extensional, then, conversely, every sage is a fixed point of 0, and thus the fixed points of 0 are precisely the fixed point combinators! Prove this.
Problem 15.4 Without the assumption of extensionality, we do not know whether or not all sages are fixed points of 0 (even assuming combinatory completeness). [This might be an interesting open problem.] However, Cs'
even without the hypothesis of extensionality, some of the sages that we have derived can be shown to be fixed points of 0. Which ones?
Problem 15.5 Sate whether the following is true or false: for any sage B, the element 90 is a sage.
Problem 15.6 By a sage producer let us mean a combinator A that for every element x, the element Ax is a sage.
Construct a sage producer from B, C, and M. [Hint: it can be constructed from P2, M2, and C.]
§16 Sages from B and W We have seen that a sage can be constructed from the one proper combiowe
nator U, but U, though proper, is not regular. In T.M.M. we left open the following two questions: (1) can a sage be constructed from just one regular combinator?; (2) if not, can a sage at least be constructed from B and just one other regular combinator? Richard Statman has answered the second question affirmatively (and, I believe, the first one negatively). He constructed a sage from B and W. Since then a host of B, W-sages have been discovered by Wos and McCune
at Argonne National Laboratories. We shall now turn to the study of Statman's sage constructed from just B and W.
Problem 16 (a) From B and W construct a combinator L1, satisfying the condition:
Lixyz = x(yyz)
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§17 n-sages
.ti
[In fact, L1 can be constructed from B and L and then reduced to an expression in B and W. Indeed, it is simpler this way.] (b) Show that Lix(Lix)(Lix) is a fixed point of x.
(c) From B and W construct a combinator M3 with the property: M3xy = xy(xy)(xy).
(d) Now construct a sage from M3 and L1 and reduce to B and W. Note 090
I also raised the problem of whether a sage can be constructed from B and M. According to Larry Wos, who is an expert on sage constructions, this particular problem has not been solved to this day (16 June, 1992), though he states that some progress may be in sight. Another problem I would like to suggest: is there an effective means of deciding which finite sets of combinators are sufficient to produce a sage?
§17 n-sages Looking at applicative systems from the viewpoint of sequential systems, to say that A has the recursion property is to say that for every element a there is an element b such that for all x, bx = ax(bx) ([b]1(x) -+ a,x, [b]1(x)). A
stronger property is that there is an element r such that for all x and y, rxy = xy(rxy). Such an element r we will call a 2-sage. Actually, the existence of a 2-sage is equivalent to the sequential system S(A) having the 1-1 Myhill property (the existence of a E-function O(x, y) such that for all x and y, q(x, y) -+ x, y, q(x, y)), hence every combinatorially complete system has a 2-sage. But we are interested in applicative systems that are not necessarily combinatorially complete.
Problem 17 (a) Suppose that every element has a fixed point and that the combinator S is present. [Sxyz = xz(yz).] Prove that there is a 2-sage. (b) Specifically, construct a 2-sage from S and L.
Problem 17.1 Show that a 2-sage can be constructed from B and W. [Tricky!]
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Problem 17.2 Suppose A is extensional. We know that there is then a combinator derivable from B, C, and W (namely, 0) whose fixed points are those and only those elements that are sages. Is there necessarily a combinator derivable from B, C, and W whose fixed points are those and only those elements that are 2-sages (assuming extensionality)?
By an n-sage we mean a combinator I' satisfying the condition: Pxl ... xn, = xl ... xn(rxl ... xn). Problem 17.3 Find two regular combinators whose presence is enough to guarantee the existence of n-sages for all n > 1.
Part III Multiple fixed points In preparation for the study of multiple fixed point properties, it will be handy to have some more proper combinators available for immediate use.
§18 Some more useful combinators Problem 18 From B, C, and W, derive combinators el, e2, S4, C1, C2, D1, D2 satisfying the following conditions.
(a) elxy = xyx (b) e2xy = yxy
(c) S4xyzwv = z(xwv)(ywv)
(d) Clzxy = z(yxy)(xyx) (e) C2zxy = z(xyx)(yxy)
(f) Dlxyzw = xz(yw)(xz) (g) Dlxyzw = xw(yz)(xw)
Double fixed points §19 The cross-point property We will say that A has the cross-point property if for any elements al and a2 there are elements bl and b2 such that albs = b2 and a2b2 = bl.
§20 The weak double fixed point property
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Problem 19 Show that if the combinators M and B are present then A has the cross-point property.
Problem 19.1 We have just, seen that the cross-point property can be obtained using two proper combinators (M and B). Actually, it can be obtained with only one proper combinator derivable from B, C, and W. What combinator works? cap
§20 The weak double fixed point property
a°;
E-+
The weak double fixed point property for an applicative system A is that for any elements a1 and a2, there are elements b1 and b2 such that a1b1b2 = b1 and a2b1b2 = b2. We will call a pair (b1, b2) of such elements a double fixed point of the pair (al, a2). For a complete applicative system, there are many interesting ways to establish the double fixed point property, as the following problems reveal. For applicative systems not necessarily complete, the different ways yield different results.
Problem 20 Show that if the combinators C1 and C2 of Problem 18 are present, then A has the weak double fixed point property.
Problem 20.1 Suppose that every element has a fixed point and that for any element a there are elements b1 and b2 such that for all x and y: bixy = a(xy)(yx) and b2xy = a(yx)(xy). Show that the system has the weak double fixed point property.
Problem 20.2 The following is a more interesting one: Show that if every element has a fixed point and if S is present, then A has the weak double fixed point property. [This of course implies that if S and L are present, then A has the weak double fixed point property.]
Problem 20.3 Show that if B and W are present then A has the weak double fixed point property.
Problem 20.4 Here is a way of deriving the weak double fixed point property using the pairing combinators V and its inverse combinators K1 and K2 (see Problem 8.2). Suppose that V, K1, and K2 are present and that every element has a fixed point and that the combinator S2 is present. [S2xyzw = xzw(yzw)
Chapter 17. Fixed point properties of combinatory logic
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- Problem 3.11.] Show that the system has the weak double fixed point property.
Problem 20.5 For those who like curiosities, suppose that there are elements al and a2 such that for all z, x, y, w: (1) alzxy = x(yy)(zzxy) (2) a2zwxy = w(yy)(zxy) Prove that A has the weak double fixed point property.
§21 The property DM o`)
The double Myhill property for an applicative system A is that there exists combinators M1 and M2 satisfying the conditions:
(1) Mlxy = x(Mlxy)(M2xy) (2) M2xy = y(Mixy)(M2xy) Such a pair (Ml, M2) will be called a DM-pair. We say that such a pair (M1, M2) is derivable from a given set of combinators if M1 and M2 are both derivable from the set. There are many different ways of deriving a DM-pair from B, C, and W. Let us consider some.
Problem 21 Derive a DM-pair from the combinators C1i C2, Dl, D2 of Problem 18.
Cdr
Problem 21.1 There exist two proper combinators A1, A2 derivable from Z.,"
B, C, and W, such that a DM-pair is derivable from Al and A2. Find such combinators Al and A2. [The ones we have found are of order four.]
---
We recall the combinator 0 satisfying the condition: Oxy = y(xy), and we know that every fixed point of 0 is a sage. This fact has a double analogue, namely that in a complete applicative system, there are combinators 01 and 02 such that every double fixed point of (01, 02) is a DM-pair. Indeed, 01 and 02 can be derived from B, C, and W. [This yields yet another method of obtaining a DM-pair from B, C, and W.]
Problem 21.2 Construct such a pair (01, 02) from B, C, and W.
§22 Nice combinators and symmetric combinators
339
§22 Nice combinators and symmetric combinators By a nice combinator we shall mean a combinator N satisfying the condition:
Nzxy = z(Nxxy)(Nyxy) By a symmetric combinator we shall mean a combinator s satisfying the condition: sxy = x(sxy)(syx)
There are several ways in which nice combinators and symmetric combinators can be derived from B, C, and W:
Problem 22 Find proper combinators n1 and n2 derivable from B, C, and W such that (a) A nice combinator is derivable from n2.
(b) All fixed points of n1 are nice combinators, and if the system is extensional, then all nice combinators are fixed points of n1.
Problem 22.1 Find proper combinators sl and s2 derivable from B, C, and W such that: (a) A symmetric combinator is derivable from s2.
(b) All fixed points of s1 are symmetric combinators, and if the system is extensional, then all symmetric combinators are fixed points of sj. Discussion
Nice combinators and symmetric combinators provide another approach to double fixed point properties. It is obvious that the presence of a nice combinator implies the weak double fixed point property, and so does the presence of a symmetric combinator together with the combinator C. Also, from a symmetric combinator s together with B and C, a DM-pair can be constructed (Problem 22.2 below), and a DM-pair can also be constructed from a nice combinator N together with C, W, and W1.
Problem 22.2 Prove the last statement.
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Discussion: Grand and multiple symmetric combinators
For any n > 2 there is a )I-combinator A (in fact one derivable from B, C, and W) satisfying the following condition:
Awzxl ... xn = z(wxlxl ... xn)(wx2xl ... xn) ... (wxnxl ... xn,) Any fixed point g of A satisfies the condition: gzxl...xn = z(9xlxl...xn)(9x2xl...xn)...(gxnxl...xn)
Such a combinator g we could call a "grand" combinator. [Thus a nice combinator is a grand combinator for n = 2.1
Of course the presence of g guarantees the weak n-fold fixed point property. But also, for each i < n, there is a combinator gi of the )Jcalculus (in fact one derivable from B, C, and W) satisfying the condition:
gill ... xn = gxnxl ... xn. And therefore we have the "uniform" n-fold fixed point property (the n-0 Myhill property) that for all i < n: 9ixl...xn = xi(91x1...xn)...(gnxl...in)
`°a
Of course for a combinatorially complete system (the AK-calculus) the existence of such combinators g, gl, ... , gn follows from earlier results on unified sequential systems, but the point of the above discussion was to indicate that all this could be done in the )J-calculus3 (in fact using just
B,C, and W). Also for each n > 2, from B, C, and W one can get a "multiple symmetric" combinator h satisfying the condition: hxl...1n = xl(hxl...xn)(hx2...xnxl)...(hxnxl...xn-1)
We leave this to the reader.
Solutions 1 (a) It is sometimes easiest to work these problems backward. We are
looking for an element D such that for all x, y, z, w: Dxyzw = (xy)(zw). Let us look at the expression (xy)(zw) and see how we can get back to Dxyzw, where D is the element to be found. Well, we look at the expression (xy) as a unit - call it A - and so (xy) (zw) is A(zw), which we recognize as BAzw, which is B(xy)zw. So the first step of this "backward" argument is to recognize (xy)(zw) as 3For the observation that my scheme works in the AI-calculus I am indebted to Henk Barendregt.
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341
B(xy)zw. Next, we look at the front end B(xy) of the expression and recognize it as BBxy, and so B(xy)zw = BBxyzw. Therefore we take D to be BB. Let us check by running the argument forward:
Dxyzw = BBxyzw, since D = BB B(xy)zw, since BBxy = B(xy) (xy)(zw) = xy(zw) (b) Since we have already found D from B, we are free to use it. In other words, in any solution for B2 in terms of B and D, we can replace D by BB, thus getting a solution in terms of B alone. Again we will work the problem backward. -ti
x(yzw) = x((yz)w) = Bx(yz)w, but Bx(yz) = DBxyz, and so Bx(yz)w = DBxyzw. And so we take B2 to be DB, which in terms of B is (BB)B (which we can also write BBB). So our solution is BBB. Let us check: DBxyzw = Bx(yz)w = x(yzw). In the remaining parts of this problem, we will just give solutions, leaving the checking to the reader.
(c) Take E = BB2 (= B(BBB))
(d) Take B3 = EB (= B(BBB)B) - also = BB2B (e) Take D1 = BD (= B(BB)) (f) Take D2 = D1B (= B(BB)B) (g) Take D3 = DD (= BB(BB))
(h) Take E = EE (= B(BBB)(B(BBB))) 1.1
Take B1 = B.
Now suppose that n is such that there
is
a combinator Bn satisfying the condition: Bnxy1 ... ynz = x(yl ... YnZ). Then take Bn+1 = BBnB. Then Bn+lxy1 ... ynyn+lz = BBn Bxy1 ... ynyn+lz = Bn(Bx)y1 ... ynyn+lz = Bx(y1 ... ynyn+l)z (because
. . ynyn+l = Bx(yi . . . ynyn+l) by inductive hypothesis), and Bx(y1i... , yn, yn+1)z = x(y1 ... ynyn+lz). Thus Bn+ixyi ... ynyn+lz
Bn(Bx)y1
.
= x(y1 ... ynyn, z). This completes the induction. And so B1 = B, B2 = BB1B2, B3 = BB2B, B4 = BB3B, and so forth.
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C'1
2 (a) Take R = BBT. [Check: BBTxyz = B(Tx)yz = Tx(yz) = yzx.] (b) Take C = RRR. [Check: RRRxyz = RxRyz = Ryxz = xzy.]
(c) Thus C = BBT(BBT)(BBT) = B(T(BBT))(BBT). (d) Given C, we can take R = CC (since CCxyz = Cyxz = yzx). 2.1
Take Q = CB. [Check: CBxyz = Byxz = y(xz).]
2.2 CB = B(T(BBT))(BBT)B = T(BBT)(BBTB) = T(BBT) (B(TB)) = B(TB)(BBT). 2.3 Take Qo = QQ(QQ). [Check: QQ(QQ)xyzw = QQ(Qx)yzw = Qx(Qy)zw = Qy(xz)w = xz(yw).]
2.4 Take Qo = BCD (= BC(BB)). [Check: BCDxyzw = C(Dx)yzw = Dxzyw = xz(yw).] We henceforth leave the checking to the reader.
2.5 (a) Take Q1 = BCB (alternatively Q(CQ)C).
(b) Take Q2 = CQ1 (= C(BCB)) - alternatively C(Q(CQ)C). (c) Take Q3 = BC(CB) - alternatively QQC. [In terms of B and T, we could simply take Q3 = BT.]
(d) Take Q4 = CQ3 (= C(BC(CB))) - alternatively C(QQC). (e) Take Q5 = BQ.
(f) Take Q6 = B(BC)Q5 (= B(BC)(BQ)).
2.6 (a) Take B = CQ. (b) Take B = QT(QQ). (c) Take C = QQ(QT).
2.7 Take G = BBC. 2.8 Take V = C(BB(CC))(CC) (alternatively BCT).
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2.9 We use the combinators Bn, of Problem 1.1 (B,,,xy1 ... y,,,z = x(y1 ... y,,z)). First we take V2 = B(BC)C. Then V3 = B3CV2; V4 = B4C
V3i...,Vn+1 = B.+iCV.. 2.10 Take C2 = BC(BC); C3 = BC(BC2), ... , Cn+1 = BC(BC,,) (n > 2).
2.11 TakeF=CV. 3 (a) We can take W' to be BMR, or BM(CC). In terms of B, M, and T, BMR is BM(BBT). We could also take Wto be B(BMB)T. (b) We take W = CW'.
(c) W = CW' = RRRW' = RW'R = BBTW'R = B(TW')R = If we take BM(BBT) for W' we get B(TW')(BBT). B(T(BM(BBT))(BBT). If instead we take B(BMB)T for W', we get B(T(B(BMB)T))(BBT), which is Rosser's expression for W in terms of B, T, and M. 3.1
Yes, it is even derivable from T and W. Take M = WT (thus WTx =
Txx = xx).
3.2 (a) We can take L = CBM, or RMB. We can also take L = B(TM)B. (b) We can take L = BWB. Then we can take CQ for B, and L becomes CQWB, which is QBW, which becomes Q(CQ)W.
(c) Take L = QM.
3.3 (a) Take M2 = BM. (b) Take M2 = Q(WQo)W - which is Q(W(QQ(QQ)))W. 3.4 Take W1 = BW; W2 = BW1i W3 = BW2,... , W.+1 = BWn. A simple induction shows that these combinators work.
3.5 (a) Take P = W3Q5 (= B(BW)(BQ)), or B(BWQ), or B(QQW).
(b) Take P1 = LQ - alternatively W(BQ). (c) Take P2 = CP1 (= C(LQ)).
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(d) Take P3 = BCP.
3.6 (a) Take So = B(BW)C (which is W2C). '+1
(b) Take S' = BSo which is B(B(BW)C).
(c) Take S = S'(BB). ,1.
(d) Then S becomes B(B(BW)C)(BB) (the standard expression). .may
3.7 We can take S = W2G, which is B(BW)G, which is B(BW)(BBC).
6U0
3.8 We can take W' to be SRR. Then we take W to be CW' = C(SRR). Taking CC for R, we get C(S(CC(CC))), which we can take for W.
3.9 (a) Take W = ST. (b) Take M = WT. (c) Thus we can take M to be STT. 3.10 Take D = B(BS)B. Then take '4 = BD31..... 4n+l = B4),,-D.
3.11 (a) Take S1 = BS. (b) Take S2 = DS. (c) Take S3 = B-D. 4 4.1
Take O = QQW. Take P = BO.
4.2 (a) W1P1
(b) LO (c) WP
(d) PM
'D2
B-D,P, 4)3 = BDA,
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Solutions
4.3 (a) W1P1 is BWP1, which is also QP1W, which is Q(LQ)W, which
is also Q(W(BQ))W. Also Q(W(BQ))W = BW(W(BQ)). Thus Q(LQ)W is a Turing combinator, and so are BW(W(BQ)) and Q(W(BQ))W.
(b) LO = L(QQW), so L(QQW) is a Turing combinator (also LO = .fir
L(BWQ)). And so Q(LQ)W and L(QQW) are two different expressions for a Turing combinator in terms of L, Q, and W.
(c) In WP, we can take BO for P, thus W(BO) is a Turing combinator. Replacing 0 by QQW, we get W(B(QQW)). Alternatively, we can take W(B(BWQ)), and so we have two other expressions for a Turing combinator in terms of B, Q, and W.
(d) PM becomes BOM (taking BO forP); alternatively we can take QMO, which in turn is QM(QQW). 5
6
(a) (b) (c) (d)
Take T = CI. Take M = WI. Take M = LI. Take M = SII.
(f)
Take O=SI. Take M=0I.
(g)
Take 0
(e)
(h)
P1.
Take M=PII.
First take Q3 to be GI. Then take C to be GGII. Then take Q =
G(CC)Q3 (which is G(CC)(GI)). Then, of course, we take B = CQ, then T = CI (or more simply GII).
7 M(BTM) is such an element X. Another is LT(LT). [A further discussion of this problem follows the solution of Problem 10.]
7.1 Obviously L is such an element.
7.2 LL(LL) is such an element. 8
Take I = SKK. Then take M = SII. Then take T = S(K(SI))K.
Then take B = S(KS)K.
8.1 (a) Suppose Kx = Ky. Then for any z, Kxz = Kyz. But Kxz = x
and Kyz=y, sox =y. (b) Suppose Kx = K. Then for any element y, Kxy = Ky, hence x = Ky
(since x = Kxy). Then for any elements yi and Y2, Ky1 = Ky2
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(since they are both equal to x), hence yl = Y2 (by (a) above). Thus any elements yl and y2 are the same, so N has only one element. (c) For each positive n, we define Kn, by the following inductive scheme:
K1 = K; K.+1 = KKn, (e.g., K4 = K(K(KK))). We show that for n 0 m, Kn, 0 Km, and hence the set {K1i K2, ... , K,,,, ...} contains infinitely many distinct elements.
It follows from the cancellation law (a) that if K,,,,+i = Kn,+1, then
Km = Kn, (because Kn,,+1 = K,,,+1 implies KKm = KKn,). By repeated applications of this, it follows that if K, = Km+k then K1 = K1+k (e.g., if K3 = K5, then K2 = K4, hence K1 = K3), but K1 = Kl+k means that K = K(Kk), which is not possible unless K is the only element of N. And so if K is not the only element of N, then infinitely many elements are derivable from K.
8.2 We first do (b). Take K1 = TK and K2 = T(KI). Then for any x, K1x = xK and K2x = x(KI). Then Ki(Vxy) = VxyK = Kxy = x; K2(Vxy) = Vxy(KI) = Klxy = ly = y. Thus Ki(Vxy) = x and K2(Vxy) = y.
Part (a) is then immediate, because if Vxy = Vzw then Ki(Vxy) _ Ki(Vzw), hence x = z. Likewise K2(Vxy) = K2(Vzw), hence y = w. 8.3 (a) Take neg = V f t.
(b) Take c = Rf (Rxyz = yzx). (c) Take d = TK.
(d) Take i = RK. (e) Take e = CSneg.
8.4 For an extensional system, we can take neg to be C. 9 We now give combinators that work, but how these were found will be explained in Part II of this chapter.
(a) M(BMM) is idempotent. (b) LM(LM) is idempotent. (c) LLL(LLL)(LLL)(LLL) is idempotent.
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Solutions
9.1
Again we give the combinators, but explanations are given in Part II.
(a) M(BKM) is voracious. (b) LK(LK) is voracious.
10 (a) We showed in Chapter 2 that if M is present and if the system has the compositional closure property that for any elements a and b there is an element c that composes a with b (i.e., is such that for all x, cx = a(bc)) then every element has a fixed point. More specifically, ono
if y composes x with M, then yy is a fixed point of x. Well, BxM composes x with M, and so BxM(BxM) is therefore a fixed point of x. Hence also M(BxM) is a fixed point of x. We can verify directly: M(BxM) = BxM(BxM) = x(M(BxM)). (b) Since for any x and y, Lxy = x(yy) then we can take Lx for y and we have Lx(Lx) = x(Lx(Lx)), and so Lx(Lx) is a fixed point of x. Discussion ova
We saw in the solution to Problem 7 that M(BTM) and LT(LT) are both elements that have the property of commuting with all elements. Also, they are both fixed points of T! Actually, if A is any fixed point of T, then A commutes with all elements, because then A = TA, hence Ax = TAx = xA. .--.
11 (a) Suppose x is a fixed point of M. Thus Mx = x. But also Mx = xx. Hence xx = x, so x is idempotent. We know by the solution to Problem 9 that for any x, M(BxM) is a fixed point of x. Hence M(BMM) is a fixed point of M, hence is idempotent. Also, LM(LM) is a fixed point of M, so LM(LM) is idempotent. [This is how we found solutions to (a) and (b) of Problem 9.]
(b) Suppose LLx = x. Then L(xx) = x (since LLx = L(xx)). Hence L(xx)x = xx. But L(xx)x = xx(xx), hence xx(xx) = xx, which means that xx is idempotent. Now, a fixed point of LL is L(LL)(L(LL)), which is also LLL(LLL) (since L(LL) = LLL). Therefore LLL(LLL)((LLL)(LLL)) is idempotent. [This is how I found a solution to (c) of Problem 9. Shorter solutions have been found by Larry Wos and others at Argonne National Laboratories. I believe a solution has been found with only seven letters.]
Chapter 17. Fixed point properties of combinatory logic
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Fm's
11.1 Suppose Kx = x. Then for any y, Kxy = xy, but Kxy = x, hence xy = x, so x is voracious. Since M(MKM) is a fixed point of K, it is voracious. Since LK(LK) is a fixed point of K, it is voracious. [This is how solutions were found for Problem 9.1.] 12 [Proof of F1] Suppose f (x,xl, ... , x,,,) is realizable; let b be an element that realizes it. If now every element has a fixed point, let a be a fixed point of b. Then since a = ba, then ax, ... xn, = baxl ... x = f (a, x1, ... , xn).
[Proof of F21 Suppose that b realizes the function f (xx, yl, ... , yn). Then bxy1 ... yn = f (xx, Y 1 ,- .. , yn) for all x, yl, ... , yn, hence bby1 ... yn = f (bb, yl, . , yn), and so we take a = bb. .
.
,.y
The following is an obvious corollary of Theorem F2. Suppose that for every realizable function f (x, yl, . , yn) the function f (xx, yl, ... , yn) is realizable. Then for every realizable function f (x, yl,... , yn) there is an element a such that for all y1, ... , yn, ayi ... yn = f (a, yl, ... , yn) Obviously every complete applicative system satisfies the hypothesis of this corollary, and also the hypothesis of Theorem F1. . .
E-+
13 (a) We have shown that M(BxM) is a fixed point of x, so we seek a combinator 9 such that 9x = M(BxM) for every x. Well, BxM =
.-,
RMBx, hence M(BxM) = M(RMBx) = BM(RMB)x. Thus BM(RMB) is a sage.
(b) Also, BxM = CBMx, hence BM(CBM) is a sage. (c) We now use the fact that Lx(Lx) is a fixed point of x, and Lx(Lx) _ M(Lx) = BMLx. Hence BML is a sage. Thus also M2L is a sage. (d) Since we can take BWB for L (Problem 3.2(b)) then BM(BWB) is a sage (as the reader can directly verify).
13.1 We again use the fact that M(LX) (which is LX(LX)) is a fixed point of x. But M(Lx) = QLMx, hence QLMx is a fixed point of x, and so QLM is a sage. But we can take L to be QM (Problem 3.2(c)) and so Q(QM)M is a sage. Actually, without bringing L into it, we can reason more directly as follows. We use the first fact that M(BxM) is a fixed point of x. Now,
M(BxM) = M(QMx) = BM(QM)x = Q(QM)Mx, so Q(QM)M is a [Alternatively, we have already shown that BM(CBM) is a sage, but BM(CBM) = BM(QM) = Q(QM)M.] sage.
349
Solutions
40H
As we are at it, let us note that if we take L to be QM, then BxM and Lx are the same element (since BxM = QMx = Lx), and so Lx(Lx) and M(BxM) (which is BxM(BxM)) are really the same.
13.2 (a) If A is an assistant sage, then WA is a sage, since then WAx = Axx, which is a fixed point of x. In fact A is an assistant sage if and only if WA is a sage.
(b) QL(QL) is an assistant sage, for QL(QL)xx = QL(Lx)x = Lx(Lx), which is a fixed point of x. And so W(QL(QL)) is a sage. (c) QoLL is also an assistant sage, for QOLLxx = Lx(Lx). Actually, QoLL
same sage as W(QL(QL)).
-per
is the same element as QL(QL), for Qo = QQ(QQ), so QoLL = QQ(QQ)LL = QQ(QL)L = QL(QL). Thus also W(QoLL) is the (d) M2 is such an X! We already know that M2L is a sage, and we can take M2 to be Q(WQ0)W, as in the solution of Problem 3.3(b). And so Q(WQo)WL is a sage. Actually, this is the same sage as the one we just got in (c), since Q(WQ0)WL = W(WQOL) = W(QoLL) (since WQOL = QoLL).
13.3 (a) GLL is an assistant sage, since also GLLxx = Lx(Lx), which is a fixed point of x. [Thus W(GLL) is a sage, which can also be expressed as W(WGL).]
coax
(b) One way is to use the fact that QL(QL) is an assistant sage and to take CB for Q, thus obtaining CBL(CBL), which can be shortened to B(CBL)L. Another way is to use the fact that GLL is an assistant sage and to take BBC for G, thus obtaining BBCLL, which can be shortened to B(CL)L. [This is more economical than B(CBL)L.]
13.4 We have seen that W(WGL) is a sage, and we can take L to be BWB, and so W(WG(BWB)) is a sage.
13.5 On the one hand, B(CBL)L is an assistant sage (Problem 13.3), so W(B(CBL)L) is a sage, and if we take BWB for L we get W(B (CB(BWB))(BWB)). But it is more economical to take the sage W(WGL) and replace G by BBC (Problem 2.9) and L by BWB, thus getting W(W(BBC)(BWB)).
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13.6 SLLx = Lx(Lx), hence SLL is a sage.
13.7 We take BWB for L in SLL, and so S(BWB)(BWB) is a sage. This can be shortened to WS(BWB), which is Curry's expression for a fixed point combinator.
13.8
(a) M(BOM) (b) LO(LO)
(c) M(PM) (d) WP(WP) (e) PII(P(PII)) (from (c), taking P11 for M). [How these were obtained will be revealed after Problem 14.] r-1
1 (a) P is such a combinator. [Pxyz = z(xyz) - Problem 3.5.] Suppose PYY = YY. Then for any element x, PYYx = YYx, but PYYx = x(YYx), hence x(YYx) = YYx, which means that YY is a sage.
(b) Suppose X has this property. Then WX(WX) = X(WX(WX)), and since X has this property, WX(WX) must be a sage. Since P has this property (by (a)) then WP(WP) is a sage, as we have already seen in the last problem.
14 Uxy = y(xxy), so UUy = y(UUy), thus UU is a sage! This is surely the simplest construction of a sage of any. Remarks
Now we shall explain how solutions to Problem 13.8 were found. In the sage UU, if we take PM for U (Problem 4.2(d)), we get PM(PM), which can be shortened to M(PM). This gives (c) of Problem 13.8. We can also take BO for P (Problem 4.1), and then M(PM) becomes M(BOM), which is (a) of Problem 13.8. As for (b) of Problem 13.8, we take LO for U (Problem 4.3(b)), and UU becomes LO(LO). As for (d), we can take WP for U in UU (Problem 4.2(c)), and we get WP(WP) (which we also got as a corollary of the solution of (b) of Problem 13.9).
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Solutions
14.1 Suppose UX = X X . Then for any element y, UXy = X X y. But also UXy = y(XXy). Hence XXy = y(XXy), so XX is a sage.
Suppose 0 is a sage and that Ax = Ox for every x. Then for every x, x(Ax) = x(Bx), but x(Ox) = Ox (since 0 is a sage), so x(Ax) = 9x, so x(Ax) = Ax (since Ox = Ax), which means that A is a sage. 15
15.1 Suppose 0 is a sage. Then for any x, 09x = x(Ox) = Ox. Thus OBx = Ox for all x, and so by the last problem (taking 00 for A), OB is a sage.
15.2 Suppose A is a fixed point of 0, i.e., OA = A. Then for any x, OAx = Ax, hence x(Ax) = Ax, hence A is a sage. 15.3 Suppose B is a sage. Then for any x, x(Ox) = Ox, thus OBx = Ox. If now the system is extensional, then 00 = 0, which means that B is a fixed point of O.
15.4 Suppose A is a combinator with the property that for any element
x, Ax = O(xx). Then A is certainly a Turing combinator, since then t-+
Axy = O(xx)y = y(xxy). Let us call such a combinator a special Turing combinator. [In an extensional system, there is only one Turing combinator,
and so there is then no difference between a Turing combinator and a special Turing combinator. But we find it of interest to consider applicative systems that are not necessarily extensional.] Now, if A is a special Turing
combinator, then AA is not only a sage, but is even a fixed point of 0 (because, since Ax = O(xx) for every x, then AA = O(AA)). It so happens that the Turing combinators W1P1i LO, WP, and PM that we studied in Problem 4.2 are even special Turing combinators. Let us verify this:.
(a) W1P1x = BWP1x = W(Pix) = W(LQx) = W(Q(xx)) _ BWQ(xx) = O(xx) (since BWQ = 0). (b) LOx = O(xx).
(c) WPx = Pxx = BOxx = O(xx). (d) PMx = BOMx = O(Mx) = O(xx). And so if U is any of these four combinators, then UU is a fixed point of O.
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15.5 It is true: BOx = 0(0O)x = x(BOx). [In fact BO = 0(00), so BO is a fixed point of 0, hence is a sage by Problem 15.2.]
Cps
15.6 We use P2i M2, and C. [P2xyz = x(yyz); M2xy = xy(xy).] We take A to be C(M2P2). Then for any x, Ox is a sage, since for all y, Oxy = C(M2P2)xy = M2P2yx = P2y(P2y)x = y(P2y(P2y)x) = y(Axy) Thus for all y, Axy = y(Oxy), so Ox is a sage.
16 (a) We can take L1 = BLB, which in terms of B and W is B(BWB)B. [Also we could take L1 = W1B2, where W1 is BW and B2 is BBB. Thus BW(BBB) could also be taken for L1, but we prefer BLB.] (b) Obvious.
(c) Take M3 = B(WW).
(d) Then M3L1 is a sage. Reduced to B and W, it is B(WW)(B(BW B)B). [Alternatively, we could take B(WW)(BW(BBB)).] 17 (a) Suppose that S is present and that every element has a fixed point. We will first show that A has the recursion property. Well, given an element a, let b be a fixed point of Sa. Then b = Sab. Hence for any x, bx = Sabx, but Sabx = ax(bx), and so bx = ax(bx). Thus A has the recursion property. And more specifically, for any element a, if b is a fixed point of Sa, then for all x, bx = ax(bx). We now take S itself for a, and so if y is any fixed point of SS, then for all x, -yx = Sx(-yx), hence -yxy = Sx(-yx)y = xy(yxy). Thus any fixed point y of SS is a 2-sage.
(b) We know that for any x, Lx(Lx) is a fixed point of x, and so L(SS)(L(SS)) is a fixed point of SS, hence by (a), it is a recursion combinator.
The combinator L(SS) - call it "a" - is an interesting one. It is a proper combinator, since axyz = L(SS)xyz = SS(xx)yz = Sy(xxy)z = yz(xxyz). Thus axyz = yz(xxyz). Taking a for x, we see that aayz = yz(aayz), and so as is indeed a 2-sage. 17.1 We have already constructed a sage B from B and W (Statman's sage, Problem 16) and if B is a sage, then BB is a 2-sage (since BOxy = B(xy) = xy(O(xy)) = xy(BBxy)). [Not so tricky, after all!]
Solutions
353
17.2 Yes, the combinator SS does the job. We have already seen in the solution of Problem 17 that every fixed point of SS is a 2-sage. Conversely,
suppose r is a 2-sage. Then SSPxy = Sx(Px)y = xy(Pxy), but xy(Pxy) _ Pxy (since P is a 2-sage), hence SSPxy = Pxy. If now A is extensional, then SSP = P, which means that r is a fixed point of SS.
17.3 B and W will work: we have seen that if 0 is a sage, then BO is a 2-sage. Actually, for any positive n, if 0 is an n-sage, then BB is an
(n + 1)-sage (as is easily verified). And so, starting with Statman's sage 0 constructed from B and W, BB is a 2-sage, B(BB) is a 3-sage, and so forth.
18 (a) Take el = WC. (b) Take e2 = Cel (= C(WC)). (c) Take S4 = BC(C42) [4)2xyzwv = x(ywv)(zwv) - Problem 3.10.] (d) Take C1 = S4e2e1. (e) Take C2 = S4e1e2.
(f) Take Dl = BQo(Bel). [Qoxyzw = xz(yw) - Problem 2.3.] (g) Take D2 = BG(Bel). [Gxyzw = xw(yz) - Problem 2.7.] 19 Since M and B are present, every element has a fixed point (Problem 10(a)). Given a1 and a2, let b1 be a fixed point of Ba2a1. Thus Ba2a1b1 =
bl, so a2(albl) = b1. Let b2 = albs. Thus a2b2 = b1 and albs = b2-
19.1 We use the combinator P3 of Problem 3.5 satisfying the condition:
P3xyz = y(xzy). We temporarily let A be the combinator LP3 - it is proper and satisfies the condition: Axyz = y(xxzy). Then the presence of A guarantees the cross-point property because for any elements a1 and a2, AAa1a2 = al(AAa2a1), and AAa2a1 = a2(AAala2), and so we take b2 = AAa1a2 and b1 = AAa2a1, and we have a1b1 = b2 and a2b2 = b1. 20
Given a1 and a2, take b1 = Clal(C2a2)(Clal) and b2 = C2a2(C1
ai)(C2a2).
Chapter 17. Fixed point properties of combinatory logic
354
20.1 Given a1 and a2, by hypothesis there are elements cl and c2 such
that for all x and y: clxy = ai(xy)(yx) and c2xy = a2(yx)(xy). Let d1 be a fixed point of cl and d2 a fixed point of c2. Then d1d2 = c1d1d2 = ai(did2)(d2di) and d2d1 = c2d2d1 = a2(did2)(d2di). And so we take bi = did2 and b2 = d2d1.
20.2 By Problem 17, the present hypothesis implies that A has the recursion property (in fact there is even a 2-sage). Then given elements a and b, let e be such that (1)
ex = bx(ex) (for all x)
Next, let c be a fixed point of Sae, so c = Saec = ac(ec). Hence: (2)
c = ac(ec)
By (1), taking c for x, we have ec = bc(ec). Thus c = ac(ec) and ec = bc(ec), so if we let d = ec, we have c = acd and d = bed. [This proof should be compared with that of Theorem 1.2, Chapter 16.]
20.3 Since B and W are present, so is BWB (which can be taken for L), so every element has a fixed point. Given a and b, let e be a fixed point
of BW(Ba). Then for all x, ex = BW(Ba)ex = W(Bae)x = Baexx = a(ex)x. And so we have: (1)
For all x, ex = a(ex)x
Now let d be a fixed point of W(Bbe). Then d = W(Bbe)d = Bbedd = b(ed)d, and so (2)
d = b(ed)d
By (1), ed = a(ed)d. We let c = ed, and we have c = acd and d = bcd.
20.4 Let e be a fixed point of S2V(S2aK1K2)(S2bK1K2). Then e
= S2V(S2aK1K2)(S2bK1K2)e = V(S2aK1K2e)(S2bK1K2e)
= V(a(Kie)(K2e))(b(Kie)(K2e))
Hence Kie = a(Kie)(K2e) and K2e = b(Kie)(K2e). And so we take c = Kie and d = K2e and we have c = acd and d = bcd.
20.5 Let j3 = alai. Then /3xy = x(yy)(,3xy) (because,3xy = alaixy = x(yy)(alaixy) = x(yy)(/3xy)).
Solutions
355
Given elements a and b, let h = a2/3ab. Then hx = a2/3abx = a(xx)(/3bx). Hence hh = a(hh)(/3bh). But also /3bh = b(hh)(/3bh) (since for every x and y, /3xy = x(yy)(/3xy)). And so we let c = hh and d = /3bh, and we have c = acd and d = bcd. 21
(D1C1C2i D2C2C1) is a DM-pair.
21.1 We need combinators A1, A2 satisfying the following conditions:
Aizwxy = x(wzwxy)(zwzxy) A2zwxy = y(zwzxy)(wzwxy) We can take Al = W2(V4D4e2e1) A2 = W3(Vs-b4ele2)
Then (A1A2A1, A2A1A2) is a DM-pair.
21.2 We need combinators 01 and 02 satisfying the following conditions:
Oizwxy = x(zxy)(wxy) O2zwxy = y(zxy)(wxy) We can take 01 = W2(V212); 02 = W3(V31'2). Now, suppose a and b are such that Olab = a and O2ab = b. Then
axy = Olabxy = x(axy)(bxy) bxy = O2abxy = y(axy)(bxy)
Thus (a, b) is a DM-pair
22 We want combinators n1, n2 satisfying the conditions:
nlwzxy = z(wxxy)(wyxy) n2wzxy = z(wwxxy)(wwyxy) We take nl = C(V24D3W(W2C))
n2 = Ln1
Then n2n2 is a nice combinator and ni satisfies (b).
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356
22.1 We want combinators s1, 82 satisfying the conditions:
siwxy = x(wxy)(wyx) s2wxy = x(wwxy)(wwyx) We take
s1 = BW(C(C(BSID2)C)) S2 = Lsi Then 8282 is a symmetric combinator and si satisfies (b).
22.2 (C(BBs)C, C(BsC)) is a DM-pair. (WN, W1(CN)) is a DM-pair.
Chapter 18
Formal combinatory logic In preparation for the fixed point theorems of the next chapter, we must turn to some basics of axiomatic combinatory logic.
Part I The axiom systems CL and CLo §1 Formal combinatory logic For formal combinatory logic we take the five symbols
SK() = together with a denumerable sequence xl, x2, ... , xn, ... of variables (or alternatively, two symbols x and ' and take our variables to be x', x", x"', ... ). The class of terms is inductively defined by the following two conditions
(1) The symbols S and K and all variables are terms. These terms are called simple terms.
(2) For any terms X and Y, the expression (XY) is a term. Thus the set of terms is the smallest set containing S, K, and all variables, which is such that for any members X and Y, the expression (XY) is also a member. Terms that are not simple will be called compound. Thus every compound term is of the form (XY), where X and Y are terms. In displaying terms we often omit parentheses with the understanding that they are to be restored with association to the left, as we informally did in the last chapter, and we will also be free to omit outer parentheses. And so, e.g., the expression SKx3(KS) is simply an abbreviation for the term (((SK)x3)(KS)).
Chapter 18. Formal combinatory logic
358
By an equation, or formula, we shall mean an expression of the form X = Y, where X and Y are terms. We now consider the following formal axiom system CL. The axioms of CL are all formulas of one of the following
three forms (where X, Y, and Z are any terms):
Al: SXYZ = XZ(YZ) A2: KXY = X
A3: X=X The inference rules of CL are the following:
R1: From X = Y to infer Y = X.
R2: From X = Y and Y = Z to infer X = Z. R3: From X = Y to infer XZ = YZ. R4: From X = Y to infer ZX = ZY. By a proof in CL is meant a finite sequence F1, . . . , Fn, of formulas such that each member of the sequence is either an axiom of CL, or is derivable from an earlier member by R1, R3, or R4, or is derivable from two earlier members by R2. A formula is said to be provable in CL, or to be a theorem
of CL, if it is a member of a sequence that constitutes a proof in CL. Equivalently, the class of theorems of CL is the least class that contains all axioms of CL and is closed under the rules R1-R4.
Two terms X and Y are called equivalent - in symbols, X - Y - if the formula X = Y is provable in CL. The system CLo
`S~'
"yam
By a combinator is meant a term without variables. Thus all combinators are built from the four symbols S, K, (,). By a sentence is meant a formula in which no variable occurs. Thus a sentence is an expression of the form X = Y, where X and Y are combinators. We let CLo be the system CL whose axioms are restricted to sentences. Thus all theorems of CLo are sentences. Our main interest is in the system CLo. It is very easy to construct an elementary formal system over the alphabet {S, K, (, ), =} in which can be represented the set of combinators, set of sentences, set of axioms, and set of theorems of CLo (Exercise 1 below). Thus CLo is a formal system, and so the set of theorems of CLo is recursively enumerable (under any admissible Godel numbering). The combinators I, B, C, W, and others derivable from them, are definable in terms of S and K, as shown in the last chapter. All results on
§1 Formal combinatory logic
359
applicative systems informally proved in the last chapter can be formally proved in CLo, for example if we take any of the numerous sages 0 constructed in the last chapter (for example (Q(QM)M), and write it down explicitly in terms of S and K, then for any combinator X, the sentence OX = X (OX) is formally provable in CLo.
Exercise 1 Show that the set of theorems of CLo is formally representable. Do the same for CL (couched in the finite alphabet {S, K, , x" }). Induction principle
To show that a given property of terms holds for all terms, it suffices to show:
(1) The property holds for all simple terms (S, K, and variables).
(2) For any terms X and Y, if the property holds for X and Y, then it holds for (XY).
To show that a given property holds for all combinators, it suffices to show:
(1) The property holds for S and K. P..
(2) For any combinators X and Y, if the property holds for X and for Y, it holds for (XY). Combinatory completeness
We stated without proof in the last chapter that any applicative system containing S and K is combinatorially complete. In terms of formal combinatory logic, this means that for any term W and any variables Y 1 ,- .. , yn that include all variables in W, there is a combinator A such that the for-
mula Ay, ... yn = W is provable in CL (and hence for any combinators X1, ... , Xn, the sentence AX1 ... Xn = W (Xl, ... , Xn) is provable, where W (X1i ... , Xn) is the result of substituting the combinators X1, ... , Xn for the variable Y 1 ,- .. , Yn respectively in W). Let us now prove this.
We take I to be SKK, and so for any term X, the formula IX = X is provable in CL. Now, let x be any variable and W any term (in which x may or may not happen to appear). We define the term W., (also written Ax W) by the following inductive scheme:
(1) If W is x itself, we take Wx = I.
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360
(2) If W is any simple term other than x itself, we take Wx = KW. (3) For a compound term WV, we take (WV)x = SWxVx. By easy induction arguments we see:
Proposition 1
(a) The variable x does not occur in Wx.
(b) All variables of Wx occur in W.
(c) The formula Wxx = W is provable in CL.
III
III
Parts (a) and (b) are obvious. To see that (c) holds, if W is a simple term, then (c) holds by (1) and (2) above. For a compound term WV, if Wxx - W and Vxx - V, then (WV)xx = SWxVxx - Wxx(Vxx) - WV. Now we wish to show the combinatory completeness of CL, in the sense that for any term W, and for any variables yl, ... , y,,, which include all the variables of W, there is a combinator A such that the formula Ay, ... yn = W is provable in CL. Well, for any term W and any variables yl, ... , yn, we write Wy,...y. for (... ((Wy1)y2) ...)yn . By mathematical induction, Proposition 1 yields:
Proposition 2 (1) None of the variables yl, ... , yn occur in Wy1 (2) All variables of Wyly2...Y.. occur in W. (3) Wyaya-l...yl yi ... Y. = W W.
If now yl,... , yn includes all the variables of W, then by (1) above, Wyn,,,yl contains no variables, hence is a combinator A, and by (3), Ay, ... yn = W. Thus CL is combinatorially complete.
§2 Consistency and models
cam
-.C
By a non-trivial applicative system is meant an applicative system in which N contains more than one element. By a model for CL is meant a non-trivial applicative system that is combinatorially complete. Now, the consistency of CL - in the sense that not all formulas are provable - has been known for some time, but this was not enough to ensure the existence of a model. The existence of models for CL was established by Dana Scott[25]. Simplified models were subsequently found, and the following one is a variant by Meyer[9] of one due to Plotkin[12] as reported in Rosser[23].
§3 Extensional combinatory logic
361
k-0
Start with a non-empty set A and extend it to the least set B such that for every element x c B and every finite subset Q of B, the ordered pair (,(3, x) is an element of B. The model consists of all subsets of B, and for any subsets C and D of B, we take (CD) to be the set of all ordered pairs ('(3, x) such that ,Q is a finite subset of D, x is an element of B, and (/3, x) E C. We then take K to be the set of all (a, ((3, x)) such that a, /3 are finite subsets of B and x E a. We take S to be the set of all (a, (/3, (-y, x))) such that a,,3, -y are finite subsets of B and x c ary(,3ry). It is not too difficult to verify that
for any subsets C, D, E of B, KCD = C and SCDE = CE(DE). We thus have a model of CL.
§3 Extensional combinatory logic No more sentences (formulas without variables) are provable in CL than in CLo, but a host of other sentences are provable if we add to CL the following "extensionality" rule:
R5 From Wx = Vx to infer W = V, where W, V are terms and x is a variable that does not occur in either W or V.
With R5 added, the resulting system - extensional combinatory logic - is abbreviated "CL-l-ext". With R5 added, one can prove, for example, C11
the sentence SKK = SKS (because SKKx = SKSx is provable in CL, since SKKx = x and SKSx = x are both provable in CL).
'C3
As we have said, our main interest is in CLo and many sentences not provable in CLo are provable in CL+ext. And now comes something amazing: there are four sentences X1, X2, X3, X4 provable in CL+ext, not provable in CL, such that if they are added as further axioms to CLo, then all sentences provable in CL+ext become provable! In fact-, if we add X1-X4
as axioms to those of CL, the resulting axiom system - which we call CL1 - is equivalent to CL+ext, in the sense that the same formulas are
awl
Ski
provable in CL1 as in CL+ext. We will not need this remarkable fact for the next chapter, but its proof is a real gem, and so we give a version of it in the appendix that follows - a version that we believe to be particularly easy to understand by those not familiar with the result.
Appendix on CL1 We let x, y, z be any distinct variables. We first observe that the following four formulas are provable in CL+ext:
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362
Q1: S(Kx)(Ky) = K(xy)
Q2: S(Kx)I = x Q3: S(S(KK)x)y = x Q4: S(S(S(KS)x)y)z = S(Sxz)(Syz)
That these are provable in CL+ext follows from the fact that the following four formulas are provable in just CL (which the reader can easily verify):
(1) S(Kx)(Ky)z = K(xy)z
(2) S(Kx)Iy = xy (3) S(S(KK)x)yz = xz (4) S(S(S(KS)x)y)zw = S(Sxz)(Syz)w By the completeness of CL, there are combinators 01, 02, I', K ' , ' 0 1 , 02 such that the following six formulas are provable in CL:
(la) O1xy = S(Kx)(Ky) (2) (3)
(4a)
(1b)
02xy = K(xy)
I'x = S(Kx)I K'xy = S(S(KK)x)y z/J1xyz = S(S(S(KS)x)y)z (4b)
02xyz = S(Sxz)(Syz)
[Specifically, we can take 01 = C(BB(BSK))K, 02 = BK, I' = C(BS K)I, K' = BS(S(KK)), 01 = BB(BS)(BS(SKS)), 02 = C(BB(B(S(B (BS)S)))). But this particular choice is inessential.]
Then by virtue of Q1-Q4i the following formulas are provable in CL+ext. Qi : 01xy = 02xy Q'2 :
I'x = Ix (since x = Ix is provable)
Q'3 : K'xy = Kxy (since x = Kxy is provable)
Q4 : 01xyz = 02xyz Then by further applications of R5 (two of which are required for Q'1 and Q3, and three of which are required for Q4) the following four sentences are provable in CL+ext. X1 01 = 02
X2I'=I
§3 Extensional combinatory logic
363
X3 K'=K X4 01 = 02
These are the "mystery" sentences X1-X4! We let CL1 be CL with X1X4 added as axioms. Since X1-X4 are provable in CL+ext, then all formulas of CL1 are provable in CL+ext. We now wish to show that everything provable in CL+ext is provable in CL1. And for this it suffices to show that CL1 is closed under rule R5, i.e., that if Wx = Vx is provable in CL1 and if the variable x does not occur in W or in V, then W = V is provable in CL1.
Since X1-X4 are provable in CL1, then Q1-Q4 are provable in CL1. The formulas Q1-Q4 are the important ones for the proof that follows.
For any terms W1 and W2, we write W1 - W2 to mean now that W1 = W2 is provable in the system CL1. For any term W and variable x, we define Wx as in the proof of combinatory completeness. Until further notice, we let x be a variable fixed for the entire discussion, and it will Cc)
be notationally convenient to write W* for Wx. Thus, if W = x, then W* = I; if W is any simple term other than the variable x, W* = KW, and for a compound term, (W1W2)* = SWiW2. Now we shall see how the formulas Q1-Q4 enter the picture. [Of course they also hold, replacing x, y, z by any terms whatsoever.]
Fact 1 If x does not occur in W, then W* - KW. Proof By induction on the structure of W: if W is a simple term distinct from x, then W* is KW, hence W* - KW. Now suppose W is a compound term W1W2i where W1 and W2 are both free from x and both such that Wi KW1 and W2 KW2. Then W* = SW1* W2 - S(KW1)(KW2), but S(KW1)(KW2) -- K(W1W2) (by Q1!). Hence W* - KW (since W is W1W2). This completes the induction
Fact 2 If x does not occur in W, then (Wx)* - W. Proof Suppose that x doesn't occur in W. Then by Fact 1, W* - KW. Now, (Wx)* = SW*x* - S(KW)x* = S(KW)I. But S(KW)I - W (by Q2), and so (Wx)* - W.
Fact 3 (KW1W2)* - W. Proof (KW1W2)* = S(S(KK)W)W2 (see note below) 1
Wi (by Q3)-
364
Chapter 18. Formal combinatory logic
Note
(KW1W2)* = S(KW1)*W2 = S(SK*W*)W2 = S(S(KK)W*)W* (since
K* = KK). Fact 4 (SW1W2W3)* - (W1W3(W2W3))*.
Proof (SW1W2W3)* = S(S(S(KS)W*)W2)W3 (see note below) =- S(S W*W3)(SW*W3) (by Q4) = S(W1W3)*(W2W3)* = (W1W3(W2W3))* Note
(SW1W2W3)* = S(SWiW2)*W3 = S(S(SW1)*W*2)W3 = S(S(SS* W*)W2)W3 = S(S(S(KS)Wi)W*)W3. Having proved Facts 1-4, the formulas Q1-Q4 (and hence the sentences X1-X4) have now served their purpose, and it will no longer be necessary to refer to them. If F is a formula W = V, by F* we shall mean the formula W* = V. We now aim to prove that if F is provable in CL1, so is F*.
Lemma 0 If F is provable in CL1 and F contains no variables, then F* 55.
is provable in CL1.
Proof Suppose W = V is provable in CL1i and W and V contain no variables. Then KW = KV is provable in CL1, but since neither W nor V contain any variables, then W* = KW and V* = KV are both provable in CL1 (by Fact 1). Hence W* = V* is provable in CL1.
Lemma 1 If F is any axiom of CL1, then F* is provable in CL1. Proof Suppose F is an axiom of CL1. If F is one of the sentences S1-S4, then F contains no variables, hence F* is provable in CL1 by Lemma 0. If F is an axiom of type A3, i.e., if F is W = W for some term W, then F* is the formula W* = W*, which of course is provable in CL1.
If F is of type A1, i.e., of the form KW1W2 = W1, then F* is
E-4
(KW1W2)* = Wi, which is provable in CL1 by Fact 3. If F is of type A2, i.e., of the form SW1W2W3 = W1W3(W2W3), then F* is (SW1W2W3)* = (W1W3(W2W3))*, which is provable in CL1 by Fact 4. This takes care of all possible cases.
§3 Extensional combinatory logic
365
Lemma 2 If Wi = W2 is provable in CL1, so are (W1V)* = (W2V)* and (VW1)* = (VW2)*.
Proof This is obvious. Suppose Wi = W2 is provable in CL1. Then so is SW1V* = SW2V*, which means that (W1V)* = (W2V)* is provable in CL1. Likewise SV*Wi = SV*W2 is provable, hence (VW1)* = (VW2)* is provable.
Now, let CL1* be the axioms system CL1, with the following axioms and rules added: For every axiom F of CL1 we take F* as an additional axiom of CL1*. Then we add the rules that from Wi = W2 we may infer (W1V)* = (W2V)* and (VW1)* = (VW2)*. By Lemma 2, these two new inference rules do not enlarge the class of provable formulas of CL1, and by Lemma 1, the new axioms of CL1* are already provable in CL1, and
E-+
therefore all formulas provable in CL1* are provable in CL1 (though a proof of a formula in CL1 will in general be considerable longer than a proof of it in CL1*). Now, suppose F1,.. . , F,,, is a sequence of formulas that constitute a proof in CL1, Then the sequence Fl*,..., F,*, is a proof in CL1*. Therefore, if F is provable in CL1, then F* is provable in CL1*, but since everything provable in CL1* is provable in CL1 we have
Theorem A,, If F is provable in CL1, so is F*. Now we easily get:
III
Theorem A [Main result] - If Wx = Vx is provable in CL1 and if x does not appear in W or in V then W = V is provable in CL1.
Proof Assume the hypothesis. Since Wx = Vx is provable in CL1, so is (Wx)* = (Vx)* (by Theorem A0). But since x doesn't appear in either W or in V, then (Wx)* = W and (Vx)* = V are provable in CL1 (by Fact 2), hence W = V is provable in CL1.
This shows that the systems CL1 and CL+ext are equivalent.
Chapter 19
A second variety of fixed point theorems There is a result in combinatory logic known as the "second fixed point theorem," whose statement, proof, and applications to undecidability will be dealt with in Part II of this chapter. It is based on methods of naming combinators within combinatory logic itself, and this fits in ideally with
the overall purpose of this volume. Part I of this chapter is devoted to adequate naming functions. The second fixed point theorem itself has a host of variants, all of which have pleasant generalizations to sequential systems, which will be dealt with in our final chapter.
CAS
Part I Self-reference in CLO As we stated in the last chapter, our main interest is in the system CLo; 0°.
ASCU
extensionality is not needed for what we will be doing. For any combinators C,3
.ti
X and Y, by X - Y we now mean that the sentence X = Y is provable in CLo.
§1 Admissible name functions Let us consider a 1-1 mapping II that assigns to each combinator X a s0.
combinator X (more often written X-') that we will regard as the name of X. We shall call the mapping admissible (as a name function) if there are combinators A and S such that for all combinators X and Y the following two conditions hold: III
C1: AXY-XY
§1 Admissible name functions
367
C2: SX=X We remark that the usual procedure to obtain an admissible name function is to first assign to each natural number n a combinator n as its representative, and then to take an effective Godel numbering of all combinators, coo
and then to take X to be n, where n is the Godel number of X. But we shall take a more direct approach, briefly comparing it later on with the coo
standard approach. Before we exhibit a specific admissible name function, we wish to establish some general principles about name functions. In preparation for this, let us recall a few things. First, we are defining t to be K and f to be KI (§8, Chapter 17) and for any combinators X and
Y, tXY - X and fXY - Y. We are taking neg to be V f t (V satisfies the condition: Vxyz - zxy), and we recall that negt = f and neg f - t. We are taking the ordered pair [X, Y] to be VXY and K1 to be TK and K2 to be TKI (T, not to be confused with t, satisfies the condition: Txy - yx), and we recall that K1 [X, Y] - X and K2 [X, Y] - Y. We recall that the combinators S and K are called simple and all other combinators are called compound.
Pre-admissible mappings We shall call the mapping X -> X pre-admissible if there are combinators Z1, Z2, Ql, 0`2 satisfying the following conditions: P1:
(1) Z1K
t, but for any X
K, Z1X - f
(2) Z2St, but foranyX4S,Z2X-f P2: Q1XYY - X and ay2XYY - Y
'C3
a,"
We call such an element Z1, a K-tester and such an element Z2 an Stester. The combinators al and 0'2 might be called predecessor combinators. [We can think of X and Y as the first predecessor, second predecessor of XY respectively.] Now comes a lovely result, whose proof uses a modification of a brilliant use of fixed points due to Alan Turing[36].
Theorem R [A recursion principle] If the mapping X -> X is preadmissible, then for any combinator a and any combinators a1 and a2, there is a combinator ,Q such that:
(1) pK=a1 and ,3S-a2 (2) For any compound combinator XY, 9XY - a(QX)(QY)
Chapter 19. A second variety of fixed point theorems
368
Proof Given a, al, a2, by the fixed point theorem (Theorem F, §12, Chapter 17) there is a combinator/d satisfying the following condition: ,3x = Zlxaj(Z2xa2(a(Q(olx))\N\o2x))))
III
III
Since Z1K = t, then it is easily seen that /.3K - a1. Since Z1S = f and Z2S = t, then ,dS = a2. Next, for a compound combinator XY, since Z1XY f and Z2XY ° f, then (3XY = a(,3(o1XY))(0(a2XY)) a(,3X) (,3Y).
As our first application of the recursion principle we have:
X11
Theorem 1.1 Suppose that the mapping X -+ X is pre-admissible and that there is a combinator A' such that for all X and Y: A'XY - XY. Then the mapping satisfies condition C2 of admissibility, i.e., there is a combinator 6 such that for all x: 6X - X.
Proof By Theorem R, taking a = A', a1 = K, a2 = SL there is a combinator S (called "0" in Theorem R) such that SK = K, SS = S, and 6XY = A'(6X)(SY). [Specifically, we can take S satisfying the condition Sx = Z1xK(Z2xS(A'(6(0'lx))(6(o2x)))).] The result then follows by inducIII
III
III
III
III
tion. We already have SK K and 6S - S. Now suppose that X and Y are such that SX - X and SY - Y. Then SXY = A'(6X)(SY) = A'XY XY. This completes the induction.
Corollary 1.2 A sufficient condition for a mapping X --> X to be admissible is that it be pre-admissible and that there are combinators A and A' such that for all combinators X and Y:
AXY = XY
A'XYXY More on admissibility The following will be useful:
Proposition 1.3 If M is admissible then for every n > 2 there is a combinator An, such that for all X 1 ,.. . , X,,,: A,zXl ... X,, - X1 ... X. Proof By induction on n > 2: we take A2 = A. Now suppose there is a combinator A,,, satisfying the condition A,,Xl ... X,,, = X1 ... X. We
§1 Admissible name functions
369
III
then take An+l satisfying the condition: An+1x1... xnxn+l - A(Anxl ... xn)xn+l. Then An+1X1... XnXn+l - A(AnXl ... XnXn+1 = AXl ... XnXn+l - Xl ... XnXn+l
Strong admissibility We shall call a mapping X --> X strongly admissible if it is admissible and if also there is a combinator H such that for every combinator X:
HX -X Theorem 1.4 If the mapping X -> X is pre-admissible then there is a combinator H satisfying the condition: HX - X.
III
Proof Suppose the mapping is pre-admissible. Then by Theorem R, taking a = I, al = K, a2 = S, there is a combinator H such that HK - K,
HS - S, and HXY = I(HX)(HY) - hence HXY - (HX)(HY). By an obvious induction, HX - X for every X.
Corollary 1.5 If the mapping X --> X is both pre-admissible and admissible then it is strongly admissible. Remarks
We do not know whether or not every admissible mapping is strongly admissible. There are certain fixed point theorems of this chapter whose hypothesis requires only admissibility, and others whose hypothesis appears to require strong admissibility. As a point of minor interest, we have:
Proposition 1.6 If the mapping X -> X is strongly admissible then there is a combinator A' satisfying the condition: A'XY - XY.
Proof Take A' such that A'xy = S(A(Hx)(Hy)). 6(A(HX)(HY)) - S(AXY) - SXY XY
Then A'XY
Full admissibility :,.
Tao
.°-
A trivial example of a strongly admissible map is the identity function. We shall now define a mapping X --> X to be fully admissible if it is both admissible and pre-admissible. A fully admissible mapping is strongly admissible by Corollary 1.5.
Chapter 19. A second variety of fixed point theorems
370
More on pre-admissibility For applications, the following proposition will prove handy: v?.
Proposition 1.7 Suppose that there are combinators al and a2 such cad
that (1) If X is simple (K or S) then a1X = t, but if X is compound, then III
a1X - f. III
(2) a2Ktand a2S- f Then there is a K-tester Z1 and an S-tester Z2. Proof Take Z1 such that Z1x = alx(a2x)f and Z2 such that Z2x
ajx(neg(a2x)) f. Then it is easily verified that Z1 is a K-tester and Z2 is an S-tester.
The above with earlier results gives the main result of this section.
Theorem I A sufficient condition for a mapping X - X to be fully admissible (and hence also strongly admissible) is that there are combinatory o , 0`2, al, a27 A, A' such that for all combinators X and Y: III
III
QjXY = X III
a2XY = Y a1X t, if X is simple, otherwise a1X = f III
III
a2K = t and a2S = f AXY - XY III
A'XY XY §1.1 Construction of a fully admissible map We now exhibit a concrete name function X -* X and show that it is fully admissible. We inductively define X as follows:
K = [t, t] (= VKK)
S = [t, f] (= VK(KI)) XY = [f, [X, Y]] (= V f (VXY)) We see that every name X is an ordered pair whose first component is t if X is simple, and f if X is compound. We must now find combinators
§2 Admissible maps via numeral systems
371
al, a2, al, a2, A, A' such that the sufficient condition for full admissibility given by Theorem I holds. We take a1, a2 such that for all x, alx = Kl(K2x) and vex = K2(K2x).
Then it is obvious that a1XY = X and 0`2XY = Y (since K2XY [X7 YD_
As for al and a2, if X is simple, then K1X = t, otherwise K1X = f, and so we take al = K1. Also, K2K = t and K2S = f, and so we take a2 = K2. We now have al, a2, al, a2, and so our mapping is pre-admissible (since by Proposition 1.7 there is then a K-tester Z1 and an S-tester Z2). As for A, we obviously take A satisfying the condition: Axy V f (Vxy). Then AXY = XY. As for A', we note XY = V f (VXY) = V f (VV f (V f (V(V f (VYX)) Y))) (as the reader can verify by direct computation). And so we take A' such as to satisfy the condition:
A'xy = Vf(VVf(Vf(V(Vf(VVx))y))) Thus our mapping is fully admissible.
Exercise 1 Show that the mapping X -* X that we have constructed is formally representable over the alphabet of CLo. ?.1
*§2 Admissible maps via numeral systems A more standard type of admissible mapping uses a numeral system together with a Godel numbering, which we now briefly sketch. By a numeral system is meant a 1-1 map that assigns to each natural number n a combinator n such that there exist combinators a and Z such that:
(1) ZO=t, but for alln#0, Zn= f (2) For every natural number n, an = n+ (n+ = n + 1) Such a combinator or is called a successor combinator and such a combinator Z is called a zero tester. A specific numeral system used in Barendregt[1] is inductively defined as follows. We take 0 = I and for any n, we take n+ = [f, n]. This is indeed a numeral system, since we can take or = V f and Z = Tt. Then obviously III
III
V fn = [f, n] = n+, and ZO = TtI = It = t, but Zn+ = Tt(V fn) _ III
V f nt = t f n- f. Thus or is a successor combinator and Z is a zero-tester. III
In this numeral system there is a crucial combinator P - called a predecessor combinator - such that for all n, Pn+ = n, namely, take
Chapter 19. A second variety of fixed point theorems
372
`ti
P= T f. Then Pn+ = T f (V f n) - V f n f- f fn - n. [Note also that PO - f.]
CAD
Now, a combinator a is said to define a numerical function f (xl, ... , xn) (with reference to a given numeral system) if for all numbers al, ... , an: act, ... an - f (a,, ... , an). And f is said to be A-definable if some combinator a defines it. The beautiful thing is that the A-definable functions are precisely the recursive functions! This is obvious in one direction. From C'1
CAD
E-+
the fact that CLo is axiomatizable (the set of provable sentences is formally representable) it easily follows that every A-definable function is recursive 0-CCl
t',
(*Exercise 2, below). As for the converse, the key facts are that primi..O
tive recursion and minimization can be captured in combinatory logic (see *Exercise 3 below). Under any of the standard Godel numberings, there are recursive func-
tions con(x, y) and numx such that for any combinators X and Y with F31
respective Godel numbers x and y, the number con(x, y) is the Godel number of (XY) and num(x) is the Godel number of the numeral x (see T.M.M. for a handy Godel numbering). Then, for any combinator X, one defines
X (more usually written X,) to be the numeral x, where x is the Godel number of X. The mapping X -* X is then admissible (in fact strongly ..cam-r
admissible'). This is the usual procedure. By contrast, our admissible mapping does not use a numeral system or Godel numbering, and the proof if its strong admissibility is relatively simple.
*Exercise 2 Show that every A-definable function is recursive. *Exercise 3 (a) Show that the set of A-definable functions is closed under composition, explicit transformations, and introduction of constants.
(bl) Suppose f (x) is A-definable by a. Let c be any number and let g(x) be the function defined by recursion from f and c, i.e., the function
`±o
satisfying the following conditions: g(O)
9(n+)
= c =
f(g(n))
CAD
By the fixed point theorem, there is a combinator a satisfying the condition
Qx - zxx(a((3(Px))) 1The proof of this is rather elaborate - see Barendregt[1], whose proof uses a 1-point base (a non-proper combinator X from which S and K are both derivable).
§3 The second fixed point theorem
373
Show that Q defines the function g(x).
(b2) More generally, suppose g(y1i... , yn), h(x, yl,... , yn, z) are )-definable functions, and that f (x, y1i ... , yn) is a function defined from g and h by the following recursion: = g(Y1, . " , yn) h(x, yl.... 7 Yn' f (x, y1, ... , yre)) f (x+, yl, ... , yn) /J (0, y1, ... , yn)
-
Show that the function f (x, Y1, ... , yn) is A-definable.
(c1) Suppose that p is a combinator such that for every natural number
n, either pn = t or pn - f. By the fixed point theorem there is a combinator H satisfying the condition Hx - pxx(H(ax)). Now oar
suppose that n is a number such that pn - t, but for every m < n, pfn = f (in other words, n is the smallest number such that pn t). III
III
Prove that HO = n.
(c2) Suppose that f (x1, ... , xn, y) is a regular function on the natural numbers, i.e., , for every x1i .... xn there is at least one y such that f (XI, ... , xn, y) = 0. Let 9(x1, ... , xn) be the smallest y such that
f (X1, xn, y) = 0. 0.1
Now suppose that a is a combinator that defines the function f . By the fixed point theorem there is a combinator H satisfying the condition: Hx1 ... xny = Z(ax1
... xny)y(Hx1 ... xn(a'y))
Let 3 be a combinator satisfying the condition ,3x1 ... xn - Hx1 ... xnO. Show that the function g is definable by p. [This proves that the set of A-definable functions is closed under minimization, which concludes the proof that all recursive functions are A-definable.]
Part II Fixed point theorems of the second type §3 The second fixed point theorem In what follows, II will be an admissible mapping X - X fixed for the discussion. [Strong admissibility will not be required till later.] Now, which of the following statements, if true, would surprise you?
Chapter 19. A second variety of fixed point theorems
374
(1) There is a combinator X such that X = X.
(2) There is a combinator X such that X XX. (3) There is a combinator X such that X = XX. III
(4) There is a combinator X such that X = XX. III
(ID
(5) There is a combinator X such that X X.
(7) There is a combinator X such that for every combinator Y, XY YX.
III
III
(6) There is a combinator X such that for every combinator Y, XY = X.
Well, all of these statements are true! They are all but special cases of the following theorem.
Theorem 3.1 [Second fixed point theorem of combinatory logic] For every combinator a there is a combinator X such that aX = X. Proof we continue to let A -and 6 be combinators such that for all X and
III
III
Y, AXY = XY and SX - X. We now let t be a combinator satisfying the condition: Ax = Ax(Sx). Then for any combinator X, AX = XX (since AX = AX(6X) = AXX = XX). Next, given a, let Y be a combinator such that for all x, Yx = a(Ax) (we could take Y = Bat). Then YY = a(MY) = a(YY). We hence take X = YY, and so X = aX. Exercise 4 Why are the statements (1)-(7) preceding the theorem consequences of it?
§4 Applications We let II be an admissible mapping fixed for the discussion, and we shall define X to be a II-fixed point of a if aX = X. Theorem 4.1 tells us that every combinator has a 11-fixed point. Before turning to some significant applications of this, let us see why statements (1)-(7) preceding the theorem are special cases of it.
For (1), let X be a 11-fixed point of the the identity combinator I. For (2), we obviously take X to be a II-fixed point of the combinator M
(Mx = xx). For (3), we take X to be a 11-fixed point of a fixed point combinator 0. Then X = OX = X(OX) = XX. For (4), we take X to
§4 Applications
375 III
be a II-fixed point of a voracious combinator a (ax = a for all x). Then "CS
aX = X, but also aX - a, hence X - a. Then, since aX = X and a- X, we have XX _= X. For (5) we take X to be a II-fixed point of S. For (6) we take X to be a II-fixed point of K, and for (7) we take X to be a II-fixed point of T.
Representability We shall say that a combinator a represents the set of all combinators
X such that aX = t. We shall say that a set M of combinators is Arepresentable if it is represented by some combinator a.
It is easy to prove that every A-representable set is formally representable (thus r.e. under every standard Godel numbering) and it can be shown from known results that every formally representable set of combinatory is A-representable. From Theorem 3.1 we get:
Theorem 4.1 For any A-representable set M of combinators there is a combinator X such that X- t F-+ X E M.
Proof Suppose a represents M. By Theorem 3.1 there is a combinator X such that aX =_ X. But also aX =_ t <--> X E M (since a represents M). Thus X - t<-->X E .M.
For any set M of combinators, by its complement .M we mean its complement with respect to the set of all combinators, i.e., M is the set of all combinators not in M. We let T be the set of all combinators a such that a =_ t and T be its complement. Thus T is the set of all a such that
a0- t. Corollary 4.2 The set T is not A-representable. Exercise 5 For any sentence X = Y of CLo, define its code to be the car
coo
combinator [X, Y] (i.e., VXY). Call a sentence a Godel sentence for a set M of combinators if the sentence is provable in CLo if its code is in M. Show that for any A-representable set M, there is a Godel sentence for M. Conclude that the set of codes of the non-provable sentences of CLo is not A-representable.
III
III
III
We shall say that a combinator a dines a set M if aX =_ t for every x c M, and aX = f for every X E M. We call M A-definable if some
Chapter 19. A second variety of fixed point theorems
376
c--
combinator a defines M. It follows from the consistency of CLo that t 0 f (because if t = f , then for every X and Y, tXY f XY, which means X = Y, which would mean thatCL0 is inconsistent). Therefore, if a defines III
III
M, then a certainly represents M (since it is immediate that X E M = aX = t, and conversely, aX = t -(aX = f) = -X E M = X E M, and thus X E M +-+ aX = t). Also, if a defines M then neg a defines M, and so M is also A-representable. Thus if M is A-definable then M and .M are both A-representable. [The converse also happens to be true, since A-representability is equivalent to formal representability and A-definability is equivalent to formal solvability.] Since T is not A-representable, as we have just seen, it of course follows that T is not A-definable. We are about to prove a theorem that generalizes this fact. We shall call a set M of combinators extensional, or closed under equivalence, if for every combinator X in M, all combinators equivalent to X are also in M. The following theorem, which generalizes Corollary 4.2, is essentially Scott's variant[24] of Rice's theorem[19]. ci-
Theorem 4.3 (a) If M1 and M2 are disjoint extensional non-empty sets, then no superset of M1 disjoint from M2 is A-definable. (b) No extensional set of combinators is A-definable, except for the empty set and the set of all combinators.
Proof (a) Suppose (M1, M2) is a disjoint pair of extensional non-empty sets,
and that M1 C M, and that M is A-definable. We show that M cannot be disjoint from M2. Let al be any element of M1 and a2 any element of M2 and let a be a combinator that defines M. Now let 3 be a combinator satisfying III
III
the condition: Ox - axa2al. Then if X E M, aX = t, hence ,3X = a2, whereas if X M, then aX = f , hence /3X = a1. Thus X E M = QX - a2, and X M = 3X - al. Now, by Theorem 3.1 we can take X such that QX = X. Suppose X M. Then X = a1 (since X = dX al) and therefore X E M1 (since M1 Jill
III
[++
III
III
is extensional and contains al). This is impossible, since M1 C_ M.
Therefore X E M. Hence 3X = a2, so X =_ a2, hence X E M2 (since M2 is extensional and contains a2). Therefore X is in both M and M2, and so M cannot be disjoint from M2.
§4 Applications
377
C':1
I-.
(b) Suppose M is extensional and neither empty nor the set of all combinatory. Then X4_ is also extensional and non-empty. The only superset of M disjoint from M is M itself, which by (a) is not A-definable. Thus M is not A-definable.
tit
Discussion
c1+
!"'(
It is immediate from (b) above that the set T is not A-definable, since T is obviously extensional and is neither empty nor the set of all combinators. Indeed, it follows from (b) of Theorem 4.3 that for no combinator X is it the case that the set of all combinators equivalent to X can be A-definable. In particular, the set of all combinators equivalent to I is not A-definable - which is essentially Church's theorem[3]. This set is one of the earliest examples of a A-representable set that is not A-definable (which under any standard Godel numbering goes over to a recursively enumerable set that is not recursive).
Part (b) of Theorem 4.3 is the combinatory analogue of Rice's theorem[19], which goes as follows. In recursion theory a set A of numbers
is called extensional if for every number n in A, all numbers m are in A such that w.,,,, = wn, - in other words, A contains with every number n, all numbers m such that m and n are indices of the same r.e. set. Rice's theorem is that no extensional set can be recursive, except the empty set and the set of all natural numbers.
Exercise 6 Using the fact that for any recursive function f (x) there is a number n such that wn = W f(n), prove Rice's theorem.
A method of simultaneously generalizing Rice's theorem and Scott's theorem is given in the next exercise. ICY
Exercise 7 Let - be an equivalence relation on a set N and let E be a collection of functions from N into N having the fixed point property that
for every f E E there is some x c N such that f (x) - x. Let S be a set of subsets of N such that for every element A E S and any elements al, a2
of N there is a function f E E such that f (x) - al for every x c A and f (x) __ a2 for every x c A. Call a set A of elements of N closed (with respect to the equivalence relation -) if A contains with every element x, all elements y such that y - x.
(a) Prove that no closed set other than N or 0 can be a member of S.
Chapter 19. A second variety of fixed point theorems
378
(b) Show that both Rice's theorem and Theorem 4.3 can be obtained as special cases of (a). coy
§5 Other fixed point theorems of the second variety We continue to assume that the mapping X -* X is admissible. The c0.
following theorem and its corollary can be proved:
Theorem 5.1 [Second double fixed point theorem] For any combinators III
a and 3 there are combinators X and Y such that aXY - X and /3XY Y.
Such a pair (X, Y) will be called a 11-double fixed point of the pair (a, a).
Corollary 5.2 For any combinators a and ,Q there are combinators X and Y such that aX - Y and ,QY - X. We shall not prove these here; they follow as corollaries of more general results to be proved in the next chapter. Likewise, the following theorems about strongly admissible mappings will be seen to be special cases of the theorems proved in the next chapter.
Theorem 5.3 [The uniform second fixed point theorem] If the mapping X --> X is strongly admissible then there is a combinator 0 - that we will call a II-sage such that for every combinator X, OX - XOX. Remark
Without the hypothesis of strong admissibility we do not know whether Theorem 5.3 holds or not. Similar remarks apply to the remaining theorems of this chapter.
Theorem 5.4 Under the same hypothesis (strong admissibility) there is a combinator 0 such that for any combinators X and Y, qXY - XYgXY. coo
Theorem 5.5 [Uniform version of Theorem 5.1 - a "double Myhill theorem of the second type] If the mapping X -i X is strongly admissible then there are combinators 01 and 02 such that for all combinators X and Y:
(1) c1XY - X01XY 02XY
§5 Other fixed point theorems of the second variety
(2) 02XY
379
Y01XY 02XY
Such a pair (01i 02) will be called a II-DM pair.
Nice combinators and symmetric combinators of the second type As the reader probably suspects, we also have: yam]
't3
Theorem 5.6 Suppose that the mapping X --> X is strongly admissible. Then
(a) There is a combinator N such that for any combinators Z, X, Y: NZXY - ZNXXY NYXY. (b) There is a combinator a such that for any combinators X and Y:
aXY - XaXY aYX. The reader might find it a profitable exercise to try proving some of the above theorems before turning to the next chapter.
Chapter 20
Extended sequential systems We now turn to a variety of fixed point theorems that simultaneously generalize the fixed point theorems of the last chapter and earlier fixed point theorems for sequential systems. .ti
We let S be a sequential system (N, E, ->) such that for every n > 2 there is an exact E-function E(xl,... , xn). And now we consider two other items. The first is an equivalence relation - on N that satisfies the following two conditions:
El If a - b then for any E-function O(x, yl, ... , yn), the equivalence 0(a, yi, ... , yn) = 0(b, yl, ... , yn) holds for all yl, ... , yn. III
E2 If a - b, then for any finite sequences 0, r of elements of N, the ''7
equivalence B, a, r -40, b, r holds.
8,°
Of course ordinary identity is such an equivalence relation. For applications to the formal system CLo of combinators,, we define a, x1,. .. , xn -> b, yl,... , Yk to mean that axl ... xn - by, ... Yk (which, we recall, means that axl ... xn = by, ... Yk is formally provable in CLo, but not necessarily that axl ... xn is the same combinator as by, ... yk). Since
for CLo, a - b implies ac - be and ca - cb, then the equivalence relation between combinators is easily seen to satisfy conditions El and E2. Our second item to be considered is a mapping M that assigns to each element x of N an element Jr- of N. W e shall say that a function fo(xl, ... , xn)
mirrors a function f (X 1, ... , xn) (with respect to M and - understood) if for all elements x1, ... , xn: AG _l, ... , xn) - f (xl, ... , xn). And now we define the mapping M to be admissible if the following two conditions hold: III
M 1 Each exact E-function E(xl,... , xn) (n > 2) is mirrored by a Efunction Eo(xi, ... , xn) [Eo(xi, ... , xn) = E(xl, ... , M2 There is a E-function 6(x) such that for all elements x: 5(x) - x.
§0 M-diagonalizers
381
C)'
We let E be the triple (S, -, M) and we call E an extended sequential system. We call E of type 1, type 2, universal if S is respectively of type 1, type 2, universal. We call E admissible if M is an admissible mapping (with respect to S and -). In applications to combinatory logic, M is to be an admissible map
II, as defined in the last chapter. By Proposition 1.3 of the last chapter, for each n > 2 there is a combinator A,, such that for all X 1 ,.. . , Xn; AnXl ... Xn - Xl ... Xn, and so we take Eo(xl,... , xn) to be Anxl, ... xn,
and since we are taking E(xl,... , xn) to be the product xl ... xn, then Eo(Xl,... , Xn) = AX, ... X. - XI ... X. = E(Xi,... , Xn), and so condition Ml holds. We of course take a combinator S satisfying the condition
Sx - x, and we then define S(x) to be 6x, and thus 6(x) - x, and so condition M2 holds. Coming back to extended sequential systems in general, let us note the
obvious, but nevertheless important fact that if we take - to be ordinary identity and M to be the identity map, then M is trivially admissible (with 6(x) the identity function and Eo(xl,... , xn) = E(xl,... , xn), where E(xl,... , xn) is an exact E-function). And so all theorems that we will prove about admissible extended sequential systems have applications to sequential systems as well as to combinatory logic with an admissible map II.
In all that follows, for each n > 2, E(xl, ... xn) is assumed to be an exact E-function.
§0 M-diagonalizers We shall call a function 0(x, yl, ... yk) (k > 0) an M-diagonalizer if for any elements x, y1, ... , yk: 0(x, yl, ... , yk) - E(x, x, yl, ... , yk), where E(x, yl, ... , yk) is an exact E-function. [For k = 0, this condition is
0(x) - E(x, x).] And now we call E diagonalizable if for each n > 1, 'CS
E contains an M-diagonalizer of n arguments. The following lemma is basic to all that follows.
Lemma A If E is admissible and of type 1, then E is diagonalizable.
, . . .
4-.
Proof Suppose E is admissible and of type 1. Given n > 1, let E(x, y, z1, ... , zk) be an exact E-function of k + 2 arguments (k = n 1) and let Eo(x, y, z 1 , zk) be a E-function that mirrors the function E(x, y, z1i ... , zk). Take a E-function 6(x) satisfying condition M2
and take 0(x, yl, ... , yk) = Eo(x, 6(x), 6(yl), ... , 6(yk)). Then A is a E-function (since E is of type 1) and we have 0(x yl, ... , yk) =
382
Chapter 20. Extended sequential systems III
III
Eo(x b(x), 6(y1), ... 6(yk)) Eo(x, x, y1 ... yk = E(x, x, y1, ... yk). Thus A is an M-diagonalizer of n arguments. [If n = 1, then k = 0, and we take 0(x) = Eo(x, S(x)), and so O(x) - E(x, x).]
III
Corollary A If lI is an admissible mapping X -> X of combinatory logic, then for each n > 1, there is a combinator An such that for all combinators X,Yl,...,Yn-1: AnXY1...Yn-1 =XXY1...f7n-1
III
Of course for combinatory logic, such a combinator An is one satisfying the condition: Onxyli ... yn_1 - Anx(Sx)(Sy1) ... (6yn-1). Such a 500
combinator On we will call a H-diagonalizer.
§1 M-fixed points We recall (Chapter 12) that we called an element b a fixed point of a if b -> a, b. We now define b to be an M-fixed point of a if b -4a, b. The following theorem generalizes the second fixed point theorem of combinatory logic (Theorem 3.1, Chapter 19).
Theorem 1.1 If £ is diagonalizable then every element has an M-fixed point.
Proof By hypothesis, E contains an M-diagonalizer O(x). Given a, let c be such that for all x: c, x -4a, O(x). [We are assuming that all sequential systems are weakly connected.] Then E(c, c) -4 c, c -> a, 0(c) --4a, E(c, c) (since 0(c) = E(c, c)). We thus take b = E(c, c) and sob -> a, b. By Theorem 1.1 and Lemma A we have:
Corollary 1.2 If £ is admissible and of type 1, then every element has an M-fixed point.
The second fixed point theorem of combinatory logic is thus a consequence of Corollary 1.2. As another corollary of Theorem 1.1 we have:
Corollary 1.3 If £ is universal and diagonalizable then for any Efunction f (x), there is an element b such that b -+ f (b).
§2 The double M-fixed point theorem
383
Exercise 1 Prove Corollary 1.3.
Exercise 2 Is the second fixed point theorem of combinatory logic also a consequence of Corollary 1.3? Quasi-M- diagonalizers
Imo
Theorem 1.1 has a strengthening to which we will now turn. Call a function g(x, yl, ... , yk) a quasi-M-diagonalizer if for every element x there is an element x1 such that for all y ,7. .. , Ilk: q(-t1, 917... , Yk) E(x, xi 7 y17 ... , yk), where E(x, z, yl, ... , yk) is an exact E-function. [For
r'!
k = 0, q(x) is a quasi-M-diagonalizer if for every x there is an element x1 such that q(xl) -> E(x, xl).] Obviously every M-diagonalizer is also a quasi-M-diagonalizer (just take xl = x). Then Theorem 1.1 can be strengthened to
Theorem 1.1# If E contains a quasi-M-diagonalizer q(x) of one argument then every element has an M-fixed point.
Proof Exercise 3 below. III
Let us note that for - the identity relation and M the identity mapping, an M-diagonalizer is simply a diagonal function (as defined in Chapter 12); a quasi-M-diagonalizer is simply a quasi-diagonal function, just as an M-
fixed point is simply a fixed point. In Chapter 12 we proved that if E contains a quasi-diagonal function of one argument then every element has a fixed point. This, then, is but a special case of Theorem 1.10 above.
Exercise 3 Prove Theorem 1.1#.
§2 The double M-fixed point theorem III
In the last chapter we stated (Theorem 5.1) that for any combinators a and a there are combinators X and Y such that aXY - X and ,39Y - Y. [-4
The following is a generalization of this.
'C3
Theorem 2.1 If S is diagonalizable - or even if E contains an Mdiagonalizer of two arguments - then for any elements al and a2 there are elements b1 and b2 such that b1 -> al, b17 b2 and b2 -> a2, b1, b2.
Chapter 20. Extended sequential systems
384
Proof Suppose that E contains an M-diagonalizer 0(x, y). Given al and a2, let cl and c2 be elements such that for all x and y:
I{-
(1) cl, x, y -' a1, 0(x, y), 0(y, x) (2) c2, x, y - * a2, A
X), A
Y)
Then E(cl, C1, c2) -4 C1, El, c2 - a,, 0(C1v C2), A(C2, Cl) - a1, E(C1,
Also, E(C2, E2, El) - a2,2,1 -f
a2, L (C1, C2), CSI
el, C2), E(C2, E2, C1).
AA, Z4) -4 a2, E(c1, C1, C2), E(C2, c2, C1). We thus take b1 = E(c1, c1, C2) and b2 = E(c2, c2, c1).
Exercise 4 Show that the conclusion of Theorem 2.1 follows from the weaker hypothesis that E contains a quasi-M-diagonalizer q(x, y). [Note that this generalizes Theorem 1.1 of Chapter 14: if E contains a quasidiagonalizer of two arguments then S has the weak double fixed point property.]
Exercise 5 Show that if the conclusion of Theorem 2.1 holds for £, then for any elements a1 and a2 there are elements b1 and b2 such that b1 -> a1, b2 and b2 --4a2, b1 (we might call this property the M-cross-point property).
Exercise 6 Suppose that E is of type 1 and contains a quasi-Man]
diagonalizer of one argument. Does S necessarily have the M-cross-point property?
Exercise 7 State and prove an n-fold generalization of Theorem 2.1.
§3 Strong M-fixed point properties .ti
As might be expected, we define S to be strongly admissible if £ is admissible and M also satisfies the following condition:
M3 There is a E-function H(x) such that for all x: H(x) - x. U]'
"L7
Such a function H(x) we will call an M-neutralizer, or just a neutralizer for short, when the mapping M is fixed for the discussion. For combinatory logic with a strongly admissible mapping X
we of course take H(x) = Hx, where H is a combinator satisfying the condition HX - X. For applications to ordinary sequential systems (which can be looked at as extended sequential systems in which - is the identity relation and M is the identity mapping) we of course take H to also be the identity mapping.
§3 Strong M-fixed point properties
385
Adequate systems
.nom
.'3
"C3
^C3 ..d ,a?
p..
And now we shall define £ to be adequate if it is of type 2, diagonalizable, and contains an M-neutralizer H(x). By Lemma A, if £ is strongly admissible and'of type 1 and type 2, then £ is adequate. The rest of this chapter will be devoted to M-fixed point properties of adequate systems. In what follows, H(x) will be an M-neutralizer fixed for the discussion. The M-recursion and M-Myhill properties
We shall say that £ has the M-recursion property if for every element a there is a E-function O(x) such that for all x: 0(x)
a, x, 0(x)
Theorem 3.1 If £ is adequate then £ has the M-recursion property. .:$
't3
Proof Suppose £ is adequate. Then E contains an M-diagonalizer 0(x, y) of two arguments. Then, for any element a, take b such that for all y and x: b, y, x -> a, H(x), A(y, x)
Then E(b, 6,x)
->
a, H(x), 0(b, x)
->
a, x,E(b,b,x)
And so we take O(x) = E(b, b, x). [Since S is assumed to be of type 2, then E(b, b, x) is a E-function of x.]
alb
We shall say that £ has the M-Myhill property if there is a E-function O(x) such that for all x, 0(x) --4x, 0(x). [This is a uniform version of the M-fixed point property.] It is obvious that if S is universal and has the M-recursion property, then £ has the M-Myhill property (just take a such that a, x, y -4 x, y). And so as a consequence of Theorem 3.1 we have:
Theorem 3.2 If £ is adequate and universal then there is a E-function O(x) such that for all x: 0 (x) -> x, 0(x)
Chapter 20. Extended sequential systems
386
III
mar,
We note more specifically that if E is adequate and universal, then there is an element b such that for all y and x: b, y, x -> H(x), 0(y, x), and then E(b, b, x) is a E-function O(x) satisfying the condition: O (x) -4 x, 0(x). For combinatory logic, this tells us that for a strongly admissible mapping II, there is a 11-sage - a combinator 0 such that for all X, we have OX - X 0X, and so the uniform second fixed point theorem of combinatory logic (Theorem 5.3, Chapter 19) is now established. We can, in fact, now see how such a combinator 0 can be obtained. We let b be a combinator satisfying the condition: byx Hx(A3y(6y)(Sx)) (so that for any combinators Y and X, bYX - XYYX) and we take 0 = bb.
There is, however, another argument, which does not appear to be
0(D
in_
generalizable to extended sequential systems, and which yields the uniform second fixed point theorem as a corollary of the second fixed point theorem.
0q9
III
III
We take a combinator a satisfying the condition: ayx - Hx(Ay(bx)). We then let 0 be a 11-fixed point of a (0 - a0), and so for any X: OX - aOX HX(A9(6X)) - X(AOX) - XOX. Thus 0 is a 11-sage. This argument reveals an additional fact of interest, namely that there
'C3
is a combinator a all of whose 11-fixed points are 1I-sages. Coming back to extended sequential systems, Theorem 3.1 (and hence also Theorem 3.2) can be strengthened as follows. For any positive n, let us say that E is n-adequate if E is of type 2 and contains a neutralizer H(x) and an M-diagonalizer A(x1i . . . , x,,,) of n arguments. Thus E is adequate if E is n-adequate for all positive n. It is obvious from the proof of Theorem 3.1 that it would still hold if we replaced "adequate" by "2-adequate." And
now let us say that E is quasi-n-adequate if E is of type 2 and contains a neutralizer H(x) and a quasi-M-diagonalizer q(xj, ... , xn) of n arguments. By a slight modification of the proof of Theorem 3.1 it can be seen that if ,F is quasi-2-adequate, then E has the M-recursion property (Exercise 8 below). This not only yields the second fixed point theorem of combinatory logic, but also generalizes Theorem 2.1 of Chapter 13 that says that if E is of type 2 and contains a quasi-diagonalizer Q(x, y) of two arguments, then S has the recursion property.
Exercise 8 Prove that if E is quasi-2-adequate then E has the Mrecursion property.
Exercise 9 Suppose that E is universal and that .6 is adequate - or even quasi-3-adequate. Prove that there is a E-function q(x, y) such that for all x and y: 0(x, y) -> x, y, 0(x, y). [Note that Theorem 5.4, Chapter 19, is a consequence of this.]
§4 Uniform double M-fixed point properties
387
§4 Uniform double M-fixed point properties Theorem 4.1 If S is adequate then for any elements al and a2 there are E-functions 01(x, y), 02 (x, y) such that for all elements x and y: (1) 01(x, y) - al, x, y, 01(x, y), 02(x, y) (2) 02(x, y) - a2, x, y, 01(x, y), 02(x, y)
Proof (sketch) Take b1, b2 such that z1, x, y) bl, z1, z2, x, y - al, H(x), H(y), A(z1, z2, x, y), A b2, zl, z2, x, y - a2, H(x), H(y), A(z2, zl, x, y), A(zl, z2, x, y)
Then take 01(x, y) = E(bl, b1, b2, x, y) and 02(x, y) = E(b2, b2, b1, x, y).
Corollary 4.2 If S is adequate then for any elements a1, a2 there are E-functions 01(x, y), 02 (x, y) such that for any x, y:
-> a1, x,
01(x, y), 02(x, y) and 02(x, y) - a2, y, 01(x, y), 02(x, 9) -
Corollary 4.3 If E is adequate, then for any elements al, a2 there are E-functions O1(x), 02(x) such that for all x: 01(x) -> a1i x, q1(x), 2() and 02(x) - a2, x, 01(x), q2() From Corollary 4.2 we have:
Theorem 4.4 If E is adequate and universal then there are E-functions 01(x, y), 02 (x, y) such that for all x, y: 01(x, y)
x, 01(x, y), 02 (x, y)
III
02 (x, y) - y, 01(x, y), 02 (x, y)
As a consequence of Theorem 4.4, we see that for combinatory logic with a strong admissible mapping X -49 there are combinators 01 and 02 such that for all X and Y, 019Y = X01XY 02(XY) and 02XY Y01XY 02(XY) (Theorem 5.5, Chapter 19). But this is also a consequence of the second double fixed point theorem of combinatory logic (Theorem 5.1, Chapter 19) together with the following:
Chapter 20. Extended sequential systems
388
Theorem 4.5 If II is a strongly admissible map X -49 for combinatory logic, then there exist combinators al, a2 such that any II-double fixed point (01, 02) of (al, a2) is a II-DM pair (i.e., if a,
01 and
02, then for any combinators X and Y, 01XY - Xqi1XY 02XY and 02XY - Y01XY 02XY.
Proof (Sketch) Take al, a2 satisfying the conditions: alzwxy alzwxy
Hx(A3z(6x)(6y))(A3w(6x)(6y)) Hy(A3z(Sx)(6y))(A3w(Sx)(6y))
§5 M-nice and M-symmetric functions Theorem 5.1 If E is adequate and universal then: (a) There is a E-function N(z, x, y) such that for all x and y: N(z, x, y) -> zN(x, x, y), N(y, x, y)
(b) There is a E-function s(x, y) such that for all x and y: x, s(x, y), s(y, x)
Proof (Sketch) (a) Take b such that b, w, z, x, y -> H(z), 0(w, x, x, y), 0(w, y, x, y), and take N(z, x, y) = E(b, b, z, x, y).
(b) Take b such that b, w, x, y -> H(x), 0(w, x, y), 0(w, y, x), and take s(x, y) = E(b, b, x, y).
Exercise 10 Consider combinatory logic with a strongly admissible mapping II.
(a) Show that there is a combinator a such that if N is any II-fixed point
of a, then for all X, Y, Z: NZYX - ZNXXY NYXY. (b) Show that there is a combinator a such that if s is any II-fixed point
of a, then for all X, Y: sXY - XsXY sYX.
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Can
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[35] Alfred Tarski. Der wahrheitsbegriff in den formalisierten sprachen. Studia Philisophie, 1:261-405, 1936.
[36] Alan Turing. The p-function and Ak conversion. Journal of Symbolic Logic, 2:153-163, 1937.
[37] Erik G. Wagner. Uniformly reflexive structures. Transactions of the American Mathematical Society, 144, 1969.
Index n-k Myhill property, 284 n-m recursion property, 284 n-adic concatenation, 133 n-adic notation, 132 n-ary predicates, 49 n-place predicates, 49 n-sage, 336 n-synchronizer, 286 n-universal, 239 n-universal element, 239 ord(II), 286 x is part of y, 152 '-'
(U), 135
t..
Ai, 182
chi-
K-tester, 367 M, 320 M-Myhill properties, 385 M-Myhill property, 385 (s,
L', 79 1I-double fixed point, 378 Ea, 124 E0-formulas, 149 E0-relation, 124 E1-formulas, 149 E1-relation, 124 A#, 50 A-representable, 375 DMk, 277 w-consistent, 174 w-inconsistency, 70 w-inconsistent, 174
x begins y, 152 x ends y, 152 xBy, 152 xEy, 152
H
Wo, 50 L1, 75
'C3
M-diagonalizers, 381 M-fixed points, 382 M-neutralizer, 384 M-nice functions, 388 M-recursion, 385 M-symmetric functions, 388 S-tester, 367 W*, 50
s-,
Jn(x1i...,x,,), 128
(C)
J(x, y), 126
CSC
pip
B!", 182 G,,,, 153
xPy, 152 (many-one) reducible, 178 CLo, 357 DM, 273 DM', 275 DMo, 273 DMk, 273 DRo, 267 DR1, 267 DR2, 267 2-sage, 335
adequate systems, 385 admissible, 92, 380 admissible functions, 67 admissible name functions, 366 admissible notation, 114
Index
393
almost a semi-D.G. predicate, 206 almost represents, 205 application, 315 applicative system, 25, 230, 315 Arithmetic, 78 arithmetic, 78 assistant sage, 332 associate, 7, 206 axiomatizable, 201
."S.
.._
:D:
208
pop
(CD
effective sequence, 210 effectively a Rosser system for sets 208
effectively inseparable, 193 effectively unconquerable, 224 effectively universal, 203 effectively universal systems, 203 elementary dyadic arithmetic, 113 elementary formal systems, 102 EM, 311 Enumeration Theorem, 142 enumeration theorem, 141 equivalence relation, 17 exact functions, 234 exact separability, 63 exactly separates, 63 ...
4,4
D.G., 187 D.G. function, 187 D.G. predicate, 205 D.U., 186 D.U. pairs of sets, 186 decidable, 48, 201 definability, 63 defines, 375 diagonal function, 50, 236 diagonalizable, 381 diagonalization, 3, 17, 50
effective representation system, 20 effective Rosser systems, 219 effective Rosser systems for sets,
(CD
193
completely represents, 64 consistent, 48 constructive definability, 122, 124 contrarepresentable, 57 creative, 181 cross-point, 19 cross-point property, 336
cm.
Church's theorem, 377 CL, 357 class abstracts, 79 co-productive, 181 combinatory completeness, 359 complete, 48 complete representability, 63 completely creative, 181 completely effective inseparable,
..1
pmt
bar fixed points, 309 bases, 326
`i-
'47
diagonalizer, 17, 236 diagonalizes, 17 double enumeration, 157, 182 double enumeration theorem, 183 double fixed points, 263, 336 double indexing, 182 double iteration theorem, 185 double Myhill property, 273 double quasi-diagonal lemma, 269 double recursion properties, 267 double universality, 182 doubly generative pairs, 187 doubly generative systems, 205 doubly indexable, 182 doubly universal, 186 doubly universal pairs, 158 duplicator, 320 dyadic arithmetic, 113 dyadic concatenation, 117 dyadic Godel numbering, 129 dyadic Godel numbers, 153
394
Index
inverse functions K2 , 128 involute, 186 involution, 186 iteration properties, 177 iteration theorem, 145
f.r. (formally representable), 104 finite existential quantification, 122 finite quantifications, 122 finite universal quantification, 122 first diagonal lemma, 50 first predecessor, 367 first-order definable, 124 fixed point, 52 fixed point combinators, 331 fixed point functions, 249 formal combinatory logic, 357 formal system, 105 formally representable over K, 104 full admissibility, 369 fully admissible, 369
Lemma N, 247 Lemma QQ, 269 lexicographical Godel numbering,
C."
explicit transformations, 106 expresses, 56, 77 expressibility, 77 expressible, 56 extended Myhill property, 311 extended recursion property, 251 extended sequential systems, 380 extension of alphabets, 111 extensional, 329, 376
Kleene pair (Kl, K2), 189 Kleene's symmetric incompleteness theorem, 68
aCV
132, 133
Godel number, 26, 48, 49, 78 Godel sentences, 26, 79, 140 Godelizing operations, 91 generative, 147, 179, 204 generative function, 147, 179 generative predicate, 204 generative sets, 179 generative systems, 204 hyperconnected, 257
284
multiple fixed points, 336 multiple symmetric, 284 Myhill property, 228
near D.G. predicate, 206 near diagonalization, 18 near diagonalizer, 18, 247 near generative predicate, 204 near normalization, 85, 89 near semi-D.G. predicate, 206 negation property, 58 neutralizer, 384 nice combinators, 339 nice functions, 281 norm, 4, 15 normal, 52 normality, 52 normalization, 4 normalization lemma, 247 normalizers, 245 .N. .N.
c.o
idempotent, 329 incomplete, 48 inconsistent, 48 index, 142 indexicals, 1 inverse functions K and L, 127
master predicate, 210 minimization, 121 mirrors, 380 modus ponens, 103 multiple fixed point properties, 263,
omniscient function, 224
Index
395
one-sided quotation, 7
off=
cad
tea)
Second diagonalization lemma, 60 second fixed point theorem of combinatory logic, 374 second predecessor, 367 self-reference in CLo, 366 semi-D.G., 190, 205 semi-D.G. function, 190 semi-D.G. predicates, 206 semi-D.U., 194 semi-reducibility, 193 semi-reducible, 193 semi-Rosser systems, 209 sentential systems, 231 separability, 60 sequential systems, 228 solvability over K, 110 solvable over K, 110 SP, 287 SR, 287 stable, 70 standard systems, 177 stem functions, 209 C))
.°.°.°
Scott, 376
CC))
quasi M-diagonalizers, 383 quasi-diagonalizer, 237 quasi-exact, 237
sage, 331 sage producer, 334 satisfies, 77 "N"
pairing function, 328 Parikh, 72 Peano notation, 115 pre-admissible mappings, 367 predecessor combinators, 367 predicate, 48 predicate of n arguments, 49 predicates of degree n, 49 primitive recursive functions, 119 productive, 181 property Bn,, 309 propositional combinators, 328 provability by stages, 69 provable sentences, 48
Rice's theorem, 376, 377 Rosser function, 208, 219 Rosser system for relations, 208 Rosser system for sets, 208 Rosser systems, 207 rule D, 102 rule of detachment, 103
C))
t;:
cad
'C3
(1)
C)R
'C3
00O
S."
113
recursively inseparable, 192 reduces, 178, 186 reducible, 147, 186 refutable sentences, 48 regular, 121 repeat, 7 represent, 49 representability, 104 representation function, 51 representation system, 48, 49
0.'
r.e., 113 r.e. sequence, 210 r.e. system, 201 recursion property, 229, 245 recursive, 113 recursive enumerability, 113 recursive functions, 118 Recursive pairing functions, 126 recursively enumerable, 113 recursively enumerable relations,
strong M-fixed point properties, 384
strong admissibility, 369 strong double Kleene property, 272 strong fixed point functions, 253 strong Kleene property, 258 strong separability, 62 strongly admissible, 369, 384
Index
396
cad
strongly connected, 253 strongly connected systems, 253 strongly equivalent, 231 strongly separates, 62 successor combinator, 371 SW, 287 symmetric combinators, 339 symmetric functions, 279, 310 synchronization, 286 synchronized fixed point properties, 286, 287 synchronized fixed point theorems, ti,
cm.
288
synchronized Myhill property, 287 synchronized recursion property, 287 .,.
synchronized sequences of func-
tions, 286 synchronized weak fixed point property, 287 synchronizer of order n, 286 Tarskification, 82 tarskification, 88
term t of (E), 103 "'O
term definable, 78 the double M-fixed point theorem, 383
the universal system (U), 135 Theorem EM, 311 truth combinators, 328 Turing sages, 333 type 2 approach, 249, 269 types 0,1,2, 233 unary notation, 115 unconquerable systems, 223 undecidable, 48, 201 uniform double M-fixed point properties, 387 uniform double iteration theorem, 184
uniformly D.U., 186 uniformly semi-D.U., 195 uniformly universal, 178 universal, 147, 178, 239 universal sets, 178 universal systems, 239 unstable, 70
very strongly connected, 256 very strongly connected systems, 278
very strongly equivalent, 231 voracious elements, 329 weak double fixed point property, 263, 337 weak double Kleene property, 266 weak fixed point property, 228 weak fixed point function, 249 weak Kleene property, 229 weakly connected, 233 weakly equivalent, 231 weakly separates, 60
zero tester, 371
