Aspects of Incompleteness (Lecture Notes in Logic 10)

Lecture Notes in Logic

Per Lindstrom

Aspects of Incompleteness

Springer

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Lecture Notes in Logic Editors: S. Buss, San Diego A. Lachlan, Burnaby T. Slaman, Chicago A. Urquhart, Toronto H. Woodin, Berkeley

10

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Per Lindstrom

Aspects of Incompleteness

Springer

Author Per Lindstrόm Department of Philosophy University of Gothenburg S-41298 Gothenburg, Sweden e-mail: [email protected]

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Lindstrό'm, Per, 1936A s p e c t s of I n c o m p l e t e n e s s / Per L i n d s t r ό m . p. cm. — (Lecture notes 1n l o g i c 10) Includes b i b l i o g r a p h i c a l references and I n d e x . ISBN 3-540-63213-1 ( s o f t c o v e r : a Ik. p a p e r ) 1. I n c o m p l e t e n e s s t h e o r e m s . 2. Recursion t h e o r y . II. S e r i e s . QA9.65.L56 1997 511.3—dc21

I.

Title.

97-25726

CIP

Mathematics Subject Classification (1991): 03F30, 03F40 ISBN 3-540-63213-1 Springer-Verlag Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. © Springer-Verlag Berlin Heidelberg 1997 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: Camera-ready by the authors SPIN: 10633635 46/3142-543210 - Printed on acid-free paper

To Eva for all she's done for me

PREFACE

By GodeΓs incompleteness theorem there is no complete axiomatization of mathematics, not even of first order arithmetic. This leads naturally to the project of investigating the family of, inevitably incomplete, arithmetical theories. In this book we present some of the results obtained in pursuing this aim. A brief summary of the contents of the book is given in the introduction. GodeΓs main idea was to translate metamathematical statements, concerning formulas, formal theories, provability in such theories etc., into arithmetic - the arithmetization of metamathematics. Combining this with the technique of self-reference, he was able to construct a (formal) sentence which is undecidable, neither provable nor disprovable, in a given theory T of arithmetic, thereby demonstrating, in an entirely novel way, that T is incomplete. The ideas developed by Godel proved fruitful far beyond their original application. Later essentially new ideas were added. Most important are the basic concepts (and results) of recursion theory - these are needed for a completely general formulation (and proof) of GodeΓs theorem - notable are also the so called partial truth-definitions, the idea of formalizing the proof of the completeness of first order logic in arithmetic, and the general concept of interpretability. In this book we shall be concerned almost exclusively with properties that are common to all sufficiently strong, axiomatizable theories. Thus, for example, investigations of particular theories such as, for example, Peano Arithmetic and its fragments, requiring special, model-theoretic or proof-theoretic, methods, fall outside the scope of the book. But even so, the choice of material in some respects reflects the author's preferences; this is particularly true of Chapter 7. Credits for results, proofs, and exercises, remarks on alternative proofs, related results, etc. are given in the notes following each chapter. Results (exercises) not attributed to anyone are either easy, folklore, or due to the author. The reader is assumed to be acquainted with (first order) predicate logic including Henkin's completeness proof. We also presuppose knowledge of the elements of recursion theory. Finally, the reader may find our presentation difficult to follow unless (s)he has previously seen a more detailed development of formal arithmetic, an explicitly defined arithmetization of metamathematics, and a full proof of GδdeΓs theorem. I am grateful to Daniel Vallstrom and Rineke Verbrugge; they have very kindly and carefully read substantial portions of this book, pointed out numerous (minor) errors and suggested many improvements. The responsibility for the blemishes that remain is, of course, mine. May 1997. P.L.

CONTENTS

CHAPTER 0. Introduction

1

CHAPTER i. Preliminaries Exercises Notes

5 20 22

CHAPTER 2. Incompleteness §1. Incompleteness §2. Consistency statements §3. Independent formulas §4. The length of proofs Exercises Notes

23 23 25 31 33 35 40

CHAPTER 3. Numerations of r.e. sets §1. Numerations of r.e. sets §2. Types of independence Exercises Notes

42 42 45 50 51

CHAPTER 4. Axiomatizations §1. Finite and bounded axiomatizability; reflection principles §2. Irredundant axiomatizability Exercises Notes

52 52 57 59 61

CHAPTER 5. Partial conservativity Exercises Notes

62 70 73

CHAPTER 6. Interpretability §1. Interpretability §2. Faithful Interpretability Exercises Notes

75 75 84 88 91

CHAPTER 7. Degrees of interpretabilty §1. Algebraic properties §2. A classification of degrees

94 94 102

x

Contents

§3. Σ! and Πx degrees Exercises Notes CHAPTER 8. Generalizations §1. Incompleteness §2. Axiomatizations §3. Interpretability Notes

104 114 117 119 119 121 122 124

REFERENCES

125

INDEX

130

NOTATION

132

0. INTRODUCTION

Let T be a sufficiently strong theory formalized in the language LA of (first order) arithmetic. Following Godel, we want to show that there is a sentence φ of LA which is true (of the natural numbers) but not provable in T. GδdeΓs idea was to achieve this by constructing φ in such a way that (*) φ is true if and only if φ is not provable in T. Then, assuming (for simplicity) that all theorems of T are true, we are done. For, suppose φ is provable in T. Then, by (*), φ is not true and so, by hypothesis, φ is not provable in T. Thus, φ is not provable in T But then, by (*), φ is true. One way to achieve (*) is to find a sentence φ which, in some sense, "says" of itself that it is not provable in T. There are then three major difficulties. First of all, the sentences of LA deal with natural numbers, they do not deal with syntactical objects such as sentences (of a formal language), proofs, etc. Secondly, even if some of the sentences of LA can, somehow, be understood as dealing with syntactical objects, it is not clear that it is possible to "say" anything about provability (in T) using only the means of expression available in LA. And, finally, even if this is possible, there may be no sentence which "says" of itself that it isn't provable. Godel, however, was able to overcome these difficulties. The first problem is solved by assigning natural numbers to syntactical expressions in a certain systematic way. This is sometimes called a Godel numbering, and the number assigned to an expression, the Godel number of that expression. Thus, the numeral of the number assigned to an expression can be regarded as a name of that expression and certain number theoretic statements can be regarded as statements about syntactical objects. (In what follows "φ is a formula", "p is a proof", etc. are short for "φ is the Godel number of a formula", "p is the Godel number of a proof", etc.) To overcome the second difficulty Godel (re)invented the primitive recursive functions (sets, relations). He showed that a number of crucial properties of (Godel numbers of) expressions, such as that of being a (well-formed) formula, are primitive recursive. In particular, Godel showed that, if the set of axioms of T is primitive recursive, this is also true of the relation PRFτ(φ,p): p is a proof of the sentence φ in T φ is provable in T, PRτ(φ), if and only if ΞpPRFτ(φ,p). This property, however, is not (primitive) recursive. Godel then went on to prove that all primitive recursive functions (sets, relations) are definable in LA. Thus, in particular, there is a formula Prfτ(x,y) of LA defining PRFT(k,m). But then Prτ(x) := 5yPrfτ(x,y) defines PRT(k). (In what follows we write Th φ for PRτ(φ) ) Godel, however, proved more and this is crucial: for every sentence φ, Th φ if and only if Th Prτ(φ). (This is the first time we use the assumption that T is sufficiently strong; but, of course, if T isn't, T is incomplete for that reason.)

2

0. Introduction

This takes care of the second difficulty. So now there remains only the problem of finding a formal sentence which "says" of itself that it is not provable in T. Gδdel solved this problem in the following way. Consider the substitution function SBST(k,m) defined by: SBST(k,m) = the formula obtained from the formula m by replacing the free variable "x" by the numeral of the number k, if m is a formula, = 0 otherwise. This function is primitive recursive and so is defined in T by a formula Subst(x,y,z) in the sense that for any formula δ(x) and any number k, Th Vz(Subst(k,δ,z) <-> z = δ(k)); in other words, for all k, δ(x), T proves that: "z satisfies Subst(k,δ,z) if and only if z is the formula δ(k)". Now consider the formula 5y(Subst(x,x,y) Λ -Prτ(y)), call it γ(x). Let θ be γ(γ). Intuitively, θ "says" that the result of replacing the variable "x" by the numeral for the number γ in the formula γ(x) is not provable in T. But this result, γ(γ), is θ itself. Thus, θ "says" that θ is not provable in T. Formalizing this argument we obtain: (**) Th θ ^ -Prτ(θ). (This is an instance of the very important fixed point lemma; Lemma 1, Chapter 1.) And now GδdeΓs proof can be completed as follows. First we show that (***) Th θ. Suppose Th θ. Then Th Prτ(θ). But then, by (**), Th - θ and so T is inconsistent (whence Th 1, where _L := ->0 = 0), contrary to assumption. Thus, (***) holds. But this is exactly what - Prτ(θ) "says". So -ιPrτ(θ) is true and consequently, by (**), θ is true. Let Conτ be the sentence ->Prτ(_L). Conτ is then a natural formalization of "T is consistent". In proving (***) we actually proved that if Thθ then Th_L and so that if T is consistent (Th_L), then Thθ. It turns out that this proof can be formalized in T provided that T is sufficiently strong. Thus, Th Conτ -» -πPrτ(θ) and so, by (**), Th Conτ -> θ. But then, since Th θ, it follows that Th Conτ; in other words, T cannot prove its own consistency. This is GόdeΓs second incompleteness theorem. This, in brief, is what Gδdel accomplished (restricted to theories in LA; generalizations to other theories containing arithmetic, for exampel set theory, are straightforward). In GδdeΓs original proofs it is assumed that the set of axioms is primitive recursive. Subsequently, when the (general) recursive functions had been defined, it turned out, however, that this assumption could, without altering the structure of the proofs, be replaced by the weaker assumption that the set of axioms is recursive. In fact, it became clear that GόdeΓs first incompleteness theorem holds for all formal systems, in the most general sense, and is actually a result belonging to recursion theory: the set of true (Π1) sentences of L A is not recursively enumerable.

0. Introduction

3

In (the above sketch of) GδdeΓs proof, and in virtually all proofs in the following chapters, the method of arithmetizing metamathematics, i.e. translating metamathematical concepts, statements, etc. into arithmetic, plays a central role. This method is based on a large number of definitions and preliminary results. In Chapter 1 we introduce the basic notation and terminology and state a number of Facts concerning these notions. These Facts will not be proved but references will be given; some of them are proved in almost every exposition of GδdeΓs theorems, others require quite extensive proofs that would be out of harmony with the rest of the book. In Chapter 1 we also prove the fixed point lemma (Lemma 1.1), the essential undecidability of Robinson's Arithmetic, Q, a very weak finite subtheory of PA, (Theorem 1.2), and the nonexistence of truth-definitions (Theorem 1.3). In Chapter 2 we present the first and most important results of the subject, GόdeΓs incompleteness theorem and his (second) theorem on the unprovability of consistency (Theorems 2.1 and 2.4). GδdeΓs results were subsequently improved in various respects and we present some of these improvements. The main result of Chapter 3 is that, assuming that T contains a minimum of arithmetic (Q), every recursively enumerable set is numerated by a (Σι) formula in T (even if not all Σ^ sentences provable in T are true) (Theorem 3.1). We also prove some refinements of this result. Given that the set of axioms of a theory T is infinite, it is natural to ask if these axioms can be replaced by a finite set of axioms. In Chapter 4 we apply the so called reflection principles to prove some negative results concerning this problem (Theorems 4.1, 4.2). For example, neither Peano Arithmetic, PA, nor any one of its consistent extensions (in LA) is finitely axiomatizable. On the other hand, every extension of PA has an axiomatization which is irredundant in the sense that none of the axioms can be derived from the other axioms (Theorem 4.6). We also prove the existence of not irredundantly axiomatizable theories (Theorem 4.7). Let Γ be a set of sentences, for example Σn or Πn. A sentence φ is Γ-conservative over T if every sentence in Γ provable in T + φ is provable already in T. Partial conservativity is studied in Chapter 5, where the basic existence theorems are proved (Theorems 5.2, 5.3,5.4); it plays an important role in Chapters 6 and 7. An interpretation of a theory S in a theory T is, roughly speaking, a function t on the set of formulas of S into the set of formulas of T such that t preserves logical form and Tht(φ) whenever Shφ. S is interpretable in T, S < T, if there is an interpretation of S in T. These concepts were introduced by Tarski. If, in addition, Sh φ whenever Tht(φ), we shall say that t is faithful and that S is faithfully interpretable inT. Interpretability was originally used as a tool in proofs of (relative) consistency and undecidability Interpretability (in arithmetical theories) is studied for its own sake in Chapter 6. The key result is that if T is an extension of PA and Cons is a sentence which (in a suitable sense) in T "says" that S is consistent, then S < T + Cons (the arithmetization of GδdeΓs completeness theorem; Theorem 6.4). From this it

4

0. Introduction

follows that S < T if and only if for every finite sub theory S' of S, Th ConS' (Lemma 6.2) and that if S, too, is an extension of PA, then S < T if and only if every Γ^ sentence provable in S is provable in T (Theorem 6.6). We also prove similar characterizations of faithful interpretability (Theorems 6.13, 6.14). Mutual interpretability is an equivalence relation; its equivalence classes will be called degrees of interpretability. Let T be a consistent extension of PA. The degrees of extensions of T, partially ordered by the relation induced by <, form a distributive lattice (Theorem 7.1). This lattice is studied in Chapter 7 both from a purely algebraic point of view and in terms of the nature of the theories belonging to a given degree. It is quite often true in the following pages that a result stated for extensions of PA actually holds for all extensions of some (much) weaker, sometimes finitely axiomatizable, subtheory of PA. We shall, however, pay little attention to facts of this type: what we are mainly interested in here are the properties shared by all theories containing a sufficient amount of arithmetic. But, if a result is (proved to be) true of Q (and its extensions), this will be explicitly noted. Almost all the results presented in this book hold in a very general setting. In spite of this we shall in Capters 1-7, for reasons of simplicity and readability, formulate (and prove) these results for theories formalized in LA only. We partly make up for this lack of generality in Chapter 8, which is devoted to generalizations, usually straightforward, to theories formalized in other languages; the most important of these is the language of (first order) set theory.

1. PRELIMINARIES

In this chapter we introduce the basic notation and terminology which will be used throughout this book. We also state a number of basic Facts. These Facts will not be proved; some of them are rather obvious (and easy to believe), others are substantial and well-known theorems; further Facts will be stated when they are needed. These Facts are sufficient for most of the proofs in this book; the chief exceptions are the proofs of Theorems 3.5 and 6.4. Finally, we prove the very important fixed point lemma (Lemma 1) and apply it to prove that Robinson's Arithmetic Q is essentially undecidable (Theorem 2) and that in extensions of Q there are no truth-definitions (Theorem 3). The language LA of elementary arithmetic can be described as follows. The alphabet consists of: the propositional connectives: -•, Λ, v, —», <-^, the quantifiers: 5, V, the equality symbol: =, symbols used to form (individual) variables: v,', parentheses: (,), the arithmetical constants: 0, S, +, x. (The intended interpretation of S is the successor function.) Thus, the alphabet is finite. The variables of LA are the expressions v, V, v", etc. We write vn for v followed by n occurrences of '. In most contexts x, y, z, u, v, w, possibly with subscripts etc., will be used for variables. The terms, formulas, and sentences of LA are defined as usual. Among the terms we distinguish the numerals 0, SO, SSO, SSSO,.... These will be written 0, 1, 2, .... Thus, we shall omit bars and other devices ordinarily used to indicate numerals (or Godel numbers) and use the same symbols for natural numbers and for formal numerals. In most cases this will cause no trouble as long as the symbols for formal variables are kept strictly apart from the symbols for natural numbers (and numerals). For the latter we use k, m, n, p, q, r, s, possibly with subscripts etc. and symbols for formulas (see below). N is the set of natural numbers. For sentences and formulas of LA we use lower case Greek letters. Sentences will be written as φ, ψ, θ, χ, etc. and formulas as α(x), β(x,y), γ(x), ξ(x), T^X^...^), p(x,x')/ τ(x), ξ, γ, etc. The variables displayed are almost always exactly the free variables of the formula. ξ(y) is obtained from ξ(x) by replacing x by y, assumed not to be free in ξ(x), and, possibly renaming bound variables in the usual way. ξ(k) is obtained from ξ(x) by replacing x by the numeral k (or, if you prefer, by the numeral for the number k). This generalizes in the obvious way to substitutions involving more than one variable. We use := to denote equality between formulas. By a theory T we understand a set of sentences (to be thought of as the (nonlogical) axioms of T). (It would be inconvenient to identify a theory T with the set of

6

1. Preliminaries

its theorems, since quite often we need to know that there is a formula binumerating (defined below) (the set of axioms of) T.) Note that, although we shall mainly be interested in theories that are reasonable from an arithmetical point of view, such reasonableness is not part of the concept theory. T + φ is the theory obtained from T by adding φ as a (new) axiom. T + X, where X is a set of sentences, is understood similarly. We assume given a fixed complete deductive calculus PL for first order logic. Referring to PL certain (finite) formal objects (sequences of sentences) are proofs (in T). A proof is a proof of its last sentence. The sentence φ is provable in T, Th φ, if there is a proof of φ in T. Th ξ(x1,...,xn), where x1,...,xn are all the free variables of ξ(x1,...,xn), is short for Th Vx1...xnξ(x1,...,xn). Th(T) is the set of theorems of, i.e. sentences provable in, T. If X is a set of sentences, we write Th X or XH T to mean that Th φ for every φe X. Thus, SH T means that S is a subtheory of T (T is an extension of S). We write h φ for 0h φ, where 0 is the empty set. Thus, h φ means that φ is provable in logic (PL). N = (N, +, x, S, 0) is the standard model of arithmetic. A sentence φ is true if it is true in N. A theory is true if all its axioms (and therefore, all its theorems) are true. There are two (true) theories PA (Peano Arithmetic) and Q (Robinson's Arithmetic) that will play a prominent role in what follows. Q is a finite theory; its axioms are (we omit the initial universal quantifiers): Ql Sx = Sy —» x = y, Q2 -Ό = Sx, Q3 -.0 = x -> 3y(x = Sy), Q4 x + 0 = x, Q5 x + Sy = S(x + y), Q6 x x 0 = 0, Q7 x x Sy = (x x y) + x. We introduce the two-place predicates < and < by means of the definitions: x < y =df 5z(z + x = y), χ

< y =df x ^ y Λ ~χ = y

With our present simplified notation certain (harmless) ambiguities arise. For example, 2 + 3 can be read as a numeral but also as an expression containing the symbol +. In Fact 1 below we have indicated that the latter is the intended reading by underscoring the relevant function symbol. But, of course, terms such as Sy, x + y, x + 3, 4 x z, etc. are unambiguous. Also, with very few exceptions, it is, in view of Fact 1, not important which way, say, 2 + 3 is understood. Fact 1. The following formulas are provable in Q for all k, m, n, (i) -i k = m for k Φ m, (ii) k ± m = k + m, (iii) k x m = k x m, (iv) x < m — > x = 0vx = l v...v x = m, (v) x < m v m < x.

1. Preliminaries

7

Q is a very weak theory. For example, the sentences Vx(0 + x = x), Vχ-<(x = Sx), Vx(x < x), cannot be proved in Q. The axioms of PA consist of the axioms of Q plus the (universal closures) of formulas of the form α(0) Λ Vx(α(x) -> cc(Sx)) -» Vxα(x). (Here α(x) may contain free variables other than x.) This is the induction scheme and is as close as we can get to the full (second order) induction axiom in first order arithmetic. From the induction scheme we can derive the least number principle: for every formula α(x) as above, PAh Ξxα(x) -> 5x(α(x) Λ Vy(y < x -> ->(x(y)). Obviously, QH PA. In PA axiom Q3 is redundant. Vxy(x < y v y < x) is provable in PA, but not in Q. In fact, this is sometimes the sole reason for writing "PAh" rather than "Qh". Godel proved that every primitive recursive function is definable in first order arithmetic. Formalizing this proof, he proved that: Fact 2. For each primitive recursive function f^ko,...,]^) there is a formula δf(xo,...,xn,y) such that for all ko,...,!^, (i) Qh δfίV ^y) y = fίVΛi)(ii) PAh δf(xo,...,xn,y) Λ δf(xo,...,xn,z) -> y = z, (iii) PAh ayδf(x0,...,xn,y). In Fact 2 (i) f(k0,...,kn) is, of course, a numeral, i.e. does not contain the symbol f. A formula δ(xo,...,xn/y) sucn tnat f°Γ a^ ko/ /kn/ Thδ(V vk n ,y)~y = f(k0.....kj will be said to define f in T. For (general) recursive functions we have the following weaker Fact (for Fact 3 (b), see below): Fact 3. (a) For every (total) recursive function f^ko,...,!^), there is a formula δf(xo,...,xn,y) defining f in Q. The formula p(x0,...,xn) numerates the relation R^,...,!^) in the theory S if for all kQ,..., kn (as usual "iff" is short for "if and only if"), RίV v k J i f f S h p t k o.....kj. Thus, ξ(x) numerates X in S if for every k, keXiffShξ(k). p(x0,...,xn) Enumerates the relation R(ko,...,kn) in S if for all ko,...,!^, not R(ko,...,kn) iff Sh In particular, ξ(x) binumerates X in S if for every k,

8

1. Preliminaries

keXiffShξ(k), teXiffShξ(k). If a formula binumerates X (R) in S, it binumerates X (R) in every consistent extension of S. If S is recursively enumerable (r.e.), any set (relation) numerated by some formula in S is r.e. and any set (relation) binumerated by some formula in S is recursive. Fact 3 (a) has the following: Corollary 1. (a) A set X (relation R) is recursive iff there is a formula binumerating X (R) in Q. (b) A set X (relation R) is r.e. iff there is a formula numerating X (R) in Q. This corollary and most of those below in this chapter are easy consequences of the relevant Facts; their proofs are, therefore, left to the reader. Note that, in view of Corollary 1, we have the remarkable fact that any set X (relation R) which is (bi)numerated by some formula in some r.e. theory, is (bi)numerated by a (possibly different) formula already in Q. We write 3x β(x)). Ξxφ. Corollary 2. If p(xl/.../xn) is PR, then Qh pίk^...,]^) iff p^,...,]^) is true. It follows that (i) every PR formula is decidable in Q, (ii) a set X (relation R) is primitive recursive iff there is a PR formula binumerating X (R) in Q. A PR formula binumerating X (R) in Q will be called a PR Enumeration ofX (R). A

1. Preliminaries

9

numeration ofX is a formula numerating X in PA. In what follows Corollary 2 will be applied without further mention. Suppose PAH T. Then, by Fact 2, if f^,...,!^) is a primitive recursive function and δf(x0/.../xn/y) is the corresponding formula, we can add the function symbol f to the language of T and add as a new axiom. (Thus, we shall be using the same function-symbol in the object language as in the metalanguage.) The resulting theory S is then a conservative extension of T in the sense that every sentence in the language of T provable in S is provable already in T. Thus, we may assume that f is a symbol in the language of T. Occasionally, the choice of the defining formula δf(xo,...,xn,y) is essential, e.g. in the proof of Theorem 3.5, but most of the time it is not. In particular, we shall use the function symbols <x,y> and (x)y for the primitive recursive functions and (k)m defined by: = 2kχ3m, (k)m = the number n such that p^ divides k, but p^+1 doesn't if k > 0, = 0 if k = 0. (Here pm is the m1*1 prime number: p0 = 2, p^ = 3, etc.) The function (k)m will be used to code finite sequences of natural numbers; namely, for each finite sequence n0,..., n^ of natural numbers, there is a number n such that (n)j = nj for i < k. The function (k)m can be used to transform an inductive definition into an explicit definition in PA in the following way. Suppose, for example, f(k) is defined by: f (0) = 0, f(n+l) = g(f(n),n)ifneX, = h(f(n))ifn*X. Suppose g(k,m), h(k), and X are formally represented by g, h, and ξ(x). Let δ(x,y) := Ξz((z)0 = 0 Λ Vu<x(ξ((z)u) -» (z)u+1 = g((z)u/u)) A Vu<x(- ξ((z)u) -> (z)u+1 = h((z)u)) A (z)x = y). Then δ(x,y) defines f in PA and PAh Vx5yVz(δ(x,z) <-> z = y). Thus, we may introduce a function symbol f by means of the definition: f(x) = y <-» δ(x,y). It is then easy to see that the formalizations of the clauses of the definition of f (k) become provable in PA: PAh f(0) = 0, PAh ξ(x) -» f(x+l) = g(f(x),x), PAh - ξ(x) -> f(x+l) = h(f(x)). We assume given a Godel numbering of the formal objects of LA (extensions of LA obtained by adding symbols for certain functions) among them all proofs. Since there is really no reason to distinguish between a formal object and its Godel number, we shall "identify" the two. (We do not really care exactly what the formulas,

10

1. Preliminaries

proofs etc. of a theory are; the only thing that matters is how they are related to one another.) Thus, on the pages of this book you will find no formulas of LA, only "formulas" referring to such formulas. (But the reader may, of course, still think of formal objects as strings of symbols.) Formulas and sentences being numbers, it follows that symbols for formulas and sentences are (symbols for) numerals. Thus, for example, ξ(η(y)) makes perfectly good sense; it is the result of replacing x in ξ(x) by (the numeral for) η(y). (Note that y is not free in ξ(η(y)).) ξ(η(k)) is obtained by first replacing y by k in η(y), giving η(k), and then replacing x by η(k) in ξ(x). The Gδdel numbering can be defined in such a way that everything, that should be primitive recursive, is. In particular, the following is true (see also Fact 4 (d) below): Fact 4. (a) The function corresponding to concatenation is primitive recursive. (b) The function corresponding to substitution of numerals for variables is primitive recursive. (c) The sets of formulas and sentences are primitive recursive. By Fact 4 (a), -«φ, φ -> ψ, etc. are primitive recursive functions of φ and ψ and so we may, and shall, use ->, —», etc. as formal symbols for these functions and write -ιχ, x-»y, etc. for -ι(x), ->(x,y), etc. As has already been mentioned, in many cases our (simplified) notation is not unambiguous. For example, x = φ -» ψ can be read in three different ways. One of these is eliminated by writing x = (φ—»ψ). But this formula is still ambiguous: does it contain the function symbol ->, or doesn't it? The answer to this and similar questions will always be clear from the context, when it matters. For example, we are allowed to add the symbol f for an (arbitrary) primitive recursive function f to the vocabulary of T only if we have assumed that PAH T, and then it doesn't really matter which way f(2+3), say, or φ-»ψ, occurring as a term, is understood. On the other hand, terms such as f (x) or y-»φ are, of course, unambiguous. In this book we shall be interested in r.e. theories only. In most contexts it is not necessary to distinguish between deductively equivalent theories. Thus, we may take advantage of the following result known as Craig's theorem. Theorem 1. For any r.e. set X, there is a primitive recursive set Y such that YHhX. Proof. If X = 0, this is trivial. Suppose X Φ 0. There is a primitive recursive function f such that X = (f(k): ke N}. For any sentence φ, let φ(°) := φ and φ(n+1) := φ(n) Λ φ.LetY = {f(k)(k):keN}. In view of Craig's theorem we may adopt the first of the following three conventions; the other two are introduced to avoid needless repetition:

1. Preliminaries

11

Convention 1. All theories denoted by single (decorated) letters, S, S0/ T, T', A, B etc. are primitive recursive. Convention 2. All theories denoted by single (decorated) letters, S, S0, T. T', A, B, etc. are consistent. Convention 3. From now on until Chapter 8, T is an extension of Q, QH T. If a theory is written as S, S', S0, etc. this is meant to indicate that, unless the contrary is explicitly assumed, the fact that this theory is formalized in LA is really irrelevant. We now define the arithmetical hierarchy of formulas (sentences) of L^, in other words, the sets 1^ and Πn in the following way. 2^ and Πn are the least sets containing PR, closed under Λ, v, and bounded quantification and such that (i) ^ u Πn C Σ^-L n Πn+1, (ii) if ξ is Σn (Πn), then -ξ is Πn (ΣJ, (iii) if ξ0 is ^ (Πn) and ξx is Πn (ΣJ, then ξ0 -* ξα is Πn (ΣJ, (iv) if ξ is 1^ (Πn) and δf(x0,...,xn,y) is as in Fact 2, then Ξz(δf(x0,..vXn/z) Λ ξ) and VzίδfCxo/.-vXn/2) -» ξ) are 1^ (Πn), (v) 1^ is closed under existential quantification, and (vi) Πn is closed under universal quantification. It follows that Σ0 = Π0 = PR. Bn is the set of Boolean combinations of T^ formulas. Let Φ be either 1^ or Πn or Bn. Then Φτ = {ξ: Ξηe Φ: Th ξ <-> η}. A formula is Δ£ if it is Πn and Σ^ or 1^ and Π^; ^ = Δ£A. In what follows Γ is either Σn+1 or Πn+1 and Γ+ is either Σn or Πn. Γd, the dual of Γ, is Σn, if Γ is Πn, and Πn, if Γ is Σ^ In writing Σn, Πn, Δ^, or Bn we almost always omit the (obvious) assumption that n > 0. The arithmetical hierarchy generalizes to formulas containing new symbols for primitive recursive functions in the obvious way: if ξ(x) is Γ+ and g^,...,^) is primitive recursive, then ξ(g(xo,...,xn)) is Γ+. In particular, Γ+ is closed under Vx Ξxπ(x). (b) For each Πn+1 formula (sentence) π, we can effectively find a Σj^ formula σ(x) such that PAh π <-> Vxσ(x). By Fact 5, if we are working in an extension of PA, we can always assume that any Σn+1 formula (sentence) is of the form 3xπ(x), where π(x) is Πn and that any Πn+1 formula (sentence) is of the form Vxσ(x), where σ(x) is Σn. Also note that it follows from Fact 5 that for each Σn+1 formula σ(xo,.-.,Xk-i) there is a PR formula X - / '

' ) such that

12

1. Preliminaries

PAh σίxo^.vXk-i) <-» 3y0Vy1...Qynp(x0/.../Xk-i/yo/ vyn)/ where Q is Ξ or V according as n is even or odd. Similarly, for each Πn+1 formula π(xo,...,Xk_ι), there is a PR formula p(xo/-/Xk-i/yo/-/yn) such that PAh πίxo/ .vXk-i) <-> Vyo a yi QynP(^> vXk-l'yo> 'Yn)/ where Q is V or 5 according as n is even or odd. (This includes the case k = 0, in which case σ and π are sentences.) Fact 3 (a) can be improved as follows: Fact 3. (b) The formula δf(xo,...,xn/y) of Fact 3 (a) can be taken to be Σ ^ Corollary 1 can now be improved as follows. Corollary 3. For every recursive set X (relation R) there is a ΣI formula and, therefore, a Π1 formula binumerating X (R) in Q. (b) For every r.e. set X (relation R) there are a Σ1 formula and a HI formula numerating X (R) in Q. A theory T is T-sound if every Γ sentence provable in T is true. For every r.e. set X, there is a primitive recursive relation R(k,m) such that X = {k: 5mR(k,m)}. Thus, from Corollary 2 (ii), we get the following: Corollary 4. Suppose T is Σ1-sound. Then for every r.e. set X, there is a Σ1 formula numerating X in T. In Chapter 3 it will be shown that the assumption that T is Σ^-sound can be omitted (Theorem 3.1). A function f(ko,...,kn) is provably recursive in T if there is a ΣI formula δfίxQ,...^^) such that (i) Th δrfV /kn/y) ~ y = f(ko,...,kn). (ii) Th δf(xo,...,xn,y) Λ δf(x0,...,xn,z) -> y = z, (iii) Th 3yδf(x0/...,xn/y). (In (i) fίko,...,!^) is, of course, a numeral.) Thus, all primitive recursive functions are provably recursive in PA. Suppose (i), (ii), (iii) are true. Then we may add the function symbol f to the language of T and add f(xo/-/*n) = 7 <-» δf(xo,...,xn/y) as a new axiom, where δf(x0,...,xn,y) is as above. The resulting theory is then a conservative extension of T. Suppose α(x) and β(y) are Σn+1 and f(k0,...,kn) is provably recursive in T. Then (1) Ξx(α(x) Λ Vy^f(x0/.../xn)β(y)) is not (necessarily) Σn+l'/ it is, however, Σ^+1, since it is, provably in T, equivalent to

1. Preliminaries

13

Ξxz(α(x) Λ δf(x0,..vXn/z) A Vy 3y u = ξ(m), Qh Subst2(k0,m0,k1,m1,η(vk(),vkl),u) <-> u = rKmQ,!^). As already mentioned, we may in any extension of PA introduce the corresponding function symbols Sbs^ and Sbst2. If vk is the only free variable of ξ, we write ξ(x) for Sbst1(k,x,ξ). In writing, for example, η(x, y) we assume that there are k and m such that η := η(vk,vm) and that η(x, y) := Sbst2(k,x,m,y,η). Note that, although x is not free in, say, ξ(η(x)), it is free inξ(η(x)). Given a formula σ(z), let

Prfσ(χ,y)

be a formula whose intuitive meaning is: "there is a v such that (y)v = x, (y)u = 0 for all u > v, and for every u < v, either (y)u is a logical axiom, satisfies σ(z), or is obtained from formulas (y)w with w < u using one of the (logical) rules of derivation"; in other words "y is a proof of the sentence x from the set of sentences satisfying σ(z)". (Thus, if there are nonsentences "satisfying σ(z)", they are simply disregarded.) The fact that there is a formula Prfσ(x,y) with the desired properties (see below) follows from Facts 2 (i) and 4 (and the details of the formalization of predicate logic.) If σ(z) is Γ+, then Prfσ(x,y) is Γ+. Let Prσ(x) := ayPrfσ(x,y), Conσ := -Prσ(l), where 1 := ~Ό = 0. Thus, the intuitive meaning of Prσ(x) is: "the sentence x is provable from the set of sentences satisfying σ(z)" and Conσ intuitively says: "the set of anc sentences satisfying σ(z) is consistent". If σ(z) is Z^, then Prσ(x) is I^+l' * Conσ is Πn+1. For any formula σ(x), let (σ I y)(x) := σ(x) Λ x < y, (σ + y)(x) := σ(x) v x = y. In what follows we shall use Prfs(x,y), Prfs+z(x,y), PrfS|z(x,y), Prs(x), Cons, etc. to

14

1. Preliminaries

denote (ambiguously) any formula Prfσ(x,y)/ Prfσ+z(x,y), Prfσ |z(x,y), Prσ(x), Conσ/ etc. where σ(x) is a PR binumeration of S. If S = 0, we assume that σ(x) := -»c = x; if S is finite and nonempty, S = {φo, ,φn}/then σ ( x ) := x = Φθv

vx

= Φn

Fact 6. h Vx(σ(x) -> σ'(x)) -> Vy(Prσ(y) -> Prσ/(y)). Consequently h Vx(σ(x) -> σ'(x)) -> (Conσ, -» Conσ). Fact 7. Suppose σ(x) numerates S in T. (a) If p is a proof of φ in S, then Th Prfσ(φ,p). (b) If Sh φ, then Th Prσ(φ). (c) Suppose PAH T. Let α(xo,...,xn-l) be any formula whose free variables are XO/-/ xn-l If Sh Ot(xo/-/Xn_i), then Th Pr^V-A-i))(d) If σ(x) binumerates S in T and p is not a proof of φ in S, then Th ->Prfσ(φ,p). Fact 8. Let σ(x) be any formula, (i) PAh Prσ(x) A Prσ(x-»y) -» Prσ(y), (ii) PAh Prσ+y(z) <-> Prσ(y->z), (iii) PAhPr σ (x)^ayPr σ|y (x). Corollary 5. Let σ(x) be any formula. (i) PAhPrσ(β(x))^Prσ(Ξxβ(x)), (ii) PAhPr σ (Vxβ(x))^Pr σ (β(x)), (iii) PAh Prσ(x) A Prσ(-x) -> -Conσ/ (iv) PAh Prσ(-ιχ) 4^ -«Conσ+x and PAh Prσ(x) ^ - Conσ+^x, (v) if PAH T, σ(x) numerates S in T, and Sh γ(x) -> δ(x), then Th Prσ(γ(x)) -^ Prσ(δ(x)), (vi) if PAH T, σ(x) numerates S in T, and Sh φ -> ψ, then Th Prσ(φ) -^ Prσ(ψ). All true Σ^ sentences are provable in Q; in fact, this is provable in PA; in other words, Q is Σ^-complete provably in PA: Fact 9. Suppose φ and δ(x0,...,xn_ι) are Σ1. (a) If φ true, then Qh φ. (b) PAh δ(x0/.../xn.1) -> PrQίδίxo,...,^)); in particular, PAh φ -» Prg(φ). By Fact 9 (a), if ψ is Γ^ and T + ψ is consistent, then ψ is true. Corollary 6. Suppose σ(x) numerates an extension of Q in PA. (a) If φ is a Σl sentence, then PAh φ -> Prσ(φ). (b) If σ(x) is Σl and τ(x) is a numeration of T (in PA), then

1. Preliminaries

15

PAhPr σ (φ)^Pr τ (Pr σ (φ)); in particular, PAh Prσ(φ) -» PrQ(Prσ(φ)). The following conditions (cf. Fact 7 (b), Fact 8 (i), and Corollary 6 (b)) are known as the Bernays-Lδb provability conditions (for PrT(x)). (BLi) if Th φ, then PAh Prτ(φ), (BLii) PAh Prτ(φ) Λ Prτ(φ-»ψ) -^ Prτ(ψ), (BLiii) PAh Prτ(φ) -> Prτ(Prτ(φ)). The construction of "self-referential" sentences and formulas will play a decisive role in what follows. Such constructions are possible in virtue of the following result, the fixed point lemma; we list a number of special cases; a completely general formulation would be needlessly complicated, φ is a fixed point of ξ(x) in T if Th φ ^ ξ(φ). Lemma 1. (a) For any Γ+ formula γ(x), we can effectively find a Γ+ sentence φ such that Qh φ <-> γ(φ). (b) For any Γ+ formula γ(x,y), we can effectively find a Γ+ formula ξ(x) such that Qhξ(x)<->γ(x,ξ). (c) For any Γ+ formulas Yo(*/y) and Yι(x,y), we can effectively find Γ+ sentences φ0 and q>! such that Qh φ0 <-» Y0((po, Yi((p0,γ(k,ξ(k)). (e) Suppose PAH T. For every Γ+ formula γ(x,y), we can effectively find a Γ+ formula ξ(x) such that PAhξ(x)^γ(x,ξ(x)). Proof. In what follows x is vm and y is vn. (a) Let δ(x) := Ξzβubstj(m,x,x,z) Λ γ(z)). We have Qh Subst^nxδΛz) <-> z = δ(δ). It follows that Qhδ(δ)^γ(δ(δ)). Thus, φ := δ(δ) is as desired. Φ (b) Let η(x,y) := ΞzίSubst^n^^z) Λ γ(x,z)).

16

1. Preliminaries

We have Qh SubstxOvη r^z) <-> z = η(x/η). It follows that Qh η(x,η) <-> γ(x,η(x,η)). Thus, ξ(x) := η(x,η) is as desired. * (c) For i = 0, 1, let δi(x,y) := az0z1(Subst2(m/x,n/y/x,z0) Λ Subst2(m,x,n,y,y,z1) Λ J We have Qh Subst2(m,δ0/n,δ1,δ0,z0) <-» z0 = Qh Subst2(m,δ0/n,δ1/δ1,z1) <-» zα = It follows that Qh δotδoA) ~ γo(δθ(δ(>δι)>δι(δθA))' Qh δjίδoA) <->γL(δ0(δ0/δ1)/δ1(δ0/δ1)). Thus, φ0 := δgίδo/δ!) and φ^ := δ1(δ0/δ1) are as desired. (d) Let η(x,y) := Ξz(Subst2(m,x/n,y,y/z) Λ γ(x,z)). We have Qh Subst2(m,k,n^^,z) <-> z = η(k,η). It follows that Thus, ξ(x) := η(x,η) is as desired. 4 (e) Let η(x,y) := γ(x,Sbst2(m/x,n,y,y)). We have PAh Sbst2(m,x,n,η,η) = η(x,η). It follows that PAh η(x,η) <-» γ(x,η(x,η)). Thus, ξ(x) := η(x,η) is as desired. The cases listed in the above formulation of the fixed point lemma do not exhaust the possibilities of self-reference. (This should be clear from the proof.) However, applications of self-reference (in what follows) not covered by these examples can be obtained by straightforward generalization. For example, let γ(x,y) be any formula and suppose we want to construct a sentence θ such that Qh θ <-» γ(θ,- θ). This can be done as follows. There is a (PR) formula v(x,y) such that for every φ, Qh v(φ,y) o y = -«φ. Let δ(x) := 3y(v(x,y) Λ γ(x,y)). By the fixed point lemma, there is a sentence θ such that Qh θ <-» δ(θ). Clearly θ is as desired. From this point on the fixed point lemma will be used without further mention. The phrase "let φ be such that Qh φ «-> ξ(φ)", where ξ(x) is Γ+, is short for "let φ be a (Γ+) sentence such that Qh φ <-^ ξ(φ)" and the same applies mutatis mutandis to

1. Preliminaries

17

all similar phrases. Applying the fixed point lemma we now prove two basic and very important theorems. A theory S is decidable if Th(S) is recursive, otherwise undecidable. S is essentially undecidable if S and all its consistent extensions are undecidable. Theorem 2. Q is essentially undecidable. This follows at once from Corollary 1 (a) and: Lemma 2. There is no formula binumerating Th(T) in T. Proof. Suppose τ(x) binumerates Th(T) in T. Let φ be such that (1) Qh φ <-» ιτ(φ). If Th φ, then Th τ(φ) and so, by (1), Th ->φ. But then T is inconsistent, contrary to Convention 2. It follows that TM φ. Since τ(x) binumerates Th(T) in T, this implies that Th ->τ(φ) and so, by (1), Th φ, a contradiction. Let U be any, not necessarily r.e., consistent extension of Q. By a truth-definition for U we understand a formula υ(x) such that for every sentence φ, (tr) Uh φ ^ υ(φ). The following result is known as the Tarski, or Gδdel-Tarski, theorem. Theorem 3. There is no truth-definition for U. Proof. The proof is almost the same as that of Lemma 2. Suppose υ(x) is a truth-definition for U. Let φ be such that Qh φ <-> -«υ(φ). This together with (tr) implies that U is inconsistent, contrary to assumption. The proof of Theorem 3 is a formal version of the so called liar paradox. In the latter one considers a sentence saying of itself that it isn't true: (*) (*) isn't true. (*) is both true and not true, a contradiction. Thus, a sentence saying, what (*) seems to say, cannot exist. Let M be any model of Q. The set X of natural numbers is defined in M by the formula ξ(x) if X = {k: ξ(k) is true in M}. X is definable in M if there is a formula defining X in M. Applying Theorem 3 to the set of sentences true in M we get the following: Corollary 7. Suppose M is a model of Q. The set of sentences true in M is not definable in M; in particular, this is true of N.

18

1. Preliminaries

Thus, there is no full truth-definition in arithmetic. We do, however, have the following partial positive fact. A partial truth-definition for Γ sentences in T is a formula Trr(x) such that for every Γ sentence φ, Th φ <-> Trr(φ). Let Γ(x) be a "natural" PR binumeration of the set of Γ sentences (cf . Fact 4 (d)). We (may) assume that if Γ c Γ', then PAh Γ(x) -> Γ'(x). Fact 10. (a) There is a Γ formula Satr(x,y) with the following properties: (i) For every Γ formula γ(x), PAh γ(x) <-» Satr(x,γ). (ii) Let Trr(x) := Satr(0,x). Then for every Γ formula γ(x), PAhγ(x)^Tr Γ (γ(x)) and so for every Γ sentence φ, PAh φ ^ Trr(φ). (iii) PAh Γd(x) Λ Γ(y) Λ Trrd(x) Λ Trr(x->y) -> Trr(y). (b) There is a \^+ι formula Satgn(x,y) such that for every Bn formula β(x), PAh β(x) Fact 10 (a) (i), (ii) can be used to justify self-referential constructions such as the following one. Let γ(x,y) be any Γ formula. There is then a Γ formula ξ(x) such that PAhξ(k)<-+ξ(k+l)vγ(k,ξ). Indeed, this is equivalent to PAhξ(k)^Tr Γ (ξ(k+l))vγ(k,ξ). Applying Fact 10 (a), we can now show that the arithmetical hierarchy, pertaining to formulas of LA/ is proper; for the corresponding result for sentences, see Corollary 2.5. Theorem 4. Suppose PAH T. There is a Γ formula which is not Π1'1. Proof. Let γ(x) := Satr(x,x). Suppose η(x) is Γ1 and Th η(x) <-* γ(x). -ιη(x) is Γ. Thus, Th -ιη(x) η(- η). But then T is inconsistent, contrary to Convention 2. In terms of the partial truth-definitions we can formulate the following: Fact 11. For every Γ, PAh Vx(Γ(x) Λ Pr0(x) -> Trr(x)). Let X I k = {ne X: n < k}. (Formulas such as Prx ( k(x) are then ambiguous, but the ambiguity is harmless.) A theory T is reflexive if Th Conτ (k for every k. T is essentially reflexive if every extension of T (in the same language) is reflexive.

1. Preliminaries

19

Corollary 8. PA is essentially reflexive. Proof. Suppose PAH T. Let k be arbitrary and let Γ be such that -^ΛTI k is Γ. By Fact 10 (a) (ii) and Fact 11, Th Pr0(-ΆT I k) -> --AT I k and so Th -1Pr0(--AT I k). But then, by Corollary 5 (iv), Th Conτ |k, as desired. From Fact 6 and Corollaries 5 (iv) and 8 we get: Corollary 9. Suppose PAH T. (a) If τ(x) binumerates T in T, then Th Conτj ^ for every k. (b) For all k and φ, Th PrT ik(φ) -^ φ. Corollary 8 will be of crucial importance, especially in Chapters 6 and 7. But it should be observed that, although many results proved in the following pages for extensions of PA do depend on Corollary 8, others, for example, most of those of Chapter 2 and all results of Chapter 5, do not. The latter results generalize to (possibly finitely) axiomatized, consistent extensions of PA, not (necessarily) formalized in LA The following elementary observations are occasionally useful. Lemma 3. Let π := Vx(α(x) -> 3y<xβ(y)), σ := 3x(α(x) A Vy<x-β(y)), Then (i) PAh π v θ, (ii) PAh-(σAχ), (iii) PAh (π A θ) -» Vχ- α(x), (iv) PAh Ξxα(x) -> (σ v χ), (v) PAh σ <-> (Ξxα(x) A θ), (vi) PAh 3xα(x) -> (χ <-> - σ).

θ := Vy(β(y) -» 3x
Proof, (i) Argue in PA: "Suppose -iπ and ->θ. Let z and u be such that α(z), -«3yz < u, impossible. Thus, π v θ." This proves (i). (ii) follows from (i). (iii) Argue in PA: "Suppose π, θ, and 3xα(x). By the least number principle, there is a smallest z such that α(z). Since π holds, there is a u < z such that β(u). But then, by θ, there is a v < u such that α(v). It follows that v < z, a contradiction." This proves (iii). (iv) follows from (iii). (v) follows from (i) and (iii). (vi) follows from (ii) and (iv).

20

1. Preliminaries

Corollary 10. Suppose α(x) and β(y) are PR. Let θ be as in Lemma 3. (a) If Ξxα(x) and Vyβ(y) are true, then PAh θ. (b) PAh (Ξxα(x) A Vy-β(y)) -> PrPA(θ). Proof. Let π be as in Lemma 3. (a) If 5xα(x) and Vχ- β(x) are true, so is -«π. Since - π is Σ lx it follows that PAh -iπ. But then, by Lemma 3 (i), PAh θ. * (b) h (5xα(x) Λ Vy-<β(y)) -> - π. Since -«π is Σl7 we have PAh ->π -> PrPA(-«π), by Fact 9 (b). By Lemma 3 (i), PAh - π -> θ. By (BLi) and (BLii), we get PAh PrPA(- π) -> PrPA(θ). Putting these together we get the desired conclusion. For the concepts and results of (elementary) recursion theory used in this book we refer to Soare (1987). A set is Π^ (Σ°) if it is defined in N by a Πn (ΣJ formula. X is a complete Π ° (Σ°) set if X is Π ° (Σ°) and for every Π ° (Σ°) set Y, there is a recursive function f(k) such that for every k, kE Y iff f(k)e X. No complete Π£ (Σ°) set is Σ° (π£). Complete Π° (Σ°) sets exist. If T is true, Theorem 4 follows directly from the fact that for each n > 0, there is a Γl£ (Σ°) set which isn't Σ£ (π£). The following notions will be needed in Chapter 7. A partially ordered set L = (L, <) is a lattice if any two members a, b of L have a least upper bound (l.u.b.) a u b and a greatest lower bound (g.l.b.) a n b. Thus, a u b < c i f f a < c and b < c; similarly, c < a n b iff c < a and c < b. It follows that a < b iff a n b = a iff a u b = b. L is distributive if for all a, b, ce L, a n (b u c) = (a n b) u (a n c); or equivalently, a u (b n c) = (a u b) n (a u c). The inequalities (a n b) u (a n c) < a n (b u c), a u (b n c) < (a u b) n (a u c) hold in all lattices. Suppose L has a minimal (maximal) element OL (IL). If a n b = 0^ and a u b = 1L, then b is a complement of a (and a a complement of b). If L is distributive, each a has at most one complement. If b = max{c: a n c = 0L}, then b is the pseudocomplement (p.c.) of a.

Exercises for for Chapter 1 1. 1. Improve Lemma 2 by showing that if U is any consistent extension of Q, not necessarily r.e., there is no formula numerating N - Th(U) in U. 2. (a) Let Y be any r.e. set of formulas decidable in T. Show that there is a recursive set X such that no member of Y binumerates X in T. (b) Suppose PAH T. Show that every Δα formula is decidable in T. Conclude that

Exercises

21

there is a recursive set which is not binumerated by any Δj formula in T (compare Corollary 3 (a)). 3. (a) Suppose T is Σ1-sound. Show that not every recursive function is provably recursive in T (compare Exercise 2.28 (b)). [Hint: Let δ0(x,y), δ1(x,y),... be an effective enumeration of all Σ^ formulas δ(x,y) provably in T defining (total) functions, i.e. such that (tot) Th VxΞyVz(δ(x,z) <-» z = y). For each m, let fm(k) be the recursive function defined by δm(x,y) in T. The function g(k) = fk(k)+1 is recursive.] (b) Suppose PAH T. Show that if T is not Σ1-sound, then every recursive function is provably recursive in T. (Thus, the restriction to Σ^-sound theories in (a) is essential.) (c) Show that for each recursive function f(k), there is a formula δ(x,y) defining f in T and such that (tot) holds. (Thus, the restriction to Σ^ formulas in the definition of "provably recursive" is essential.) 4. Show that there is no formula α(x) such that for all φ, ψ, (i) Th α(φ) -> φ, (ii) Th α(α(φ)->φ), (iii) if h φ, then Th α(φ), (iv) Th α(φ) and Th α(φ-»ψ), then Th α(ψ). (This improves the Godel-Tarski theorem; see also Exercise 4.4.) [Hint: Let χ := AQ. Let φ be such that Qh φ <-> ~ια(χ->φ). It follows that

i- (α(χ->φ) -> (χ -»φ)) -> (χ -> φ)

Show that Th φ and Th - φ.]

5. Suppose PAH T. (a) Show that Trr(x) is not Π1^ (compare Theorem 4). (b) Show that there is a Δn+1 formula which is not B^ (compare Theorem 4). 6. (a) Show that if T is Σj^-sound, then T is Πn+1-sound. (b) Suppose PAH T and T is true. Let φ be such that PAh φ <* 3z^(z) A Prτ+φ(z) Λ -Tr^z)). (These sentences will reappear in Chapter 5.) Thus, φ "says" that T + φ is not Σn-sound. φ is Σn+1. Show that φ is false. Conclude that Σj^-soundness does not imply Σn+1-soundness.

22

1. Preliminaries

Notes for Chapter 1. The background material on formal arithmetic and Godel numberings presupposed in this book can be found in several textbooks, for example, Kleene (1952a), Mendelson (1987), Smoryήski (1985), Kaye (1991), Boolos (1979), (1993), Hajek and Pudlak (1993). We follow Feferman (1960) in identifying formal expressions with their Godel numbers. Facts 2 and 4 are due to Godel (1931). When, somewhat later, the recursive functions were defined, the proof of Fact 3 presented no new difficulties. For proofs of these Facts and Fact 1, see any one of the textbooks just mentioned. The terms "numerate" and "binumerate" are due to Feferman (1960). Theorem 1 is due to Craig (1953). The exact definitions of PR, Σ^, Πn/ ^ vary from one author to another, depending on the intended applications; for example, other authors often use ΔQ to denote the set of bounded formulas, i.e. formulas all of whose quantifiers are bounded (cf. e.g. Kaye (1991) and Hajek and Pudlak (1993)). The present definitions have the advantage that the concepts are easy to work with (in the present setting) and that the sets PR, Σ^ Πn are primitive recursive (Fact 4 (d)) (Δ^ n > 0, is not recursive; see Exercise 2.4 (d)). The formulas Prfσ(x,y), Prσ(x), Conσ were introduced by Feferman (1960). Fact 9 is due to Feferman (1960). The Bernays-Lδb provability conditions are due to Lob (1955), simplifying the original conditions due to Bernays (cf. Hubert and Bernays (1939)). Part (a) of the fixed point lemma is implicit in Godel (1931); it was first stated explicitly by Carnap (1934) (see also Godel (1934)). The more general versions (b) - (e) were subsequently obtained by Ehrenfeucht and Feferman (1960) and Montague (1962). Lemma 2 and Theorem 2 first appeared in Tarski, Mostowski, Robinson (1953); for a stronger result, see Exercise 2.3. Theorem 3 was first published by Tarski (1933) (see also Godel (1934)). The application of (partial) truth-definitions goes back to Hubert and Bernays (1934,1939); a full proof of Fact 10 is given in Kaye (1991). For a slightly different proof of Theorem 4 and a related result, see Exercise 5. Fact 11 is essentially due to Kreisel and Wang (1955) (see also Mostowski (1952a)); for a sketch of a proof of a related result, which can easily be turned into a proof of Fact 11, see Kaye (1991), p. 140. Corollary 8 is due to Mostowski (1952a). Exercise 4 is due to Montague (1963).

2. INCOMPLETENESS

The methods of arithmetization and self-reference were originally used to prove incompleteness theorems for arithmetical theories. In this chapter we present the most important theorems of this type. A sentence φ (in the language of S) is undecidable in S if Sl^ φ and SI/ - φ. S is complete if no sentence is undecidable in S, otherwise incomplete.

§1. Incompleteness. We begin with the first and most important result of the whole subject, GόdeΓs incompleteness theorem (for theories in LA). Theorem 1. Let φ be a Π1 sentence such that (G) Qh φ <-» -Prτ(φ). Then φ is true and TI/ φ. Thus, if T is Σ1~sound, then also TI/ -«φ. Proof. Suppose Th φ. Then, by Fact 7 (b), Qh Prτ(φ). But then, by (G), Qh ->φ and so Th -»φ. It follows that T is inconsistent, contrary to Convention 2. Thus, Ί\f φ. By (G), φ is true. Thus, -«φ is a false Σ^ sentence and so TI/ - φ if T is Σ1-sound. Notice the close similarity between the proofs of Theorem 1, Lemma 1.2, and Theorem 1.3 (the liar paradox). To derive the conclusion that TI/ -tφ in Theorem 1, we needed the assumption that T is Σ1-sound. We can now see that this is stronger than mere consistency: T + -ιφ is consistent but not Σ1-sound. (Note that it does not follow from Theorem 1 that T + -iφ is incomplete.) Thus, the question arises if, assuming consistency only, there is a (Π^) sentence which is undecidable in T. Our next result, known as Rosser's theorem, shows that the answer is affirmative. Theorem 2. Let θ be a Πj sentence such that (R) Qh θ o Vz(Prfτ(θ,z) -> au
24

2. Incompleteness

Next we prove that Th ->θ. Suppose Th - θ and let p be a proof of ->θ in T. Then TH θ and so, by Fact 7 (d), Qh - Prfτ(θ,q) for every q. By Fact 1 (iv), it follows that Qh z < p -> -Prfτ(θ,z), whence, by Fact 1 (v), (2) Qh Prfτ(θ,z) -» p < z. By Fact 7 (a), Qh Prfτ(- θ,p). Hence, trivially, Qh p < z -* au
(or Th ψ <-^ ξ(- ψ)), then ψ is undecidable in T. If PAH T, the above proof of Theorem 2 can be replaced by the following argument. Suppose Th θ. Then TI/ ->θ. By (R), it follows that ->θ is true and so, -»θ being ΣI, Th -iθ, by Fact 9 (a), a contradiction. (This part does not use the assumption that PAH T.) Next suppose Th -iθ. Then TM θ. But then, by Corollary 1.10 (a) and (R), Th θ, again a contradiction. That T is incomplete also follows from Theorem 1.2, since every complete r.e. theory is decidable. This proof, however, does not (directly) yield an example of a sentence undecidable in T. Furthermore, the present proof of Theorem 1 is needed in the proof of Theorem 4, below. That every complete r.e. theory U is decidable is seen as follows: If U is inconsistent, decidability is trivial; thus, suppose U is consistent. Let φ be any sentence of U. To decide whether or not φe Th(U), generate, in some effective way, all proofs in U. If a proof of φ is found, conclude that φG Th(U); if a proof of - φ is found, conclude that φ£Th(U). Conversely, Theorem 1.2 follows from Theorem 2. Indeed, suppose U is a consistent, decidable extension of Q. There is then a complete, recursive, consistent extension U' of U. U' is an extension of Q. Hence, by Craig's theorem (Theorem 1.1), there is a complete, consistent primitive recursive extension of Q. This, how-

§1. Incompleteness

25

ever, contradicts Theorem 2. That any consistent, decidable theory U has a complete, consistent, decidable extension can be seen as follows: Let φ0, (p^,... be an effective enumeration of all sentences of the language of U. Define Un by: U0 = U, Un+1 = Un + φn if Un \f -»φn, Un+1 = Un + -ιφn otherwise. Let U' = LJ{Un: neN}. Then U' is complete and consistent. By assumption, it can be effectively decided whether U n h -ιφn or not. It follows that U' is decidable. Theorem 2 can be strengthened as follows. A family {Tk: ke N} of theories is r.e. if the binary relation φe Tk is r.e. Theorem 3. If {Tk: ke N} is an r.e. family of theories, there is a Hi sentence which is simultaneously undecidable in all the theories T^. We derive Theorem 3 from the following slight improvement of Theorem 2. Let us say that a set X of sentences is monoconsistent with T if T + φ is consistent for every φe X. Thus, for example, if φ is undecidable in T, then {φ, -iφ} is monoconsistent with T. Also, if X and Y are monoconsistent with T, so is X u Y. Let φ° := 1 φ and φ := - φ. Lemma 1. If X is r.e. and monoconsistent with Q, then there is a Γ^ sentence θ such that θ i ίX / i = 0,1. Proof. The proof is almost the same as the proof of Rosser's theorem. Let R(k,m) be a primitive recursive relation such that X = {k: ΞmR(k,m)} and let p(x,y) be a PR binumeration of R(k,m). Let θ be such that (1) Qh θ <-> Vz(p(θ,z) -> Ξu m < z. Now Qh p(-ιθ,m) and so Qh Vz(p(θ,z) -> 3uθ, which is impossible, since θeX. Thus, we have derived the desired contradiction and the proof is complete. Proof of Theorem 3. The set LJ{Th(Tk): keN} is r.e. and monoconsistent with Q. Now use Lemma 1. §2. Consistency statements. Most arguments carried out in this book can be formalized in PA. In particular this is true of the proof of Theorem 1. This leads to a

26

2. Incompleteness

proof of the following very important result, GόdeΓs second incompleteness theorem (for theories in LA). (Recall that a numeration of a set X is a formula numerating X in PA.) Theorem 4. (a) Suppose PAH T. Let φ be as in (G). Then PAh Conτ -> φ and consequently ΊV- Conτ. (b) If τ(x) is any Σ^ numeration of T, then TI/ Proof, (a) We follow closely the proof of Theorem 1 (a). By (BLiii), (1) PAhPrT(φ)^PrT(PrT(φ)). By (G) and (BLi), PAh Prτ(Prτ(φ) -> - φ) and so, by (BLii), PAhPrτ(Prτ(φ))->Prτ(-πφ). But then, by (1), PAh Prτ(φ) -» Prτ(-φ), whence, by Corollary 1.5 (iii), PAh Prτ(φ) -> -»Conτ and so, by (G), PAh Conτ -> φ. But then, assuming that Th Conτ, we get Th φ, contradicting Theorem 1 (a). It follows that ΊV Conτ. 4The proof of (b) is obtained from the above by replacing Prτ(x) by Prτ(x). In Theorem 4 (b) it is sufficient to assume that τ(x) is Σ} and numerates T in some theory S such that PAH SH T; but the assumption that τ(x) is Σ^ cannot be omitted; see Theorem 7, below. In applying Theorem 4 to an extension S of PA, we often show that there is a PR binumeration (Σ^ numeration) σ(x) of S such that Sh Conσ and conclude that S is inconsistent. A somewhat shorter proof of Theorem 4 (a) is as follows. By (G), PAh -iφ -» PrT(φ). By provable Σ1-completeness (Fact 9 (b)), PAh -iφ -> Prτ(->φ). But then, by Corollary 1.5 (iii), PAh - φ -» -«Conτ and so PAh Conτ -» φ. A similar proof yields Theorem 4 (b). Combining Theorem 4 and Corollary 1.8, we get. Corollary 1. If PAH T, then T is not finitely axiomatizable. Proof. Suppose T is finitely axiomatizable. Then there is a k such that TH T I k. Also, by Corollary 1.8, Th Conτ (k, whence T I kh Conτ ,k. But, since PAH T I k, this contradicts Theorem 4. Corollary 1 will be strengthened in Chapter 4 (Corollary 4.1) and Chapter 6 (Theorem 6.3). The proof of Theorem 4 can also be formalized in PA yielding:

§2. Consistency statements

27

Corollary 2. If PAH T, then PA + Conτh Conτ+^ConT. Proof. Let φ be as in (G). By Theorem 4 (a), (1) PAh Conτ -» φ. But then, by (BLi) and (BLii), PAh Prτ(Conτ) -> Prτ(φ) and so, by (G) (2) PAh Prτ(Conτ) -> -.φ. From (1) and (2) we get PAh Prτ(Conτ) -> -ιConτ which, by Corollary 1.5 (iv), yields the desired conclusion. The proof of our next result is another exercise in formalization, in this case of the proof of Theorem 2. Theorem 5. Let θ be a Rosser sentence for T. Then PA + Conτh -ιPrτ(θ) Λ - Prτ(-ιθ). Proof. We follow closely the above proof of Theorem 2. By Corollary 1.5 (iii), (1) PA + Conτh Prτ(θ) -> ->Prτ(- θ). It follows that PA + Conτh Prfτ(θ,z) -> - Prfτ(->θ,u) and so (2) PA + Conτh Prfτ(θ,z) -» Vu γ(z). By Fact 9 (b), we have, PAh γ(z) -» Prτ(γ(z)). Combining this with (3) yields PA + Conτh Prfτ(θ,z) -* Prτ(γ(z)), whence, by Corollary 1.5 (i), (4) PA + Conτh Prτ(θ) -> Prτ(3zγ(z)). By (R), Th Ξzγ(z) -+ -.0. But then, by (BLi) and (BLii), PAh Prτ(5zγ(z)) -» Prτ(-.θ). Combining this with (4), we get PA + Conτh Prτ(θ) -^ Prτ(-ιθ). But then, by (1), (5) PA + Conτh - Prτ(θ), as desired. Next we prove that (6) PA + Conτh-ιPrτ(-θ). From (1), we get PA + Conτh Prfτ(-.θ,u) -> - Prfτ(θ,z) and so PA + Conτh Prfτ(- θ,u) -^ Vz
28

2. Incompleteness

PA + Conτh Prτ(- θ) -> Prτ(5uδ(u)). (R) easily implies that Th Ξuδ(u) -> θ. But then (6) follows, by an argument almost the same as the proof of (5). If PAH T, this proof of Theorem 5 can be replaced by the formalization of the above short proof of Theorem 2. By (R), PAh Prτ(θ) Λ - Prτ(-θ) -> -«θ. Since ->θ is Σα, PAh -.0 -» Prτ(-ιθ). It follows that PAh Prτ(θ) -> Prτ(- θ) and so, by Corollary 1.5 (iii), PAh Conτ -> -πPrτ(θ). Next, by Corollary 1.10 (b), (R), (BLi), and (BLii), PAh Prτ(--θ) Λ --Prτ(θ) -> Prτ(θ), whence PAh Prτ(--θ) -> Prτ(θ) and so, by Corollary 1.5 (iii), PAh Conτ -> -.Prτ(- θ). Combining Theorem 5 and Corollary 1.5 (iv) we get: Corollary 3. Let θ be as in (R). Then PA + Conτh Conτ+θ Λ The sentence φ in (G) above says of itself that it is not provable in T. Let us now consider a sentence χ saying of itself that it is provable in T, i.e. such that Qh χ <-> Prτ(χ). Is χ provable in T? In this case no simple argument in terms truth will yield an answer, not even if T is true. Nevertheless, it turns out that Th χ provided that PAH T. This follows from our next result, known as Lob's theorem. Theorem 6. Suppose PAH T and let φ be any sentence such that Th Prτ(φ) —» φ. Then Th φ. Proof. Let θ be such that (1) PAh θ <-> (Prτ(θ) -> φ). From this, (BLi), and (BLii), we get (2) PAh Prτ(θ) -> (Prτ(Prτ(θ)HPrτ(φ)). By (BLiii), (3) PAh Prτ(θ) -> Prτ(Prτ(θ)). From (2) and (3) it follows that (4) PAh Prτ(θ) -> Prτ(φ). Since, by hypothesis, Th Prτ(φ) -> φ, this implies that (5) Th Prτ(θ) -> φ. But then, by (1), Th θ, whence, by (BLi), PAh Prτ(θ). Finally, this together with (5) yields Th φ, as desired. There is a semantic paradox related to the above proof in somewhat the same way as the liar paradox is related to the proof of Theorem 1. Let (**) If (**) is true, the earth is flat.

§2. Consistency statements

29

"Prove", by considering (**), that the earth is flat. Theorem 6 is a strengthening of Theorem 4: let φ := _L But Theorem 6 can also be derived from Theorem 4 as follows. Suppose Th Prτ(φ) —» φ. Then T + -«φh - Prτ(φ), whence, by Corollary 1.5 (iv), T + -iφh Conτ+_,φ. But then, by Theorem 4, T + -iφ is inconsistent and so Th φ. By slightly modifying the proof of Theorem 6 we can derive the stronger result that for every sentence φ, (L) PAh Prτ(Prτ(φHφ) -> Prτ(φ). In fact, from (4) we get PAh (Prτ(φ) -> φ) -> (Prτ(θ) -> φ). But then, by (1), PAh (Prτ(φ) -> φ) -» θ, whence, by (BLi) and (BLii), PAh Prτ(Prτ(φ)->φ) -> Prτ(θ). Finally, (L) follows from this and (4). Theorem 4 is sometimes informally expressed by saying that if T is as assumed, then T does not prove that T is consistent. That this must be interpreted with some care is clear from the following result. Theorem 7. Suppose PAH T. Let τ(x) be any formula binumerating T in T and let τ*(x) := τ(x) Λ Conφ. Then (i) τ*(x) binumerates T in T and (ii) PAh Con,.*. The following intuitive proof of Theorem 7 (ii) (formalizable in PA) is probably easier to understand than the formal argument below, but its formalization would be somewhat longer: "Any proof p from the set X defined by τ(x) Λ Con^ contains a greatest sentence φe X. Since φ satisfies Conτ|x, it follows that the set of members of X occurring in p is consistent. Thus, p cannot be a proof of L." Proof of Theorem 7. Note that x is free in Cort^. (i) If ke T, then Th τ(k). By Corollary 1.9 (a), Th Con^. Thus, Th τ*(k). If, on the other hand, keT, then Th -ιτ(k) and so Th - τ*(k). (ii) Trivially h τ*(x) -» τ(x). Hence, by Fact 6, (1) h Con,- -> Corv Since PA is reflexive, we have PAh Con^. (We assume that 0 is not a formula.) Also, by Fact 8 (iii), PAh VzCoriφ —> Cor^.. By the least number principle, it follows that (2) PAh -iConτ -> Ξz(-.Conτ|z+1 Λ Conψ). By Fact 6, PAh -'Conτ|z+1 -> (Conτjx -> x < z). Hence, by the definition of τ*(x), PAh ~ Conτ|z+1 -> (τ*(x) -^ τ(x) Λ x < z). Hence, again by Fact 6, PAh - Conτ|z+1 A Conτ|z But then, by (2),

30

2. Incompleteness

PAh -"Coriτ -» Conτ* and so, by (1), PAh Corv,.*, as desired. If τ(x) is PR, then τ*(x) is Π1. By Theorems 4 and 7, τ*(x) is not provably in T equivalent to a Σ^ formula. The formula τ*(x) may seem like a mere curiosity, but certain closely related formulas are actually of crucial importance in connection with interpretability (see the proof of Lemma 6.2.). By Theorems 4 and 7, there are formulas τ0(x) and τ ^x) binumerating T in T such that ConTQ and Conτ are not provably equivalent in T. We now show that this is so even if we restrict ourselves to PR formulas. Theorem 8. Suppose PAH T. Let τ(x) be any PR binumeration of T. (a) There is a PR binumeration τ'(x) of T such that (i) Th Con,. -> (ii) Ί\f- Coiv -> (b) Let π be a true Γ^ sentence such that Th π -> Conτ. There is then a PR binumeration τ'(x) of T such that Th π <-» Cor^/. Proof, (a) Let τ'(x) be such that PAh τ'(x) <-» τ(x) Λ Vy<χ-PrfT(Coιv -^Cor^ ,y). By Fact 6, (i) holds. Suppose (ii) is false, i.e. (1) Th Coiv -> Conτ. Let p be a proof of Coiv -> Cor^ in T. Then, by Fact 7 (a) and Fact 1 (v), PAh Vy<χ-'PrfT(Conτ/^Conτ /y) -» x < p and so PAh τ'(x) -» τ(x) Λ x < p. By Fact 6, it follows that, (2) PAh Conjip -> Conτ,. But Th Conψ, by Corollary 1.9 (a). Hence, by (1) and (2), Th Cor^ , contradicting Theorem 4 (a). This proves (ii). Finally, by (ii), Fact 1 (iv), and Fact 7 (d), τ'(x) is a PR binumeration of T. φ (b) By Fact 5 (b), we may assume that π := Vxδ(x), where δ(x) is PR. Let τ'(x) := τ(x) v 3y<χ-ιδ(y). Since π is true, τ'(x) is a PR binumeration of T. Clearly 7 T + πh τ (x) -» τ(x). Thus, by Fact 6, T + πh Cor^ -» Cor^/ and so Th π —» Cor^/. To show that the converse implication is provable in T we use the fact that evidently PAh 3y- δ(y) -» -Coiv. But then, by Fact 6, T + ->πh -"Co^/ and so Th Con^ —» π. Suppose τ(x) is a PR binumeration of T. Then, by Theorem 4, it may be true that Th -iCor^. However, from Theorem 8 (a) it follows that we can always choose τ(x) so that this does not hold: Corollary 4. If PAH T, there is a PR binumeration τ(x) of T such that TI/

§3. Independent formulas

31

§3. Independent formulas. A formula ξ(x) is independent over T if the only prepositional combinations of sentences of the form ξ(k) provable in T are the tautologies. f k This, of course, is the same as saying that T + (ξ(k) ( ): ke N} is consistent for any The following result is a strengthening of Theorem 2. Theorem 9. There is a Πj formula which is independent over T. Proof. Let R(k,i,γ,p) be the primitive recursive relation: there is a binary sequence s such that sk = i (so i = 0 or i = 1) and p is a proof in T of -(γ(0)so Λ...Λ γ(k)sk). Let p(x,y,z,u) be a PR binumeration of R(k,i,γ,p). Let μ(x) be such that Qh μ(x) <-> Vz(p(x,l,μ,z) -> 3u 0 and leave the case n = 0 to the reader.) Case 1. sn = 0. Then (2) Th μ(0)so Λ...Λ μ(n-l)sn-ι -> -,μ(n), (3) Th p(n,0,μ,p), (4) Th -ιp(n,l,μ,q) for q < p. From (3) and (4) we get Th μ(n) as in the proof of Rosser's theorem. But then, by (2), (5) Th -(μ(0)so Λ...Λ μ(n-l)sn-ι), contrary to the fact that n is minimal. Case 2. sn = 1. Then s (6) Th μ(0) o Λ...Λ μίn-l)^-! -» μ(n), (7) Th p(n,l,μ,p), (8) Th -πp(n,0,μ,q) for q < p. From (7) and (8) we get Th - μ(n) and so, by (6), we again get (5), again contrary to the minimality of n. Theorem 9 can be improved as follows; Theorem 10 will be used in Chapter 6 (proof of Lemma 6.8). Theorem 10. For any 1^ formula δ(x), there is a I^+l formula η(x) such that for any f, ge 2N, if Tf = T + (δ(k)f(k): ke N} is consistent, so is Tf + (η(k)δ(k): ke N}. Proof. For every fe 2N, let Rf (k,i,γ,ρ) be the relation: there is a binary sequence s such that (s)j,, = i and p is a proof in Tf of -.(γ(0)(s)o Λ...Λ γ(k)(s)k) (compare the relation R(k,i,γ,p) defined in the proof of Theorem 9). Using the for-

32

2. Incompleteness

mula δ(x), we are going to define a formula p*(x,y,z,w) such that for every f, f (1) p*(x,y,z,w) binumerates R (k,i/γ,p) in Tf. (Thus, p*(x,y,z,w) behaves in relation to Tf, in the same way as the formula p(x,y,z,u) in the proof of Theorem 9 behaves in relation to T.) Let R+(k,i,γ,t,n,p) be the following primitive recursive relation, where t is a binary sequence: there is a binary sequence s such that (s)^ = i and p is a proof of s s -.(γ(0)( )o Λ...Λ γ(k)( )k) in T + δ(0)«o +...+ δ(n)Wn. Then (2) Rf(k,i,γ,p) iff Ξnt 2χ3χ...χpn, where pn is the nth prime mumber. Let p+(x,y,z,u,v,w) be a PR binumeration of R+(k,i,γ,t,n,p). By Fact 2, there is a PR formula σ(x,z,u) such that Qh σ(k,m,u) <-» u = (k)m. Let β(x,y) := Vz σ(x,z,0)) A (-δ(z) -> σ(x,z,l))). Then for every n and every t, (3) T f h β(t,n) <* A{(t)m = f(m): m < n}. In view of (2) and (3), the obvious definition of p*(x,y,z,w) is now: p*(x,y,z,w) := Ξuv<w(β(u,v) Λ p+(x,y,z,u,v,w)). To prove (1), suppose first Rf(k,i,γ,p). By (2), there are then n, t < p such that (t)m = f(m) for all m < n and R+(k,i,γ,t,n,p). But then Th p+(k,i,γ,t,n,p). By (3), it follows that T f h β(t,n) and so that T f h p*(k,i,γ,p). Next suppose -«Rf(k,i,γ,p). Then, by (2), -ιR+(k,i,γ,t,n,p) for every n < p and every t < p such that (t)m = f(m) for all m < n. It follows that Th - p+(k,i,γ,t,n,p) for all such n and t. Also, by (3), T f h ~-β(t,n) for all t such that (t)m Φ f(m) for some m = z <-> (x = (z)0 A y = (z)^. Theorem 11. Suppose PAH T. Then there is a Γ (Δn+1) formula γ(x) such that T +

§4. The length of proofs

33

Vx(γ(x) <-> 6(x)) is consistent for every Γ (Bn) formula δ(x). Proof. Suppose first Γ = 2^; the case Γ = Πn follows by taking negations. Let S(k,m,n) be a primitive recursive relation such that δ(x)e Σj, & Th -Vx(η(x) <-> δ(x)) iff ΞnS(η,δ,n). Let σ(x,y,z) be a PR binumeration of S(k,m,n) and let σ*(x,y,z) := σ(x,y,z) Λ Vy'z'( < -» -xr(x,y',z')). Finally, let γ(x) be such that (1) PAh γ(x) ^ ayz(σ*(γ,y,z) A Sat^y)). Suppose there is a 1^ formula δ(x) such that (2) Th-Vx(γ(x)^δ(x)). For each such formula, there is an n such that S(γ,δ,n). Now pick δ(x) and n so that <δ,n> is minimal. Then, by (+), PAh σ*(γ,y,z) <->y = δΛZ = n. Hence, by (1) and Fact 10 (a) (i), PAh Vx(γ(x) <-> δ(x)), contradicting (2). Thus, (2) is false for all Σ^ formulas δ(x), as desired. To obtain a Δn+1 formula as desired, replace Sat^^y) by SatBn(x,y) in (1). For extensions T of PA, Theorem 9 follows at once from Theorem 11. Theorem 11 has the following: Corollary 5. Suppose PAH T. There is a Γ (Δn+1) sentence not in Π*'T (B^). Proof. Let γ(x) be as in Theorem 11 and let φ := γ(0).

§4. The length of proofs. We begin by showing that the length of proofs of sentences φ is not bounded by any recursive function of φ. Theorem 12. Let f(k) be any recursive function. There is then a Πj sentence φ such that Th φ and the least proof of φ in T is > f(φ). Proof. Let δf(x,y) be a Σ1 formula defining f in Q (cf. Fact 3 (b)). Let φ be such that Qh φ <-> Vy(δf(φ,y) -> Vz y = f(φ) and, by Fact 7 (a), Qh Prfτ(φ,p), it follows that Qh ->φ and so Th -»φ, a contradiction. Thus, φ has no proof p < f(φ) in T. But then, by Fact 1 (iv) and Fact 7 (d), Qh Vz
34

2. Incompleteness

Suppose ΊΫ φ. Then T + φ is stronger than T not only in the sense that it proves more theorems but also in the sense that there are infinitely many theorems of T which have "much shorter" proofs in T + φ; more exactly: Theorem 13. Suppose W φ. Let f be any recursive function. There is then a sentence θ such that Th θ and there is a proof q of θ in T + φ such that θ has no proof < f(q) in T. Proof. We may assume that f is increasing. Let δf^y) be a formula defining f in Q (cf. Fact 3 (a)). Let ψ be such that Qh ψ <-» 3yz(Prfτ+φ(φvψ,y) Λ δf(y^z) Λ Vu 3y(Prfτ(Prτ(φ),y) Λ Vz Vz(Prfτ(φ,z) -> 3y
Exercises

35

Then T + Prτ(Prτ(φ)) + -Prτ(φ)h φ*. Clearly, Th φ* -> φ and so, by (BLi) and (BLii), Th Prτ(φ*) -> Prτ(φ). Since φ* is Σl7 we have T + φ*h Prτ(φ*). It follows that T + Prτ(Prτ(φ))hPrτ(φ). The rest of the proof is now the same as above, except that we observe that, since φ is HI and Th φ, φ must be true (Fact 9 (a)). Λ For any sequence p of formulas and any formula θ, let p θ be p followed by θ. Λ If p is a proof of Prτ(θ) in T we may think of p θ as an R-proof of θ in T, i.e. a proof in T in which we are allowed to use the rule R. Now, let h be any primitive recurΛ sive function and let g(θ,p) = h(p θ). Then g is primitive recursive. Let φ and q be Λ as in Theorem 14 and let r = q φ. Then r is an R-proof of φ in T and φ has no proof
Exercises for Chapter 2. In the following Exercises we write Prf(x,y), Pr(x), Con for Prfτ(x,y), Prτ(x), Conτ, respectively. 1. Suppose T is true. Show that T is not complete by using the fact that Th(T), being r.e., is definable in N together with Corollary 1.7. 2. Let U be a (not necessarily r.e. or true) consistent extension of Q. Suppose there is a formula υ(x) binumerating U in U. Show that U is not complete. 3. Let Ref(T) = {φ: Th -.φ}. Let X be any set such that Th(T) £ X and Ref(T) n X = 0. Show that there is no formula binumerating X in T. (This improves Lemma 1.2.) Conclude that Th(T) and Ref(T) are recursively inseparable, i.e. there is no recursive set Y such that Th(T) £ Y and Ref(T) n Y = 0. (This implies Theorem 1.2.) 4. (a) Suppose T is Σj-sound. Use the fact that there is an r.e. nonrecursive set to show that there is a (true) Γ^ sentence not provable in T. (b) Let XQ and X1 be disjoint r.e. sets. Let pj(x,y) be a PR formula such that Xj = {k: ΞmQh pi(k,m)}, 1 = 0,1. Let ξ(x) := 3y(p0(x,y) Λ Vz
36

2. Incompleteness

σ be a Σn formula which is not Δ^. Let Xj and pj(x,y) be as in (b). Suppose XQ and X1 are recursively inseparable. Let η(x) := 3y(Po(*/y) A Vz -ιPr(Con). (Thus, there is a sentence, Con, satisfying (G) not constructed using self-reference.) We also have PAh Pr(- φ) —> Pr(->Con) (compare the last part of Theorem 1). (b) Suppose PAH T and T is Σ1-sound. Show that ΊV Con -+ - Pr(- φ) (compare Theorem 5). 7. T is co-consistent iff for every formula α(x), if Tl—>α(k) for every k, then ΊV Ξxα(x). (c) Show that if T is co-consistent, then T is Γ^-sound. (d) Suppose T is true. Show that there is a false Σ3 sentence φ such that T + φ is ω-consistent. Conclude that co-consistency does not imply Σ3-soundness. [Hint: Let φ be a sentence "saying" that T + φ is not co-consistent.] (e) Suppose PAH T and T is true. Show that for every n, there is an extension S of T which is Σj^-sound but not co-consistent. [Hint: Let δ(x) be a Πn formula such that PAh φ ^ Ξxδ(x), where φ as in Exercise 1.6 (b). Let S = T + Ξxδ(x) + bδ(k): keN}.] 8. Suppose PAH T. Let θ be a Γ^ Rosser sentence for T and let ψ := Vu(Prf(- θ,u) -> 5z Con, W ψ -> Con, and PAh θ Λ ψ -» Con. Conclude that ΎV- ψ. [Hint: Use Theorem 4 and Corollary 3.] 9. Suppose PAH T. Suppose φ is undecidable in T. Show that there is a (Σl7 ny sentence ψ such that Tl^ φ -> ψ, PAh Pr(φ) -> Pr(ψ). [Hint: Construct ψ is such a way that PAh Pr(φ) Λ -tPr(φ-^ψ) -> Pr(ψ).] 10. Strengthen Lemma 1 in the following way Suppose X is r.e. and monoconsistent with Q. Show that there is a Γ^ formula η(x) such that the only prepositional combinations of sentences of the form η(k) which are members of X are the tautologies.

Exercises

37

11. Suppose PAH T. Show that there is a na formula κ(x) such that T> κ(k) for every k, but Th κ(k) v κ(m) whenever k Φ m. (This can also be obtained as a special case of Theorem 3.5.)- [Hint: Let κ(x) be such that PAh κ(x) <-> Vy(Prf(κ(x),y) -» azu(«x,y> Λ Prf(κ(z),u))).] 12. Suppose PAH T. Let f(k) be any recursive function. Show that there is a Πj sentence θ such that Th θ and Tlf(θ)M θ. (This improves Corollary 1; also compare Exercise 4.5.) [Hint: Let δf(x,y) be a Σl formula defining f in Q (cf. Fact 3 (b)) and let θ be such that Qhθ^Vy(δ f (θ,y)^-Pr T | y (θ).] 13. Prove Lob's theorem by considering a sentence θ such that PAh θ <-» Pr(θ->φ). (This is essentially the proof of Theorem 6 using Theorem 4 mentioned in the text.) Show that this proof can be formalized in PA. 14. Suppose PAH T Show that there is a PR formula δ(x) such that Th VxPr(δ(x)) and TI/ Vxδ(x). [Hint: Let φ be as in the proof of Theorem 1 and let δ(x) := -Prf(φ,x).] 15. Suppose PAH T. Let τ(x) be a PR binumeration of T and let τ*(x) be as in Theorem 7. (a) Let ψ be such that PAh ψ <-> ->Prτ*(ψ). Show that ψ is undecidable in T. (b) Show that Th Conτ*+_,conr [Hint: Let φ be as in Theorem 1. Then Tl— φ —> Prτ*(- φ) and so Th -Kp -> ~ Prτ*(φ). It follows that Th Prτ(φ) -> -^Prτ*(φ). Also Th -"Prτ(φ) -> -«Prτ*(φ) and so Th - φ.] 16. Suppose PAH T. Prove the following strengthening of Corollary 4. Suppose X is r.e. and monoconsistent with T. There is then a PR binumeration τ(x) of T such that - Conτ£ X (see Exercise 6.6 (b)). [Hint: Let τ'(x) be a PR binumeration of T, let p(x,y) be a PR binumeration of a relation R(k,m) such that X = {k: ΞmR(k,m)}, let φ be such that PAh φ o Conτ,(x)ΛVy<x_np(^φ/y), and set τ(x) := τ' 17. Suppose PAH T. Let τ0(x) and τ-^x) be PR binumerations of T. (a) Show that there is a PR binumeration τ(x) of T such that Th Con,. <-> Con. Λ Con^. [Hint: Let τ(x) := τ0(x) v ay<xPrfτι(_L,y). See also Theorem 8 (b).] (b) Show that there is a PR binumeration τ(x) of T such that Th Con,.
38

2. Incompleteness

(c) Suppose Ί> φ. Show that there is a PR binumeration τ(x) of T such that TI/ Coriτ —» φ (compare Theorem 8 (a)). (d) Suppose TV φ and TV - ψ. Show that there is a PR binumeration τ(x) of T such that TI/ Con,. -> φ and TY ψ -^ Co^. [Hint: Use Lemma 1 and Theorem 8 (b).] 18. Suppose PAH T Let α(x), β(x) be PR formulas and let α < β mean that there is a primitive recursive function g such that PAh Vx(Prfα(l,x) -> Prfp(JL,g(x)). (α < β implies PAh Conβ —> Conα, bur not conversely.) Let α = β mean that α < β < α. Let τ(x) be a PR binumeration of T and let α(x) be such that PAh τ(x) -> α(x). Show that there is a Γ^ (ΣI) sentence φ such that α = τ + φ. [Hint: In the Γ^ case let φ be such that PAh φ <-> Vx(Prfα(_L,x) -> 3y<xPrfτ+φ(_L,y)). Use the fact that to every PR formula δ(x), there is a primitive recursive function h such that PAh δ(x) -> Prfτ(δ(x),h(x)).] 19. Prove the following strengthening of Theorems 3 and 9. If {Tk: ke N} is an r.e. family of theories, there is a Π^ formula which is simultaneously independent over all the theories Tk. Strengthen Theorems 10 and 11 in the same way. 20. (a) Derive Theorem 9 for extensions of PA from Theorem 11. (b) Formulate and prove a generalization of Theorem 11 which implies Theorem 10 for extensions of PA. 21. Suppose PAH T. Let σ be any Σ-^ sentence. Show that there is a ΣI sentence χ such that PAh (σ v Pr(_L)) «-» Pr(χ). Conclude that (i) for every Σ^ sentence σ such that Th Pr(_L) -> σ, there is a ΣI sentence χ such that Th σ <-> Pr(χ) and so for any sentences φ, ψ, there is a Σj sentence χ such that Th Pr(χ) <-> Pr(φ) v Pr(ψ), and (ii) for every Γ^ sentence π such that Th π -> Con, there is a Π1 sentence θ such that Th π <-^ Conτ+θ (compare Theorem 8 (b)). [Hint: Let δ(y) be a PR formula such that σ := 3yδ(y). Let χ be such that PAh χ ^> Ξy(δ(y) Λ Vz
Exercises

39

able in T. Let σ be such that Qh σ <-» 3z((Prf(-.σ,z) v δ(z)) Λ Vuξ(φ),z)) v (-Trr(φ) Λ 3z(Prf(φ^ξ(φ),z) A Vy f(φ) (compare Theorem 12). (b) Show that there is a Γ^ formula ξ(x) such that for every n, Th ξ(n) and the least proof of ξ(n) is > f(n). 25. Suppose PAH T and let g(k) be any recursive function. Show that there are Π^ sentences ψ0, Ψi provable and a proof p of ψ0 v ^ such that neither ψ0 nor \\f^ has a proof < g(p). [Hint: Let δg(x,y) be a Σ1 formula defining g(k) in T. Let Prf'(x,y) := Prf(x,y) Λ VzPrf(x,z) and let ψj be such that Th ψi ^VyzίPrf'tψoVψ^y) Λ 3v(δg(y,v) Λ z < v) Λ Prf(ψi,z) -> 3uι) is true. Show that g(k) is not provably recursive in T even if we restrict ourselves to Σj sentences φ0, φlβ [Hint: Suppose not. Assume that Th g(y) = 0 v g(y) = 1. Let ψj be such that Th Ψi <-> 3y(Prf/(Pr(ψ0)vPr(ψ1),y) Λ g(y) = 1-i), where Prf '(x,y) := Prf(x,y) Λ Vz
40

2. Incompleteness

27. Suppose PAH T, T is Σ1-sound, and g is primitive recursive. (a) Show that there are true Σα sentences σ0/ GI and a proof p of Pr(σ0) v Pr(σα) such that neither Pr(σ0) nor Pr^) has a proof < g(p). [Hint: Show that there is a primitive recursive function h such that h is provably increasing in T and if Th σ <-» 3z(Prf(Pr(χ),z) Λ Vu
Notes for Chapter 2. Theorem 1 is due to Godel (1931). (However, Godel assumed that T is co-consistent (see Exercise 7) but then applied this assumption only to the formula (corresponding to) Prfτ(φ,x).) For a quick proof of what is the essential content of GδdeΓs theorem, namely: truth and provability in arithmetic are not equivalent (or: the set of true sentences of L^ is not r.e.), see Exercise 1; this also follows from each of the Exercises 1.2 (a), 1.3 (a), and 1.6 (b). Theorem 2 is due to Rosser (1936). Theorems 1 and 2 can be strengthened and generalized in a number of different directions as indicated in Exercises 1, 2, 3, 4 (see also Chapter 8). However, these "directions" lead away from the central theme of this book and so will not be pursued further; but see, for example, Kleene (1952a), Mostowski (1952b), (1961), and Kreisel and Levy (1968). Lemma 1 is due to Lindstrδm (1979). Theorem 3 is due to Mostowski (1961); for a stronger result also due to Mostowski (1961), see Exercise 19. Theorem 4 is essentially due to Godel (1931); the present general formulation is due to Feferman (1960). Corollary 1 is due to Mostowski (1952a) and RyllNardzewski (1952); this result is strengthened in Chapter 4 (Corollary 4.1) and Chapter 6 (Theorem 6.3). Theorem 6 is due to Lob (1955). Lob's theorem or, more exactly, (L), is one of the keys to the modal logic of provability (cf. Boolos (1979), (1993), Smorynski (1985), Lindstrόm (199?)). Theorem 7 is due to Feferman (1960).

Notes

41

Theorem 8 (a) (with a different proof) is due to Feferman (1960); Theorem 8 (b) is due to Orey (see Feferman (I960)). Theorem 9 is due to Mostowski (1961); for a stronger result also due to Mostowski (1961), see Exercise 19. Theorem 10 is due to Scott (1962). Theorem 11 (with a different proof) is due to Montagna (1982). For Theorem 12 with Γ^ replaced by Σlr see Exercise 24 (a). A result similar to Theorem 13 was first obtained by Gδdel (1936) (cf. also Mostowski (1952b)); for a stronger result, see Exercise 3.3. Theorems 12 and 13 can also be derived from the fact that the set of (Π^ sentences provable in T (T + φ) is not recursive (cf. Exercise 4 (c) and Theorem 1.2). Theorem 14, improved as in Exercise 28 (a), is due to Parikh (1971); the present proof was pointed out to me by Christian Bennet; see also de Jongh and Montagna (1989); a more general result has been proved by Montagna (1992); cf. also Hajek, Montagna, Pudlak (1992); for related results, see Exercise 5.15. Exercise 1 is implicit in Tarski (1933) (see Gδdel (1934) and Mostowski (1952b)). Exercise 6 (a) is a special case of a general result, the fixed point theorem of provability logic due to Dick de Jongh (unpublished) and Sambin (1976) (cf. also Boolos (1979), (1993), Smorynski (1985), Lindstrδm (199?)). Exercise 11 (with a different proof) is due to Kripke (1963). Exercise 13 is due to Kreisel (see Smorynski (1985)). Exercise 15 (b) is due to Feferman (1960); it was used by him to prove Theorem 6.8. Exercise 17 is due to Hajkova (1971); her papers contain many related results. Exercise 18 is due to Bennet (1986). Exercise 19 is due to Mostowski (1961). Exercise 21 is due to Warren Goldfarb. The equivalence of (i), (ii), (iii) in Exercise 22 is due to Jensen and Ehrenfeucht (1976) and Guaspari (1979); for similar results, see Exercise 5.2. Exercise 27, improved as in Exercise 28 (a), is due to Shavrukov (1993) (with different proofs).

3. NUMERATIONS OF R.E. SETS

Any set numerated in T is r.e. The question arises if the converse of this is true, in other words, if every r.e. set can be numerated in T. If T is Σ j-sound, then, of course, the answer is "yes" (Corollary 1.4). If T is not Σ1-sound, the answer is still "yes" although this is not so obvious. This is the first and most important result of this chapter. We also prove some refinements of this result. Beginning in this chapter we omit most references to the Lemmas, Facts, and Corollaries of Chapter 1. To avoid too much repetition, proofs are sometimes left to the reader.

§1. Numerations of r.e. sets. Let X be any r.e. set. Our first task is to show that X can be numerated in T even if T is not Σ1-sound. We have already solved a similar problem in generalizing GδdeΓs incompleteness theorem to non Σ^-sound theories (Theorem 2.2). A similar construction will suffice for our present problem. Theorem 1. Let X be any r.e. set. There is then a Σ^ (Π^) formula ξ(x) which numerates X in T. Proof. There is a primitive recursive relation R(k,m) such that X = (k: 5mR(k,m)}. Let p(x,y) be a PR binumeration of R(k,m). Let ξ(x) be such that (1) Qh ξ(k) ~ 3y(p(k,y) A VzPrfτ(ξ(k),p) for every p. It follows that (3) Qh Vz<m-Prfτ(ξ(k),z). Combining (2) and (3) we get Qh 3y(p(k,y) A Vz
§1. Numerations of r.e. sets

43

Next let ξ(x) be such that Qh ξ(k) <-> Vy(Prfτ(ξ(k),y) -> Ξz
The proof is again left to the reader. Lemma 1 also follows from Lemma 2 (a), below. We now ask if there are (Σ1) formulas ξ(x) which not only numerate X in T but also satisfy additional conditions in terms of provability or nonprovability of (propositional combinations of) sentences of the form ξ(k) with kg X. The following result is a first step in that direction. Theorem 2. Let X0 and X± be disjoint r.e. sets. There is then a ΣI formula ξ(x) such that ξ(x) numerates XQ in T and - ξ(x) numerates X1 in T. Proof. Let R^m) be a primitive recursive relation such that Xi = {k: i = 0, 1. Let Pi(x,y) be a PR Enumeration of R^m). Let ξ(x) be such that (1) Qh ξ(k) ~ 3y((Po(k,y) v Prfτ(-ξ(k),y)) A Vz
44

3. Numerations of r.e. sets

Th -πay<m((p0(k,y) v Prfτ(-ξ(k),y)). But then, by (1), Th -ιξ(k), contrary to assumption. Thus, Th - ξ(k). Next suppose Th ->ξ(k) and let p be such that Th Prfτ(-ιξ(k),p). We also have and TV ξ(k) and so Th -.Prfτ(ξ(k),m) for all m. Suppose now keX^ Then Th -ip^m) for all m. It follows that Th Prfτ(-ξ(k),p) Λ Vz
§2. Types of independence

45

Next let ξ(x) be such that (4) Qh ξ(x) ^ Vu(σ(ξ,u) -* Ξz
§2. Types of independence. By a type (of independence) we understand a consistent r.e. set F of propositional formulas P in the propositional variables pn, ne N, closed under tautological consequence. Let <φk: k<ω > be a sequence of sentences. Let P(<φk: k<ω >) be obtained from P by replacing pk by φk for each k. If ξ(x) is a formula, let P(ξ) = P(<ξ(k): k<ω >). <φk: k<ω > is of type F over T if F = {P: Th P(<φk: k<ω >)}. ξ(x) is of type F over T if this is true of <ξ(k): k<ω >. Theorem 4. For each type F, there is a primitive recursive sequence <φk: k<ω > of B! sentences of type F over T. Proof. In what follows p£ is pk, if i = 0, and ^pk, if i = 1. Let s be a sequence of O's and 1's, s = . Then Ps is p^o Λ...Λ p^k. Assuming that (p0,...,(pk have been defined, we define φs in a similar manner. We now define φ0, (Pi, 92/--- It will be clear that the sentences φk are B1 and that

46

3. Numerations of r.e. sets

the sequence <φ^: k<ω > is primitive recursive. In addition to this it is sufficient to >

guarantee that for every k and every s = pm)E F}, X* = {m: (P* -> -pm)e F}. Next let ξs(x) be a Σl formula defined as in the proof of Theorem 2 with X0 and Xl replaced by XQ and X^ and T replaced by T + φs. Then (2) if F + Ps is consistent, then XQ and Xf are disjoint, (3) if T + φs is consistent and XQ and X^ are disjoint, then X* = {m: T + φsh ξs(m)}, X* = {m: T + φ*h -ξs(m)}. Let s0,...,sq be all sequences of O's and Γs of length n+1. Finally, set Ψn+l := (ΦS°A ξsj/n+l)) v-v (ΨS(1 Λ ξsq(n+1))

Then (4) T To complete the induction, we now have to show that (5) F + PSΛ pn+1 is consistent iff T + φsΛ φn+1 is consistent, (6) F + PSΛ ~"pn+i is consistent iff T + φsΛ ~"9n+i is consistent. To prove (5), suppose first F + PSΛ pn+1 is consistent. Then n+lg Xj . Moreover, F + Ps is consistent and so, by (2), XQ and X^ are disjoint and, by the inductive assumption, T + φs is consistent. It follows, by (3), that T + φsM - ξs(n+l) and so, by (4), T + φsΛ φn+1 is consistent. Next suppose T + φsΛ φn+1 is consistent. Then F + Ps is consistent. Hence, by (2), (3), (4), n+l£ Xf and so F + PSΛ pn+1 is consistent. This proves (5). The proof of (6) is similar. From Theorem 4 and Fact 10 (b) we get: Corollary 1. Suppose PAH T. Then for each type F, there is a Δ2 formula of type F overT Suppose T is Σ1-sound, ξ(x) is Σl7 and ξ(x) is of type F over T. Then F is positively prime (p.p.) in the sense that for all propositional variables Pn0/-/Pnk/ if PΠQ v v pnkE F, there is an i < k such that pn.E F. (A formula P is p.p. if the set of tautological consequences of P is p.p.) We now prove that, for extensions of PA, the converse of this is true. Theorem 5. Suppose PAH T. Then for each p.p. type F, there is a Σl formula of type F over T. The proof of Theorem 5 is different from the other proofs in this book. We shall have to rely on the reader's ability to formalize (fairly simple) intuitive arguments in PA (or willingness to believe that these arguments can be so formalized). It will

§2. Types of independence

47

be essential to distinguish between the claims (i): for every k, PA proves: ...k... and (ii): PA proves: for every k, ...k... Here (ii) is the stronger claim; it may very well be the case that (i) is true and (ii) is false. We are going to define a certain primitive recursive function f(k,m,n); the details of the definition will be crucial. The (inductive) definition of the function f(k,m,n) is given in the metalanguage and the task of formalizing this definition is left to the reader. The numbers 0,1 will be thought of as the truth-values falsity and truth, respectively. A function tG2 N can then be regarded as a truth-value assignement: t assigns truth to pj iff t(i) = 1. We always assume that t(i) = 0 for all but finitely many i. Thus, t is essentially a finite object and can be coded by, and treated as, a natural number. t[P] is the truth-value assigned by t to the formula P; for example, t[pj = t(i)

By induction on the length of P, it is easy to show that for every P, PA proves: if for every i such that pj occurs in P, ξ(i) iff t(i) = 1, then P(ξ) ifft[P] = l. Suppose g, he 2N. Then g precedes h in the lexiographic ordering if g Φ h and g(k) < h(k), where k is the least number such that g(k) Φ h(k). We shall also use the following partial ordering of 2N: g « h iff g(k) < h(k) for all k. (1)

Lemma 3. (a) Suppose F is p.p. Then there is a primitive recursive function Q(s) such that (i) F is tautologically equivalent to (Q(s): se N}, (ii) for every s, Q(s) is p.p., (iii) PA proves: for all s, s7, if s < s7, then Q(s') -> Q(s) is a tautology (we may assume that Q(0) is a tautology), (iv) PA proves: for all i, s, if pj occurs in Q(s), then i < s. (b) If P is p.p. and consistent, then there is a «-least t such that t[P] = 1. (c) Let ts be the «-least t such that t[Q(s)] = 1. For every s, ts « ts+1. Proof, (a) There is a primitive recursive function QQ(S) such that F = (Qo(s): seN}. Let Qι(s,n) be the conjunction of the set of tautological consequences of Λ{Q0(s'): s'< s} which contain no propositional variables other than pi for i < n. Next let r(s) = max{n<s: Qι(s,n) is p.p.}. Finally, let Q(s) = Q1(s,r(s)). Then Q(s) is primitive recursive and (i) - (iv) are satisfied. * (b) Let t0,...,tn be all assignments t such that t[P] = 1 and t(i) = 0 for every PJ not in P. Let p^,...,?^ be all propositional variables pj such that P —> pj is a tautology. Let t'(i) = 1 iff ie {i0,-,im}. Then t'« tk for k < n. Suppose t'[P] = 0. Then for every k < n, there is a jk£ {io,...,im} such that tk(jk) = 1. But then P —» pj v...v p; is a tautology and so the same is true of P -> PJ for some k < n, a contradiction. Thus, t'[P] = 1. * (c) This is clear, since, by (a) (iii), ts+1[Q(s)] = 1. Proof of Theorem 5. Let f(s,m,i) be the primitive recursive function defined below; m will always be assumed to be a formula η(x), the value of f(s,m,i), when m is not a formula, is irrelevant and we may set f(s,m,i) = 0. Now let ξ(x) be such that

48

3. Numerations of r.e. sets

(2) PAh ξ(x) <-> 3z(f(z,ξ,x) = 1) and let h(s,i) = f(s,ξ,i). Also, let hs be such that for all i, hs(i) = h(s,i). hs(i) may be thought of as the truth-value assigned to pi at stage s. It will be clear from the definition of f that for fixed η and i, f(s,η,i) is nondecreasing in s. Thus, informally, ξ(i) is true iff the truth-value eventually assigned to pj is 1. Our goal is to define f in such a way that the following two claims can be established; Q(s) is as in Lemma 3 (a). Claim 1. For every s, PAh (Q(s))(ξ). Claim 2. For every P, if Th P(ξ), then Pe F. By Lemma 3 (a) (i), Theorem 5 follows from Claims 1 and 2. Cases 1.1 and 2 of the definition of f are designed to ensure the validity of Claim 2: If Th P(ξ) and, for a suitable s, Q(s) -> P is not a tautology, Case 1.1 applies at Stage s+1 and so hs+1[P] = 0. Also Case 2 applies at all later stages and so hs/ = hs+1, whence hs,[P] = 0, for all s'> s. This is provable in PA. It follows, by (1), that PAh -«P(ξ), contradicting the assumption that Th P(ξ). We now define f(s,η,i) and at the same time an auxiliary function g(s,η) as follows: Stage 0. f(0,η,i) = g(0,η) = 0. Stage s+1. Case 1. g(s,η) = 0. Case 1.1. s = , m is a proof of P(η) in T, and there is a t such that (3) t[Q(s)] = 1, (4) t[P] = 0, (5) f(s,η,i) < t(i) for i < s. 7 Let t be the lexicographically least such t. Set g(s+l,η) = 1 and Case 1.2. Not Case 1.1 and there is a t such that (5) holds and (6) t[Q(s+l)] = 1. Let t' be the lexicographically least such t. Set g(s+l,η) = 0 and f(s+l,η,i) = f (i). Case 1.3. Otherwise. Set g(s+l,η) = 0 and Case 2. g(s,η) > 0. Set g(s+l,η) = 1 and Inspection of the above definition in conjunction with Lemma 3 (a) (iv) shows that hs(i) = 0 whenever i > s; in fact, this can easily be proved in PA, in other words: (7) PA proves: for all i and s, if i > s, then hs(i) = 0. Furthermore, (8) if s<s', thenh s «h s ,. For s' = s+1, this can be seen by inspection; the full result follows by induction. Using (7), the proof of (8) can be formalized in PA and so we have (9) PA proves: for all s, s7, if s < s7, then hs « hs/.

§2. Types of independence

49

Next we show that (10) for every s, g(s,ξ) = 0; in other words, if η := ξ, Case 1.1 never applies. Suppose not and let s7 be the least number such that g(s',ξ) = 1. Then Case 1.1 applies at Stage s'. Thus, s7-! = , m is a proof of P(ξ) in T, whence (11) Th P(ξ), and hs/[P] = 0. Let V = hs,. For η := ξ Case 2 now applies at every s > s' and so hs = t' for every s > s'. By (8), hs « V for s < s'. It follows that t'(i) = 1 iff there is an s such thath(s,i)=l. Using (9), this argument can be formalized in PA and so PAh Ξz(h(z,x) = 1) <-> t'(x) = 1. But then, by (2), PAh ξ(x) <-> t'(x) = 1 and so, by (1), PAh P(ξ) <-> t'[P] = 1. But PAh t'[P] = 0. It follows that PAh -tP(ξ), contradicting (11). This proves (10). We now show that for all s, (12) hs = ts, where ts is as in Lemma 3 (c). Since Q(0) is a tautology, this holds for s = 0. Suppose (12) holds for s. Then, by Lemma 3 (c), hs « ts+1. Since ts+1[Q(s+l)] = 1, either Case 1.1, Case 1.2 or Case 2 applies at s+1. By (8), Cases 1.1 and 2 don't and so Case 1.2 does. Also, the lexicographically least t' mentioned in Case 1.2 with η := ξ is ts+1. It follows that hs+1 = ts+1. This proves (12). From (7) and (12) it follows that (13) for every s, PA proves: hs = ts. Next we show that (14) for every s, PA proves: for every s'> s, hs/[Q(s)] = 1. Argue in PA: "For s'= s we have hs/ = ts, by (13), and so hs/[Q(s)] = 1. Suppose s'> s and the statement holds for s'. If Case 1.3 or Case 2 applies at s'+l, then hs/+1 = hs/ and so, by the inductive assumption, hs/+1[Q(s)] = 1. If Case 1.1 or Case 1.2 applies at s'+l, then hs,+1[Q(s')] = 1 or hs/+1[Q(s'+l)] = 1 and so, by Lemma 3 (a) (iii), hs/+1[Q(s)] = 1. Now the desired conclusion follows by induction." (Since this argument takes place in PA, Cases 1.1 and 2 cannot be ruled out.) This proves (14). Proof of Claim 1. Fix s. Argue in PA: "By (2) and (9), there is an s'> s such that for every i < s, hs,(i) = 1 iff ξ(i). By (14), hs/[Q(s)] = 1. By Lemma 3 (a) (iv), no p{ with i > s occurs in Q(s). Thus, by (1), (Q(s))(ξ)."φ Proof of Claim 2. Let m be a proof of P(ξ) in T. Let s = . By Lemma 3 (a) (i), it is sufficient to show that Q(s) —> P is a tautology. Suppose not. Let t be such that t[Q(s)] = 1 and t[P] = 0. Then ts « t. By (13), hs = ts and so hs « t. But then Case 1.1 applies at s+1 and so g(s+l,ξ) = 1, contrary to (10). Thus, Q(s) —> P is a tautology. Finally, Q(s)e F and so Pe F. + This concludes the proof of Theorem 5. For PAH T, Theorems 1, 2, 3 are, of course, special cases of Theorem 5.

50

3. Numerations of r.e. sets

Exercises for Chapter 3. 1. Suppose QH T0H Tx. Show that for every r.e. set, there is a Σx formula which numerates X in both TQ and T^. 2. We write SHpT to mean that S is a proper subtheory of T. (a) Suppose QH lΌHpIV Let XQ and Xl be r.e. sets such that XQ c X χ . Show that there is a formula ξ(x) numerating \ in Ti7 i = 0,1. [Hint: Let θ be such that TQ!^ θ and ΊI\- θ. There exist a formula ξ1(x) numerating X x in T0 and in Ί± and a formula ξo(x) numerating XQ in T0 + -»θ. Let ξ(x) := ξ1(x) Λ (θ v ξo(x)).] (b) Suppose QH T0Hp... HpTn. Let Xi/ i < n, be r.e. sets such that Xj c χ.+1 for i < n. Show that there is a formula ξ(x) numerating Xj in T^ for i < n. (c) Suppose QH T0 H Tj and suppose there is a formula σ(x) which numerates Th(S) in S for every S such that T0H S H Tα. Show that Tα H T0. [Hint: Suppose T x h θ and let φ be such that Qh φ ^-> -ισ(φvθ). Show that T0h ~ φ.] (d) Suppose QH T0/ QH T l7 and TQ and T^ are incomparable (with respect to H). Let X0 and \ι be any two r.e. sets. Show that there is a formula ξ(x) which numerates Xi in TJ, i = 0,1. 3. Suppose QH T0HpT1. Show that there is a formula ξ(x) such that for every recursive function f, the set {n: TQ|- ξ(n) & there is a proof p of ξ(n) in T1 such that ξ(n) has no proof < f(p)inT 0 ) is infinite, in fact, nonrecursive (this improves Theorem 2.13). [Hint: Let X be an r.e. nonrecursive set and let ξ(x) be a formula numerating X in TQ and N in T^.] 4. Let X0 and \ι be r.e. sets. Let ξ0(x) be a Σ^ formula numerating X0 in T. Show that there is a Σ^ formula ξ1(x) numerating X x in T such that ξ0(x) v ξχ(x) numerates X0 u Xι in T. (If n = 1 and T is Σ1-sound, this is trivial.) [Hint: Let p(x,y) be a PR formula such that Ξyp(x,y) correctly numerates X± in T, let ξ(x) be such that Qh ξ(k) ^ 3y(p(k,y) A Vz η(k): kg X} is consistent. (This improves Theorem 3.) 6. (a) Suppose PAH T and T is not Σ1-sound. Show that the sentences φk in Theorem 4 can be taken to be Δ^. [Hint: Use Lemma 1.3 (vi).] (b) Suppose QH S. Show that there are primitive recursive enumerations φ0, over S is the same as the type of <ψn: n<ω> over T.

Notes

51

7. (a) Let pj(y), i = 0,1, be PR formulas. Let φ be such that Qh φ <-> Ξy((Prfτ(-φ,y) v Po(y)) Λ Vzp1(z)) is true, Th ->φ iff 3z(p1(z) Λ Vy
Notes for Chapter 3. Theorems 1 and 2 are essentially due to Ehrenfeucht and Feferman (1960) and Putnam and Smullyan (1960), respectively; the present proofs are due to Shepherdson (1960). Lemmas 1 and 2 are due to Lindstrδm (1979), (1984a). Theorem 4 follows from a result of Pour-El and Kripke (1967) restricted to theories in LA (see Exercise 6 (b)); the proof is just an "effective" version of the proof that every denumerable Boolean algebra is embeddable in every denumerable atomless Boolean algebra. Theorem 5 is new; the proof is an adaption of a proof of Solovay (1985); the result solves Problem 32 of Friedman (1975); an interesting special case of Theorem 5 is proved in Montagna and Sorbi (1985). Exercise 3 is due to di Paola (1975). Exercise 6 (b) is a result of Pour-El and Kripke (1967) restricted to theories in LA. Exercise 7 (a) is the so called Shepherdson-Smoryήski fixed point theorem (see Smoryήski (1980) and Hajek and Pudlak (1993)); a more general result is proved in Smoryήski (1981a).

4. AXIOMATIZATIONS

S is an axiomatization of T if SHhT. Suppose SH T. S + X is an axiomatization ofΊ over S if X is r.e. and THh S + X. In this chapter we discuss some important properties of axiomatizations: finiteness, boundedness, and irredundance.

§1. Finite and bounded axiomatizability; reflection principles. We shall say that T is a finite extension ofS if there is a sentence φ such that THh S + φ. T is essentially infinite over S if no consistent extension of T is finite over S. T is essentially infinite if T is essentially infinite over the empty theory (logic). We already know that PA is essentially infinite (Corollary 2.1). By the local reflection principle for S we understand the set Rms = (Prs(φ) -» φ: φ any sentence of LA}. Thus, Rfns is a piecemeal (local) way of saying that every sentence provable in S is true. (The latter statement, the full (global) reflection principle for S, cannot be expressed in T, since, by the Gδdel-Tarski theorem, truth is not definable.) Clearly PA + Rfhjh Cony (let φ := J_). Also note that T is essentially reflexive iff Th Rfn T ! k for every k (cf. Corollary 1.9 (b)). We now use the local reflection principle to construct an essentially infinite extension of a given theory S. Note that RfnsH T implies SH T. Theorem 1. If RfnsH T, then T is essentially infinite over S. Proof. Suppose TH S + θ. We are going to show that S + θ is inconsistent. Let ψ be such that (1) Qhψ^Prs+θ(ψ). By hypothesis, Th Prs(θ->ψ) -> (θ -> ψ). From this and (1) it follows that Th θ -> ψ. But then (2) S + θh ψ. It follows that Qh Prs+θ(ψ) and so, by (1), Qh -.ψ. But QH S + θ and so, by (2), S + θ is inconsistent. If PAH T, the conclusion of Theorem 1 can be strengthened; see Corollary 2, below. There is a stronger principle, the uniform reflection principle, which is a better approximation than Rfns of the full reflection principle for S, namely, RFNS = (Vx(Γ(x) Λ Prs(x) -» Trr(x)): Γ arbitrary}. Clearly T + RFNsh Rfns provided that PAH T. Applying the uniform reflection principle we can derive a stronger conclusion than in Theorem 1. A set X of sentences is bounded if X c Γ for some Γ. Let Prfs Γ(x,y) :=

§1. Finite and bounded axiomatizability; reflection principles.

53

az(Γ(z)ΛTr Γ (z)ΛPrf s+z (x,y)) and let Prs/Γ(x) := ΞyPrfs/Γ(x,y). Lemma 1. For every φ, PA + RFN s hPr s/Γ (φ)-*φ. d

Proof. Suppose φ is Γ . Argue in PA + RFNS: "Suppose Prs Γ(φ). There is then a Γ sentence ψ such that Trr(ψ) and Prs(ψ->φ). By RFNS/ Vz(Γd(z) Λ Prs(z) -> Trrd(z)). Since ψ -» φ is Γd, it follows that Trrd(ψ-^φ). But Trr(ψ). Consequently, by Fact 10 (a) Trrd(φ) and so φ, as desired/' Theorem 2. Suppose PAH T and Th RFNS. If X is any bounded (not necessarily r.e.) set of sentences such that TH S + X, then S + X is inconsistent. Proof. Let Γ be such that X c Γ. Suppose TH S + X. We are going to show that S + X is inconsistent. Let ψ be such that (1) PAhψ^-Pr s/Γ (ψ). By Lemma 1, Th Prs/Γ(ψ) -* ψ. From this and (1) it follows that Th ψ and so (2) S + Xh ψ. But then there is a conjunction θ of members of X such that S + θh ψ. It follows that T + θh Trr(θ) Λ Prs+θ(ψ) and so T + θh Prs/Γ(ψ), whence, by (1), T + θh ->ψ and so S + Xh -πψ. Thus, by (2), S + X is inconsistent. Note the obvious analogy between the proofs of Theorems 1 and 2, on the one hand, and the proof of GodeΓs theorem (Theorem 2.1), on the other. Note also that if T is Σj-sound, then X = {->Prτ(φ): TI/ φ} is a (non-r.e.) set of Πj sentences such that T + RfnτH T + X and T + X is consistent. Since PAh RFN0 (Fact 11), we have (a) of the following corollary, improving Corollary 2.1. Corollary 1. (a) There is no consistent bounded set X such that PAH X. (b) If PAH T, there is no bounded set X such that T + RFNTH T + X and T + X is consistent. If PAH S, the above proof of Theorem 2 can be replaced by the following simple argument; the proof of Theorem 1 can be simplified in a similar way. Let χ:=Vx(Γ(x)ΛPr s (x)-*Tr Γ (x)). Now let θ be any Γd sentence such that S + θh χ. Then S + θh ->Prs(^θ), whence S H- θh Cons+θ and so S + θ is inconsistent, by Theorem 2.4. This argument and (a somewhat more detailed version of) the above proof of Theorem 2 can be looked at from a different point of view which will be further

54

4. Axiomatizations

elaborated in Chapter 5: Let φ be any Γ sentence such that S + ->χh φ. Then S + ->φh χ and so Sh φ. Thus, -<χ is Γ-conservative over S in the sense that if φ is any Γ sentence and S + ->χh φ, then Sh φ. Next we show that if PAH T, no bounded extension of T is essentially infinite over T (and a bit more). Theorem 3. Suppose PAH T, let X be an r.e. set of Γ sentences, and let Y be any r.e. set of sentences such that T + XI/ ψ for every ψe Y. There is then a Γ sentence θ such that T + θh X and T + θ^ ψ for every ψe Y. Proof. By Craig's theorem, we may assume that X and Y are primitive recursive. Let ξ(x) and η(x) be PR binumerations of X and Y, respectively. Case 1. Γ= Πn. Let θ be such that PAh θ ^ Vy(ξ(y) A Vzu -πPrfτ+θ(z,u)) -> TrΠn(y)). Suppose ψeY and T + θh ψ. Let p be a proof of ψ in T + θ and let q = max{p,ψ}. Then (1) PAhVzu -.Prfτ+θ(z,u)) for all r. It follows that T + θh X, as desired. Case 2. Γ= Σn. Let θ be such that PAh θ <-> Ξy(Ξzu Tr^z))). The verification that θ is as desired is left to the reader. From Theorem 1 and Theorem 3 with Y = {_!_} we get the following: Corollary 2. Suppose PAH T. If X is any bounded r.e. set of sentences such that RfnτH T + X, then T + X is inconsistent. Suppose T is Σ^-sound. We have already mentioned that PA + Rfn τ h Conτ. By Theorem 1, T + Conτl^ Rfnτ. Clearly PA + RFNτh Rfnτ. It has been pointed out that T + {-ιPrτ(φ): TI/ φ} is a consistent, bounded extension of T + Rfnτ. Thus, by Theorem 2, if PAH T, then T + Rfnτl^ RFNT. These observations can be strengthened as follows. We define the sentences Con(n,S), for n > 0, by: Con(l,S) := Cons, Con(n+l,S) := Con(l,S + Con(n,S)). Let Con£ = (Con(n,S): n > 0}. The proof of the following lemma is straightforward and left to the reader. Lemma 2. (a) If k > m > 0, then

§1. Finite and bounded axiomatizability; reflection principles.

55

PAh Con(k,S) -> Con(m,S). (b) For all k, m > 0, PAh Con(k,S + Con(m,S)) <-> Con(k+m,S). The sets Rfn(n,S) are defined as follows: Rfn(0,S) = 0, Rfn(l,S) := RfnS, Rfn(n+l,S) := Rfn(l,S + Rfn(n,S)). Next let Rfng = U{Rfn(n,S): neN}. We write SHpS' to mean that S is a proper subtheory of S'. Theorem 4. Suppose PAH T and T is Σ1-sound. (a) T + Con^HpT + Rfnτ. (b) T + Rfn^HpT + RFNT. Lemma 3. (a) PA + Rfn τ h Rfnτ+Con . (b)PA+RFN T hRFN T+Rfnr Proof, (a) Let φ be any sentence. PA + Rfnτh Prτ(Conτ -^ φ) -» (Conτ -> φ). But, as we have already observed, PA + Rfn τ h Conτ. It follows that PA + Rfnτh Pr

τ+Conτ(φ) -» Φ^ as desired. Φ (b) We give an informal proof using the fact that Fact 10 (a) is provable in PA. We assume, as we may, that the PR binumeration p(x) of Rfnτ implicit in the notation RFNj+Rfn is such that PA proves that every sentence satisfying p(x) is of the form Prτ(θ) -> θ. Suppose Σα c Γ. Now argue in PA + RFNT: "Let ψ be any Γ sentence provable in T + Rfhτ and let PrT((pj) -> φir for i < n, be the members of Rfnτ occurring in the proof. We may assume that -iPr-j^), for i < n, since those Prτ(φ) -» φ for which Prτ(φ) are provable in T and we may add the proofs of them to the original proof. Since -iPrT((pj) —> (Prτ(φj) —»9^) is (trivially) provable in T, it follows that θ := -.Prτ(φ0) Λ...Λ - Prτ(φn) -^ ψ is provable in T. By RFNT, Trr(θ). But, by Fact 10 (a) (ii), Trrd^Pr^cpi)), for i < n. Hence, by Fact 10 (a) (iii), Trr(ψ), as desired." Proof of Theorem 4. (a) In view of Lemma 3 (a), it follows, by induction, that T + Rfn τ h Conίj!. T + Conίj! is consistent, since T is Σ1-sound, and Con^ is an r.e. set of Π1 sentences. Thus, by Corollary 2, T + Con^ \f- Rfhτ. Φ (b) By Lemma 3 (b), T + RFNτh Rfn£ Let Xk = {- Prτ+Rfn(k/τ)(φ): T + Rfn(k,T)l^ φ}. Then, by induction, T + U{Xk: k < n}h Rfn(n+l,T). Let X = U{Xk: keN}. Then X is a (non-r.e.) set of true Π^ sentences, whence T + X is consistent, and T + Xh Rfnίj?. Thus, by Theorem 2, T + Rfn^ \t- RFNT. If T is Σ1-sound then, by Theorem 4 (a), T + Conίjί is a proper subtheory of T + Rfhτ. In our next result we show that if we restrict ourselves to Π1 sentences, this is no longer true.

56

4. Axiomatizations

We write SHΠlS' to mean that S is a Uγ-subtheory of S', i.e. every Γ^ sentence 7 provable in S is provable in S . Theorem 5. If PAH T, then T + Rfn τ H Πl PA + In the proof we use the following observation. Lemma 4. If QH S, then SHΠlPA + Cons. Proof. Let π be a Πj sentence such that Sh π. Then PAh Prs(π). Since - π is Σ l 7 we also have, PAh - π -> Prs(-ιπ). It follows that PAh ->π -> -«Cons and so PA + Consh π. Proof of Theorem 5. Let φ0/ (pi, 92,-. be all sentences of LA. For every theory S, let It is sufficient to show that for every n, there is a k such that TnHΠ PA + Con(k,T). By Lemma 4, TnHΠ PA + ConTn and so we need only prove that PA + Con(k,T)h ConTn. First we note that (1) for any sentence φ, PA + Con(2,S)h Cons+Pr ^ _> φ. Argue in PA: "Suppose -iCons+Pr / φ j _> φ/ in other words, S + Prs(φ) —» φh J_. Then Sh Prs(φ) and Sh - φ. But then Sh Prs(- φ) and so Sh - Cons/ whence -τCon(2£)" This proves (1). We now show that for every n, (2) for every extension S of PA, PA + Con(2n+1,S)h ConSn. For n = 0 this holds, by (1). Suppose (2) holds for n = k. Let S be any extension of PA. Then (3) PA proves: PA + Con(2k+1,S) h ConSk, (4) PA proves: if Con(2k+1,S + ConSk), then (S + ConSk)k is consistent. Now argue in PA: "Suppose -ιConSk+1/ in other words, Sk + Prs(φk+1)->φk+1h_L Then, since SH Sk, But then, by (1), Sk + ConSkh _L and so S + ConSk + A{Prs(9i) ^Φi: i < k}h 1. It follows that S + ConSk + A{Prs+ConSk(9i) -> Φi: i < k}h 1, in otiier words, (S + ConSk)kh _L. But then, by (4), (5) -Con(2k+1,S + ConSk). By (3), we also have PA + Con(2k+1,S)h ConSk. From this and (5) we get

§2. Irredundant axiomatizability k+1

57 k+1

-πCon(2 ,S + Con(2 ,S)), k 2 and so, by Lemma 2 (b), -<:on(2 + ,S)." Thus, we have shown that (2) holds for n = k+1. It follows that (2) holds for all n. For S = T, this yields the desired conclusion. For completeness we mention, but do not prove, that PA + RFNT is not a Πj-subtheory of T + Rfn^; for example, PA + RFNτh Conτ+R£nω.

§2. Irredundant axiomatizability. A set X of sentences is irredundant over T if for every φeX, T + (X - {φ})h φ. An extension S of T is irredundantly axiomatizable (i.a.) over T if there is an axiomatization T + X of S such that X is irredundant over T. In this case we shall also say that T + X is irredundant over T. If S is a finite extension of T, then S is i.a. over T. A theory is irredundantly axiomatizable (i.a.) if it is i.a. over the empty theory (logic). If T is i.a. over a finite theory, then T is i.a. Theorem 6. If PAH T, then T is i.a. Lemma 5. Suppose X is recursive and S + X is not a finite extension of S. Then S + X is i.a. over S iff there is a recursive function f(n) such that for every conjunction χ of members of X, S + Xh f(χ) and Sh χ -> f(χ). Proof. "If". Let f(n) be as assumed. Let φ0, cpi/ cp2/ be an effective enumeration of X. Let χn := φ0 Λ...Λ φn. We may assume that SI/ φ0. We effectively define sentences ψn in the following way. Let ψ0 := (po Suppose ψn has been defined and S + Xh ψn. We can then effectively find an m such that S + χmh ψn. Let ψn+1 := χm Λ f(χm). Then S + XHh S + (ψn: ne N}, h ψn+1 -> ψn, and Sh ψn -> ψn+1 for every n. Next let Θ0 := ΨQ and θn+1 := ψn -> ψn+ι. Again we have S + XHh S + {θn: neN}. For every n, S+-«θn is consistent. Also I— θn —> θk for every k Φ n. It follows that S + {θk: k * n}h θn. Thus, S + {θn: ne N} is an axiomatization of S + X which is irredundant over S. "Only if". Let S + Y be an axiomatization of S + X which is irredundant over S. Let χ be a conjunction of members of X. Given χ, we can effectively find a conjunction ψ of members ψo/ /Ψk ofγsucntnat S + ψh χ. Since S + X is not finite over S, we can now effectively find a sentence ΘE Y - {ψ0,. .,Ψk} Let f(χ) = θ; if n is not a conjunction of members of X, let f(n) = 0. Since S + Y is irredundant over S, it follows that f(n) is as desired. Proof of Theorem 6. Let φ be as in Theorem 2.1 with T = Q + χ. If Th χ, then Q + χ is consistent and so Q + χh φ. By Theorem 2.4, PA + CoriQ+χh φ. Set f(χ) = φ. Then y.χ _> f(χ). Also, by Corollary 1.8, Th CoriQ+χ and so Th f(χ). The desired conclusion now follows from Lemma 5 with S = 0 and X = T. To prove the existence of non-i.a. theories we borrow the following lemma from recursion theory.

58

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Lemma 6. There is a coinfinite r.e. set H such that for every recursive function h(n) (such that h(n) < h(n+l) for every n), there is a number m such that {k: h(m) < k < h(m+l)} c H. (It follows that H is not recursive.) Theorem 7. There is a Πα (Σj) formula η(x) such that T + fη(k): ke N} is not i.a. over T. Proof. Let H be as in Lemma 6. By Theorem 3.3, there is a Π1 (Σj) formula η(x) numerating H in T and such that if kg H, then (1) T + {η(m):m*k}*η(k). Let S = T + (η(k): kEN}. By (1) and since H is coinfinite, S is not finite over T. Suppose S is i.a. over T. Let φn := η(0) Λ...Λ η(n). By Lemma 5, there is then a recursive function f(n) such that for every n, Sh f(φn) and ΊV φn -» f(φn). There is a recursive function g(n) such that for every n, Th φg/n\ -» f(φn). It follows that ΊV φn —> But φg(n). Let h(0) = 0 and h(n+l) = g(h(n)). Then for every n, ΊV φh(n) -» φh(n+i) Th η(k) for ke H. It follows that {k: h(n) < k < h(n+l)} ς£ H for every n, contradicting Lemma 6. Corollary 4. If T is finite, there is a Πx (Σj) formula η(x) such that T + (η(k): keN} is not i.a. Let S = T + (φk: ke N}. Suppose S is i.a. over T. By the proof of Lemma 5, there are conjunctions ψm of the sentences φk such that if Θ0 := Ψo/ θ m +l := Ψm "^ Ψm+l/ tnen T + {θk: ke N} is an axiomatization of S which is irredundant over T. However, irredundance has been obtained at the price of a slight increase in complexity: supposing that the sentences φ^ are Γ, it does not follow that this is true of the sentences 6^. Thus, we may ask if irredundance can always be achieved without raising complexity. By our next result, the answer is negative. Let us say that S is irredundantly Γ-axiomatizable (i. Γ-a.) over T, if there is an r.e. set Z c: Γ such that T + Z is an axiomatization of S which is irredundant over T. Theorem 8. If PAH T, there is a Πn formula ξ(x) such that T + (ξ(k): ke N} is i.a. over T but not i. Πn-a. over T. The proof of Theorem 8 uses methods which will be developed in Chapter 5; it is given at the end of that chapter.

Exercises for Chapter 4. 1. (a) Show that PA + φ + Rfnsh Rfhs+φ,

Exercises

59

P A + φ + RFNshRFNs+φ. (b) Let R£ns(Γ) = (Prs(φ) -> φ: φ is Γ}, RFNS(Γ) := Vx(Γ(x) Λ Prs(x) -> Trr(x)). (i) Improve (a) by showing that if φ is Γ, then PA + φ + Rfns(Γd)h Rfns+φ(Γd), if φ is Γ, then PA + φ + RFNs(Γd)h RFNs+^P1). (ii) Show that if QH S, then PA + Consh RίnsίΠ^, PA + RFNs(Σn)hRFNs(Πn+1). (iii) Suppose PAH T. Show that if X c Γ is r.e. and T + Xh Rfnτ(Γd), then T + X is inconsistent, if X e Γ and T + Xh REN^Γ1), then T + X is inconsistent. Define the sets Rfh®(Γ) and RFN^Γ) in the natural way. Suppose S and T are true. ^ D Conclude that

T + Rfn^ RFN^),

T + Rfn^ΣJ^ Rfnτ(Πn) for n > 2,

2. Suppose PAH T. Let RFNτ = (Vx(Γ(x) Λ Prτ(x) -» Trr(x)): Γ arbitrary}. Let φ be any sentence such that W φ. Show that there is a PR binumeration τ(x) of T such that T + RFNτ^ φ. 3. Suppose PAH T and T is Σ1-sound. (a) Show that T + RfnT(Γ) is not essentially infinite over T. (b) Let S be such that T + Rfnτ(Σ1)H SH T + Rfnτ. Show that S is infinite over T. [Hint: Use (the proof of) Theorem 5 and Theorem 2.4.] 4. (a) Suppose the formula α(x) is such that for every φ, if Th φ, then Th α(φ). Show that there is a sentence ψ such that TV α(ψ) ^ ψ. [Hint: Use Exercise 1.4.] (b) Suppose there is a formula α(x) such that for every φ, if h φ, then Th α(φ), Th α(φ) —> φ. Show that T is not finitely axiomatizable. (This also follows by the proof of Theorem 1 with S = 0.) 5. T is reducible to S if there is a recursive function g(n) such that for all sentences φ, (i) Th g(φ) and (ii) if Th φ, then Sh g(φ) -» φ. If T is a finite extension of S, T = S + θ, then T is reducible to S: let g(φ) = θ for every φ. Prove the following result, a strengthening of Theorem 1: if RfnsH T, T is not reducible to S. [Hint: Suppose T is

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reducible to S and let g(n) be the relevant recursive function. Let δ(x,y) be such that for every sentence φ, Qh δ(φ,y) <-> y = (g(φ)->φ) (cf. Fact 3). Let ψ be such that Qhψ^Ξy(δ(ψ,y)Λ-Pr s (y)). Show that Th ψ and Qh - ψ.] 6. (a) Suppose SΫ φ and S + iφ + Z is non-i.a. over S + ->φ. Show that S + {φ v ψ: ψeZ} is non-i.a. over S. (b) Suppose THpT'. Show that (i) there is a theory S such that TH SH T' and S is not i.a. over T, (ii) if T is finitely axiomatizable, there is a theory S such that TH SH T7 and S is not i.a. 7. Suppose PAH T. Let X and Y be any r.e. sets of Γ sentences such that if φe X and ψe Y, then Th φ -> ψ. Show that there is a Γ sentence θ such that if φG X and ψe Y, then Th φ -^ θ and Th θ —» ψ. [Hint: Suppose Γ = Πn. Suppose X and Y are primitive recursive. Let ξ(x) and η(x) be PR binumerations of X and Y. Let θ := Vx(η(x) A Vy<x(ξ(y) -> -TrΠn(y)) ^ TrΠn(x)).] 8. Suppose PAH T and T is Σ^-sound. (a) Show that there is a Πj formula β(x) such that for every m, T + β(m) is consistent and T + β(m)h Conτ+β(m+1). (Note that if T is true, so are the theories T + β(m).) [Hint: Let the primitive recursive function f be defined (in T and in the real world) as follows; we assume that δ(x) is a PR formula: f(δ,ξ,0) = 0, f(δ,ξ,n+l) = m if m > f(δ,ξ,n), Vz<m->δ(z), n is a proof in T of -iξ^m), if there is such a number m, = f(δ,ξ,n) otherwise. If the value of f(k,m,n) is not determined by these conditions, it is irrelevant and we may set f(k,m,n) = 0. Next let γ(z,x) be such that PAhγ(z,x)^Vy(f(z,γ,y)<x). Letg(k,s) = f(k,γ,s). Claim. If Ξlxδ(x) is true, then for every n, g(δ,n) = 0. Proof. Let k be the least number such that δ(k) is true. Then for every n, g(δ,n) < k. Thus, if the claim is false, there is a largest n such that g(δ,n) Φ g(δ,n+l). Let m = g(δ,n+l). Then n is a proof of - γ(δ,m). It follows that ^y(g(δfy) < m) is provable

Notes

6i

and so is true, a contradiction. Let δ'(x) be a PR formula such that PAh Ξxδ'(x) o Prτ(-Vy(g(δ',y) = 0)). Letβ(x):=Vy(g(δ',y)<x).] (b) Show that with each rational number a > 0, we can effectively associate a Γ^ sentence θa such that T + θa is consistent and if a < b, then T + θah Conτ+θb. [Hint: Define a function g in much the same way as in case (a) except that g may, in a sense, take rational numbers > 0 as values.]

Notes for Chapter 4. Theorems 1 and 2 are due to Kreisel and Levy (1968). The formula Prs Γ(x) and the present formulation of the proof of Theorem 2 are due to Smoryήski (1981b). Corollary 1 (a) is due to Montague (1961) and Rabin (1961). What we have called the uniform reflection principle RFNS is not quite what is usually referred to by that term, but for theories containing PA the difference is negligible. Theorem 3 is due to Lindstrom (1984a). Corollary 2 is due to Kreisel and Levy (1968). Theorem 4 (b) is a weak form of a result of Feferman (1962). For (partial) improvements of Theorems 1,2,4 and Corollaries 1,2, see Exercise 1. Theorem 5 is due to Goryachev (1986) (with a different proof); the bound 2n+1 obtained in the proof is far from optimal; using methods not explained here, it can be shown that n+2 will do (cf. also Beklemishev (1995)). More information on (transfinite) iterations of consistency statements and reflection principles, a rather technical subject which falls outside the scope of this book, can be found in Feferman (1962) and Beklemishev (1995). What we have called an irredundant axiomatization is usually called an independent axiomatization. Theorem 6 is due to Montague and Tarski (1957). Lemma 5 is due to Tarski (cf. Montague and Tarski (1957)). For a proof of the existence of an r.e. set as described in Lemma 6, a so called hypersimple set, see Soare (1987). The idea of using hypersimple sets to construct non-i.a. theories is due to Kreisel (1957). Theorem 7 is related to a result of Pour-El (1968) and Corollary 4 is Pour-EΓs result restricted to theories in LA. Theorem 8 is new; Theorem 8 with Πn replaced by Σ^ and restricted to Σn-sound theories is also true but seems to require a quite different proof. Exercise 3 (b) is due to Beklemishev (199?). Exercise 4 is due to Montague (1963). Exercise 5 is due to Kreisel and Levy (1968). Exercise 8 (a) was proved by Harvey Friedman, Smoryήski, and Solovay, independently, answering a question of Haim Gaifman; for a different proof, due to Friedman, see Smoryήski (1985), p. 179. Exercise 8 (b) is due to Alex Wilkie (with a different proof); see Simmons (1988). The present proof can be modified to yield much stronger conclusions.

5. PARTIAL CONSERVATIVITY

A sentence φ is Γ-conservatΐve over T if for every Γ sentence θ, if T + φh θ, then Th θ. In this chapter we study this phenomenon for its own sake. Results on Γ-conservativity are, however, also very useful in many contexts, in particular in connection with interpretability (see Chapters 6 and 7). Our task in this chapter is to develop general methods for constructing partially conservative sentences satisfying additional conditions such as being nonprovable in a given theory. We assume throughout that PAH T. The results of this chapter do not depend on the assumption that T is reflexive. A first example of a Π^-conservative sentence is given in the following: Theorem 1. ->Conτ is Π1-conservative over T. Proof. Suppose θ is Γ^ and (1) T + -ιConτh θ. From (1) we get PAh Prτ(--θ) -> Prτ(Conτ), whence (2) PAh Prτ(-θ) -* -Conτ^Conτ. By provable Σ1-completeness, (3) PAh -θ -» Prτ(-ιθ). By Corollary 2.2, (4) PA + ConτhConτ^Conr Combining (2), (3), (4) we get PAh -πθ -» - Conτ and so by (1), Th θ. By Corollary 2.4, Theorem 1 provides us with an example of a (Σ1) sentence φ which is Π^-conservative over T and nontrivially so, i.e. such that TM φ, even if T is not Σ1-sound. 1 If φ is Γ-conservative over T and ψ is Π , then clearly φ is Γ-conservative over T + ψ. Also note that if T is Σ1-sound and π is Πj, then π is Σ1-conservative over T iff π is true iff T + π is consistent. Let us now try to construct a sentence φ which is nontrivially Γ-conservative over T. Thus, given that (1) T + φh θ, where θ is Γ, we want to be able to conclude that Th θ. This follows if (1) implies that (2) T + - θh φ. The natural way to ensure that (1) implies (2) is to let φ be a sentence saying of itself that there is a false Γ sentence (namely θ) which φ implies in T. Thus, let φ be such that (3) PAh φ ^ 5u(Γ(u) Λ Prτ+φ(u) A -Trr(u)), where Γ(x) is a PR binumeration of the set of Γ sentences. Then (1) implies (2).

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63

It is, however, not generally true that T> φ. This holds if T is true, since φ is then false. But, for example, T + -ιConτh φ, and so if Th ->Conτ, then Th φ. To prevent this from happening, we redefine φ as follows: let φ be such that PAh φ <-> ay5uvψ is pd-conservative over T; in fact, that is exactly what is done in the proof of Theorem 4.2. This also follows from the fact that (3) with φ replaced by -«ψ and Γ by Γd is true. Let [Γ]s(x,y) := Vuv Trr(u)). This formula is constructed to yield the following: Lemma 1. [Γ]τ(x,y) is a Γ formula such that (i) PAh [Γ]τ(x,y) A z < y -> [Γ]τ(x,z), (ii) T + φ h [Γ]τ(φ,m) for all φ and m, (iii) if ψ is Γ and T + φh ψ, there is a q such that PA + [Γ]τ(φ,q)h ψ. Proof, (i) is clear, (ii) Let GQ,...^ be all Γ sentences < m provable in T + φ and whose proofs are < m. Then PAh Vuv<m(Γ(u) Λ Prfτ+φ(u,v) -^ u = Θ0 v...v u = θk). Also clearly, by Fact 10 (a) (ii), T + φh u = ΘQ v...v u = θk -> Trr(u). It follows that T + φ h [Γ]τ(φ, m). (iii) Suppose ψ is Γ and T + φh ψ. Let p be a proof of ψ in T + φ and let q = max{ψ,p}. Then PA + [Γ]τ(φ,q)h Trr(ψ) and so PA + [Γ]τ(φ,q)h ψ. S is a Γ-subtheory of T, SHΓ T, if every Γ sentence provable in S is provable in T. We write [Γ](x,y) for [Γ]τ(x,y). Lemma 2. Suppose χ(x,y) is Π1. There is then a Π* formula ξ(x) such that for all k and m, (i) T + ξ(k)hχ(k,m), (ii) T + ξ(k)H Γ T + {χ(k,q):qeN}. Proof. Case 1. Γ = Σ^. Let ξ(x) be such that (1) PAh ξ(k) <-> Vy([Σj(ξ(k)/y) -> χ(k,y)). Then (i) follows from Lemma 1 (ii) and (1). To prove (ii), suppose ψ is Σj^ and T + ξ(k)h ψ. By Lemma 1 (iii), there is a q such that

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Hence, by Lemma 1 (i), PA + Vy χ(k,y)) and so, by (1), PA + Vy ΞyHΠn](ξ(k),y) A Vz
Theorem 3. (a) There is a Γ sentence φ such that φ is Γ -conservative over T and -«φ is Γ-conservative over T. (b) If X is r.e. and monoconsistent with T, there is a Γ sentence φ such that φ is Γd-conservative over T, - φ is Γ-conservative over T, and φ, -iφe X. We derive Theorem 3 from: Lemma 4. Suppose %o(x,y) is Γd and Xι(x,y) is Γ. Then there is a Γ formula ξ(x) such that for i = 0,1,

5. Partial conservativity

(i) (ii)

65

T + ξi(k)h Vy<mχi(k,y) -> X^^m), d 1 1 if ψ is Γ and T + ξi(k)h ψ , then T + (χ^^^q): qe N}h ψ .

Proof. We need only prove this for Γ = Σ^. Let ξ(x) be such that (1) PAh ξ(k) *-> Ξy(HΠn](ξ(k),y) v -χ0(k,y)) A Vzm. It follows that T + ξ(k) + Vy<mχ0(k,y)h HΠn](ξ(k),y) v -χ0(k,y)) -» y > m. But then, by (1), T + ξ(k) + Vy<mχ0(k,y)h χι(k,m), as desired. (ii) Suppose ψ is Πn and (2) T + ξ(k)hψ. By Lemma 1 (iii), there is a q such that T + [Πn](ξ(k),q)h ψ and so (3) T + iψh -[Πn](ξ(k),q). By Lemma 1 (ii), for every m, (4) T + -ξ(k)h[Σn]K(k),m). By (3), (4), Lemma 1 (i), and (1), it follows that T + -ψ + -ξ(k) + Vyφe X} = {k: 3mR1(k,m)}. Let ξ(x) be as in Lemma 4 with Xi(x,y) := ~»Pi_i(x,y). Let φ be such that PAh φ <-> ξ(φ). Suppose φe X or -iφe X. Let m be the least number such that Ro(φ,m) or R1(φ,m). Suppose Ri(φ,m). Then not R1_i(φ,n) for n < m. (We may assume that Ro(k,n) implies not R1(k,n).) But then, by Lemma 4 (i), Tl—'ξHψ)/ whence Tl—'(p*. But this is impossible, since q^eX. It follows that φ, -«φ^ X. But then, by Lemma 4 (ii), φ is Γd-conservative over T and -<φ is Γ-conservative over T. Let Prf'τ/Γ(x,y) := ΞuvPrf jpd(->x,y). Then, for example, formula (1) in the proof of Lemma 4 becomes: (Sm) PAh ξ(k) <-> Ξy((Prf'τ/ΣnK(k),y) v -χ0(k,y)) A Vz
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Our next result is related to Theorem 4.3; it will be used several times, in some cases indirectly, in Chapters 6 and 7. S is a Γ-conservative extension of T if TH SHpT. By Theorems 4.4 (a) and 4.5, T + Rmτ is a Π1-conservative extension of PA + Con^l. Theorem 4. (a) Let X be an r.e. set of Γ sentences. There is then a Γ sentence θ such that T + θ is a Γ^-conservative extension of T + X. (b) Let γ(x,y) be any Γ formula. There is then a Γ formula η(x) such that for every k, T + η(k) is a Γ^-conservative extension of T + (γ(k,m): me N}. Proof, (a) By Craig's theorem, we may assume that X is primitive recursive. Let η(x) be a PR Enumeration of X. Then for every q, (1) PA + Xhη(q)->Tr Γ (q). By Lemma 2 with (Γ replaced by Π1 and) χ(x,y) := η(y) -> Trr(y), there is a Γ sentence θ such that for all φ, (2) T + θhη(φ)^Tr Γ (φ), (3) T + ΘHpd T + (η(q) -> Trr(q): qe N}. From (2) it follows that T + θh X and from (1) and (3) it follows that T + ΘHpd T + X. Φ (b) Left to the reader. So far there has been no indication that the properties of Σ^ and Πn, n > 1, in terms of partial conservativity may be different, but we shall now show that they are. Let ψ0 and ψj be Γ sentences. If (1) ThΨo v Ψl, then, trivially, (2) ψj is pd-conservative over T + -ιψι_j, i = 0, 1. If Γ = Πn, the converse of this is true. This follows from our next: Lemma 5. Let ψ0 and ψj be any Πn sentences. There are then Πn sentences Θ0 and G! such that (i)

ThθQVθ!,

(ii) (iii)

Th Ψi -> θj, i = 0, 1, Th Θ0 Λ Ql -> ψ0 A ψχ.

Proof. By Fact 5, we may assume that ψi := Vxδ^x), where δj(x) is Σ^. Let θj := Then (i), (ii), (iii) are easily verified (cf. Lemma 1.3). From (ii) and (iii) of Lemma 5 it follows that T + -ΊI^ + ψι_ih -iθj. Hence, assuming (2), T + -^ψih -«θj. It follows that Th Θ0 v θj^ -» ψ0 v ψ1 and so, by Lemma 5 (i), we get (1). We now prove that if Γ = Σ^, then (2) does not imply (1).

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Theorem 5. (a) There are 1^ sentences ψ0/ ψi such that (i) Th -.(Ψo ΛΨl), (ii) ΊV ψ0 v ψl7 l (iii) ψi is Πn-conservative over T + ~ ψι_j, i = 0, 1. (b) Suppose X is r.e. and monoconsistent with T. Then there are Σn sentences ψ0, ψ! such that (i) and (iii) hold and (iv) ψQ We derive this theorem from: Lemma 6. Let X be an r.e. set. There are then Σ^ formulas ξo(x) and ξι(x) such that for i = 0, 1, (i) Th^ξoMΛξiM), (ii) ifkeX,thenTh-.ξi(k), (iii) if kg X, then ξj(k) is Πn-conservative over T + -iξ^k). Proof. Let p(x,y) be a PR formula such that X = {k: 3mPAh p(k,m)}. For i = 0, 1, let ζi(x)/ λi(x)/ δj(x,y) be, respectively, 2^, Σ^, and Πn_ι formulas such that (1) PAh Xi(k) <-> ay(-[Πn](ξi(k),y) Λ Vz χ^x). Now (ii) follows from this and (3). Finally, to prove (iii), suppose kg X. Now suppose ψ is Πn and By (i), it follows that (4) T + ξi(k)h ψ. But then, by Lemma 1 (iii), there is a q such that T + [Πn](ξi(k),q)h ψ. Also Th - p(k,m) for all m. By (1), it now follows that T + -«ψh χ^k). Thus, by (2), T + -iψh Ξyδi(k,y). But then Combining this with (4) we get T + ^i_i(k)h ψ. This proves (iii). Proof of Theorem 5. (a) follows from (b). φ (b) We may assume that if ψe X and Th ψ —» θ, then θe X. Let ξ^x) be as in Lemma 6. Let φ be such that

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PAh φ <-> ξ0(φ) v ξ^φ). Set ψi := ξj(φ). If φeX, then, by Lemma 6 (ii), Th -^(φ) for i = 0, 1, and so Th ->φ, impossible. Thus, φg X and so (iv) holds, (i) and (iii) follow from Lemma 6 (i) and (iii), respectively. Theorem 5 (b) will be used in the proof of Theorem 7.7 (b), below. Note that, by Theorem 5, Lemma 5 with Πn replaced by 1^ is false. We can now partially improve Corollary 2.5 as follows:

Corollary 1. There are 1^ sentences ψ0, Ψi, such that Th ψ0 -> ~«ψι and there is no Δn sentence φ for which Th ψ0 -» φ and Th φ -> -iψj. Proof. Let ψ0, ψj be as in Theorem 5 (a). Suppose φ is Δ^ Th ψ0 —> φ, and Th φ —> -•ψ!. Then Th --ψ! -^ φ and Th -ιψ0 -» -«φ and so Th ψ0 v ψlx a contradiction. Let Cons(Γ,T) be the set of sentences Γ-conservative over T. It is clear from the definition of Cons(Γ,T) that it is a Π^ set. We now show that this classification is correct. Our next lemma follows at once from Lemma 3.2 (b) but has a simpler direct proof which we leave to the reader. Lemma 7. Let R(k,m) be any r.e. relation. There are then formulas Po(*/y) and p1(x,y) such that pg(x/y) is Σ lx Pι(x,y) is n l x po(*/y) numerates R(k,m) in T, PAh Pθ(k,m) —> p1(k,m), and if not R(k,m), then Th Theorem 6. (a) Cons(Γ,T) is a complete Π^ set. (b) If Γ Φ ΣI, then Π1 n Cons(Γ,T) is a complete n° set. Proof. Let X be any Π^ set and let R(k,m) be an r.e. relation such that X = (k: VmR(k,m)}. Let p(x,y) be a formula numerating R(k,m) in T, which is ΣI if Γ = Σn and Π^ if Γ = Πn. Let ξ(x) be as in (the proof of) Lemma 2 with χ(x,y) := p(x,y) To prove (a) it is now sufficient to show that (1) kEXiffξ(k)eCons(Γ,T). By Lemma 2, (2) T+ ξ(k)hp(k,m), (3) T + ξ(k)H Γ T + {p(k,q):qEN}. If keX, then Th p(k,q) for every q and so, by (3), ξ(k)eCons(Γ,T). If kgX, there is an m such that Th p(k,m) and so, by (2), ξ(k)£ Cons(Γ,T) (in fact, ξ(k) is not Σα- or not Π1-conservative over T, as the case may be). Thus, (1) holds.This proves (a). If Γ is Σj^ or Πn with n > 2, then ξ(x) is Γd as claimed in (b). Finally, suppose Γ = Πj. Let p0(x,y) and Pι(x,y) be as in Lemma 7. Let p(x,y) := Po(x/y) Then ξ(x) is Σα. By Lemma 7, ξ(k)£ Cons(Πl7T) if k^ X. Thus, (b) holds in this case, too. Suppose T is Σ1-sound and θ is Π^ Then θ is Σ1-conservative over T iff θ is true. Thus, H! n Cons(Σl7T) is Π°.

5. Partial conservativity

69

We conclude this chapter with a proof of Theorem 4.8. We derive this result from the following lemma; a refinement of this lemma (for n = 1) will be proved in Chapter 7 (Lemma 7.22). Lemma 8. There is a Πn formula ξ(x) such that for every k, (i) Th ξ(k), (ii) Thξ(k+l)->ξ(k), (iii) ξ(k) is ^-conservative over T + ->ξ(k+l). Proof. In a first attempt to prove Lemma 8 it is natural to let ξ(x) be such that PAh ξ(k) ^ ξ(k+l) v Vv([ΣJK(k+l)Λξ(k),v) -> -Prfτ(ξ(k),v)). But then (i) does not follow and so we have to proceed in a more indirect way. Let δ(u) be any formula. Let κ(z,x,y) be a Πn formula such that (1) PAh -κ(z,x,0), (2) PAh κ(δ,k,y+l) ^ κ(δ,k+l,y) v Vv([ZJ(^5(k)Ak(k),v) ^ -Prfτ(ξδ(k),v)), where ξδ(x) := Vu(δ(u) -+ κ(δ,x,(u ^ x) + 1)), ηδ(x) := Vu(δ(u) -* κ(δ,x+l,u ^x)). (-i is the function such that k-im = k - m i f k > m and = 0 otherwise.) In (2) set y = u - k. Then, since neither y nor u is free in the second disjunct of (2), by predicate logic, we get (3) PAh ξδ(k) ^ ηδ(k) v Vv([ΣJ(^δ(k)Λξδ(k)Λ) -* -Prfτ(ξδ(k),v)). It follows that (4) ifThξ δ (k),thenThη δ (k). For let p be a proof of ξδ(k) in T. By Lemma 1 (ii), T + -ηδ(k)Λξδ(k)h-Prfτ(ξδ(k),p), whence T + ξδ(k)h ηδ(k) and so Th ηδ(k). Clearly (5) if Th δ(u) -+ u > k, then Th ηδ(k) <-> ξδ(k+l). Suppose now δ(u) is PR. Then (6) if 3uδ(u) is true, then Th ξδ(0). Suppose Ξuδ(u) is true and Th ξδ(0). Let m be the least number such that δ(m) is true. Then Th δ(u) -> u > m. By (4) and (5), it follows that Th ηδ(m). But also Th δ(m) and so, by (1), Th -<ηδ(m), a contradiction. Thus, (6) is proved. Now let δ7(u) be a PR formula such that (7) PAhΞuδ'(u)<-+Prτ(ξδ,(0)). If 3uδ7(u) is true, then, by (6), Prτ(ξδ,(0)) is false and, by (7), it is true. Thus, 3uδ'(u) is false, whence, by (7), Prτ(ξδ,(0)) is false and so Th ξδ,(0). Let ξ(x) := ξδ,(x) and η(x) := ηδ/(x). Then Th ξ(0). Hence, by (3) and (5) with δ(u) := δr(u), we get (i) and (ii). (iii) can be verified as follows. Suppose (8) T + πξ(k+l) + ξ(k)h σ,

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where σ is Σ^. Then, by (5), T + -"η(k) + ξ(k)h σ. Hence, by Lemma 1 (iii), there is a q such that But then, by (i), (3), and Lemma 1 (i), T + - σh ξ(k), whence T + - ξ(k)h σ and so, by (8), T + - ξ(k+l)h σ, proving (iii). Proof of Theorem 4.8. Let ξ(x) be as in Lemma 8. By Lemma 8 (i) and (iii), TI/ ξ(k) -> ξ(k+l). It follows that T + ξ(0) + (ξ(k) -> ξ(k+l): keN} is an axiomatization of T + (ξ(k): ke N} which is irredundant over T. Let π^, ke N, be Πn sentences such that T + {%: ke N} is an axiomatization of T + (ξ(k): ke N}. Let r be arbitrary. By Lemma 8 (ii), there is an m such that T + ξ(m)h πr. Let s be such that T + π0 Λ...Λ π s h ξ(m+l). We may assume that s > r. It follows that T + ξ(m) Λ -«ξ(m+l)h ->(π0 Λ...Λ πτ_ι Λ πr+1 Λ...Λ πs). But then, by Lemma 8 (iii), T + π0 Λ...Λ πτ.ι Λ πr+1 Λ...Λ π s h ξ(m+l). It follows, by Lemma 8 (ii), that T + {πk: k Φ r}h πr. Thus, T + {π^: keN} is not irredundant over T. We have actually proved more than is stated in Theorem 4.8. First of all, for every r, T + (πk: k Φ r}h πr; in fact, for every m, T + {πk: k > m}h πr. Secondly, this holds for all, not necessarily r.e., sets {πk: ke N} of Πn sentences such that T + (π^: keN} Hh T + {ξ(k): keN}. The theory T + (η(k): keN} constructed in the proof of Theorem 4.7, on the other hand, is deductively equivalent to T + (η(k): kg H} and (η(k): kgH} is irredundant over T. (The set (η(k): kgH} is not r.e. (cf. Lemma 4.6).)

Exercises for Chapter 5. In the following exercises we assume that PAH T. 1. Let θ be a Γ^ Rosser sentence for T. Show that -iθ is not Π1-conservative over T (compare Exercise 2 (c)). 2. Suppose T is not Σ^-sound. (a) Show that Conτ is not Σ1-conservative over T. [Hint: Let δ(y) be a PR formula such that 3yδ(y) is false and provable in T. Let χ be as in Exercise 2.21. Then W χ and T + -χh Prτ(χ) A Prτ(-χ).] (b) Improve (a) by showing that if TI/ -«Conj, there is a Σ^ sentence σ such that T + Conτh Prτ(σ) and Tb6 Prτ(σ). (c) Improve (a) by showing that if θ is a Π1 Rosser sentence for T, θ is not Σ1-conservative over T. [Hint: Let ψ := 3u(Prfτ(-«θ,u) Λ Vz
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4. φ is a self-prover in T if Th φ -> Prτ(φ). Every Σ1 sentence is a self-prover. (a) Show that φ is a self-prover in T iff there is a sentence θ such that Th φ <-> (θ Λ Prτ(θ)). (b) Show that for every n > 0, there is a 1^ (Πn+1) self-prover in T which is not

5. (a) Show that Lemma 2 (ii) can be replaced by if PAH S H T, then S + ξ(k)HΓ S + (χ(k,q): qe N}. (b) φ is hereditarily Γ-conservative over T if φ is Γ-conservative over S for every S such that PAH SH T. Show that in Lemma 3 and Theorem 2 we can replace "Γd-conservative over T" by "hereditarily Γ^-conservative over T". (c) Show that in Theorem 3 we cannot in general replace "Γ- (P^-) conservative" by "hereditarily Γ- (Γ1-) conservative". [Hint: Let φ be a Σl sentence and ψ a Π L sentence such that PA -H φ Λ ψ is consistent and PAI/ φ v ψ. Let T = PA + φ Λ ψ.] 6. (a) Show that there are sentences φ and ψ such that, T + φl/ ψ, T + ψl/ φ, φ is Πn-conservative over T + ψ, and ψ is ^-conservative over T + φ. (b) Improve (a) by showing that there are sentences φ and ψ as in (a) such that φ is ΣΠ and ψ is Πn. [Hint: Let Th φ ^ 3zHΠn]τ+ψ(φ,z) Λ Vu -Prfτ(ψ,z)). Use Exercise 5 (b).] 7. Show that there are 1^ sentences ψ0/ Ψi as in Theorem 5 satisfying the additional condition that -"ψj is ^-conservative over T, i = 0, 1. 8. (a) S is a proper Γ-subtheory of T if Shr T/1Γ S. Suppose AH B/l Πl A. Show that there is a sentence χ such that A is a proper Γ^-subtheory of A + χ1 and A + χ1 HΓ B, i = 0, 1. (b) Show that there are sentences φ0, (pi such that φ0, (pi, ^(po v -«φ1 are Γ-conservative over T and -ι(p0, -«(pi, (po Λ (pj are not Π1-conservative over T. [Hint: Use Lemma 4.] 9. (a) Show that there is a Δn+1 sentence φ such that φ and -«φ are Πn-conservative over T. [Hint: Let φ be such that PAh φ <-> ΞyHΠn](φ,y) Λ Vz
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Th V{Λ -.α: X c {0,..vn} & X has exactly k elements}.] jeX

I

10. Let X0 and X± be disjoint r.e. sets. (a) Show that there is a Σ^ formula ξ(x), such that ξ^x) numerates Xi in T, i = 0, 1, and if kg XQ u X l7 then ξ(k) is Πn-conservative over T and - ξ(k) is ^-conservative over T. (b) Show that there is a formula ξ(x) such that (i) if keX 0 / then Th ξ(k), (ii) if keX l Λ then Th ~ ξ(k), (iii) if Y0 and ΎI are any disjoint finite subsets of (X0 u X^0, then Λ{ξ(k): ke Y0} Λ Λ{-iξ(k): ke Y α } is Γ-conservative over T. [Hint: First define a formula η(k) such that all the sentences (- )η(O) Λ...Λ (~ι)η(k) are Γ-conservative over T. Then let ξ(x) := (ξo(x) v η(x)) Λ -.ξ^x) for suitable ξ0(x), ξι(x).] 11. (a) Let X and Y be r.e. sets of Γ and Γd sentences, respectively, such that if φe X and ψeY, then Th φ v ψ. Show that there is a Γ sentence θ such that T + θ is a Γd-conservative extension of T + X and T + ->θ is a Γ-conservative extension of T + Y. (b) Let Θ0, QI, Θ2,... be a recursive sequence of Γ sentences such that Th -«(θk Λ θm) for k Φ m. Let XQ and \ι be disjoint r.e. sets. Show that there is a sentence φ such that X0 = {k: Th θk -> φ} and Xl = {k: Th θk -> - φ}. 12. Suppose T is not Σ1-sound. Show that Π^ n Cons^/Γ) is a complete Π^set. [Hint: Let R(k,m) and S(k,m,n) be an r.e. and a primitive recursive relation such that X = {k: VmR(k,m)} and R(k,m) iff 3nS(k,m,n). Let σ(x,y,z) be a PR binumeration of S(k,m,n). Let γ(x) be a PR formula such that Ξxγ(x) is false and provable in T. Let p0(x,y), Pι(x,y), and δ(x,y,z) be such that PAh p0(x,y) <+ Vz(Prfτ(p1(x/y)/z) -+ Ξu: Φie Cons(Πn/ T + -φ^), i = 0, 1} is a complete Π^ set.

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14. (a) Suppose φ is Σn, and Πn-conservative over T. Let ψ be any Πn sentence which is ^-conservative over T + φ. Show that T + ->φh ψ. Conclude that no Πn sentence is nontrivially ^-conservative over T + φ and T + ->(p. [Hint: Let φ := Ξxγ(x) and ψ := Vxδ(x), where γ(x) and δ(x) are Π^ and Σ^, respectively. Then T + φ + ψh 3x(γ(x) A Vy<xδ(y)).] (b) Show that there is an r.e. family of consistent extensions of PA such that for no Γ does there exist a Γ sentence which is nontrivially Γd-conservative over every member of the family. [Hint: Let φ be a Π^ sentence undecidable in PA. Then {PA + - θ: PAh θ -» φ} u {PA + θ: PAh φ -> θ} is an r.e. family of extensions of PA. Suppose θ is Πn and nontrivially Σn-conservative over all members of this family. Then PA + φl/ θ. θ is Σj^-conservative over T + -ι(θ Λ φ). It follows that PA + φh θ, a contradiction. The dual case is similar.] 15. This exercise may be compared with Theorems 2.13, 2.14. (a) For each Γ, there is a primitive recursive function f such that for every Γ sentence φ, f(φ) is a proof in PA of φ <-» Trr(φ). Use this to show that there is a Γ sentence θ and a primitive recursive function g(k) such that θ is Γ^-conservative over T and if ψ is any P* sentence and q a proof of ψ in T + θ, then g(q) is a proof of ψ inT. (b) Let f be any recursive function. Show that there are sentences φ, ψ such that φ is Γ-conservative over T, ψ is Γ, Th ψ, and there is a proof p of ψ in T + φ such that q > f(p) for every proof q of ψ in T.

Notes for Chapter 5. The general concept Γ-conservative is due to Guaspari (1979). Theorem 1 is due to Kreisel (1962). Lemma 2 is due to Lindstrδm (1984a). Lemma 3 and Theorem 2 with X = Th(T) are due to Guaspari (1979); for somewhat stronger results, also due to Guaspari (1979), see Exercise 5 (b). The proofs of Lemma 3 and Theorem 2 are from Lindstrδm (1984a). Lemma 4 is due to Lindstrόm (1984a). (Lemmas 2 and 4 and their proofs are similar to and were inspired by results of Guaspari (1979), Solovay (cf. Guaspari (1979)), and Hajek (1971); for further applications, see e.g. Hajek and Pudlak (1993).) Theorem 3 less the references to the set X is due to Solovay (cf. Guaspari (1979); see also Jensen and Ehrenfeucht (1976); the full result is proved in Smoryriski (1981a) and Lindstrδm (1984a). The formula Prf'τ/Γ(x,y) was introduced by Smoryriski (1981a); (Sm) and the fixed point mentioned in Exercise 3.7 (a) are special cases of a very general construction due to Smoryriski (1981a); however, in the proof of his main theorem Smoryriski has to assume that the formulas %i(x,y) are PR. Theorem 4 is due to Lindstrδm (1984a). Lemma 6 and Theorem 5 are due to Bennet (1986), (1986a). Corollary 1 with Σ^ replaced by Πn is false (Exercise 3). Theorem 6 for Γ = Γ^ and for Γ = Πn+1 are essentially due to Solovay (cf. Hajek (1979)) and Hajek (1979), respectively, (in both cases with different proofs);

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Theorem 6 for Γ = Σ^ n > 1, is due to Quinsey (1980), (1981) (with a different proof); the present proof is due to Lindstrom (1984a). For more information on Cons(Γ,T) and related sets, see Exercises 12 and 13. Lemma 8 is due to Lindstrom (1993); Lemma 8 with Πn and 1^ interchanged and restricted to Σ^-sound theories is also true but the proof is quite different. An alternative concept of partial conservativity has been introduced and studied by Hajek (1984). Exercise 2 (a) is due to Smoryriski (1980); Exercise 2 (c) is due to Svejdar (cf. Hajek and Pudlak (1993)). Exercise 4 is due to Kent (1973). Exercise 5 (b) is due to Guaspari (1979). Exercise 7 is due to Bennet (1986). Exercise 10 (a) is due to Smoryriski (1981a). Exercise 12 is due to Quinsey (1981); the suggested proof is due to Bennet. Exercise 13 (c) is due to Bennet (1986). Exercise 14 is due to Misercque (1983).

6. INTERPRETABILITY

Let S and S7 be arbitrary theories. S7 is interpretable in S if, roughly speaking, the primitive concepts and the range of the variables of S7 are definable in S in such a way as to turn every theorem of S' into a theorem of S. If, in addition every nontheorem of S7 is transformed into a nontheorem of S, then S7 is faithfully interpretable in S. In this chapter, we assume that PAH T. Thus, T is essentially reflexive.

§1. Interpretability. Let S and S7 be arbitrary theories. By a translation (of the language of S' into the language of S) we understand a function t on the set of formulas (of S') into the set of formulas (of S) for which there are formulas TJO(X)/ %(*/y)/ η+(x,y,z), ηx(x,y,z) and a formula μt(x) such that t satisfies the following conditions for all formulas φ, ψ, ξ(x): (*) t(x = y) := x = y, t(x = 0) := η0(x), t(Sx = y) := ηs(x,y), t(x + y = z) := η+(x,y,z), t(x x y = z) := ηx(x,y,z), t(- φ) := -t(φ), t(φ Λ ψ) := t(φ) Λ t(ψ), t(3xξ(x)):=3x(μt(x)Λt(ξ(x))). (Here x, y, z are arbitrary variables.) We assume that V and the connectives v, ->, <-» are defined in terms of 3, -•, Λ. Note that t, on the formulas for which it is defined by the above conditions, is uniquely determined by its values on atomic formulas together with the formula μt(x). So far t(φ) is only defined provided that φ is written in a certain "normal form". For example, t is not defined on the formula x + 0 = y. But this formula is equivalent to 3z(z = 0 Λ x + z = y) and t is defined on this formula so we can set t(x + 0 = y) := t(3z(z = 0 A x + z = y)). Similarly, for any formula φ not already on "normal form", replace φ in some canonical way by φ* on "normal form" (logically equivalent to φ) and set t(φ) := t(φ*). It follows, for example, that t(Vxξ(x)) is equivalent to Vx(δ(x) -> t(ξ(x))). Clearly t is a primitive recursive function. The translation t is an interpretation in S iff (**) Sh 3xμt(x), Sh 3x(μt(x) Λ Vy(μt(y) -> (η0(y) <-> y = x))), Sh Vx(μt(x) -» 3y(μt(y) A Vz(μt(z) -> (ηs(x,z) <-> z = y)))), Sh Vxy(μt(x) Λ μt(y) -> 3z(μt(z) Λ Vu(μt(u) -> (η*(x,y,u) <-> u = z)))), * = +, x. Thus, t is an interpretation in S iff Sh t(φ) for every logically valid sentence φ. t is an interpretation o/S7 in S, t: S7< S, iff Sh t(φ) for every φ such that S 7 h φ. S7

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is interpretable in S, S'< S, if there is an interpretation of S' in S. S'< S means that S'< 7 S^S . Trivially, if S'H S, then S'< S. The reader should check that < is a transitive rela7 tion. Also note that if S'< S, then every finite subtheory of S is interpretable in a finite subtheory of S. 7 If S'< S and S is consistent, so is S'. For suppose S is not consistent. Let φ be any sentence. Then S'h φ Λ -<φ. But then Sh t(φ Λ - φ). But t(φ A -<φ) := t(φ) Λ -ιt(φ), whence Sh t(φ) Λ -«t(φ) and so S is inconsistent. Since every translation t is a primitive recursive function, we may in (extensions of) PA use t as a function symbol, t can always be defined such that the following Fact holds and the argument in the preceding paragraph can be formalized in PA. Fact 12. Suppose t: S'< S. (a) The conditions (*) and (**) are provable in PA. (b)PAhPr 0 (x)^Pr s (t(x)). This Fact has the following: Corollary 1. Suppose t: S'< S and S' is finite. Then PAh Prs/(x) ^ Prs(t(x)) and consequently PAh Cons -> ConS'. The assumption that S' is finite in Corollary 1 cannot be omitted: S'< S may be true but not provable in PA (see Corollary 5 and Theorem 12, below). But we do have the following weaker result. (Recall that a numeration of a set X numerates X in PA.) Theorem 1. Suppose SQ < S^ and let σ1(x) be a Σ^ numeration of S^. There is then a Σ! numeration σ0(x) of S0 such that PAh Conσι -» ConσQ. Proof. Suppose t: S0 < Sj. Let σ(x) be a PR binumeration of S0 and let σ0(x) := σ(x) Λ Prσ (t(x)). Then σ0(x) is a Σl numeration of S0 and (1) PAhPrσ()(x)->Prσι(t(x)). To prove this, we reason (informally) in PA as follows: "Suppose φ is derivable from formulas satsifying σ0(x). Then there are ψ0/ /ψn °f formulas satisfying σQ(x), such that Λ{ψk: k < n} -> φ is provable in logic. But then, by Fact 12 (this chapter), t(Λ{ψk: k < n}) -> t(φ) is provable from the set defined by σχ(x). But t(A{ψk: k < n}) := Λ{t(ψk): k < n}). Also, by the definition of σQ(x), each t(ψk) is derivable from the set defined by σα(x). But then so is A{t(ψk): k < n}). It follows that t(φ) is derivable from the set defined by σ1(x).// This proves (1). From (1) we easily get the desired conclusion. Theorem 1 in combination with GόdeΓs second incompleteness theorem (Theorem 2.4) yields the following strengthening of GodeΓs result. For a different

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77

improvement of Theorem 2.4, see Theorem 8, below. Theorem 2. T + Conτ £ T. Proof. Suppose T + Conτ < T. Then, by Theorem 1, there is a Σx numeration τ'(x) of T + ConT such that T + Conτh Coiv By Theorem 2.4 it now follows that T + Conτ is inconsistent. But then, since T + Conτ < T, T is inconsistent, contrary to Convention 2. Since Conτ is Πj, Theorem 2 is also a direct consequence of Theorem 2.4 and the following: Lemma 1. If π is a Π1 sentence and Q + π < T, then Th π. Proof. There is a k such that Q + π < T I k. So, by Corollary 1, Th Conτ |k -* Cong+π. It follows that Th CoriQ+π. Since -iπ is Σl7 we have, by provable Σ1-completeness, Th -.π -> -»ConQ+π. It follows that Th π. Note that we have actually proved that Q + Con^ ^ T. In Chapter 2 (Corollary 2.1) we proved that PA is essentially infinite (in fact, PA is essentially unbounded; Corollary 4.1). This can now be improved as follows: Theorem 3. T is not interpretable in any finite subtheory of T. Proof. Let S be a finite subtheory of T and suppose T < S. By Theorem 1, there is then a Σ1 numeration τ(x) of T such that PAh Cons -» Cor^. Since, by Fact 11, T is reflexive, we have Th Cons and so Th Cor^, contradicting Theorem 2.4. Most positive results on the existence of interpretations in the sequel are applications of the following fundamental result, the arithmetization of GδdeΓs completeness theorem. Theorem 4. Let σ(x) be a formula numerating S in T. Then S < T + Conσ. Proof (informal outline). A full proof of this result would be quite long and we shall be content to give a fairly detailed sketch. The main idea is to show that (the denumerable case of) the Henkin completeness proof for first order logic can be formalized in PA. (The reader is assumed to be familiar with that proof.) We begin with an outline of Henkin's proof. Let S be a (countable) set of sentences (theory) assumed to be consistent. Let cn, ne N, be new individual constants. Let L be the language obtained from Ls by adding the constants cn. Let α^x^, ne N, be a primitive recursive enumeration of all formulas of L with one free variable. We can then form a primitive recursive set Z = {Ξx n α n (x n )^α n (c Jn ):neN} such that

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(1) for every sentence θ of S, if S + Zh θ, then Sh θ. It follows that S + Z is consistent. Now let θn/ ne N, be a primitive recursive enumeration of all sentences of L. The sentences φn are then inductively defined as follows: (2) φn = θn if S + Zh Λ{φm: m < n} -> θn/ = -«θn otherwise. (Here A{φm: m < 0} := 0 = 0.) φn is not in general a recursive function of n. Let X = {φn:nGN}. Then (3) Th(S) c X and, since S + Z is consistent, (4) X is Henkin complete in the sense that X is complete and consistent and for every formula α(x) of L with the one free variable x, if 3xα(x)e X, there is a constant ck such that α(ck)e X. We can now define a model M M M M M = (M, S , + , χ , 0 ) of X in the following way. The domain M of the model is the set {cn: neN}. (Here we ignore the minor difficulty that X may contain sentences of the form ck = cm with k Φ m and so the members of M cannot in general be the constants themselves but must instead be certain "equivalence classes" of these constants or, in the present context, members of such equivalence classes. If we disregard the trivial case where S has only finite models, this can be avoided by defining Z in a slightly different way.) M 0 = c i , CM = cn, n/Cn): Ck + Cm = Cne X},

XM = {(ck,cm,cn): ck x cm = CΠG X}, where q is the (uniquely determined) constant such that 0 = q e X. Finally, it can be shown, by induction and using the fact that X is Henkin complete, that for every sentence φ of L, (5) φ is true in M iff φe X. This is true, by the definition of M, if φ is atomic. Finally, Th(S) c X and so M is a model of Th(S). We can now transform this into a proof that S < T + Conσ in the following way. th We first define in PA a primitive recursive function c(x) (= the X new individual constant). By a c-formula we understand a formula obtained from a formula of LA by replacing each free variable v by c(v). (Thus, the c-formulas are the counterparts of the sentences of L.) Let ζ(x) be a suitably defined PR Enumeration of Z, where Z is defined as above except that we now use the function symbol c. Then (the reader will hopefully believe that) for every sentence φ of S, (6) PAhP Γσvζ (φ)^Pr σ (φ). (compare (1)). It follows that (7) PAh Conσ ->

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The inductive definition of φn can, using methods available in PA, be turned into an explicit definition. Let χ(x,y) be a suitable formalization of this explicit definition (cf. Chapter 1, p. 9). Let ξ(x) := Ξyχ(x,y). (Thus, intuitively, ξ(x) means "x is a member of X".) Then (compare (3)) (8) PAhPr σ (x)^ξ(x). Let Hcmξ be the sentence saying that the set defined by ξ(x) is Henkin complete. Thus, for all c-formulas α, β, (9) PA + Hcmξh ξ(- α) <-> -ιξ(α). (10) PA + Hcmξh ξ(α) Λ Pr0(α-»β) -» ξ(β). Moreover, for every formula α(x) such that Ξxα(x) is a c-formula, (11) PA + Hcmξh ξ(3xα(x)) -> 3uξ(α(c(ύ))). The (inductive) proof of (4) does not use any means of proof beyond those available in PA. Thus, we get PAh Conσvς —» Hcmξ and so, by (7), (12) PAh Conσ -» Hcmξ. We can now define a translation t, corresponding to the model M, as follows. Let μt(x) := 3u(x = t(Sx = y) := 3uv(x = c(u) A y = c(v) A ξ(Sc(ύ) = c(v))), t(x + y = z) := 3uvw(x = c(u) Λ y = c(v) Λ z = c(w) Λ ξ(c(ύ) + c(v) = c(w))), t(χ x y = z) := 3uvw(x = c(u) Λ y = c(v) Λ z = c(w) Λ ξ(c(ύ) x c(v) = c(vί))). These equations uniquely determine t. The proof corresponding to the proof of (5) now yields for every formula P(XO/— /XH-I) of LA containing no free variables other than XQ/ /XR-I/ (13) PA + Hcmξh μt(xQ) Λ...Λ μt(xn_ι) -> (t(β(xo/.../Xn-i)) ** 3u0,...,un_1(x0 = c(u0) Λ...Λ xn-1 = c(un_x) Λ ξ(β(c(ύ)0,...,c(ύ)n_1)))). By the definition of t, this holds for atomic β(x0,...,xn_1). The inductive steps dealing with -i and Λ follow easily, by (9) and (10). Let us consider the step dealing with 3. For simplicity, let n = 1 and write x for x0. Let α(x,y) be such that β(x) := 3yα(x,y). Then t(β(x)) := 3y(μt(y) Λ t(α(x,y)). By the inductive hypothesis, PA + Hcmξh μt(x) A μt(y) -> (t(α(x,y)) <-> 3uv(x = c(u) A y = c(v) A ξ(α(c(ύ),c(v)))). By (10) and (11), PA + Hcmξh 3vξ(α(c(ύ),c(v))) *-> ξ(3yα(c(ύ),y)). But then it is fairly easy to see that PA + Hcmξh μt(x) -> (3y(μt(y) A t(α(x,y)) <-> 3u(x = c(u) A ξ(3yα(c(ύ),y)))), as desired. This proves (13). From (12) and (13), we get for every sentence φ, (14) PA + Con σ ht(φ)^ξ(φ). Finally, let φ be any sentence provable in S. Then Th Prσ(φ). Hence, by (8), Th ξ(φ) and so, by (14), T + Conσh t(φ). It follows that t: S < T + Conσ.

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This concludes our sketch of the proof of Theorem 4. If we don't insist on mimicking every detail of Henkin's proof, we can instead use the simpler interpretation t' defined in the following way: MO := x = x/

t'(x = 0) := ξ(0 = c(x)), t'(Sx = y):=ξ(Sc(x) = c(y)), f (x + y = z) := ξ(c(x) + c(y) = c(z)), t ' ( x x y = z):=ξ(c(x)χc(y) = c(z)), Thus, MX)istrivial anc*can be omitted. (This is true as long as we are dealing with theories of (elementary) arithmetic; it is not true in general.) It is via the following lemma, the (Feferman-)Orey-Hajek Lemma (and Theorem 6, below) that Theorem 4 becomes such a powerful tool in the theory of interpretability (of arithmetical theories; see also Lemma 8.4). Lemma 2. S < T iff Th Cons ( k for every k. To prove this we need the following lemma whose proof is essentially the same as that of Theorem 2.7. Lemma 3. Suppose Th Cong \k for every k. Let σ(x) be any formula binumerating S in T and let σ*(x) := σ(x) Λ Conσ)x. Then (i) σ*(x) binumerates S in T and (ii) PAh Conσ*. Proof of Lemma 2. Suppose first S < T. Let k be arbitrary. There is then an m such that SI k < T I m. By Corollary 1, PAh Conτ |m -» Cons |k. But Th Conτ ( m and so ThCon s , k . Next suppose Th Con$( k for every k. Let σ(x) be a PR binumeration of S and let σ*(x) := σ(x) Λ Conσ|x. Then, by Lemma 3, σ*(x) binumerates S in T and PAh Conσ*. Hence, by Theorem 4, S < T. There are alternative notions of interpretability more general than the one defined here. For example, we may "interpret" the equality symbol = of one theory S as a certain relation definable in another S' (and having, provably in S7, the required properties) or we may "interpret" the individuals of S as finite sequences of individuals of S' etc. It turns out, however, that if S is "interpretable" in T in some such more general, and reasonably natural, sense, then, by Lemma 2, S < T (and conversely). Thus, in the present context, there is no reason to consider these more general "interpretations". From Lemmas 2 and 3 and Theorem 4 we get the following: Corollary 2. S < T iff there is a formula σ(x) (bi)numerating S in T such that Th Conσ.

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From Lemma 2 we also obtain the following result known as Orey's compactness theorem. Theorem 5. S < T iff SI k < T for every k. In the following we use A, B, etc. to denote (consistent, primitive recursive) extensions of T. Recall that AH Γ B means that every Γ sentence provable in A is provable inB. Theorem 6. A < B iff AH Πl B. Proof. Suppose first AH Πι B. Now, Ah ConA |k for every k. It follows that Bh ConA ik for every k. But then, by Lemma 2, A < B. Suppose next A < B. Let π be any Π1 sentence such that Ah π. By Lemma 1, Bh π, as desired. By Theorem 6, A + φ < A iff φ is Π1-conservative over A. Theorem 6 has the following immediate: Corollary 3. If A < B and σ is any Σ± sentence, then A + σ < B + σ. Combining Theorem 6 and Theorem 4.5 we get: Corollary 4. T + Rfnτ < PA + Con^. In fact, this follows directly from Lemma 2 and the fact, established in the proof of Theorem 4.5, that PA + Con^h ConTn for every n. Theorem 6 can also be used to prove the following model-theoretic characterization of interpretability: Theorem 7. A < B iff for every model M of B, there is a model M' of A such that M is (isomorphic to) an initial segment of M'. Proof (sketch). "If". Let θ be any Πj sentence such that Ah θ. We show that θ holds in all models of B. Let M be any model of B. By hypothesis, there is a model M' of A such that M is isomorphic to an initial segment of M'. θ holds in M'. Since θ is Πl7 it follows that θ holds in M. Thus, θ holds in all models of B and so Bh θ. We have shown that AH Π B and so A < B, by Theorem 6. "Only if". Let t: A < B. Let M be any model of B and let M' be the structure defined by t in M. M' is a model of A. Since induction holds in M, we can in M define a function f on M satisfying the following conditions: f(0M) = 0M', f(SM(a)) = SM (f(a)). f maps M isomorphically onto an initial segment of M'. Given Theorem 6, we can now derive Theorems 8-12 below as corollaries to

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results from Chapter 5. Like Theorem 2 the following result is a sharpening of GodeΓs second incompleteness theorem. Theorem 8. T + ->Conτ < T. Proof. This follows from Theorem 5.1 and Theorem 6. A more direct proof of Theorem 8 is as follows. We need the following: Lemma 4. If S < S'+ φ0 and S < S'+ φα/ then S < S'+ φ0 v q>r Thus, if S + φ < S + -ιφ, then S + φ < S. Proof. Suppose tji S < S'+ φ^ i = 0,1. Let t be the translation which coincides with t0 if φ0 and with tj if -i(pQ Λ φ1. Thus, for example,

μt(χ) := (ΦQ Λ MtoM)v (""ΦoA ΦiA m^*))-

It follows that for all φ, (1) S' + φ 0 ht(φ)^t 0 (φ), (2) S/ + -φ 0 Λφ 1 ht(φ)<->t 1 (φ). Now, suppose Sh φ. Then S' + φQh t0(φ) and so, by (1), S' + φQh t(φ). Also S' + φ1h t^φ) and so, by (2), S' + - φ0 Λ φ1h t(φ). It follows that S'+ φ0 v q^h t(φ). Thus, t: S < S'+ φ0 v φ1, as desired. By Corollary 2.2, T + Conτh Conτ+_,ConT. But then, by Theorem 4, T + - Conτ < T + Conτ and so, by Lemma 4, T + -«Conτ < T, as desired. (In this proof of Theorem 8 it is not necessary to assume that T is (essentially) reflexive.) Theorem 9. Suppose X is r.e. and monoconsistent with T. There is then a Σ1 sentence φ such that T + φ < T and φ£ X. Proof. This follows from Theorem 5.2 and Theorem 6. Corollary 5. Let τ(x) be a formula numerating T in T such that Tl^ -"Co^. There is then a (Σ{) sentence φ such that T + φ < T and Tl^ Conτ —»Conτ+φ. Proof. Let X = {ψ: Th Coi^ -» Con^} and use Theorem 9. Theorem 10. Suppose X is r.e. and monoconsistent with T. There is then a sentence φ such that T + φ < T, T + -.φ < T, φ£X, -κp£X. Proof. By Theorem 5.3 we can take φ to be, say, a Σ2 sentence such that φ is Π2-conservative and ->φ is Σ2-conservative over T. Now use Theorem 6. A sentence φ such that T + φ < T, T + -«φ < T is known as an Orey sentence for T. Clearly, any Orey sentence for T is undecidable in T.

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The intended applications of Theorems 9 and 10 are as follows. There are consistent finitely axiomatized extensions U of T in languages extending LA. In fact, U may chosen to be a conservative extension of T in the sense that for every sentence φ of LA, Uh φ iff Th φ. Thus, U and T are equivalent in terms of provability of sentences of LA. So it is natural to ask if U and T are (ever) equivalent in terms of interpretability of sentences of LA in the sense that for every sentence φ of L A/ T + φ < T iff U + φ < U. (We assume the reader can extend the defintion of "interpretation" and "interpretable in" to the case where the theories need not be formalized in LA.) The answer is a resounding "no" (see also Corollary 8.8). To prove this we need the following essentially trivial lemma whose proof is left to the reader. Lemma 5. Let V be any r.e. theory, not necessarily in LA. Then the set {φ: U + φ < V} is r.e. Corollary 6. There is a ΣI sentence φ such that T + φ < T and U + φ ^ U. Proof. The set {φ: U + φ < U} is clearly monoconsistent with T and, by Lemma 5, it is r.e. Now apply Theorem 9. By a similar proof, but using Theorem 10 in place of Theorem 9, we get: Corollary 7. There is a sentence φ such that T + φ < T, T + -«φ < T, U + φ ^ U , and U + -φ i U. As we saw in Chapter 4, speaking in terms of provability, we have to distinguish between finite, infinite, and unbounded extensions of a given theory T. In terms of interpretability the situation is quite different. We write S = S' to mean that S < S'< S. Theorem 11. (a) If AH B, then there is a sentence φ such that A + φ = B. (b) Let X be an r.e. set of Σj sentences. Then there is a Σ^ sentence σ such that

Proof, (a) Let X = Th(B) n nx. Then, by Theorem 6, A + X = B. By Theorem 5.4 (a), there is a sentence φ such that A + φ is a Γ^- conservative extension of A + X. By Theorem 6, A + φ Ξ A + X and so A + φ = B. 4 (b) This follows from Theorems 5.4 (a) and 6. Finally, we have a result which proves the claim made earlier that the fact that, for example, A + φ < B does not imply that this is provable in PA, or in any other preassigned consistent axiomatizable theory. From the definition of < it is clear that the set {φ: A + φ < B} is Σ^. From Theorem 6, it follows, however, that {φ: A + φ < B} is Π^. That this cannot be improved follows from:

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Theorem 12. Suppose A < B. Then the set (ψE Σα: A + φ < B} is a complete Π^ set. Proof. For A = B, this follows from Theorem 5.6 and Theorem 6; we leave the proof of the general case to the reader. A translation t is given by a finite amount of information which can certainly be coded by a natural number; thus we may "identify" t with that number. Let IntAB be the set of interpretations of A in B. Corollary 8. If A < B, then IntA/B is n° but not Σ°. Proof. Clearly IntA B is Π2. Suppose it is Σ 2 . Evidently A + φ < B iff Ξte IntA B(Bh t(φ)). It follows that {φ: A + φ < B} is Σ!?, contradicting Theorem 12. π In the next § we are going to prove that Int AB is, in fact, a complete Π2 set (Corollary 12).

§2. Faithful interpretability. Let t: S'< S. t is a faithful interpretation of S' in S, t: S'^ S, if for every sentence φ, if Sh t(φ), then S'h φ. S' is faithfully interpretable in S, S'^ S, if there is a t such that t: S'^3 S. Most of the differences between < and ^ are explained by the following lemma; for example, it is not true in general that if SH T, then S ^ T. Lemma 6. If QH S ^ T, then THΣιS. Proof. Suppose t: S ^ T. Let σ be any ΣI sentence such that Th σ. Clearly t: Q + -*σ < T + - t(σ). But then, by Lemma 1, T + - t(σ)h ->σ, and so Th t(σ). Since t is faithful, it follows that Sh σ. Our main aim in this § is to prove the following characterizations of ^. Theorem 13. S ^ T iff S < T and for every φ, if Th Pr0(φ), then Sh φ. Theorem 14. A ^ B iff AH Πl BH Σι A. Corollary 9. (a) S ^ T iff for every k, Th Cons |k and for every φ, if Th Pr0(φ), then Shφ. (b) If T is Σ^sound, then S ^ T iff S < T. (c) If S < TH S, then S ^ T. Proof, (a) and (b) follow at once from Theorem 13 and Lemma 2. 4 (c) Suppose Th Pr0(φ). Then, since T is essentially reflexive (Fact 11), Th φ and so, by assumption, Sh φ. Now use Theorem 13.

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85

By Corollary 9 (c), Theorems 8, 9,10 remain true when < is replaced by ^. Theorem 13 will be derived from the following two lemmas: Lemma 7. Let σ'(x) be a (Σj) formula binumerating S in T. There is then a (Σ1) formula σ(x) binumerating S in T and such that (i) h σ(x) —> σ'(x), whence h Conσ/ -> Conσ/ (ii) for every sentence φ, if Th Prσ(φ), then there is a q such that Th Prs \ q(φ). Lemma 8. Suppose σ(x) numerates S in T and Th Conσ. There is then an interpretation t: S < T such that for every φ, if Th t(φ), then Th Prσ(φ). Proof of Lemma 7. For simplicity we assume, as we clearly may, that if p is a proof of φ in T, then φ < p. Let σ(x) be such that PAh σ(x) o σ'(x) Λ Vyz<x(Prfτ(Prσ(y),z) -> PrσΊz(y)). Then (i) is trivial. We now show that (1) if p is a proof of Prσ(φ) in T, then Th Prσ,,p(φ). Let p and φ be as assumed. Then, since φ < p, Th -Prσ, i p(φ) -> (σ(x) -> σ'(x) A x < p). It follows that Th -πPrσ, i p(φ) -» (Prσ(φ) -> Prσ,,p(φ)). But then, since Th Prσ(φ), we get Th Prσ, |p(φ), as desired. Since σ'(x) binumerates S in T, it follows from (1) that (ii) holds. To show that σ(x) binumerates S in T it suffices to show that for all φ and p, Th Prfτ(Prσ(φ),p) -> Prσ, ,p(φ). But this, too, follows at once from (1). Proof of Lemma 8. The following proof is a modification of the proof of Theorem 4. The interpretation t constructed in that proof does not necessarily have the additional property that (1) Th t(φ) implies Th Prσ(φ). To achieve this we proceed as follows. The function c, the set Z, and the formula ζ(x) are the same as before, but the definition of φn is different. Here we put (2) φn := θn if S + ZhΛ{φm: m < n}->θn or (S + ZhΛ{φm: m < n}-> - θn & ne Y), := -»θn otherwise, where Y is any set of natural numbers. As before let X = {φn: ne N}. Either θn or -«θn is put in X. We put θn in X if putting - θn in X would make X inconsistent, and similarly for -«θn. Otherwise we put θn in X iff nE Y. The idea is to achieve (1) by letting Y be formally represented by a sufficiently independent formula η(x). Let γ(x) := σ(x) v ζ(x). Let η(x) be as in Theorem 2.10 with δ(x) := Prγ(x). Next, as in the proof of Theorem 4, let χ(x,y) be the formalization of the result of turning the

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inductive definition of φn into an explicit definition using η(x) to represent Y. Let Letξ(x):=Ξyχ(x,y). As in the proof of Theorem 4 we can now define an interpretation t of S in T such that (3) Tht(φ)^ξ(φ). It remains to be shown that (1) holds. Suppose ΊV Prσ(φ). We must then show that ΊV t(φ). We have ΊV PΓγ(φ) (see (6) in the proof of Theorem 4). For any fe 2N, let Y f = (PΓγ(n)f(n): ne N}. Now let f(n) be such that f(φ) = 1 and (4) T + Yf is consistent. Next we define ψn as follows (compare (2)). ψn := θn if Prγ(Λ{ψm:m-.θn)ί Y f & Pr^)* Yf), := -«θn otherwise, where λ^ := A{ψm: m < n} Λ θn -> φ. Let ge 2N be such that and set Y f/g = Yf+{η(n)S( n ):neN}. Then, by (4) and the choice of η(x), (5) T + Yf^g is consistent. Recalling the definition of χ(x,y), we can now show, by induction, that for every n, T + Y f / g h χ(ψn,n) and so (6) T + Y f/g hξ(ψ n ). Next we show, by induction, that for every n, (7) PΓγ( Λ{ψm:mθk)E Y f or PΓγ(Λ{ψm:mφ)£Yf. In the latter case (7) holds for n = k+1. In the former case we have Prγ( Λ{ψm:mψk)e Y f and so (7) for n = k+1 follows from the inductive assumption. Case 2. ψk := -«θk. Then (8) PΓγ(λk)GYf. For suppose Prγ(λk)ί£Yf. If PΓγ( A{ψm:mθ)e Yf for every θ and so, in particular, Prγ(λk)e Yf, contrary to assumption. So PΓγ(Λ{ψm:m
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Then t: S < T. Let φ be any sentence of S such that Th t(φ). Then, by Lemma 8, Th Prσ(φ) and so there is a q such that Th Prs | q(φ). It follows that Th Pr0(AS I q-xp) and so, by hypothesis, Sh AS I q -> φ, whence Sh φ. Thus, t is faithful. "Only if". Suppose S < T. Then S < T. Let φ be such Th Pr0(φ). Suppose t: S < T is faithful. Let K be the sentence saying that t is an interpretation of 0 (logic) in T. Then, by Fact 12 (b), PAh Pr0(φ) -> Pτ0(κ->t(φ)). But then Th Pr0(κ-»t(φ)). Since T is essentially reflexive, it follows that Th K -> t(φ). But Th K and so Th t(φ). But then, t being faithful, Sh φ, as desired. Proof of Theorem 14. Suppose first AH Πι BH Σl A. Then, by Theorem 6, A < B. Suppose Bh Pr0(φ). Then, Pr0(φ) being Σα, it follows that Ah Pr0(φ). Since A is essentially reflexive, this implies that Ah φ. Hence, by Theorem 13, A ^ B. Next suppose A ^ B. By Theorem 6, AH Π B, and, by Lemma 6, BHΣ A. The analogue of Theorem 11 (a) for ^ now follows at once from Theorem 14 and Theorem 5.4 (a) with, say, Γ = Π2. We write A - B to mean that A ^ B ^ A. Corollary 10. If AH B, there is a sentence φ such that A + φ - B. The analogue of Theorem 11 (b), on the other hand, is clearly false. (Let σ^ be Σ^ sentences such that T + {σ^: k < n}h σn for every n and let X = (σ^: ke N}. Let σ be any Σ^ sentence such that T + σ ^ T + X. Then, by Lemma 6, T + σh X, whence T + XI/ σ and so, again by Lemma 6, T + X ^ T + σ.) If S is finite, then {φ: S < T + φ} is r.e., but if < is replaced by ^ this is no longer true: Corollary 11. Suppose QH S ^ T. Then X = {φ: S < T + φ} is a complete Π^ set. Proof. By Theorem 13, X is 0°. Let Y be any Π^ set. By the proof of Theorem 5.6 (a), for Γ = ΣI, there is a formula ξ(x) such that (1) if ke Y, then ξ(k) is Σ1-<:onservative over T, (2) if kg Y, then there is a Σl sentence σ such that T + ξ(k)h σ and Sl^ σ. It is now sufficient to show that (3) Y={k:ξ(k)eX}. Suppose first ke Y. Let ψ be any sentence such that T + ξ(k)h Pr0(ψ). Then, by (1), Th Pr0(ψ). But then, by Theorem 13, Sh ψ. Using Theorem 13 once again, we get S Next suppose kέY Let σ be as in (2). Since σ is Σα, PA + σh PrQ(σ) and so PA + σh Pr0(ΛQ-»σ). It follows that T + ξ(k)h Pr0(ΛQ-»σ). On the other hand -> σ. Hence, by Theorem 13, ξ(k)eX. Finally, we improve Corollary 8 as follows.

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Corollary 12. If A < B, then IntA B is a complete Π^ set. Proof. Let X = {k: VmR(k,m)}, where R(k,m) is r.e., be any Π^ set. By Theorem 3.1, there is a formula p(x,y) numerating R(k,m) in B. Let α(x) be a formula binumerating A in B. Let σ(x,y) := α(x) Λ Conα ix Λ Vz<xp(y,z). Then, by Lemma 3, for every k, (1) PAh Conσ(x/k). By Lemma 2, for every n, Bh ConA |n. It follows that (2) if ke X, then σ(x,k) binumerates A in B. Also, clearly, (3) if k£ X, there is an m such that BI/ Ξx(m < x Λ σ(x,k)). By (1) and the proof of Lemma 8, we can for each k, effectively find a translation tk such that (4) t k :{φ:Bhσ(φ,k)} Ξx(m < x Λ σ(x,k)). But then, by (3), W Prσ(x/k)(θ) and so, by (5) and (7), tkg IntA/B. This proves (6) and so the proof is complete.

Exercises for Chapter 6. In the following exercises we assume that PAH T and that A, B, C are extensions of T. 1. Show that there is a Γ^ sentence φ such that Q + φ ^ S and Q + - φ ^ S (compare Theorem 8.2). 2. (a) Suppose AH B ^ A. Show that there is a C such that AH CH B and B ^ C ^ A. [Hint: There is a sentence θ such that Bh θ and {θ} ^ A. The sets {φ: {θ} < A + -«φ} and {φ: Q + θ v φ < A} are r.e. and monconsistent with Q.] (b) Suppose A < B. Show that there is a C such that A < C < B.

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3. The proof of Theorem 4 actually yields the following stronger result: There is a finite subtheory PAσ of PA such that S < PAσ + (σ(φ): φe S} + Conσ. Use this to prove the following: (a) If τ(x) numerates T in a finite subtheory of T, then T + Cor^ ^ T (compare Theorem 2). (b) T is interpretable in a bounded subtheory of T (compare Corollary 4.1 (a) and Theorem 3). 4. (a) Suppose σ0, σj are Σ± sentences such that T + σ^ < T, i = 0, 1. Show that T + σ0 Λ σ1 < T. (b) Show that there is a Π^ set X of Σl sentences such that T + Y < T for every finite (and so for every r.e.) subset Y of X and T + X ^ T (compare Theorem 5). [Hint: Let τ(x) be a PR binumeration of T. Let p(x,y) be a PR formula such that {k: ΞmPAh p(k,m)} is not recursive. Let γ(x,y) := τ(x) Λ Vz<χ-«p(y,z). Let X = {-Coπγ(x/k): T + -^oityxjc) < T}.] 5. Improve Corollary 3 by showing that IntA B £ IntA+σ/ B+σ. 6. (a) Use Exercise 2.15 (b) to give an alternative proof of Theorem 8. (b) Use Exercise 2.16 and Theorem 8 to give another proof of Theorem 9. 7. Suppose PAH Sj. Prove the converse of Theorem 1: If for every ΣI numeration σ1(x) of Sj, there is a ΣI numeration σ0(x) of S0 such that PAh Conσ —» Conσ , then SQ ^ Sj. [Hint: Use Theorem 5 and Exercise 2.16.] 8. Show that there is a (Πl7 Σ±) sentence θ such that Th ConT -> Conτ+θ and T + θ

9. Let θ be a Hi Rosser sentence for T and let ψ := Vu(Prfτ(- θ,u) -> Ξz Vz(Con T | z+φ -> -*p(φ,z)). Show that T + φ < T and φ^ X. (b) The sentence φ in (a) is Π2. This can be improved. Let χ be such that PAh χ <-> Ξz(^Conτ(z+χ Λ Vu
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11. (a) Let φ be such that PAh φ <-> Vz(Conτ (z+φ -^ Conτ t z+-,φ). Show that φ is an Orey sentence for T (compare Theorem 10). (b) Suppose AH B. Let φ be such that PAh φ <-» Vz(ConA |z+φ -> ConB,z). Show that A + φ = B (compare Theorem 11 (a)). (c) Let φ be such that PAh φ <-> Vz(ConA,z+φ -> ConB,z+^φ). Show that A + φ = B + -iφ. 12. (a) Show that PAh Vx(ConS(),x -» ConSl,x) <-> (ConSl v Ξx(-ConS(),x A Vy<xCon Sl , y)). Conclude that the sentences φ of Exercise 11 are Δ2 (compare Exercise 5.9 (a) and Theorem 7.8). In particular, there is a Δ2 Orey sentence for T. (b) Show that no Orey sentence for T is B^. 13. Let τ*(x) be as in Theorem 2.7. In Theorem 4 let σ(x) := τ*(x) and S = T. Next let ξ(x) be as in (14) of the proof of Theorem 4. Let φ be such that PAh φ <-» -»ξ(φ). Show that φ is an Orey sentence for T. 14. Let τ(x) be any formula binumerating T in T. Let φ be such that PAh φ <-> - Prτ(φ). Show that (ii) for every n, T + φ < T + PrT | ^Cor^). Let τ(x) be the formula τ*(x) mentioned in Theorem 2.7. Conclude that φ is then an Orey sentence for T. 15. Suppose T < S. Show that there is a Σ^ formula ξ(x) such that {k:T + ξ(k)< S } is a complete Π^ set (compare Theorem 12). [Hint: Let R(k,m) be an r.e. relation such that {k: VmR(k,m)} is a complete itf set. There is a Σ1 formula p(x,y) such that if R(k,m), then Qh ρ(k,m), if not R(k,m), then Q + p(k,m) i S (Lemma 3.1). Let ξ(x) be such that PAh ξ(k) <* Ξz(-Conτlz+ξ(k) Λ Vu
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T + φ ^) T, then g(φ) is a Πx sentence such that T + φh g(φ) and ΊV- g(φ). Show that g(k) cannot be recursive. 17. Show that there are sentences φ0, cpi such that T + φ^ < T, T + φ0 Λ cp} ^ T, T + -iφi ^ T, T + -iφo v -.φ! < T, i = 0,1. [Hint: Use Exercise 5.8 (b).] 18. Show that, even if T is not Σ1-sound, there is a Σj formula τ(x) binumerating T in T such that Prτ(x) numerates Th(T) in T (by Exercise 2.22 (iv), τ(x) cannot be PR). 19. Show that if A < B, then {φe Σ^ A + φ ^ B} is a complete Π^ set. 20. (a) Show that if S is finite and QH S ^ T, then the set of faithful interpretations of S in T is a complete Π^ set. [Hint: First show that there is a sentence θ such that SI/ θ and S + θ < T.] (b) Suppose A ^ B. Show that the set of faithful interpretations of A in B is a complete Π^ set. 21. S is X-faithfully interpretable in S', S ^XS', if there is an interpretation t: S < S' which is X-faithful in the sense that for every φe X, if S'h t(φ), then Sh φ. Show that (i) S ^XT iff S < T and for every φe X, if there is an m such that Th Prs |m(φ), then Sh φ, (ii) if S < T, then S ^XT, where X = {φ: S (iii) if S %T and S < T'H T, then S *3X (iv) if S < T and SH S'< T'H T, then S'^ T7, (v) ^ cannot be replaced by ^x in (iv), (vi) A%BiffA+(Th(B)nΣ 1 )H x BH Π l A, (vii) A ^ B iff A ^ΣlB iff A ^{φ)B for every (Σx) sentence φ. 22. Show that AHΠ B iff there is a t: A < B such that for every Πn sentence ψ, Bh t(ψ) -> ψ (compare Theorem 6; note that for every t: A < B and every Hi sentence ψ, Bh t(ψ) —> ψ, by Lemma 1). [Hint: "Only if". For every k and every Πn sentence φ, Bh PrA i k(φ) -> φ. Use this to construct a formula α(x) binumerating A in B and such that PAh Conα and Bh χ -> α(χ) for every 1^ sentence χ.]

Notes for Chapter 6. The general concept (relative) interpretation due to Tarski (cf. Tarski, Mostowski, Robinson (1953); in keeping with recent usage we omit "relative"); it is an important tool in proofs of (relative) consistency and (un)decidability. The investigation of interpretability for its own sake was initiated by Feferman (1960). Theorems 1, 2, 3 are due to Feferman (1960); concerning the (im)possibility of improving Theorem 3, see Exercise 3 (b). Theorem 4 is due to Feferman (1960) building on

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work of Bernays (Hubert and Bernays (1939)) and Wang (1951); for a strengthening of Theorem 4, see Exercise 3. Lemma 2 is implicit in Feferman (I960), all but explicit in Orey (1961), and fully explicit in Hajek (1971). Corollary 2 is due to Orey (1961). Theorem 5 is due to Orey (1961) (cf. also Feferman (I960)). Theorem 6 was first stated by Guaspari (1979) and Lindstrom (1979); for a more general result, due to Guaspari (1979), see Exercise 22. Corollary 4 is due to Goryachev (1986). Theorem 8 is due to Feferman (1960) (with a different proof; see Exercise 6 (a)). Lemma 4 is due to Svejdar (1978). Theorem 9 and Corollary 6 are essentially due to Hajek (1971) (cf. also Hajkova and Hajek (1972)) (with a different proof; see Exercise 10 (a)); for yet another proof, see Exercise 6 (b). Theorem 10 less the references to the set X is due to Orey (1961); the full result is proved in Lindstrom (1979), (1984a); related results, for certain nonreflexive theories, requiring methods not explained here, can be found in Hajek and Pudlak (1993). For more information on Orey sentences, see Exercises 11 (a), 12 (b), 13,14. Corollary 5 has also been pointed out by Guaspari (1979); for a related result, see Exercise 8. The result on finite conservative extensions mentioned just before Lemma 5 is due to Kleene (1952b) (cf. also Kaye (1991)). Theorem 11 is due to Lindstrom (1979) (see Exercise 11 (b)) and (1984a); by Exercises 11 (b) and 12, the sentence φ in Theorem 11 (a) can be taken to be Δ2 (cf. also Theorem 7.8). Theorem 12 is essentially due to Solovay (cf. Hajek (1979)) (with a different proof); the present proof is from Lindstrom (1984a) (see also Exercise 15). The concept faithful interpretation was introduced in Feferman, Kreisel, Orey (1960). They observed that if QH S ^ S' and S is Σ^sound, so is S' (see Lemma 6). Theorems 13 and 14 are due to Lindstrom (1984c); see also Exercise 21. Corollary 9 (b) is due to Feferman, Kreisel, Orey (1960). Lemma 7 is due to Lindstrom (1984c); the present proof is an instance of a general argument described in Lindstrom (1988). Lemma 8 is due to Lindstrom (1984c), but the main idea of the proof, to introduce the set Y and represent Y by a sufficiently independent formula, is taken from Feferman, Kreisel, Orey (1960). Corollaries 10, 11, 12 are due to Lindstrom (1984c); for related results, see Exercises 19 and 20; Exercise 7.8, below, is an improvement of Corollary 10. An alternative notion of interpretability,/^zs/We interpretability, has been studied by Verbrugge (1992), (1994). For any formal entity q, formula, proof, etc., let Iql be the length of q, i.e. the number of (instances of) symbols occurring in q. S is feasibly interpretable in T, S
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11 (a) is due to Lindstrom (1979) and Svejdar (1978). Exercise 12 was pointed out to me by Franco Montagna (compare Theorem 7.8). Exercise 13 is due to Orey (1961). Exercise 16 (a) is due to Jeroslow (1971b). Exercise 17 can be substantially improved using results on the modal logic of (provability and) interpretability, due to Berarducci (1990), Shavrukov (1988), and Strannegard (1997). Exercise 22 is due to Guaspari (1979).

7. DEGREES OF INTERPRETABILITY

Suppose PAH T. We shall use A, B, etc. for extensions of T. (Thus, T, A, B, etc. are essentially reflexive.) The relation < of interpretability is reflexive and transitive. Thus, the relation = of mutual interpretability (restricted to extensions of T) is an equivalence relation; its equivalence classes will be called degrees (of interpretability) and will be written a, b, c, etc. Dτ is the set of degrees of extensions of T. A is of degree a if Ae a and d(A) is the degree of A. The relation < among degrees is the relation induced by the relation < among theories: d(A) < d(B) iff A < B. Dτ = (Dτ,<), the partially ordered set of degrees defined in this way, will be studied in some detail in this chapter.

§1. Algebraic properties. In this § we restrict ourselves to purely algebraic properties of D p. First we define the theory Aτ and the operations >l and T on theories as follows. AT = T + {Con A)k :keN}, AiB = T + {ConA ik v ConB |k: ke N}, ATB = T + {ConA ik Λ ConB |k: ke N}. From Lemma 6.2 and Theorem 6.6, we get the following: Lemma 1. (a) A < B iff ATH B. Thus, A τ = A and A < B iff ATH Bτ. ( b ) A < B , C i f f A θ. Instead of AlB it is sometimes convenient to use the theory AvB defined by AvB = {φ v ψ: φe A & ψe B}. Th(AvB) = Th(A) n Th(B). Evidently, AiB H AvB and, by Lemma 2, AvB H Πl AlB. But then, by Theorem 6.6, that AvB < A^B and so AvB = AiB. It follows that for every sentence φ, (A + φ)l(A + -«φ) < A. From Lemma 2 and Lemma 6.1 we get: Lemma 3. For every E^ sentence π, T + π < A?B iff A?Bh π iff there are Γ^ sentences φ, ψ such that Ah φ, Bh ψ, and T + φ Λ ψh π. For Ae a and Beb, let a n b = d(AlB) and a u b = d(AΪB). By Lemma 1, n and u

§1. Algebraic properties

95

are well-defined, a n b is the g.l.b. of a and b and a u b is l.u.b. of a and b. Thus, we have proved part of the following: Theorem 1. Dτ is a distributive lattice. To prove distributivity we need the following lemma whose proof is left to the reader. Lemma 4. (a) For every k, there is an m such that PAh Con(AvB) im -> ConA,k v ConB,k. (b) For every k, there is an m such that PAh ConA,m v ConB,m -> Con(AvB),k. Proof of Theorem 1. Let D = Aτv(BΪC) and E = (AvB)T(AvC). To prove that Dτ is distributive, it is, by Lemma 1, sufficient to show that D Hh E. Let k be arbitrary. By Lemma 4 (a), there is an m such that PAh Con(AvB) i m -> ConA,k v ConB,k, PAh Con(AvC) im -> ConA,k v Conc,k. But then Eh ConA ik v (ConB,k Λ Conc,k). It follows that DH E. The proof that Dh E is similar. Dτ has a minimal element Oτ = d(T) and a maximal element lτ, the common degree of all inconsistent theories. In our next result we answer a number of standard questions concerning Dτ; in particular, it follows that D-p is dense. Theorem 2. Suppose a < b < lτ, d0 ^ a, and b ^ d1. There are then degrees c0, q such that a < q < b, d0 ^ q ^ d^, i = 0,1, CQ n c1 = a, and CQ u c1 = b. We derive this from: Lemma 5. Suppose X is r.e. and monoconsistent with PA. Let θ be any true Π^ sentence. There are then Π^ sentences Θ0, &ι such that (i) PAhθovθi, (ii) PAh Θ0 Λ 0! -> θ, (iii) θ/eX, i,j = 0,l. Proof. We may assume that if φeX and PAh φ —> ψ, then ψeX. Let θ := Vxγ(x), where γ(x) is PR. Let R(k,m) be a primitive recursive relation such that X = {k: ΞmR(k,m)} and let p(x,y) be a PR binumeration of R(k,m). Finally, let Θ0 and θj be such that (1) PAhθ 0 ^ Vy((p(θ0,y)

96

7. Degrees of interpretability

(2) PAhθ! <-> Vz(p(θ1/Z) -» 3yΘ0, again a contradiction. Thus, Θ 0 £X and Θ 1 ^X. Finally, if -*&£ X, then, by (i), θ^e X. It follows that - θ0e X and -.θ^ X. Proof of Theorem 2. Let Aea, Beb, D^ed^ By Orey's compactness theorem (Theorem 6.5) there are sentences ψ, χ such that Bh ψ, ψ ^ A, D0h χ, χ ^ A. By Theorem 6.6, there is a Π1 sentence π such that Bh π, Ah π, and Djh π. Let X0 = {φ: Ah φ v π}, X: = (φ: ψ < A + -iφ}, X2 = {φ: χ < A + -.φ}, X3 = {φ: D x h φ v π}. Let X = X0 u \ι u X2 u X3. Then X is r.e. (cf. Lemma 6.5). It is also easy to verify that X is monoconsistent with PA. By Lemma 5, there are then Π1 sentences Θ0, &ι such that (1) PAh Θ0 v θx, (2) PAh Θ0 Λ θα -» ConB, (3) Θ^X, i,j = 0,l. Let eA = d(A+θi), i = 0, 1. Then a < e^ b ί eir since -*&& Xχ d0 ^ e{, since -O^ X2. Let q = e^ n b. Then q < b and d0 ^ Cj. If Cj < a, then, since θj is Πj, Ah θj v π, contradicting the fact that θ^ X0. Thus, Cj ^ a and so a < q. Similarly, q ^ d l r since 0^X3. By (1), c0 n G! = a. By (2), Theorem 6.4, and Lemma 3, e0 u C L > b, whence, by distributivity, c0 u QI = b n (e0 u e^ = b. From Lemma 5 we can also derive the following: Corollary 1. T is not Σ^-sound iff there are degrees a0, a^ < ly such that a0 u a± = lτ (and a0 n a1 = Oτ). Proof. Suppose T is Σ1-sound. Let a, b < lτ, Ae a, BG b. Then A?B is consistent and so a u b < lτ. Next, suppose T is not Σ1-sound. There is then a true Πj sentence θ such that Th --Θ. Let θ^ be as in Lemma 5 with X = Th(T). Let a{ = d(T+θi). Then a^ < lτ, by Lemma 5 (iii), and a0 n a^ = Oτ, by Lemma 5 (i). Finally, by Lemma 3, (T + Θ0)T(T + θ x )h θ. Since Th -.θ, it follows that (T + Θ0)T(T + θα) is inconsistent and so a0 u a^ = lτ. By Corollary 1, if PAH S and S is Σ^sound but T is not, then Ds and Dτ are not isomorphic. But suppose S and T are both Σ1-sound. It is an open problem if this implies that Ds and Dτ are isomorphic. Given that there are c0, q > a such that c0 π q = a, we may ask if any b such that a < b < l τ caps to a in the sense that there is a c> a such that b n c = a. (Dually, b cups to a if there is a c < a such that b u c = a.) In our next result this question and its dual are answered in the negative. We write a «n b to mean that a < b and b does not cap to a. Dually, a <<^ b means that a < b and a does not cup to b.

§1. Algebraic properties

97

Theorem 3. (a) Suppose Oτ < a ^ c. There is a b such that Oτ < b «^ a and b ^ c. (b) Suppose c ^ ) a < lτ. There is a b such that a «^ b < l τ and c ^ b. Proof, (a) Let Ae a and CE c. There is a Γ^ sentence θ such that Ah θ and CH θ. Let X = Th(C + - θ). X is r.e. and (mono)consistent with T + -iθ. By Theorem 5.2, there is a Π L sentence ψg X such that ψ is Σ1-conservative over T + -iθ. Let B = T + ψ v θ and b = d(B). Then Oτ < b ^ c and b < a. Suppose b u d = a. Let De d. Then, by Lemma 6.2, there is an m such that T + ψ + Con D(m h θ and so T+-»θ+ψh -«ConD i m. Since ψ is Σ1-conservative over T + -iθ, it follows that T+ -i θ h - ConD (m and so Dh θ. Thus, d > b and sod = b u d = a. * The proof of the following lemma from Lemma 6.2, Theorem 6.6, and Lemma 2 is straightforward. Lemma 6. The following conditions are equivalent: (i) AlBConB (k for every k. (iii) A < C -i- -iθ for every Uι sentence θ such that Bh θ. Let σ be any Σ1 sentence. By Corollary 6.3, the degree d(A+σ) is uniquely determined by σ and d(A). Thus, we may denote the former by d(A) + σ. A degree of the form a + σ will be called a Σγ-extension of a. If X is an r.e. set of Σ^ sentences, then, by Theorem 6.11 (b), d(A+X) is a Σ^^-extension of d(A). Lemma 7. The following conditions are equivalent: (i) a «^ b. (ii) a < b and for every Σ^^-extension c of a, if b < c, then c = lγ. Proof. Suppose (i) holds. Let Aea and Beb. Let σ be Σα and such that b < a + σ. Then B>l(A + ->σ) < (A + σ)i(A + -<σ) < A. Hence, by assumption, A + -XT < A, whence Ah ->σ and so a + σ = lτ. Thus, (ii) holds. Next suppose (ii) holds. Let c be such that b n c = a. Let Aea, Beb, Cec. Let θ be any Π1 sentence provable in C. It suffices to show that Ah θ. By Lemma 6, B < A + -«θ. But then, by assumption, Ah θ, as desired. Lemma 8. If π is Π^ A < B + π, and -iπ is E^-conservative over A, then d(A) «n d(B + π). Proof. Suppose B + π < A + σ. Then, by Lemma 6.1, A + σh π, whence A + -<πh -«σ and so Ah -<σ, in other words, A + σ is inconsistent. Now use Lemma 7. Proof of Theorem 3 (b). Let Ae a, Ce c. By Theorem 6.5, there is a sentence ψ such that Ch ψ ^ A. Let X = (φ: ψ < A + - φ}. Then, by Theorem 5.2, there is a Σl sentence χ£ X such that χ is Γ^-conservative over A. Let B = A + -iχ and b = d(B). Then c ^

98

7. Degrees of interpretability

b < lχ. Finally, by Lemma 8, a <<γ>,b. From Theorem 6.4 and Lemma 8, and Theorem 5.1, we get the following (compare Theorem 6.2): Corollary 2. d(A) «^ d(T+ConA). Theorem 3 (a) leads to the problem if for any a < lτ, there is a b such that a «^, b < lτ. (The dual of this is false: if Oτ < b < a and not Oτ <^ a, then not b «^ a.) We now show that the answer is negative. a is a cupping degree if a < l τ and a cups to every b such that a < b < lτ. Let CONT = {a < lτ: a = d(T+Cor4) for some PR binumeration τ(x) of T}. By Corollary 2.4, CONT * 0. Theorem 4. Every member of CONT is a cupping degree. Proof. Suppose a = d(T+Conτ) < 1T, where τ(x) is a PR binumeration of T. Let b be any degree such that a < b < 1T. Let Beb. We want to define a degree d such that d Jί a and au d>b. The obvious way to try is to let d = d(T+θ), where θ := Vu(PrfB(_L,u) -> Ξz 3z Prτ(φ). Since -«φ is Σl7 we also have, by provable Σ1-completeness, PAh -ιφ -» Prτ(-ιφ). Thus, (4) PAh Coriτ -> φ. Let d = d(T+ψ). Then, since ψ and Cor^ are n lr it follows from (3), (4), Lemma 3, and Theorem 6.4 that a u d > b. Suppose a < d. Then T + ψh Conτ. But then, by (2) and (4), Th φ, contradicting (1). Thus, a ^ d. Let c = d n b. Then c < b. Finally, a u c = (a u d) n (a u b) = b. Thus, a is cupping. Theorem 14', below, is an improvement of Theorem 4. A set G of degrees is cofinαl in DT if for every degree a < lτ, there is a degree be G such that a < b < lτ.

Lemma 9. CONT is cofinal in Dτ.

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99

Proof. Suppose b < lτ. By Corollary 2.4, even if T is not Σj-sound, there is a PR binumeration β(x) of a theory of degree b such that T + Conβ is consistent. By Theorem 6.4, b < d(T+Conβ). By Theorem 2.8 (b), there is a PR binumeration τ(x) of T such that Th Con,. ^ Conβ. Let a = dCΓ+ConJ. Then b < ae CONT. Let P be a property of degrees. We shall say that there are arbitrarily large degrees having property P if the set of degrees having P is cofinal in Dτ. Every sufficiently large degree has P if for every degree a < lτ, there is a b such that a < b < l τ and every degree c such that b < c < l τ has P. If a is cupping and a < b, b is cupping. Thus, from Theorem 4 and Lemma 9 we get: Corollary 3. Every sufficiently large degree is a cupping degree. By Corollary 1, if T is Σ j-sound, no degree, except Oj and 1 j, has a complement whereas if T is not Σ1-sound, some do. Also, of course, if Oχ «^ a < lτ, then a has no complement. But, even if a has no complement, it may still have a pseudocomplement (p.c). For example, if Oτ «^ a, Oτ is the p.c. of a. By Lemma 6, if π is Πj, then d(T+->π) is the p.c. of d(T+π). On the other hand we have the following: Theorem 5. There is a degree which has no p.c. The proof of this (and more) will be given in § 3 (Theorem 17). In addition to the usual (finitary) distributive laws, Dj also satisfies the following infinitary distributive laws. Let G be a set of degrees. LJG (OG) is then the l.u.b. (g.l.b.) of G, if it exists. Theorem 6. (a) If UG exists, then UG n b = U{a n b: aeG}. (b) If ΠG exists, then ΠG u b = Π{a u b: aG G}. By Theorem 6 (a), if a has no p.c., then {b: b n a = Oτ} has no l.u.b. In Lemma 23, below, we give a nontrivial example of a set G which has no g.l.b. To prove Theorem 6 (b) we need the following: Lemma 10. The following conditions are equivalent: (i) AίB>C (ii) For all (Σ ^ sentences χ and all m, if A τ + -«Conc \m H Σj T + χ, then Bh -«χ. Proof. Suppose (i) holds. Let χ and m be such that Aτ + -iCon^ | m Hχ T + χ. There is a k such that A τ + ConB |k h Conc (m. It follows that T + χh - ConB |k, whence Bh -iχ. Thus, (ii) holds. To prove that (ii) implies (i), suppose (i) fails, i.e. AtB Jί C. There is then an m such that for every k, A τ + ConB ( ^ Conc |m. But then, by Theorem 4.3, there is a

100

7. Degrees of interpretability τ

ΣI sentence χ such that A + -Cone |m~\^ + χ and T + χV- - ConB |k for every k. Since - χ is Ulf it follows, by Lemma 2, that BI/ - χ. Thus, (ii) is false, as desired. Proof of Theorem 6. (a) Let c = LJG. Clearly c n b is an upper bound of {a n b: aeG}. Suppose d is any upper bound of {a n b: aeG}. It is then sufficient to show that c n b < d. Let Be b, Ce c, De d. Then AlB < D for every A such that d(A)G G. But then, by Lemma 6, A < D + ->ConB |k for every such A and every k. It follows that for every k, C < D + -«ConB ( k for every k, whence, by Lemma 6, CiB < D and so c n b < d. Φ (b) Let c = ΓΊG. Clearly c u b is a lower bound of {a u b: ae G}. Suppose d is any lower bound of {a u b: aeG}. It is then sufficient to show that d < c u b. Again let Beb etc. Then D < A?B for every A such that d(A)e G. But then, by Lemma 10, for every such A, every m and every Σ1 sentence χ, if Bτ + -«ConD ( m^T + χ, then Ah -ιχ. It follows that for every m and every Σ^ sentence χ, if Bτ + ->ConQ | m Hj\T + χ, then Ch -ιχ. Hence, again by Lemma 10, D < CίB and so d < c u b. Suppose a < b. Let [a,b] be the interval {c: a < c < b}. (We also write [a,b) for {c: a < c < b} etc.) A natural (global) question concerning Dτ is if all intervals [a,b], where a < b < lτ, are isomorphic (in the obvious sense). The answer is negative. If c < d, let [d,c] = ([c,d], >). Another natural question is, under what conditions [a,b] is isomorphic to [d,c], where a < b and c < d. Theorem 7. (a) There are degrees a, be (0^,1^) such that the intervals [O^a] and [Oj,b] are not isomorphic. (b) Suppose a < b and c < d. Then [a,b] is not isomorphic to [d,c]. Theorem 7 (a) follows at once from our next two lemmas. The interval [a0,a1], where a0 < a l7 is said to satisfy the reduction principle if for any bg, b^e [a0,aι], if bg u b^ = a lr there are q < bj, i = 0,1, such that CQ n c^ = ap and c0 u QI = a1. A degree a is r.p. if [Oτ,a] satisfies the reduction principle. Lemma 11. If a = d(T+θ), where θ is Π l7 then a is r.p. Proof. Let b0, bl be such that b0 u b1 = a. There are then Π1 sentences ψ0, ψi such that dίT+ψi) < bi and T + ψ0 Λ ψ α h θ. By Lemma 5.5, there are ^ sentences Θ0, θα such that Th Θ0 v Θ l7 Th ψj -» Θi7 i = 0,1, Th Θ0 Λ θα -> ψ0 Λ ψx. Let q = d(T+θi), i = 0,1. Then q < bj, c0 n QI = Oτ, and, by Lemma 3, c0 u QI = b0 u b^ = a. Lemma 12. There is a degree a < l τ which is not r.p. Proof. Let π be a U^ sentence undecidable in T. In case T is not Σ1-sound we also need to assume that π is Σ1-conservative over T (cf. Theorem 5.2). We now effectively define r.e. sets Xk of ^ sentences such that (1) T + Xk + π1 is consistent, i = 0,1,

§1. Algebraic properties (2)

101

X k £ Xk+1,

(3) T + Xk + πV Xk+1/ (4) T + Xk + - π < T + Xk+1. Let X0 = 0. Then (1) holds for k = 0. Now suppose (1) holds for k = n. By (the proof of) Lemma 2.1, we can effectively find a Π^ sentence ψn such that [ (5) T + Xn + κ + -i\)4 is consistent, i = 0,1. Let Tn =df T + Xn + -.π + ψn. It follows that (6) there is no Πx sentence θ such that Tnh θ and T + θh - π. For suppose T + θh -iπ. Then T + πh ->θ and so Th -»θ, whence, by (5), Tnl/ θ. Let Xn+1 = Th(Tn) n Πα. Let k = n+1. Then (1) is satisfied for i = 1 and, by (6), (1) is satisfied for i = 0. Moreover (2) and (4) hold for k = n. Finally, T + Xn+il~ ψn and so, by (5), (3) holds for k = n. Let a0 = d(T+LJ{Xk:ke N}), a^ = d(T+π), and a = a0 u aα. Since a0 < lχ and π is Σ1-conservative over T, we have a < 1 j. We now show that a is not r.p. Let b0 and bι be such that bg ^ ag, b^ < a^, bg n bj = Op, and bp u b^ > a^. It is then sufficient to show that b0 u b1 j! a0. Let θi/k be Π1 sentences such that bj = d(T+{θ^k:ke N}), i = 0,1. We may assume that T + θi k+ιh θjk for i = 0,1 and all k. By Lemma 3, there is then an m such that T

+ Θ0,m Λ Q1#J~ π d(T+θO,m) ^ bO - aO Thus' b Y (2)'there is an n such that T + Θ0/m < T + Xn. Since b0 n bα = Oτ, for every k, T + Θ0/k v π < T + Θ0/k v (Θ0/m Λ Θ1/m)

< T + Θ0/m. It follows that T + Θ0/k v π < T + Xn, whence T + Θ0/k < T + Xn + -iπ (cf. Corollary 6.3) and so, by (4), T + Θ0/k < T + Xn+ι But this holds for all k, whence b0 < d(T+Xn+1). Next, by (3), b0 u ^ < b0 u al < d(T+Xn+1+π) 5ί a0. It follows that bg u b^ ^ a0 and so the proof is complete. Proof of Theorem 7 (b). Let Ae a and let π be a Πj sentence such that Al^ π. Then [a,d(Aτ+π)] satisfies the reduction principle (see the proof of Lemma 11). It follows that in [a,b] there is a degree e > a such that [a,e] satisfies the reduction principle. Thus, it is sufficient to show that the dual of the reduction principle is false in [c,d] whenever c < d. Let Ce c and De d, and let π be such that O π and Dhπ. Then, by Theorem 5.5 (b) with X = Th(CT + - π), there are Σl sentences σt such that Cτ+ ^σ{ = Cτ + aM, τ T T i = 0,1, and C + -ισ0 Λ ^ \f π. Let q = d(C +Gi) = d(C +-'G1_i). Then c0 n cx = c and c0 u c1 ^ d. Let dj = q n d. Then d0 n dj = c and d0 u dj < d. Suppose now dj < ej < d, i = 0,1, and e0 n eα = c. We have to show that e0 u e-^ < d. Let E0e e0. q n e0 = cι n d n e0 = d x n e0 < eα n e0 = c. It follows that (Cτ + ->σ0)lE0 < Cτ. But then, by Lemma 6, for every Γ^ sentence θ, if E0h θ, then Cτ + - σ0 < Cτ + -«θ, whence Cτ+σ0l- θ. It follows that e0 < CQ and so e0 = d0. Similarly, e^ = d^ Hence e0 u eι = d0 u d^ < d and the proof is complete. Theorem 7 (a) leads to the problem of determining the exact number of nonisomorphic intervals of Dτ. This problem remains open. We have actually proved more than is stated in Theorem 7. Let L = {<, n, u, 0,1} be the language of the theory of lattices with a bottom and a top element.

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Formulated in L, the reduction principle is an V5 sentence. Hence, by the proof of Theorem 7 (a), there are degrees a, be (Oτ,lτ) and an V5 sentence of L which holds in [Oτ,a] but not in [Oτ/b]. (This is, so far, the only known way of proving that two intervals of Dτ are nontrivially nonisomorphic.) Similarly, the proof of Theorem 7 (b) shows that if a < b and c < d, there is an ΞVΞ sentence which is true in [a,b] and false in [d,c].

§2. A classification of degrees. When there is no risk of confusion we shall use φ and X in place of T + φ and T + X. Thus, d(φ) is d(T+φ), X < φ means that T + X < T + φ, φ = ψ that T + φ = T + ψ, etc. We also write a «b to mean that a «^b. A «B means that d(A) « d(B). σ, σ0, etc. will be used to denote ΣI sentences and π, π0, etc. to denote Γ^ sentences. A degree a is Φ if there is a Φ sentence φ such that a = d(φ). By the proof of Theorem 6.11 (a), it is clear that every degree is Π2 and Σ2. This can be somewhat improved: Theorem 8. Every degree is Δ2. Proof. Let a be any degree. There is a primitive recursive set X of Γ^ sentences such that a = d(X). Let ξ(x) be a PR binumeration of X and let φ be such that PAh φ <-> VzdΠJίqvz) -> (ξ(z) -» TrΠl(z))). Then φ is Π2 and T + φ is a Π^-conservative extension of T + X (cf. the proof of Theorem 5.4 (a)). It follows that a = d(φ). Using Lemma 5.1 (i) and Lemma 1.3 (v) (applied to - φ), we get: PAh φ <-> Vz(ξ(z) -> TrΠl(z)) v az(-,[Π1](φ/z) A Vu TrΠl(u))). Thus, φ is Δ2 By Theorem 8, in terms of the arithmetical hierarchy, the only interesting (proper) subsets of Dτ are the sets of B! degrees, Σ1 degrees, Γ^ degrees, and degrees which are both ΣI and Πj. (If T is not Σ1-sound, there are also Δ^ degrees other than Oτ and lτ; see e.g. the proof of Corollary 1.) The object of the rest of this § is simply to show that these sets are different and that there is a non-B} degree. More detailed information about the ΣI and the Π} degrees will be given in the next §. Our next lemma is a restatement of Theorem 6.11 (b). Lemma 13. If X is an r.e. set of ΣI sentences, then d(X) is Σ1. The following lemma is occasionally useful. Lemma 14. There exist a (primitive) recursive sequence <σk>k<ω and a sentence σ such that (i) T + σk+1h σk, for all k, (ii) σk < σk+1, for all k, (iii) σ = {σk: keN}.

§2. A classification of degrees

103

This follows at once from (the proof of) Lemma 2.1 (applied to the sets {φ: Q + φ < T + σk}; σ0 := 0 = 0) and Lemma 13. Theorem 9. (a) There is a ΠJL degree which isn't Σ1. (b) There is a Σl degree which isn't Π^ (c) There is a degree other than Oτ and lτ which is both Σα and Πj. (d) There is a B! degree which is neither ΣI nor Γ^. (e) There is a degree which isn't B1. Proof, (a) Let π be such that - π is Π1-conservative over T and TI/ - π. Then, by Lemma 8, Oτ « d(π) and so, by Lemma 7, d(π) is not Σj. * (b) Let <σk>jc<ω and σ be as in Lemma 14. Suppose d(σ) is Π1 and let π be such that σ Ξ π. Then π = (σk: ke N} and so, by Lemma 14 (i), there is an m such that T + σ m h π. But then (σk: ke N} < σm, contradicting Lemma 14 (ii). Thus, d(σ) isn't Πα. » (d) The easiest way to prove this is to define π as in the proof of (a) and then σ as in the proof of (b), except that T is replaced by T + π. Then d(πΛσ) is neither ΣI nor Π1. Details are left to the reader. + Theorem 9 (c) will be derived from the following lemma, which will also be used later. Lemma 15. There are Πj sentences θj, i = 0,1, such that (i) W θi, (ii) Thθovθi, (iii) Th Θ0 Λ θx -^ Conτ, (iv) T + Conτh ^Prτ(θi), (v) T + Conτh θt, (vi) θiEE-θ^. Proof. Let Θi7 i = 0,1, be such that PAhθi <-> Vz(Prfτ(θi,z) -> 5u Oτ. Finally, by Lemma 15 (i) and (ii), a < lτ. Φ To prove Theorem 9 (e) we need the following:

104

7. Degrees of interpretability

Lemma 16. Suppose X is r.e. and for every k, X I k « X. Then if φ is BI and X < φ, then X «φ. Thus, a fortiori d(X) is not B1. Proof, φ can be written in the form (π0 Λ σ0) v...v (πn Λ σn). It is easily checked that for any degrees a, b, c, if a «b and a «c, then a « b n c. Thus, it is sufficient to show that if X < π Λ σ, then X « π Λ σ. Let χ be a Σ1 sentence such that π Λ σ < X + χ. Then, by Lemma 7, it suffices to show that T + Xh -iχ. By assumption, there is a k such that T + X I k + χh π. Hence T + π Λ σ H T + Xlk + ( χ Λ σ ) and so X < X I k + (χ Λ σ). But then, since X I k « X, by Lemma 7, T + Xh -.(χ Λ σ). But X < π Λ σ. It follows that T + π Λ σh -iχ, whence T + X + χh -iχ and so T + Xh -.χ, as was to be shown. Proof of Theorem 9 (e). By (the proof of) Theorem 5.2, we can effectively construct sentences 7^ such that -iT^ is Π1-conservative over but not provable in T + {πk: k < n}. Let X = {πk: ke N}. Then, by Lemma 8, X I k «X for all k. So, by Lemma 16, d(X)isnotB 1 .

§3. ΣI and Πα degrees. This § is devoted to a discussion of the ΣI and E^ degrees and the relations between them. The l.u.b. of two Hi degrees is Γ^ and the g.l.b. of two Π1 (Σ{) degrees is Γ^ (Σ1). Let us say that a is high if a » Oτ, low otherwise. Thus, by Lemma 7, a is low iff there is a ΣI degree b such that a < b < lτ. By Lemma 8, if -~*π is Π1-conservative over T, d(π) is high. By Corollary 2, every member of CONT is high. The following lemma is sometimes useful. Lemma 17. Suppose a is high. Then for any b, [a n b,b) contains no ΣI degree; in fact, if c is Σx and a n b < c, then b < c. Proof. Let Aea, Beb, and c = d(σ). Suppose A^B < T + σ. Then, by Lemma 6, A < T + σ Λ -«ConB i k for every k. Since a is high, it follows that T + σh ConB | ^ for every k, and so B < T + σ. Theorem 10. (a) The set of Γ^ degrees is cofinal in Dτ. (b) The set of Σ1 degrees is not cofinal in Dτ; in fact, for every degree a > Oτ, there is a degree b < a such that [b,a) contains no Σ^ degree. (c) There is a low Π1 degree which is not Σj. Proof, (a) Since all members of CONT are Πl7 this follows from Lemma 9. Φ (b) By Theorem 3 (b), there is a high degree c such a ^ c. Let b = c n a. Then, by Lemma 17, b is as desired. Φ (c) Let a be any low Yl^ degree > Oτ. By Theorem 3 (b), there is a high Γ^ degree

§3. Σ1 and HI degrees

105

c 5? a. Let b = a n c. Then b is low and and Π1. Finally, by Lemma 17, b is not Σj. Using Theorem 2 we can now prove the following corollary. 7

Corollary 4. (a) Suppose a is not na and ae (b,c). There are then degrees b', c such that ae (b',c') c (b,c) and [b',c'] contains no E^ degree. (b) Suppose a is not ΣI and ae (b,c). There are then degrees b', c' such that ae (b',c') c (b,c) and [b',c'] contains no Σ^ degree. Proof, (a) By Theorem 2, there are degrees b0, bx such that b < b{ < a, i = 0,1, and b0 u bl = a. Either [b0,a] or [blra] contains no Γ^ degree. If not, then a would be the l.u.b. of two HI degrees and therefore Πj. Suppose [bixa] contains no Γ^ degree and let b' = bi By Theorem 2, there are degrees c0, q such that a < q < c, i = 0,1, and C n c =a 0 l Either [b',c0] or [b',^] contains no Hi degree. For suppose d^e [b',q] and dj is Π1/ i = 0,1. Then d0 n d x e [b',a] and d0 n d1 is Πl7 a contradiction. Suppose [b',Cj] contains no U^ degree and let c' = c. Then b7 and c' are as desired. 4 (b) By a slight modification of the proof of Theorem 10 (b), which we leave to the reader, there is a degree b' such that b < b' < a and [b',a] contains no ΣI degree. The rest of the proof is the same as the proof of (a). Theorem 10 (b) leads to the question if there are arbitrarily small Σ^ degrees. By our next result, the answer is affirmative; later we shall prove a stronger result (Theorem 15). Theorem 11. If Oτ < a, then there is a ΣI and Γ^ degree be (Oτ,a). To prove this we need a lemma on partial conservativity. Lemma 18. Let X be an r.e. set. There is then a PR formula η(y,x,z) such that for all k and θ, (i) if ke X, then T + θh -.Ξzη(θ,k,z), (ii) if kgX, then 5zη(θ,k,z) is Γ^-conservative over T + θ. The proof of Lemma 18 is similar to the proof of Lemma 5.3 (for Γ = Γ^) and is left to the reader. Proof of Theorem 11. Let Vuδ(u), where δ(u) is PR, be a Πj sentence such that Oτ < d(Vuδ(u)) < a. By Lemma 18, there is a PR formula γ(x,z) such that (1) if Th φ, then Th -3zγ(φ,z), (2) if Tl^ φ, then 3zγ(φ,z) is Π1-conservative over T + φ. Let θ be such that (3) PAh θ Vu(-δ(u) -> 3z
106

7. Degrees of interpretability

(4) PAh σ <-> 3zγ(θ,z) Λ θ, (5) PA + θ + -π3zγ(θ,z)h Vuδ(u). It follows that (6) ΊV θ. For suppose not. Then, by (1), Th - Ξzγ(θ,z) and so, by (5), Th Vuδ(u), contrary to the choice of δ(u). By (3), θ < Vuδ(u) and so d(θ) < a. By (6), Oτ < d(θ). Finally, by (4), (6), (2), σ = θ. Thus, b = d(σ) is as claimed. It is natural to ask if Dτ is " generated" by some "small" set of degrees, for example, the set of ΣI degrees. We prove two negative results, Theorems 12 and 13, and one partial positive result, Theorem 14 (and 14') Let Eτ be the set of l.u.b.s of (finite) sets of Σl degrees. Note that Eτ is closed under n. By Lemma 15, CONT C Eτ. Theorem 12. There is a Πj degree not in Eτ. This is an immediate consequence of the following two lemmas. Lemma 19. If ae Eτ, there is a smallest Σ^ degree > a. Proof. Suppose a = d(σg) u...u d(σn). Then d(σ0Λ...Λσn) is the smallest Σ^ degree > d(σ0) u. .u d(σn). This can be seen as follows. Suppose d(σ0) u...u d(σn) < d(σ). Let π be such that T + σ0 A...Λ σnh π. Then T + σ0l- σj Λ...Λ σn -> π. Now, σj Λ...Λ σn —» π is a HI sentence. It follows that T + σh σj Λ...Λ σn —» π. But then T + σjh σ Λ σ2 Λ...Λ σn —> π and so T + σh σ2 Λ...Λ σn —> π. Continuing in this way we eventually get T + σh π, as desired. Lemma 20. There is a Γ^ degree a for which there is no smallest ΣI degree > a. Proof. Let <σk>k<ω and σ be as in Lemma 14. Let a = d(-ισ). Then a is Π^ Now let χ be any Σ^ sentence such that a < d(χ). Then T + χh-iσ and so T + σh-.χ. It follows that there is a k such that T + σ^h-iχ and so (1) T + χh-,σk. Since σ^ < σ, there is a sentence π such that T + σh π and T + σ^l/ π. It follows that (2) T + -iπh-iσ, (3) T + -,πl*-ισk. But then, by (2), a < d(-ιπ) and, by (1) and (3), χ i -.π. Thus, d(χ) is not the smallest Σ1 degree > a. A strengthening of Lemma 20 will be proved later (Lemma 23). Let Fτ be the set of l.u.b.s of (finite) sets of Σl and Γ^ degrees. By Theorem 12, F T SE T .

§3. Σj and Π^ degrees

107

Theorem 13. Fτ Φ Dτ. We need the following definition: A <« B iff A < B and for every set X of Σ^ sentences, if BHΠ A + X, then A + X is inconsistent. (Here X need not be r.e.) We write 7 a <«b to mean that A <« B where Ae a and Be b. (If A Ξ A' and B = B , then A <« B iff A'<« B'.) By Lemma 7, A <« B implies A « B. As will become clear, the converse of this is not true. But if a is Π1 and high, then Oτ <« a. Lemma 21. Suppose ae Fτ and for all π, if d(π) < a, then d(π) « a. Then Oτ <« a. Proof. By assumption there are π, σ0/...,σn such that a = d(π) u d(σ0) u...u d(σn). Also d(π) « a. Let Ae a. Then (1) T + at < A for i < n. Moreover, π «π Λ σ0 A...Λ σn and so, by Lemma 7, T+ πh -.σ0 v...v -ισn. But Ah π and so (2) Ah-,σ0v...v-,σn. Let X be any set of Σj sentences such that (3) AH Πl T + X. Then, by (2), T + Xh -ισ0 v...v -ισn, whence there is a kg such that T + σ0h -iΛX I ko v —<Jι v...v —ισn, and so, by (1) and (3), T + Xh —\GI v...v -ισn. Continuing in this way we eventually obtain the conclusion that T + X is inconsistent. Proof of Theorem 13. We effectively construct sentences ψ0, ψj,... such that if An = T + (ψk: k < n} and A = T + {ψk: ke N}, then (1) An«An+1, (2) notT<«A. Let a = d(A). Then for all π, if d(π) < a, there is an n such that d(π) < d(An). Also d(An) « d(An+1) < a and so d(π) « a. Thus, by (2) and Lemma 21, ag Fτ. There is an r.e. relation S(n,k,p,q) such that (not T + ψ «T + ψ + φ) iff 5pVqS(ψ,φ,p,q). By Lemma 3.2 (b), there are a E^ formula σ(x,y,z,u) and a Σ^ formula G'(x,y,z,u) such that (3) if S(n,k,p,q), then Th σ'(n,k,p,q), (4) Th σ'ίn.k^q) -» σ(n,k,p,q), (5) T + Y is consistent where Y = {-ισ(n,k,p,q): not S(n,k,p,q)}. Let AO = T. Suppose An has been defined and set θn := A{ψk: k < n}. Then (6) not An« An + φ iff apVqS(θn,φ,p,q). By (3) and Lemma 5.2, there is a Σx formula pn(x,y) such that (7) A n h ρn(φ,p) -> σ'(θn,φ,p,q), (8) if VqS(θn,φ,p,q), then pn(φ,p) is Π1-conservative over An. By Theorem 5.4 (b), there is a formula ηn(x) such that (9) An + ηn(φ) is a Π1-conservative extension of An + {-ιpn(φ,p): pe N}. Finally, let ψn be such that

108

7. Degrees of interpretability

(10) Thψ n ^η n (ψ n ). The formulas ρn(x,y), ηn(x) and the sentences ψn can be found effectively in n. To prove (1) assume it is false. Then, by (6), there is a p such that VqS(θn/ψn,p,q). But then, by (8), pn(ψn/p) is Π1-conservative over An. By (9) and (10), An+1h -»pn(ψn,p). But, by Lemma 8, this implies that An « An+1, a contradiction. This proves (1). Next we prove (2). Let Y be as in (5). Then T + Y is consistent. To prove that AH Πl T + Y we first show that (11) Suppose An+1 + Yh π. Then there is a k such that (12) A n + 1 h - , Λ Y I k v π . By (1) and (6), for each p there is a qp such that (13) not S(θn,ψn/p,qp). By (12), (9), and (10), An + Hpn(ΨrvP): P^NJh -,ΛY I k v π. By (13), (4), (7), An + Yh - pn(ψn,p) for every p. It follows that An + Yh π. This proves (11). Since (11) holds for all n, it follows that AH Πl T + Y. This proves (2) and so the proof is complete. Let Gτ be the set of degrees obtained from the Σ1 and the Γ^ degrees by closing under n and u. It is an open problem if Gy Φ Dτ. The degree mentioned in Lemma 20 cannot be arbitrarily large: if a is high, there is a smallest ΣI degree > a, namely 1 j. Similarly, the degree a defined in the proof of Theorem 13 cannot be arbitrarily large; it is not >» Oτ. This is explained, at least partially, by the following surprising: Theorem 14. (a) Every sufficiently large degree is the l.u.b. of a ΣI degree and a Π1 degree. (b) Every sufficiently large degree is the l.u.b. of two ΣI degrees. Proof. We may assume that d(Conτ) < lτ. By Lemma 9, it is sufficient to consider degrees a such that d(Conτ) < a < lτ. Let πn := Vuδn(u), where δn(u) is PR, be Γ^ sentences such that a = d({πn:nEN}). We may assume that for all n, (1) Thπ 0 -»Con τ , (2) Thπ^^π,. (a) We define Γ^ sentences φn and ψn in the following way: (3) Th φn <-> Vz(Prfτ( V{φk:k Ξu
§3. ΣI and Πj degrees

109

(7) T + -φnh Prτ(V{φk:kφn is Σ lx we have T + -"(pnl- Prτ(-«φn). Also, by (3), T + -φnh Prτ(V{φk:kPrτ(V{φk:k<m}) and so, by (7), φm. Finally, by (5), -ϊPrτ(V{φk:k<m+l}), as desired." Thus, (9) holds for n = m+1. This proves (9) and so we have proved (8). Next we show that for all n, (10) T + Λ{ψk: k < n} + Conτh φn. We first show that (11) T + Conτh φ0, (12) T + ψn + φnh φn+1. (11) follows from (7) with n = 0. (12) follows from (5) and (7). Now (10) follows from (11) and (12). Let a0 = d({-χpk:keN}), a1 = d(Conτ), a2 = d({φk:keN}). Then, by Lemma 13, a0 is Σ L. a1 is Πi By (8) and Orey's compactness theorem, a0 < a and, by hypothesis, a x < a. But then a0 u a^ < a. By (4) and (5), a0 u a2 > a. By (4) and (10), a0 u a^ > a2. It follows that a0 u a^ > a and so a0 u a± = a, as desired. Φ (b) Let θi, i = 0,1, be as in Lemma 15. We define Γ^ sentences φn and ψn in the following way: (13) Th φn <-» Vz(Prfτ(θ0W{φk:k Ξu az
HO

7. Degrees of interpretability

Now (19) follows from (20) and (21). Let a0 = d(-<θ0+bφk:kE N}), al = d(θ0). Then a0 is 1^. By Lemma 15 (vi), aα is Σx. By (18), a0 < a and, by Lemma 15 (v), al < a. But then a0 u al < a. By Lemma 15 (ii), (14), (19), and (15), a0 u a x > a. Thus, a0 u al = a, as desired. The proof of Theorem 14 actually yields the following stronger result; Theorem 14' is also an improvement of Theorem 4. Theorem 14'. (a) Suppose ae CONT and a < b < lτ. There is then a Σx degree c such that a u c = b. (b) Suppose ae CONT. There are then degrees a0, ai such that (i) a0 and ai are both Σx and Γ^, (ii) a0 n al = Oτ, (iii) a0 u aα = a, (iv) for every degree b > a, there is a Σ1 degree bj such that a^ u bj = b, i = 0,1. One way to strengthen Theorem 12 would be to show that there is a Πj degree a > Oτ such that no Σx degree cups to a. This, however, is not the case: Theorem 15. For every Πi degree a > Oτ, there is a ΣI (and Πi) degree which cups to a. Proof. The following proof is similar to that of Theorem 11. Let π be such that a = d(π) and let δ(u) be a PR formula such that π := Vuδ(u). By Lemma 18, there is a PR formula η(x,y,z) such that for all φ, ψ, (1) if T + φh π, then T + ψh - Ξzη(φ,ψ,z), (2) if T + (pi**- π, then Ξzη(φ,ψ,z) is Π1-conservative over T + ψ. Next let θ and χ be such that (3) Th θ <-> Vu(-δ(u) -» 3z Vz(η(χ,θ,z) -» Ξu Oτ, there is a Σx degree which cups to a

§3. ΣI and Γ^ degrees

111

remains open. By Theorem 14, this is true of every sufficiently large degree. Our next task is to show that the result of interchanging Σj and Γ^ in Theorem 15 is false. Theorem 16. There is a Σ} degree a > Oτ such that no Γ^ degree cups to a. Let ξ(x) be as in Lemma 5.8 with n = 1 and let a = d({ξ(k):keN}). Then a > Oτ and no Yli degree cups to a (see the proof of Theorem 3 (a)). To obtain a Σ1 degree satisfying these conditions we first prove the following refinement of Lemma 5.8 (for n = l). Lemma 22. There are Γ^ formulas ξ(x), η(x) and Σj sentences χk such that (i) W ξ(k) (ii) Thη(k)^ξ(k), (iii) Thξ(k+l)^η(k), (iv) ξ(k) is Σ1-conservative over T + -^(k), (v) {ξ(k):kEN} Ξ {χ k :kEN}. Proof. We combine the ideas of the proofs of Lemma 5.8 and Theorem 11. By Lemma 18, there is a PR formula γ(x,z) such that for all φ, (1) if Th φ, then Th ->3zγ(φ,z), (2) if Th φ, then Ξzγ(φ,z) Π1-conservative over T + φ. Let δ(u) be an arbitrary PR formula. Let κ(z,u,x,y) and v(z,u,x,y) be Γ^ formulas and μ(z,u,x,y,v) a PR formula such that (3) Th κ(z,u,x,y) <-> Vvμ(z,u,x,y,v), (4) Th -v(z,u,x,0), (5) Th κ(δ,u,k,y) ^ v(δ,u,k,y) v VvίpJ^kKg^v) -» -Prfτ(ξδ(k),v)), (6) Th v(δ,u,k,y+l) ^ Vv(-μ(δ,u,k+l,y,v) -^ Ξz<max{u,v}γ(ηδ(k),z)), where ξδ(x) := Vu(δ(u) -^ κ(δ,u,x,u ^x)), ηδ(x) := Vu(δ(u) -^ v(δ,u,x,u - x)). As in the proof of Lemma 5.8, (5) implies that (7) Th ξδ(k) ~ ηδ(k) v Vvί^K-ηgίkjΛξδίkXv) ^ -Prfτ(ξδ(k),v)). Let η^(x) := Vu(δ(u) -> v(δ,u,x,(u ^ (x+1)) + 1)). Then, by (6), (8) Th η^(k) <-> Vuv(δ(u) Λ -ιμ(δ,u,k+l,u ^ (k+l),v) ^ 3z<max{u,v}γ(ηδ(k),z)). Let χδ/k := 3z(γ(ηδ(k),z) Λ Vuv
112

7. Degrees of interpretability

and so, by (8), (10) Th χδ/k <-> Ξzγ(ηδ(k),z) A η^(k). We now show that (11) Thη δ (k)-+ξ δ (k), (12) ifThξ δ (k),thenThη δ (k), (13) Thξ δ (k+l)->η^(k), (14) if Th δ(u) -» u > k, then Th η^(k) <-> ηδ(k), (15) if Th δ(u) -> u > k and Th η§(k), then Th ξδ(k+l). (11) follows from (7). (12) follows from (7) by the same argument as in the proof of Lemma 5.8. (13) follows by predicate logic from (3) and (8). (14) is obvious. To prove (15), assume Th δ(u) -» u > k and Th ηδ(k). Then, by (14), Th η^(k). Also, by (1), Th -Έzγ(ηδ(k),z). By (8), it follows that Th Vuv(δ(u) -> μ(δ,u,k+l,u ± (k+l),v)) and so, by (3), Th ξδ(k+l). This proves (15). It can now be shown that if Ξuδ(u) is true, then ΊV ξg(0). The proof of this from (4), (12), (15) is the same as that of (6) in the proof of Lemma 5.8. As in the proof of Lemma 5.8 we can now find a PR formula δ'(x) such that Ξuδ'(u) is false and TV ξg,(0). Let ξ(x) := ξδ/(x), η(x) := ηg,(x), χk := χg,/k. The verification of (i) - (iv) is now straightforward or much the same as in the proof of Lemma 5.8; this is where (13) is needed. To prove (v), we first note that (ξ(k): keN} < {χk: keN} follows from (10), (14), (11). Next suppose T + (χk: ke N}h π. There is then an m such that T + χQh Xl Λ-Λ %m -> π BY «and (ϋ),w η(0) Hence, by (2), 3zγ(η(0),z) is I^-conservative over T + η(0). But then, by (10) and (14), T + η(0)h χl Λ...Λ χm -> π. By (10), (14), (ii), (iii), T + χ 1 h η(0). It follows that T -H χ1 Λ...Λ χ m h π. Continuing in this way we eventually get T + η(m)h π and so, by (iii), T + ξ(m+l)h π. This shows that {χk: keN} < (ξ(k): keN} and so (v) is proved. Proof of Theorem 16. Let ξ(x) and η(x) be as in Lemma 22. Let a = d({ξ(k):ke N}). Then, by Lemma 22 (i), a > Oτ. Also, by Lemma 22 (v) and Lemma 13, a is Σx. By Lemma 22 (ii), d(ξ(k)) < d(η(k)) for every k. That d(ξ(k)) doesn't cup to d(η(k)) now follows from Lemma 22 (iv). Suppose b is Πj and b < a. Then, by Lemma 22 (ii) and (iii), b < d(ξ(k)), for some k, and d(η(k)) < a. Since d(ξ(k)) doesn't cup to d(η(k)), it follows that b doesn't cup to a. Note that if a is as in Theorem 16, then a does not cup to any Γ^ degree. Indeed, let b be Π1 and > a. If a cups to b, there is a Πj degree c < a which cups to b. But then c cups to a, contrary to assumption. Finally, we prove Theorem 5 (and a bit more). We have already observed that d(-ιπ) is the p.c. of d(π). Thus, every Γ^ degree has a p.c. It follows that, in terms of our classification of degrees, the following result is the best we can do.

§3. Σ! and Π1 degrees

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Theorem 17. There is a Σ^ degree which has no p.c. This is a consequence of the following strengthening of Lemma 20. Lemma 23. There is a sentence σ such that {b > d(-ισ): b is Σ-J has no g.l.b. To prove this, we need another: Lemma 24. Suppose {πk: keN} is r.e. and let G = {d(πk): keN}. Suppose there is no finite subset H of G such that OH is a lower bound of G. Then G has no g.l.b. Proof. Let X = {π: T + π k h π for every k}. X is not r.e. This can be seen as follows. Let R(k,m) be a primitive recursive relation such that Y = {k: VmR(k,m)} is not r.e. and let p(x,y) be a PR binumeration of R(k,m). We may assume that Z = {πk: ke N} is primitive recursive; let ζ(x) be a PR binumeration of Z. Finally, let η(x) := Vz(-p(x,z) -> 3uπk. Then T < T + πk v ψ < T + πk. Let σk := g(πkvψ) and πk+1 := f(πkvψ). Then ΊV πk+1. For every k, (1) π k + 1 <σ k <π k . By Theorem 5.4 (a), there is a sentence σ such that (2) T + σ is a Π1-conservative extension of T + {-ιπk: ke N}. By (1) and (2), (3) -,σ < σk. Moreover (4) if b is Σ! and b > d(-.σ), there is a k such that b > d(πk). For suppose b = d(χ) > d(-ισ), where χ is Σx. Then T + χh -,σ, whence T + σh -.χ. But then, by (2), there is a k such that T+-,πkh-πχ, whence T+χh πk and so b > d(πk).

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7. Degrees of interpretability

Let G = (d(πk): keN}. If {b > d(-ισ): b is Σα} has a g.l.b. c, then, by (1), (3), (4), c is the g.l.b. of G. But from (1) it follows that no d(πk) is a lower bound of G. Hence, by Lemma 24, G has no g.l.b. and so {b > d(-.σ): b is Σ^ has no g.l.b. Proof of Theorem 17. Let σ be as in Lemma 23. By Lemma 6, for all B, (T + σ)iB < T iff B < T + χ for all Σl sentences χ such that T + χh -πσ. But then the p.c. of d(σ), if it had one, would also be the g.l.b. of {b > d(-.σ): b is Σj}. Thus, by Lemma 23, d(σ) has no p.c. Every ΣI degree is the p.c. of some degree. It is an open problem if the converse of this is true. If it is, the Σ1 degrees can be characterized in a purely algebraic way as those degrees that are p.c.s.

Exercises for Chapter 7. In the following exercises we assume that PAH T and that A, B, etc. are extensions ofT 1. Suppose G £ Dτ. G is independent if for any disjoint finite subsets G0 and G^ of G, OG0 ^ LJG^ (O0 = lτ, LJ0 = Oτ.) (Thus, for example, 0 is independent and {a} is independent iff Oj < a < lτ.) Show that for every finite independent set G, there are degrees b0, bj such that G u {bj is independent, i - Ό, 1, and b0 n b1 = Oτ. Conclude that every finite independent set is included in 2 κ o many maximal independent sets. 2. Suppose a < b. (a) c cups to b above a if there is a d such that a < d < b and c u d = b. Show that there is a ce (a,b] which doesn't cup to b above a. (b) c caps to a below b if there is a d such that a < d < b and c n d = a. Show that there is a ce [a,b) which doesn't cap to a below b. 3. Suppose a < b and b < l τ if T is Σ1-sound. For ce [a,b], let c* be the complement of c in [a,b] if it exists, i.e. c n c* = a and c u c* = b. (Complements are unique.) Let Cpla/b be the set of degrees in [a,b] having complements in [a,b]. (a) Show that Cpla/b is closed under n, u, and *. Let Cplab = (Cpla/b, π, u, *, a, b). Then Cplab is a Boolean algebra. (b) Show that if c, de Cpla^b and c < d, there is an ee Cplab such that c < e < d. (It follows that the Boolean algebras Cpla/b are (denumerable and) atomless and therefore isomorphic.) (c) Show that if a < c < d < b, there is an ee [c,d) such that Cpl ab n [e,d) = 0. [Hint:Cpl a/b n[c,d]CCpl c/d .] (d) Show that i f a < c < e < d < b and e£ Cpla/b, there are c', d' such that c < cr < e < d r < d and Cpla/b n [c'^7] = 0.

Exercises

115

4. Suppose a is Σ1. (a) Show that if a < b < l τ/ then a caps to Oτ below b. (b) Show that if a < b and b is high, then a «b. Conclude that if bt > a, i = 0,1, and b0 n bx = a, then b0 and b1 are low. ((a) and (b) are true of every a which is the p.c. of some degree,) 5. Show that for every low degree a, there is a low Γ^ degree > a. [Hint: Let B = T + σ and σ := Ξxδ(x), where δ(x) is PR, be such that a < d(B) < lτ. We may assume that BV - 3z χ.] 7. (a) Show that there is an r.p. degree a which is not Π^ (compare Lemma 11). [Hint: Let κ(x) be as in Exercise 2.11 and let a = d({κ(k):keN})]. (b) Improve (a) by showing that there is a non-Πj Σ^ degree a which is r.p. (compare Exercise 16 (c)). [Hint: Define πk and σk so that TI/ πk, Th κ(k) -> πk/ where κ(x) is as in (a), and KQ Λ...Λ πk_ι Λ σk Ξ= πg Λ...Λ πk. Let a = d({σk:ke N}).] 8. Suppose AH B. Show that there is a Δ2 sentence φ such that A + φ - B (compare Corollary 6.10 and Theorem 8). 9. Show that there is a Σ^ sentence σ such that Oj < d(σ) < lj and for every Σ^ sentence χ, if σ < χ, then T + χh σ. [Hint: Let σ be such that d(-^σ) is Σj.] 10. Let <σk>k<ω and σ be as in Lemma 14. Show that every Γ^ degree < d(σ) caps to Of below d(σ) (compare Exercise 26 (c)). 11. (a) Show that for every Γi^ sentence π, d(π) u d(- π) is high; in fact, if b is the p.c. of a, then a u b is high. (b) Let a = d(σ) u d(- σ). a is high. Let <σk>k<ω and σ be as in Lemma 14. Show that if b is HI and b < a, then b is low. [Hint: Use Exercise 4 (a).] 12. Show that there is a degree of the form d(σvπ) which is neither Σα nor Πj. 13. Show that there is a Πα degree a such that for every Πα degree b, if a n b = Oτ, then a u b is low (compare Exercise 18 (c)).

116

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14. Suppose a < b < lτ. Show that (a) there is a degree c < l τ such that for every d, if b n d = a, then d < c, (b) there is a degree c> Oτ such that for every d, if a u d = b, then d > c. 15. (a) Verify that in any distributive lattice, for any a, b, the intervals [a n b,a] and [b,a u b] are isomorphic (b) Show that there are degrees a, b, c, d such that a «b, c < d, not c «d, and [a,b] and [c,d] are isomorphic. [Hint: Use Exercises 4 (b) and 11 (a).] 16. (a) Verify that in any distributive lattice, if a < b < c and [a,c] satisfies the reduction principle, so does [b,c]. (b) Show that for each degree a < lγ, there is a b such that a < b < 1 j and [a,b] does not satisfy the reduction principle. (c) The non-r.p. degree a defined in the proof of Lemma 12 is high (cf. Exercise 11 (a)). Show that there is a Σ^ degree which is not r.p. Conclude from Exercise 7 (b) that there are non-Γ^ Σ± degrees such that [Oτ,a] and [Oτ,b] are not isomorphic. [Hint: Use Theorem 14' (a).] 17. (a) Suppose φ and X are as in Lemma 16. Show that if φ < X, then φ «X. (b) Suppose a < b. Show that there are c, d such that a < c < d < b and [c,d] contains no B! degree. 18. (a) Show, by combining the proofs of Theorem 4 and Lemma 15, that there are cupping degrees a0 and a^ which are Σ1 and Γ^ and such that a0 n al = Oτ. Conclude that there are low cupping degrees. (This also follows from Theorem 14' (b) ) (b) Show that there is a high (Γ^) degree a which is not cupping. [Hint: Suppose d(Conτ) < lτ. Let a = d(π) where π is Σ1-conservative over T + -»Conτ and -«π is Γ^-conservative over T + -iConp.] (c) Show that there is a low (Π1) degree a such that for every degree b, if a n b = Oτ/ then a u b is not cupping (compare Exercise 13). [Hint: Let d(π) be as in (b). Define a sentence σ such that d(σ) > Oτ and d(σ) u d(- σ) < d(π); use Theorem 11. Let a = d(- σ).] 19. Show that there are degrees a, b such that a is Σ l7 b is both Σx and n l7 and a u b is not B L. 20. Prove Lemma 15 by letting Θ0 be a Γ^ Rosser sentence for T and θ x := Vu(Prfτ(-θ0,u) -> Ξz Oτ. Show that there is a degree b < a such that [b,a] c

Exercises

117

Eτ. [Hint: We may assume that a ^ d(Conτ). Let b = a n d(Conτ) and use Theorem 14' (b). (By Lemma 17, no member of [b,a) is Σ^.] (b) Suppose there is a Σ1 degree which cups to a. Show that there is a b < a such that for every CE [b,a], there is a Σj degree which cups to c. 22. (a) Let E^be the set of degrees obtained from Oτ by taking l.u.b.s, g.l.b.s, and Σ1-extensions. Show that if ae E^ there is a least Σj degree > a. Conclude that there is a Π1 degree not in E^ (This improves Theorem 12.) (b) Let Ej,be the set of degrees obtained from E^ and the Γ^ degrees by taking l.u.b.s and Σ1-extensions. Show that the degree defined in the proof of Theorem 13 is not in F^ Conclude that there is a degree which is not the l.u.b. of a finite set of degrees of the form d(πΛσ). (This improves Theorem 13.) 23. Show that for any a, if there is a member of Gj which cups to a, then there is a ΣI degree which cups to a. (This improves Theorem 15.) 24. (a) Show that not all non-Γ^ ΣI degrees are as stated in Theorem 16. (b) Improve Theorem 16 by showing that for every degree b > 0^, there is a Σ^ degree a such that Oτ < a < b and no Γ^ degree cups to a. [Hint: By Theorem 11, there are sentences π and σ such that Oj < d(π) = d(σ) < b. Let C = T + ->π. By the proof of Lemma 22, with T replaced by C, there are Γ^ formulas ξ(x), η(x) and ΣI sentences χ^ such that (i) - (iv) hold with T replaced by C and C + (ξ(k): ke N} = C + {χk: keN}. Let a = d({ξ(k)vπ:keN}).] 25. Show that in contrast to Lemma 24 we have the following: There is a set G = {d(σ^): kE N} of Σ! degrees, where {σk: ke N} is (primitive) recursive, such that OH > Oj for every finite subset H of G and OG = Oτ. [Hint: Let a be high and such that there is no high Π^ degree < a (cf. Exercise 11 (b)). Let AE a and let σk := -«ConA j ^.] 26. (a) Show that there is a PR formula δ(u) such that if θ is defined as in the proof of Theorem 11, then d(- θ) isn't Πx. (b) Let θ be as in (a). Show that d(->θ) has a p.c. Conclude that there is a non-Γ^ ΣI degree which has a p.c. (c) Let θ be as in (a). Show that there is a Γ^ degree < d(- θ) which does not cap to Of below d(-ιθ) (compare Exercise 10).

Notes for Chapter 7. The lattice Dτ was introduced by Lindstrόm (1979), (1984b); a related lattice V τ (degrees of finite extensions of T) has been defined by Svejdar (1978) (see also Jeroslow (1971a)). (By Theorem 6.11 (a), V τ and Dτ are isomorphic.) Theorem 1 is due to Lindstrom (1979), (1984b) and (for Vτ) to Svejdar (1978). Corollary 1 is,

118

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modulo Theorem 6.6, a restatement of the equivalence of Exercise 2.22 (i) and (ii). The proof of Theorem 4 was suggested by the proof of a related result in Hajkova II (1971). Theorem 7 is new; the term "reduction principle" is borrowed from descriptive set theory and recursion theory (cf. Soare (1987)). (The only way of showing that intervals are isomorphic known so far is given in Exercise 15 (a) and works in all distributive lattices.) The remaining results of § 1 are due to Lindstrom (1979), (1984b). In connection with the proof of Theorem 4, see Exercise 6. Lemmas 11 and 12 lead to the question if there is a non-Πj r.p. degree; this question is answered in Exercise 7. Theorem 8 (with a slightly different proof; see Exercise 6.12 (a)) is due to Montagna (cf. Lindstrom (1993)). Theorem 9 is due to Lindstrom (1979), (1984b), (1993); (a) and (c) were also proved by Svejdar (1978); for a different proof of Theorem 9 (d), see Exercise 12. Theorem 10 is due to Lindstrom (1979), (1984b); (a) and the first half of (b) were also proved by Svejdar (1978). Theorems 14 and 16 are new, they were announced in Lindstrom (1993), where a weaker form of Theorem 16 is proved; Theorem 16 leads to the question if there is a Σ1 degree a such that no Π^ degree caps to a; this is answered negatively in Exercise 5; in connection with Theorem 16, see also Exercise 24. The remaining results of § 3 are due to Lindstrom (1984b), (1993). The definition of the sentences φn and ψn in the proof of Theorem 14 (a) and the observations concerning these sentences, except (8), were first used by Misercque (1982) in a different context. For improvements of Theorems 12, 13, 15, and 16, see Exercises 22 (a), 22 (b), 23, 24 (b). Theorem 17 leads to the question if no non-Γ^ Σl degree has a p.c.; this question is answered in Exercise 26 (b). For a proof of Exercise 26 (a), see Lindstrom (1993).

8. GENERALIZATIONS

So far our results have been explicitly stated (and proved) only for theories of first order arithmetic. But, as mentioned in the introduction, they hold, after suitable reformulation, in a much more general setting. Needless to say, we are not going to show this in every detail. In fact, we shall skip Chapters 3,5, 7 altogether and concentrate on some of the main results of Chapters 2,4, and 6. These examples should enable the reader to generalize (most of) the results of the preceding chapters. In this chapter he theories S, T, etc. are no longer arithmetical theories, but they are still consistent and primitive recursive and we assume that the languages of these theories are always finite. Lτ is the language of T. T is a pure extension of S if SH T and Lτ = Ls. Lower case Greek letters are now used for formulas of Lτ as well as for formulas of LA. We assume that the reader can extend the definition of t: S < T to the present more general setting. Let t~a(T) = {φ: Th t(φ)}. Then t-1(T)h ψ iff Th t(ψ). Since Ls is finite, t is primitive recursive. The following lemma is immediate. Lemma 1. (a) t: S < T iff SH Γ^T). (b) t: f XT) < T and so t-1(T) < T; in fact, t: t-1(T) ^ T; it follows that t-1(T) is consistent. (c) t-\Ί+ t(φ))Hh Γ'CΓ) + φ.

§1. Incompleteness. Our first result, GodeΓs incompleteness theorem, is a straightforward generalization of Theorem 2.1; 6t(x,y) is a formula defining t as in Fact 2. Theorem 1. Suppose t: Q < T. Let φ be such that (Gt) Qh φ ^ -3y(δt(φ,y) Λ Prτ(y)). Then φ is a true Πj sentence such that TI/ t(φ). Hence if t~\Ί) is Σ1-sound, then also Th - t(φ). By Theorem 1, for each t: Q < T, there is a true Γ^ sentence φt such that Th t(φt). By a similar generalization of Rosser's theorem, we obtain a Γ^ sentence θt such that W t(θt) and TI/ - t(θt). This result can be improved by showing that there is a single Πα sentence ψ such that W t(ψ) and W - t(ψ) for every t: Q < T: Theorem 2. There is a (true) Γ^ sentence ψ, such that Q + ψ i T and Q + - ψ i T.

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Proof, {φ: Q + φ < T} is r.e. (Lemma 6.5) and monoconsistent with Q. Now use Lemma 2.1. Our next result, GόdeΓs second incompleteness theorem, is a generalization of Theorem 2.4 (a). Since each t is primitive recursive, in PA we may use t as a function symbol. Theorem 3. Suppose t: PA < T. (a) Th t(Conτ). (b) If τ(x) is any Σ^ numeration of T, then Th ^Cor^). Proof. We prove (a); the proof of (b) is almost the same. Let φ be as in the proof of Theorem 1. Then PAh ->φ -> PrT(t(φ)). Moreover, -ι

t(φ). But then, by Theorem 1, Th t(Conτ), as desired. We shall say that t is a reflexive interpretation of S in T, t: S <Γ T, if t: S < T and for every k, Th t(Conτ (k). S is reflexively interpretable, S <Γ T, if there is a t such that t: S <Γ T. t is an essentially reflexive interpretation of S in T, t: S <er T, if t: S ConPA. (This follows from the fact that ts is a "natural" interpretation of PA in a finite subtheory of ZF.) Since PAh ConPA, it follows that PAh ConZF|k. Since t is faithful, this implies that ZFh t(ConZF (k ) and so t is not reflexive.

§2. Axiomatizations

121

§2. Axiomatizations. In this § we shall restrict ourselves to generalizing Theorem 4.2. We shall need the following generalization of part (a) of the fixed point lemma; the proof is left to the reader. Lemma 2. Suppose t: PA < T and let vn(x) := t(x = n). Let γ(x) be any formula of Lτ. There is then a sentence φ such that Thφ^3y(v φ (y)Aγ(y)). We assume given a hierarchy H = (H0/ HI,...) of the formulas of Lτ satisfying closure conditions similar to those satisfied by the hierarchy (Σ0/ Σ-L,...). Thus, each Hk is a primitive recursive set of formulas, Hk c Hk+1, and ^{Hk: ke N} is the set of all formulas of T. Let Hk(x) be a PR binumeration of Hk. We assume that for each k, there is an Hk partial truth-definition for Hk in T i.e. an Hk formula Trk(x) such that for every Hk sentence φ, (Trk) Thφ^3x(v φ (x)ATr k (x)). We also assume that the formulas Trk(x) are mutatis mutandis as in Fact 10 (a). A set X of sentences of T is said to be H-bounded if there is a k such that X £ Hk. Let RFN^= (Vx(t(Hk(x)) Λ t(Prs(x)) -> Trk(x)): keN}. Theorem 4. Suppose t: PA < T, t^) c H0, and Th RFN^ If X is any H-bounded set of sentences such that TH S + X, then S + X is inconsistent. Proof. This proof is essentially the same as the proof of Theorem 4.2. Let n be such that X c Hn. Let ψ be such that (1) Th ψ ^ 3y(vψ(y) A Vxz(t(Hn(x)) A Trn(x) A t(z = (x->y)) -> -t(Prs(z)))). By assumption, we have (2) Th Vy(vψ(y) -> Vxz(t(Hn(x)) A t(z = (x->y)) A t(Prs(z)) -> (Trn(x) -> ψ))). (1) and (2) imply that (3) Th ψ. Suppose TH S + X. By (3), there is then a conjunction θ of members of X such that S + θh ψ. It follows that (4) Th 3z(vθ^ψ(z) A t(Prs(z))): Also, by (1) and (3), S + Xh -3z(vθ^ψ(z) A t(Prs(z))). But then, by (4), S + X is inconsistent. Theorem 4 can be applied to set theory. We define Σ|.T and Πkτ as follows. Let Σ^ = Π^τ be the set of formulas of L$τ all of whose quantifiers are bounded, i.e. of the form 3xe y or VXG y. Σ^^ and Π^ aι<e then the least sets closed under bounded quantification such that Σ^τ c Π^, and Π|τ c Σ^, Σ^ is closed under existential quantification and Π^,^ is closed under universal quantification. A set X of sentences of Lsτ is bounded if X c Σkτ for some k. We then have the following:

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8. Generalizations

Fact 14. The assumptions of Theorem 4 are satisfied when t = ts/ Hk = Σ^17 T = ZF, and S = 0. From Theorem 4 and Fact 14, we get: Corollary 3. There is no bounded and consistent set X of sentences of Lsτ such that ZFHX.

§3. Interpretability. In this § we show that the relevant results of Chapter 6 generalize quite easily to the present more general setting. In using results from Chapter 6 we shall take advantage of the fact that in these results the theories S, S0, etc. need not be formalized in L^ Theorem 5. If t: PA
§3. Interpretability

123 ts

and so every Π^ sentence true in M* is true in M . From Lemma 1 (b) and Theorem 6.1, we get: Lemma 3. There is a Σ1 numeration τ'(x) of t'^T) such that PAh Conτ -» Cory. Theorem 6.2 can now be generalized as follows: Theorem 6. Suppose t: PA Cory It follows that Th t(Conτ jm) —» tίCorij). Also, by assumption, Th t(ConT|m). But then Th ^ConJ, contradicting Theorem 3 (b). Corollary 7. If T is a pure extension of ZF, then T is not interpretable in any finite subtheory of T. The Orey-Hajek lemma in the present setting reads as follows: Lemma 4. Suppose t: PA <Γ T. Then S < T iff Th t(Cons ( k ) for every k. Proof. By Lemma 6.2, S < Γ^T) iff Γ^T)!- Cons,k for every k. As in Chapter 6 we get, from Lemma 4, the following version of Orey's compactness theorem. Theorem 8. Suppose PA
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8. Generalizations

We conclude by generalizing Theorems 6.8 and 6.9; the generalization of Theorem 6.10 is left to the reader. er

Theorem 10. If t: PA < T, then T + - t(Conτ) < T. Proof. Let τ'(x) be as in Lemma 3. By Theorem 6.8, Γ^T) + --Cory < Γ^T). It fol-1 lows that t (T) + -ιConτ < Γ^T). But then, by Lemma 1 (c) and Theorem 5, T + α -1 l -ιt(Conτ) < Γ (T + - t(Conτ)) < t (T) + - Conτ < t~ (Ί) < T and so T + - t(Conτ) < T, as desired. er

Theorem 11. Suppose t: PA < T and X is r.e. and monoconsistent with T. There is then a sentence φ such that T + φ < T and φg X; φ can be taken to be of the form t(ψ), where ψ is Σ^. -1

Proof. Let Y = {θ: t(θ)εX}. Then Y is r.e. and monoconsistent with t (T). By Theorem 6.9, there is a Σα sentence ψ such that Γ^T) + ψ < t'^T) and ψeY. By Lemma 1 (c) and Theorem 5, T + t(ψ) < T. Clearly, t(ψ)<sX. Theorem 11 has the following application, where GB is Godel-Bernays (finite) set theory (compare Corollary 6.6). Corollary 8. There is a ΣI sentence φ such that ZF + ts(φ) < ZF and GB + ts(φ) ^ GB.

Notes for Chapter 8. Theorems 1 and 3 are, of course, (essentially) due to Gδdel (1931), (1934) (cf. also Feferman (I960)). Theorem 2 is due to Montague (1957), (1962) (compare Exercise 6.1). For a definition of the standard interpretation ts of arithmetic (theory of finite ordinals) in set theory, see, for example, Mendelson (1987) or Cohen (1966). Fact 13 and Corollary 2 are due to Montague (1961). Fact 14 is due to Levy (1965). Theorem 4 and Corollary 3 are due to Kreisel and Levy (1968), improving earlier work of Montague (1961). Corollaries 2 and 3 are given here as examples of applications of the corresponding general results; for a detailed discussion of similar, more general, and stronger results, see Kreisel and Levy (1968). Lemma 4, Theorems 6, 7, 8, 9, 10, 11, and Corollary 8 are straightforward generalizations of the corresponding results in Chapter 6. The question if there is a sentence φ such that GB + φ < GB and ZF + φ ^ ZF, raised in Hajek (1971), was answered affirmatively by Solovay; for this and related results, cf. Hajek and Pudlak (1993).

REFERENCES

Only works mentioned in the text (Notes) have been included among the references; for a more comprehensive bibliography, see Hajek and Pudlak (1993). Beklemishev, L. D. (1995). Iterated local reflection versus iterated consistency, Annals of Pure and Applied Logic 75, 25 - 48. Beklemishev, L. D. (199?). Notes on local reflection principles, to appear. Bennet, C. (1986). On some orderings of extensions of arithmetic, Thesis, Dept. of Philosophy, Univ. of Goteborg. Bennet, C. (1986a) Lindenbaum algebras and partial conservativity, Proc. Amer. Math. Soc. 97, 323 - 327. Berarducci, A. (1990). The interpretability logic of Peano arithmetic, J. Symb. Logic 55,1059 -1089. Boolos, G. (1979). The Unprovability of Consistency, Cambridge University Press, Cambridge, USA. Boolos, G. (1993). The Logic of. Provability, Cambridge University Press, Cambridge, USA. Carnap, R. (1934). Logische Syntax der Sprache, Springer. Craig, W. (1953). On axiomatizability within a system, J. Symb. Logic 18, 30 - 32. Davis, M. (ed.) (1965). The Undecidable, Raven Press. Ehrenfeucht, A. and Feferman, S. (1960). Representability of recursively enumerable sets in formal theories, Arch. Math. Logic 5, 37-41. Feferman, S. (1960). Arithmetization of metamathematics in a general setting, Fund. Math. 49, 35 - 92. Feferman, S. (1962). Transfinite recursive progressions of axiomatic theories, J. Symb. Logic 27, 259-316. Feferman, S., Kreisel, G., Orey, S. (1960). 1-consistency and faithful interpretations, Arch. Math. Logic 6, 52 - 63. Friedman, H. (1975). One hundred and two problems in mathematical logic, J. Symb. Logic 40,113 - 129. Godel, K. (1931). Uber formal unentscheidbare Satze der Principia Mathematica und verwandter Systeme I, Monatsh. Math. Physik. 38,173 -198 (English translation in Davis (1965)). Godel, K. (1934). On undecidable propositions of formal mathematical systems (mimeographed lecture notes by S. C. Kleene and J. B. Rosser), Princeton (reprinted in Davis (1965)). Godel, K. (1936). Uber die Lange von Beweisen, Ergebnisse eines math. Koll, 7, 23 - 24 (English translation in Davis (1965)). Goryachev, S. N. (1986). On the interpretability of some extensions of arithmetic, Mat. Zametki 40, 561 - 572; English transl. in Math. Notes 40.

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Guaspari, D. (1979). Partially conservative extensions of arithmetic, Trans. Amer. Math. Soc. 254, 47 - 68. Hajek, P. (1971). On interpretability in set theories, Comment. Math. Univ. Carol. 12, 73 - 79. Hajek, P. (1979). On partially conservative extensions of arithmetic, in: Logic Colloquium '78 (eds. M. Boffa et al), North-Holland, Amsterdam, 225 - 234. Hajek, P. (1984). On a new notion of partial conservativity, in: Computation and Proof Theory, Springer Lecture Notes in Math. 1104, 217 - 232. Hajek P., Montagna R, Pudlak P. (1992). Abbreviating proofs using metamathematical rules, in: Proof Theory and Computational Complexity (eds. P. Clote and J. Krajίcek), Oxford University Press, 197 - 221. Hajek, P. and Pudlak, P. (1993). Metamathematics of First-Order Arithmetic, Perspectives in Mathematical Logic, Springer-Verlag. Hajkova, M. (1971). The lattice of bi-numerations of arithmetic I, II, Comment. Math. Univ. Carol. 12,81 - 104, 281 - 306. Hajkova, M. and Hajek, P. (1972). On interpretability in theories containing arithmetic, Fund. Math. 76,131 - 137. Hubert, D. and Bernays, P. (1934, 1939). Grundlagen der Mathematik I, II, Springer-Verlag, Berlin. Jensen, D. and Ehrenfeucht, A. (1976). Some problems in elementary arithmetics, Fund. Math. 92, 223 - 245. Jeroslow, R. G. (1971a). Consistency statements in formal theories, Fund. Math. 72, 17 - 40. Jeroslow, R. G. (1971b). Non-effectiveness in S. Orey's compactness theorem, Zeitschr. Math. Log. Grundl. Math. 17, 285 - 289. de Jongh, D. and Montagna, F. (1989). Much shorter proofs, Zeitschr. Math. Log. Grundl. Math. 35, 247 - 260. Kaye, R. (1991). Models of Peano arithmetic, Oxford Logic Guides 15, Clarendon Press, Oxford. Kent, C. F. (1973). The relation of A to Prov'A' in the Lindenbaum sentence algebra, J.Symb. Logic 38, 295-298. Kleene, S. (1952a). Introduction to metamathematics, van Nostrand. Kleene, S. (1952b). Finite axiomatizability of theories in the predicate calculus using additional predicate symbols, Memoirs Amer. Math. Soc. 10, 27 - 66. Kreisel, G. (1957). Independent recursive axiomatization, J. Symb. Logic 22, 109 (abstract). Kreisel, G. (1962). On weak completeness of intuitionistic predicate logic, J. Symb. Logic 27,139 - 158. Kreisel, G. and Levy, A. (1968). Reflection principles and their use for establishing the complexity of axiomatic systems, Zeitschr. fur math. Logik 14, 97 - 142. Kreisel, G. and Wang, H. (1955). Some applications of formalized consistency proofs, Fund. Math. 42,101 - 110.

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Kripke, S. (1963). "Flexible" predicates in formal number theory, Proc. Amer. Math. Soc. 13, 647 - 650. Levy, A. (1965). A hierarchy of formulas in set theory, Memoirs Amer. Math. Soc. 57. Lindstrδm, P. (1979). Some results on interpretability, in: Proceedings of the 5th Scandinavian Logic Symposium 1979, Aalborg University Press, Aalborg, 329 - 361. Lindstrδm, P. (1984a). On partially conservative sentences and interpretability, Proc. Amer. Math. Soc. 91, 436 - 443. Lindstrom, P. (1984b). On certain lattices of degrees of interpretability, Notre Dame J. Formal Logic 25,127 - 140. Lindstrom, P. (1984c). On faithful interpretability, in: Computation and Proof Theory, Springer Lecture Notes in Math. 1104, 279 - 288. Lindstrδm, P. (1988). Partially generic formulas in arithmetic, Notre Dame J. Formal Logic 29,185 - 192. Lindstrδm, P. (1993). On Σj and Γ^ sentences and degrees of interpretability, Annals of Pure and Applied Logic 61,175 - 193. Lindstrδm, P. (199?). Provability logic, to appear in Theoria. Lob, M. (1955). Solution of a problem of Leon Henkin, J. Symb. Logic 20,115 - 118. Mendelson, E. (1987). Introduction to Mathematical Logic (3rd ed.), van Nostrand. Misercque, D. (1982). The non homogeniety of the E-tree - answer to a problem raised by D. Jensen and A. Ehrenfeucht, Proc. Amer. Math. Soc. 84, 573 - 575. Misercque, D. (1983). Answer to a problem by D. Guaspari, in: Open days in Model Theory and Set Theory (eds. W Guzicki et al), Poland, 181 - 183. Montagna, F. (1982). Relatively precomplete numerations and arithmetic, J. Phil. Logic 11,419 - 430. Montagna, F. (1992). Polynomially and superexponentially shorter proofs in fragments of arithmetic, J. Symb. Logic 57, 844 - 863. Montagna, F. and Sorbi, A. (1985). Universal recursion theoretic properties of r.e. preordered structures, J. Symb. Logic 50, 397 - 406. Montague, R. (1957). Two theorems on relative interpretability, Summer institute for Symbolic Logic, Cornell, 263 - 264. Montague, R. (1961). Semantical closure and non-finite axiomatizability I, in: Infinitistic Methods, Warsaw, 45 - 69. Montague, R. (1962). Theories incomparable with respect to relative interpretability, J. Sym. Logic 27,195 - 211. Montague, R. (1963). Syntactical treatments of modality, with corollaries on reflection principles and finite axiomatizability, in: Proceedings of a Colloquium on Modal and Many-Valued Logics, Acta Philosophica Fennica, Helsinki, 153 -167. Montague, R. and Tarski, A. (1957). Independent recursive axiomatizability, Summer institute for Symbolic Logic, Cornell, 270.

128

Mostowski, A. (1952a). On models of axiomatic systems, Fund. Math. 39,133 -158. Mostowski, A. (1952b). Sentences undecidable in formalized arithmetic, North-Holland. Mostowski, A. (1961). A generalization of the incompleteness theorem, Fund. Math. 49, 205 - 232. Orey, S. (1961). Relative interpretations, Zeitschr. fur math. Logik 7,146 - 153. di Paola, R. (1975). A theorem on shortening the length of proof in formal systems of arithmetic, J. Symb. Logic 40, 398 - 400. Parikh, R. (1971). Existence and feasibility in arithmetic, J. Symb. Logic 36, 494 - 508. Pour-El, M. (1968). Independent axiomatization and its relation to the hypersimple set, Zeitschr. fur math. Logik 14, 449 - 456. Pour-El, M. and Kripke, S. (1967). Deduction preserving recursive isomorphisms between theories, Fund. Math. 61,141 - 163. Putnam, H. and Smullyan, R. (1960). Exact separation of recursively enumerable sets within theories, Proc. Amer. Math. Soc. 11, 574 - 577. Quinsey, J. (1980). Some problems in logic, Thesis, Oxford University. Quinsey, J. (1981). Sets of Σk-conservative sentences are Π^ complete, J. Symb. Logic 46, 442 (abstract). Rabin, M. (1961). Non-standard models and independence of the induction axiom, in: Essays on the Foundations of Mathematics, Jerusalem, 287 - 299. Rosser, B. (1936). Extensions of some theorems of Gδdel and Church, J. Symb. Logic 1, 87 - 91 (reprinted in Davis (1965)). Ryll-Nardzewski, C. (1952). The role of the axiom of induction in elementary arithmetic, Fund. Math. 39, 239 - 263. Sambin, G. (1976). An effective fixed point theorem in intuitionistic diagonalizable algebras, Studia Logica 35, 345 - 361. Scott, D. (1962). Algebras of sets binumerable in complete extensions of arithmetic, in: Recursive Function Theory (ed. J. Dekker), Providence, 117 - 121. Shavrukov, V. (1988). The logic of relative interpretability over Peano arithmetic, Steklov Mathematical Institute, Moscow. (Russian) Shavrukov, V. (1993). Subalgebras of diagonalizable algebras of theories containing arithmetic, Dissertationes Mathematicae CCCXXIΠ, Warsaw. Shepherdson, J. (1960). Representability of recursively enumerable sets in formal theories, Arch. Math. Logic 5,119 - 127. Simmons H. (1988). Large discrete parts of the E-tree, J. Symb. Logic 53, 980 -984. Smoryrίski, C. (1980). Calculating self-referential statements, Fund. Math. 109, 189-210. Smoryriski, C. (1981a). Calculating self-referential statements: Guaspari sentences of the first kind, J. Symb. Logic 46, 329 - 344. Smoryriski, C. (1981b). Fifty years of self-reference in arithmetic, Notre Dame J. Formal Logic 22, 357 - 374.

129

Smoryriski, C. (1985). Self-Reference and Modal Logic, Springer-Verlag. Soare, R. (1987). Recursively enumerable sets and degrees, Perspectives in Mathematical Logic, Springer-Verlag. Solovay, R. (1985). Infinite fixed-point algebras, Proceedings of Symposia in Pure Mathematics 42, Providence, 473 - 486. Strannegard, C. (1997). Arithmetical realizations of modal formulas, Thesis, Dept. of Philosophy, University of Goteborg. Svejdar, V. (1978). Degrees of interpretability, Comment. Math. Univ. Carol. 19, 789 - 813. Tarski, A. (1933). Der Wahrheitsbegriff in den formalisierten Sprachen, Studia Philosophica 1, 261 - 405. Tarski, A., Mostowski, A. and Robinson, R. (1953). Undecidable Theories, North-Holland, Amsterdam. Verbrugge, L. C. (1992). Feasible interpretability, in: Arithmetic, Proof Theory and Computational Complexity (eds. P. Clote and J. Krajίcek), Oxford Univ. Press, 387 - 428. Verbrugge, L. C. (1994). The complexity of feasible interpretability, in: Feasible Mathematics II (eds. P. Clote and J. Remmel), Birkhauser, Boston, 429 - 447. Wang, H. (1951). Arithmetical models of formal systems, Methodos 3, 217 - 232.

INDEX

arithmetical hierarchy 11 axiomatization 52 Bn (formula, sentence) 11 Bn (degree) (see Φ (degree)) Bernays-Lob provability conditions 15 binumerate 7 binumeration 8 bounded (quantifier) 8,121 bounded (set of sentences) 52,121

cap 96 caps to ... below ... 114 cofinal 98 complement (in lattice) 20,114 complete (theory) 23 complete (Π°, Σ^ set) 20 conservative (partially) 62 correctly numerates 43 Craig's theorem 10 cup 96 cupping 98 cups to ... above ... 114 Δn (degree) (see Φ (degree)) Δn (formula, sentence) 11 decidable (in T, formula, sentence) 8 decidable (theory) 17 define (formula defines function) 7 definable (set in model) 17 degree (of interpretability) 94 distributive (lattice) 20 dual 11 essentially infinite (over) 52 essentially reflexive (theory) 18

essentially reflexive (interpretation) 120 essentially undecidable (theory) 17 Φ (formula, sentence) 11 Φ (degree) 102 faithful (interpretation) 84 faithfully interpretable 84 finite extension 52 fixed point 15 fixed point lemma 15,121 Γ (degree) (see Φ (degree)) Γ (formula, sentence) 11 Γ-conservative 62 Γ-conservative extension 66 Γ-sound 12 Γ-subtheory 63 GB (Gδdel-Bernays set theory) 124 g.l.b. (greatest lower bound) 20 GδdeΓs (first) incompleteness theorem 23,119 GδdeΓs second incompleteness theorem 26,120 Gόdel-Tarski theorem 17 greatest lower bound (g.l.b.) 20 H-bounded 121 Henkin complete 78 hereditarily Γ-conservative 71 high 104 hypersimple 61 i.a. (irredundantly axiomatizable) 57 i.Γ-a. (irredundantly Γ-axiomatizable) 58 independent (formula) 31

132

independent (set of degrees) 114 independent (on X over T) 44 interpretable 76 interpretation 75 interval 100 irredundant (over) 57 irredundantly axiomatizable (i.a.) 57 irredundantly Γ-axiomatizable (i.Γ-a.) 58 lattice 20 least upper bound (l.u.b.) 20 liar paradox 17 Lob's theorem 28 low 104 l.u.b. (least upper bound) 20 monoconsistent

Q 6

Robinson's Arithmetic 6 Rosser sentence 24 Rosser's theorem 23 Σn (degree) (see Φ (degree)) Σn (formula, sentence) (see Γ (formula, sentence)) Σ L-complete 14 Σn-conservative (see Γ-conservative) Σ-L-extension 97 Σn-sound (see Γ-sound) self-prover 71 Shepherdson-Smoryrίski fixed point theorem 51

25 Tarski theorem 17 theory 5 translation 75 true (sentence, theory) 6 truth-definition 17 type of independence 45

N 5 N 6 numerate 7 numeration 9

OMionsistent 36 Orey compactness theorem 81,123 Orey-Hajek lemma 80,123 PA 6 Πn (degree) (see Φ (degree)) Πn (formula, sentence) (see Γ (formula, sentence)) Πn-conservative (see Γ-conservative) Πn-sound (see Γ-sound) partial truth-definition 18,121 p.c. (pseudocomplement) 20 Peano Arithmetic 6 positively prime (p.p.) 46 p.p. (positively prime) 46 PR (primitive recursive formula)

undecidable (sentence) 23 undecidable (theory) 17 ZF (Zermelo-Fraenkel set theory) 120

8

NOTATION

LA 0, S, +, x

N := T + φ, T + X

KH Th(T) N

PA

Q x < y, x < y

δfίxo^.Λn/y)

5 5 5 5 6 6 6 6 6 6 6 7

Ξx
Pm

9 9

T (convention) Σn,Πn Bn,Γ,Γ+,Γd T Φ/Φ Δn

Sbstk Substk ξ(x), η(x,y)

Prfσ(χ,y) Prσ(x), Conσ 1

σ 1 y, σ + y

11 11 11 11 11 11 13 13 13 13 13 13 13

Prfs(x,y), Prs(x), Cons/ Prfs+z(x,y),

Prfs,z(x,y)

Xlk Πn°,Σn°

φί Ref(T)

X

c

P(ξ) Hp Rfns

13

45 50, 55 52 52

Rfn(n,S), Rfn g

Hπ,

[Πs(χ,y) ΊΓ [Γ](χ,y) Cons(Γ,T) t, μt(x) t: S'< S ^, < Hciriξ

= Int

A,B

^ -

<, d(A), Dτ, Dτ Aτ, i, T, v n, u Oτ, lτ-

CONT

u,n

[a,b], [a,b), etc. d(φ), d(X)

« σ, σo,..., n, n0,... Σn)

ET FT <«

Gτ

Prfs/Γ(x,y), Prs/Γ(x)

«rv «υ a +σ

96 97 98 99 100 102 102 102

Φ (Bn, V Γ, Πn,

F, P, P(«pk: k < ω>),

RFNτ

S, S0, T, T', A, B etc.

(convention)

SatBn(x,y)

Con(n,S), Cong

<x,y> and (x)y/ and (k)m

Satr(x,y), Trr(x)

RFNg

8 8

18 18 18 18 20 25 35 44

Γ(x)

52, 53

54 55 55 59 63 63 63 68 75 75 76 79 83 84 84 87 94 94 94 95

r^T)
LST,ZF ts vn(χ)

Hk, Hk(x)

Trk(x) RFN^

Σf,Πf GB

102 106 106 106 108 119 120 120 120 121 121 121 121 121 124

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