Recursive Number Theory: A Development of Recursive Arithmetic in a Logic-Free Equation Calculus (Studies in Logic and the Foundations of Mathematics, 20)

RECURSIVE NUMBER THEORY A DEVELOPMENT OF RECURSIVE ARITHMETIC IN A LOGIC-FREE EQUATION CALCULUS

BY

R. L. GOODSTEIN Professor of Mathematics University of Leicester

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NORTH·HOLLAND PUBLISHING COMPANY AMSTERDAM

Oopyright 1957

No part of this book may be reproduced in any form, by print, photoprint, microfilm or any other means without written permission from the publisher

PRINTED IN THE NETHERLANDS DRUKKERIJ HOLLAND N.V., AMSTERDAM

PREFACE The discovery of the reflexive paradox, that the class of all classes which are not members of themselves is a self-contradictory concept, gave rise to three new developments in mathematics. The first of these was Russell's theory of types, one part of which segregated objects into types (classes of objects of one type forming the objects of the next higher type) and prohibited the formation of classes of mixed type. This theory led to considerable complications in the construction of arithmetic since it excluded not only paradoxes but also constructions fundamental to the theory of real numbers, like that of the least upper bound of a bounded class of numbers, and the rehabilitation of these constructions necessitated the introduction of an axiom which associated with a propositional function (propositional form with a variable), whose argument ranged over objects of a given type, a propositional function with the same truth values whose argument ranged over objects of the first type. In more recent formulations of set theory alternatives to type theory (and the reducibility axiom) have been proposed; these alternative treatments depend upon some restriction of the right of membership, fortified (in the case of Quine's system) by what may be called a potential typing, according to which the object symbols in every valid formula in a type-free symbolism must admit an assignment of numbers in such a way that each object receives a number which is one less than the number of the class to which it belongs. The second development which was initiated by the discovery of the reflexive paradox was Brouwer's 'intuitionistic' logic and arithmetic, the most novel feature of which was the denial of the tertium-non-datur, the principle of logic which asserts that every proposition is either true, or false, no third possibility presenting itself. The rejection of the tertium-non-datur eliminates the reflexive paradox since the paradox rests on the assumption that every class

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PREFACE

either is, or is not, a member of itself, but it also invalidates the familiar interpretations of a considerable part of arithmetic (although Oodel has shown that intuitionistic arithmetic includes the whole of classical arithmetic, in the sense that to any formula provable by classical logic corresponds a formula provable by intuitionistic logic). The third system which was developed to escape the reflexive paradox was Skolem's recursive arithmetic. Skolem observed that he could avoid the paradox without recourse to the restrictions of type theory and without the rejection of any rule of classical logic if he did not take existence as one of the primitive notions of logic. In a calculus which expresses universality only by means of free variables this has the effect of preventing the application of the tertium non datu« in those cases where it might lead to paradox, since it rules out the negation of universal propositions. The sacrifice of existence as a primitive notion deprived Skolem of the classical method of function definition and in its place he introduced definition by recursion. A function f(n) is said to be defined by recursion if, instead of being defined explicitly, (that is, as an abbreviation for some other expression), only the value of f(O) is given, and f(n+ 1) is expressed as a function of f(n). In other words a recursive definition does not define f(n) itself, but provides a process whereby the values of f(O), f(l), f(2), f(3) and so on, are determined one after the other. In the following account of recursive arithmetic we show that logic and arithmetic may be constructed simultaneously in a free variable equation calculus in which the only statements are equations of the form a = b, where "a" and "b" stand for function signs. By means of this equation calculus the ab initio construction of logic and arithmetic may be presented in full rigour and detail at a far more elementary level than has hitherto been possible, and it is hoped that the first half of the book will prove to be suitable for the mathematical specialist in his first year at the University. In this part a great deal of the smaller detail has been separated from the text and presented in example form (with complete solutions at the end of the book) both to make the text lighter to read and to help the reader to acquire a new technique in easy stages.

PREFACE

I desire to express my warmest thanks to Professor HEYTING for the very kind interest he has taken in the preparation of this book, from the first manuscript draft to the finished typescript; to Mr. JOHN HOOLEY for preparing the index and for generous help in reading the proofs; and to the Compositors of the North-Holland Publishing Company for the excellence of their work.

Leicester, England

R. L. GOODSTEIN

INTRODUCTION The Nature of Numbers The question "What is the nature of a mathematical entity?" is one which has interested thinkers for over two thousand years and has proved to be very difficult to answer. Even the first and foremost of these entities, the natural number. has the elusiveness of a will-of-the-wisp when we try to define it. One of the sources of the difficulty in saying what numbers are is that there is nothing to which we can point in the world around us when we are looking for a definition of number. The number seven, for instance, is not any particular collection of seven objects, since if it were, then no other collection could be said to have seven members; for if we identify the property of being seven with the property of being a particular collection, then being seven is a property which no other collection can have. A more reasonable attempt at defining the number seven would be to say that the property of being seven is the property which all collections of seven objects have in common. The difficulty about this definition, however, is to say just what it is that all collections of seven objects really do have in common (even if we pretend that we can ever become acquainted with all collections of seven objects). Certainly the number of a collection is not a property of it in the sense that the colour of a door is a property of the door, for we can change the colour of a door but we cannot change the number of a collection without changing the collection itself. It makes perfectly good sense to say that a door which was formerly red, and is now green, is the same door, but it is nonsense to say of some collection of seven beads that it is the same collection as a collection of eight beads. If the number of a collection is a property of a collection then it is a defining property of the collection, an essential characteristic. This, however, brings us no nearer to an answer to our question 1

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INTRODUCTION

"What is it that all collections of seven objects have in common?" A good way of making progress with a question of this kind is to ask ourselves "How do we know that a collection has seven members?" because the answer to this question should certainly bring to light something which collections of seven objects share in common. An obvious answer is that we find out the number of a collection by counting the collection but this answer does not seem to help us because, when we count a collection, we appear to do no more than 'label' each member of the collection with a number. (Think of a line of soldiers numbering off.) It clearly does not provide a definition of number to say that number is a property of a collection which is found by assigning numbers to the members of the collection.

The Frege-Russell Definition To label each member of a collection with a number, as we seem to do in counting, is in effect to set up a correspondence between the members of two collections, the objects to be counted and the natural numbers. In counting, for example, a collection of seven objects, we set up a correspondence between the objects counted and the numbers from one to seven. Each object is assigned a unique number and each number (from one to seven) is assigned to some object of the collection. If we say that two collections are similar when each has a unique associate in the other, then counting a collection may be said to determine a collection of numbers similar to the collection counted. Since similarity is a transitive property, that is to say, two collections are similar if each of them is similar to a third, it follows that in similarity we may have found the property, common to all collections of the same number, for which we have been looking, and since similarity itself is defined without reference to number it is certainly eligible to serve in a definition of number. To complete the definition we need only to specify certain standard collections of numbers one, two, three, and so on; a collection is then said to have a certain number only if it is similar to the standard collection of that number. The numbers themselves may be made to provide the required standards

INTRODUCTION

3

in the following way. We define the property of being an empty collection as the property of not being identical with oneself, and then the number zero is defined as the property of being similar to the empty collection. Next we define the standard unit collection as the collection whose only member is the number zero, and the number one is defined as the property of being similar to the unit collection. Then the standard pair is taken to be the collection whose members are the numbers zero and unity and the number two is defined as the property of being similar to the standard pair, and so on. This is, in effect, the definition of number which was discovered by Frege in 1884 and, independently, by Russell in 1904. It cannot, however, be accepted as a complete answer to the problem of the nature of numbers. According to the definition, number is a similarity relation between collections in which each element of one collection is made to correspond to a certain element of the other, and vice-versa. The weakness in the definition lies in this notion of correspondence. How do we know when two elements correspond? The cups and saucers in a collection of cups standing in their saucers have an obvious correspondence, but what is the correspondence between, say, the planets and the Muses? It is no use saying that even if there is no patent correspondence between the planets and the Muses, we can easily establish one, for how do we know this, and, what is more important, what sort of correspondence do we allow? In defining number in terms of similarity we have merely replaced the elusive concept of number by the equally elusive concept of correspondence.

Number and numeral Some mathematicians have attempted to escape the difficulty in defining numbers, by identifying numbers with numerals. The number one is identified with the numeral 1, the number two with the numeral 11, the number three with 111, and so on. But this attempt fails as soon as one perceives that the properties of numerals are not the properties of numbers. Numerals may be blue or red, printed or handwritten, lost and found, but it makes no sense to ascribe these properties to numbers, and, conversely,

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INTRODUCTION

numbers may be even or odd, prime or composite but these are not properties of numerals. A more sophisticated version of this attempt to define numbers in terms of numerals, makes numbers, not the same thing as, but the names of the numerals; this escapes the absurdities which arise in attempting to identify number and numeral but it leads to the equally absurd conclusion that some one notation is the quintessence of number. For if numbers are the names of numerals then we must decide which numerals they name; we cannot accept the number ten for instance as both the name of the roman numeral and the arabic numeral. And if it is said that the number ten is the name of all the numerals ten then we reach the absurd conclusion that the meaning of a number word changes with each notational innovation. The antithesis of "number" and "numeral" is one which is common in language, and perhaps its most familiar instance is to be found in the pair of terms "proposition" and "sentence". The sentence is some physical representation of the proposition, but cannot be identified with the proposition since different sentences (in different languages, for instance) may express the same proposition. If, however, we attempt to say just what it is that the sentences express we find that the concept of proposition is just as difficult to characterise as the concept of number. It is sometimes held that the proposition is something in our minds, by contrast with the sentence, which belongs to the external world, but if this means that a proposition is some sort of mental image then it is just another instance of the confusion of a proposition with a sentence, for whatever may be in our minds, whether it be a thought in words, or a picture, or even some more or less amorphous sensation, is a representation of the proposition, differing from the written or spoken word only because it is not a communication. In the same way we see that the view that number is indefinable, being something which we know by our intuition, again confuses number with numeral, that is confuses number with one of its representations.

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Arithmetic and the Game of Chess The game of chess, as has often been observed, affords an excellent parallel with mathematics (or, for that matter, with language itself). To the numerals correspond the chess pieces, and to the operations of arithmetic, the moves of the game. But the parallel is even closer than this, for to the problem of defining number corresponds the problem of defining the entities of the game. If we ask ourselves the question "What is the king of chess 1" we find precisely the same difficulties arise in trying to find an answer which we met in our consideration of the problem of defining the concept of number. Certainly the king of chess, whose moves the rules of the game prescribe, is not the piece of characteristic shape which we call the king, just as a numeral is not a number, since any other object, a matchstick or a piece of coal, would serve as well to play the king in any game. Instead of the question "What is the king of chess 1" let us ask "What makes a particular piece in the game the king pieee !" Clearly it is not the shape of the piece or its size, since either of these can be changed at will. What constitute a piece king are its moves. That piece is king which has the king's moves. And the king of chess itself? The king of chess is simply one of the parts which the pieces play in a game of chess, just as King Lear is a part in a drama of Shakespeare's; the actor who plays the King is King in virtue of the part which he takes, the sentences he speaks and the actions he makes, (and not simply because he is dressed as king) and the piece on the chess board which plays the king-role in the game is the piece which makes the king's moves. Here at last we find the answer to the problem of the nature of numbers. We see, first, that for an understanding of the meaning of numbers we must look to the 'game' which numbers play, that is to arithmetic. The numbers, one, two, three, and so on, are characters in the game of arithmetic, the pieces which play these characters are the numerals and what makes a sign the numeral of a particular number is the part which it plays, or as we may say in a form of words more suitable to the context, what constitute

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INTRODUCTION

a sign the sign of a particular number are the transformation rules of the sign. It follows, therefore, that the OBJECT OF OUR STUDY IS NOT NUMBER ITSELF BUT THE TRANSFORMATION RULES OF THE NUMBER SIGNS, and in the chapters which follow we shall have no further occasion to refer to the number concept. But just as the rules of chess are currently formulated in terms of the entities of chess, so that we say, for instance, the king of chess moves only one square at a time (except in castling), instead of the completely equivalent formulation "the piece playing the part of king (or simply the king-piece) is moved only one square at a time (except in castling)" so we shall continue, in purely descriptive passages, to formulate the operations of arithmetic in terms of arithmetical entities instead of arithmetical signs. For instance, we may speak of "the sum of the numbers two and three" rather than confine ourselves to the object formulation "2 + 3", where + is the sign whose role in arithmetic is what is called addition, and "2" and "3" are numerals whose roles are those of the numbers two and three. To put it another way, in defining the part played by a. sign like +, in arithmetic, we shall say that what we are defining is the sum function, but the definition itself will refer only to operations for transforming expressions which contain the sign +.

Number Variables The parallel between chess and arithmetic breaks down when we contrast the predetermined set of pieces in the game of chess with the licence granted to arithmetic to construct numerals at will. In this respect arithmetic more closely resembles a language which places no limit, in principle, upon the length of its words. A familiar notation for numerals expresses them as words spelt with the 'alphabet' "0", "1" and" +"; each 'word' has an initial "0" followed by a succession of pairs" + 1". Thus, for instance, we form in turn "0", "0 + 1", "0 + 1 + 1", "0 + I + I + 1". The formation of numerals may be fully characterised by means of two operations, as follows. We extend the alphabet by the introduction of a new sign, "x", and form 'words' by writing either "0" or "x + I" for "x"; for example we may form in turn, "x", "x+ I", "x+ 1 + 1",

INTRODUCTION

7

"x+ 1 + 1 + 1", "0+ 1 + 1 + 1", the last of which is a numeral. This new sign we call a 'numeral variable'. The rules permitting the substitution of "z -} 1" or "0" for "x" in effect allow the substitution of any numeral for x; the object of the formulation we have adopted is that it serves to define the concept of any numeral and the concept of a numeral variable simultaneously. In the sequel, not only the letter x, but other letters, too, will be used as numeral variables. The numeral formed by writing some numeral for "x" in "x+ 1" is called the successor of that numeral. For instance, writing "0+ 1 + 1" for "x" in "z -}1" we obtain "0+ 1 + 1 + 1", the successor of "0 + 1 + 1". For this reason "x + 1" is called the (sign of the) successor function. The definite article is somewhat misleading, however, since we may write, in place of x, any other letter which is being used as a numeral variable; in a system in which x, y and z are all numeral variables, each of "x + 1", "y + 1", "z + 1" is a sign of the successor function. Nevertheless we shall talk of the successor function, the uniqueness in question being the uniqueness of the sign which results when we write some definite numeral for the variable, be it denoted by x, y or z, For purposes of standardisation of notation we shall have occasion to introduce, instead of the 'alphabet' "0", "1" and "+" for writing numerals, the 'alphabet' "0", "S" in which the numerals become "0", "SO", "SSO", "SSSO" and so on. In this notation the sign of the successor function is "Sx" and the transformation rules for a numeral variable x are (i) Sx may be written for x, (ii) 0 may be written for x. Another notation in current use employs "x'" for the successor function, so that the numerals are written "0", "0"', "0"", "0"''' and so on.

Definition of Counting No theory of the natural numbers is complete which does not also take into account the part which numbers play outside arithmetic. It is not only a property of the number nine that it is a square but also that it is the number of the planets, and this

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INTRODUCTION

latter property is not a consequence simply of the laws of arithmetic. According to the Frege-Russell definition of number, the number of a collection is found by testing it for similarity with the standard unit, pair, trio, and so on, in turn, this testing being carried out by the process of counting, but as we have proposed a definition of number which does not rest upon the undefined concept of a similarity correspondence we cannot accept counting, in the Frege--Russell sense, as a means of finding the number of a class, without readmitting this undefined concept. There is, however, an entirely different interpretation of the process of counting, which makes counting available to us as a means of recording the number of a collection, without transcending the limitation we imposed upon ourselves of expressing the properties of numbers in terms of the transformation rules of the number signs. We start by separating two distinct stages in the process of counting. The first of these is what we shall call "using a collection as a numeral" which consists in overlooking the individual 'idiosyncrasies' of the elements of the collection and regarding them as being all alike (but not identical) for the purpose in hand. This is simply a (perhaps rather extreme) form of a treatment of signs familiar in all acts of reading, writing or speaking; the letters "a" on a printed page, for instance, have their several differences and, subject to sufficiently close scrutiny, are as different as say the soldiers in a platoon, but for the purposes of reading we ignore these differences and treat the various a's as being the same sign. And so too, in speaking, we treat as the same a variety of slightly different sounds. In a different context, signs which we would accept as the same for reading purposes, are carefully distinguished, as, for example, when we test the quality of printing. The process of overlooking some differences, but not others, is fundamental in language; it is the process by which we subsume objects with a 'family likeness' under a generic name and the process which makes possible the use in language of universal words. Without it, the concept of the number of a class could never have arisen. The second stage in the process of counting consists in a transformation from one number notation to another by means of the rules "one

INTRODUCTION

9

and one is two", "two and one is three", "three and one is four" and so on. It is the recitation of these rules (in an abbreviated form in which each 'and one' is omitted, or replaced by pointing to, or touching, the object counted) which gives rise to the illusion that in counting we are associating a number with each of the elements counted, whereas we are in fact making a translation from the notation in which the number signs are "one", "one and one", "one and one and one", and so on, to the notation in which the signs are "one", "two", "three", and so on. The true nature of counting is perhaps most clearly brought out if we re-introduce the older process of making a tally. Making a tally of a collection consists in some formalised representation of the elements of the collection, say by means of dashes on a sheet of paper, so that in making a tally we are copying a number sign in some standard notation - finding the number of the collection, by treating it as a number sign and copying this sign. Thus a tally of the planets consists in the row of dashes 111111111 If we now proceed to transform this sign by means of the trans-

formation rules 11= 2, 21 = 3, 31 = 4, 41 = 5, 51 = 6, 61 = 7, 71 = 8, 81= 9we obtain in turn II III l l l l = 21111111 =3111111 =411111 = =51111=6111=711=81=9, which completes the transformation. In counting as we teach it today, the processes of tally making and sign transformation are carried out simultaneously, thus avoiding the repeated copying of the 'tail' of the number sign in transforming to an arabic numeral. It is important to realise that counting does not discover the number of a collection but transforms the numeral which the collection itself instances from one notation to another. To say that any collection has a number is just to say that any collection may be used as a number sign.

Formalisation of Counting Counting may be formalised in a system of signs by formulating the transformation rules of a counting operator "N". We represent the objects in the collections to be counted by letters a, b, e, ... ,

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INTRODUCTION

and collections by conjunctions like a & b, a & b & e; a single object being regarded also a collection. The letter l we use as a variable for an object, that is, a letter for which any object may be written; the capital letter L serves as a variable for a collection and may, in any context, be replaced by a definite collection or by "L & l", The numerals of the system are the signs (without x) obtained from l, x and the successor function x + 1 by substitution. Then we define N(l)=l,

N(L & l)=N(L)+1.

These equations suffice to determine the number of any collection. For instance, substituting "a" for the variable-sign "l", in the first, we obtain N(a) = 1, and then, substituting "a" for "L" and "b" for "l" in the second, we obtain N(a & b)=N(a)+ 1

and so, N(a & b) = 1 + 1. Next, substituting "a & b" for "L" and "e" for "l" we find N(a &b &c)=N(a &b)+1=1+H-1,

and so on. We observe that the definition of N(L) is by recursion, that is to say, N(L) is not simply an abbreviation for some other expression, as, for instance, when we define 2 = 1 + 1, the sign "2" may be replaced by "1 + 1" for which it is merely an abbreviation, but N(L) is determined only step by step, by introducing the members of the class to be counted one at a time (or shedding them one at a time). We may express this by saying that for the variable L, N (L) itself is undefind, only the result of substituting a definite class (like a & b & c) for L being defined by the recursive definition. The recursive definition is, so to speak, a schema or mould from which the definition (value) of N(a & b & ... & k) may be found by substitution for any particular class a & b & ... & k, Evolution of the Concept of a Formal System In the following chapters we shall set up arithmetic as a formal system. The idea of a formal system is one which derives from Euclid's presentation of geometry, but the notion has undergone

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considerable development during the past century. Euclid's intention in the "Elements" was to deduce the whole body of geometrical knowledge of his time from a few self evident truths (called axioms) by purely logical reasoning. Euclid did not, however, specify the nature of 'logical reasoning' and the first attempt to do so was made by George Boole, in 1847, in his Mathematical Analysis of Logic. Boole constructed a symbolic language, in which the 'laws of thought', formulated as axioms, may be studied by mathematical techniques. In the complete development of the notion a formal system is an assemblage of signs separated into various categories, their usage bound by various conventions (the axioms and transformation rules) the object of the system being to arrange sequences of formulae (which are themselves sequences of signs with certain specified formation rules) in certain relationships to one another to form a particular pattern called proof. A formal system may contain both mathematical and logical signs (the distinction is an arbitrary one), and mathematical and logical axioms; its essential feature, qua formal system, is that its operation does not presuppose any knowledge of the significance of the signs of the system than is given by the axioms and transformation rules. The mathematical axioms are no longer "self evident truths" but arbitrary initial positions in a game, and the logical axioms express, not the "laws of thought" but arbitrary conventions for the use of the logical signs. In the formal system with which we shall first be concerned in this book, the equation calculus, the only signs are signs for functions and numeral variables, and the equality sign. There are no axioms except the introductory equations for function signs, and there is no appeal to 'logic', the operation of the system being specified simply by the transformation rules for the mathematical signs. It is shown that a certain branch of logic is definable in the equation calculus and logical signs, and theorems, are introduced as convenient abbreviations for certain functions and formulae. This branch of logic is characterised by the fact that it can assert the existence of a number with a given property only when the number in question can be found by a specifiable number of trials.

CHAPTER I

DEFINITION BY RECURSION Variables We have already had occasion, in the definition of a numeral, to refer to the use of a letter as a sign for a variable. In terms of two operations (1) replacing x by x+ 1, (2) replacing x by 0, we defined numerals to be the signs constructed from x by the repeated application of the operation (1) followed by an application of the operation (2). Starting with the sign x the process of constructing numerals may be regarded as a process of eliminating x using only the operations (1) and (2). For instance the numeral 0+ 1 + 1 + 1 is constructed from x by three applications of operation (1) followed by an application of operation (2). The defining property of a numeral variable x is that it may be replaced by zero or x + 1. Any letter may of course serve as a numeral variable, but in this chapter only the letters x, y, z and w will be used. By means of variables we can make general statements about numbers, statements which hold true when any particular numeral is substituted for the variable. 1.

1.1 ADDITION The fundamental operation of arithmetic is addition. Addition is the operation of joining together two numerals by the addition sign '+'. For instance, joining the two numerals 0 + 1 + 1 + 1 + 1 and 0+ 1 + 1 we obtain (omitting the introductory part 0+ in the second numeral) the numeral 0 + 1 + 1 + 1 + 1 + 1 + 1 which is called the swm. of 0 + 1 + 1 + 1 + 1 and 0 + 1 + 1. To say that addition is the operation of joining together two numerals is not, however, a mathematical definition of the operation 13

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DEFINITION BY RECURSION

for we have simply replaced the word addition by the undefined term joining. There are obviously many ways in which we might join together two numerals, but only one of these ways gives what we mean by addition. A mathematical definition of addition must be formulated in terms of numeral variables, and the addition sign +, alone; no extraneous elements may be introduced. A child first learns to add by the simple conjunction of numerals, for instance to find the sum of 3 and 4 he combines and .... to form and 'counts the dots', i.e, transforms into 7. Later he learns to proceed from 3 to 7 by the repeated addition of 1, and this far more economical procedure is the basis of the following formal definition:

A

x+O=x x+(y+ I)=(x+y)+ I

The sign '=' here signifies that the expressions which stand on either side of it are equivalent so that either may replace, or be replaced by, the other; that is to say, Al and A2 express transformation rules by which one sign pattern may be transformed into another. (There is of course another entirely different use of the equality sign '=' in mathematics to which we shall have occasion to refer later). The variable x in equation Al may, by definition, be replaced by 0 or by x + 1 ; it follows that x may be replaced by any numeral for numerals are constructed from x by these same substitutions. And similarly both x and y in Az may be replaced by any numeral. To illustrate the application of the transformations A we evaluate the sum of 5 and 4. Substituting 5 for x in Al and Az, and 0, 1,2,3, in turn for y in Az we obtain 5+0=5, 5+ 1=6 5+2=(5+ 1) + 1 =6+ 1 =7 5+3=(5+2)+ 1 =7 + 1 =8 5+4=(5+3}+ I =8+ I =9. Each partial result after the first is obtained by means of the

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DEFINITION BY RECURSION

previous result; for instance we pass from (5+ 2) + 1 to 7 + 1 by means of the earlier equation 5 + 2 = 7. .1.11 Addition of three or more numerals is defined in terms of the ~epeated addition of pairs. Thus for instance we define

A' A"

x+y+z=(x+Y)+z x+y+z+ w=«x+ y)+z)+ w

and so on. These definitions, like definition A, treat the numerals added unsymmetrically. In definition A, x and yare not on the same footing and it is by no means obvious that x+y=y+x. In A' the three terms x, y, z are added in that order, and so too in A". We shall prove in the next chapter that the sum of two numerals is independent of the order in which they are added; from this it follows that, for instance,

y+x+z= (y+x) +z= (x+y) +z=x+y+z. The equality of x + y + z and y + z + x however requires independent proof and the general proof that addition of any number of terms is completely symmetrical is much more difficult. The definition of the sum of a variable number of terms is a problem to which we shall return later; there is however a special case which we shall consider here, and that is the repeated addition of a single number. 1.2

MULTIPLICATION

Repeated addition of the same number is called muUiplication; the sums x+x, X+X+x, x+x+x+x, and 80 on, are denoted respectively by 2· x, 3· x, 4· x, and so on, the first term of the pair denoting the number of repetitions of the second term. The formal definition of multiplication is:

M

o·x=o (Y+ I)·x=(y·x)+x

M1 ~

Substituting 0, 1, 2 and so on in turn for y in ~ (and using M1 ) we verify in turn l·x=O+x=x, 2·x= l·x+x=x+x, 3·x= 2·x+x=x+x+x and so on.

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DEFINITION BY BECURSION

x· Y is called the product of x and y; when the context renders the sign free from ambiguity we shall denote the product of x and y by xy, omitting the dot between the terms.

1.21 Products of three or more numbers are defined in terms of products of pairs. Thus

x.y.z=(x·y)·z x·y·z·w=(x·y·z)·w Multiplication like addition is a symmetrical operation, but the proof of this must also be postponed until we have set up the necessary proof processes. 1.3

EXPONENTIATION

Repeated multiplication of the same number is called exponentiation; the products x· x, z- z- x, x· z- x· x and so on, are denoted respectively by x 2 , Xl, x4, and so on, the 'index' denoting the number of repetitions of the base. The formal definition of x'/l is E

xo= I X(lI+ll =

(Xli) • X

E1 ~

Substituting 0, 1, 2 and so on, in turn for y in E 2 (and using E l ) we find in turn

and so on. Exponentiation is not a symmetrical operation, that is to say, Xli and y'" are, in general, unequal. Just as we passed from addition to multiplication and from multiplication to exponentiation by the iteration of the same number, so we can define further operations by iteration. These operations have no established names so that we shall call them simply tetration, pentation, hexation, heptation and so on. Denoting the exponentiations XX, f, xF and so on, by 2X, sx, 4X and so on, respectively we define '/Ix by T

°x=1 (lI+ll X = X,'IIlll)

17

DEFINITION BY RECURSION

so that, taking y = 0, 1, 2 in turn we verify that lX=x, 2X = x"', 3x = x(r )= xf" and so on.

Similarly if x, "'x, "''''x, ..., are denoted by we may define

IX,

~'3X,

•••

respectively,

p 1.4

SUBTRACTION

There remains to be defined one more fundamental operation of the aiithmetic of natural numbers, namely subtraction, the converse of addition. We introduce first the converse of the operation + 1, which we denote by .z: I, with the formal definition 0-=--1=0 (x+ I)-=--I =x and then the difference x-=-- y is defined by x-=--O=x x-=--(y+ 1)=(x-=--y)-=--1.

S

The use of the modified subtraction sign .z; instead of the familiar - marks an important difference between the foregoing definition of subtraction and that in elementary arithmetic. If x is not less than y then x-=-- y has its familar meaning, and, as we shall subsequently show, y+(x-=--y)=x; but if x is less than y, definition S gives x-=--y the value zero, so that y+(x-=--y)=y. Finally we define lx, YI, the positive difference between x and y by the equation

lx, yl = (x-=-- y) + (y-=--x). 1.5

FUNCTIONS

Expressions constructed from variables, like x 3 , x+Y or x·y·z are called functions; more exactly a function is an operation on numbers defined by means of variables. The several notations in which we have expressed functions so far are of considerable historical and technical importance, but their diversity tends to

18

DEFINITION BY RECURSION

conceal their fundamental structure, and we shall introduce a standardised notation. We shall denote a function by prefixing to the variables concerned the name of the operation (abbreviated where necessary). Thus the successor function x + I will be denoted by 8(x) (and so the numerals by 0,8(0), 8(8(0», 8(8(8(0))) and so on), the sum function x+y by Sum (x, y), the product x. y by Prod (x, y) and the exponential by Exp(x, y); the brackets in these signs, like the comma between the variables, are not essential parts of the sign, but are included only for typographical convenience and will be omitted when this does not lead to ambiguity. In standardised notation the definitions A, M, E, T and Stake the forms A

Sum(x, O)=x Sum(x, Sy)=S(Sum(x, y»

M

Prod(x, 0) =0 Prod(x, Sy) = Sum(x, Prod(x, y»

E

Exp(x, 0) = I Exp(x, Sy) = Prod(x, Exp(x, y»

T

Tet(x, 0)= I Tet(x, Sy) = Exp(x, Tet(x, y»

S

Dif(x,O)=x Dif(x, Sy)=P(Dif(x, y»

where P(x) is x -'- 1 defined by P(O)=O PSx=x· The common feature of all the definitions A to S is apparent; each definition takes the form

I

F(x, 0) = CJ(x) F(x, Sy) = b(x, F(x, y»

where F(x, y) is the function defined and a(x), b(x, y) are functions previously defined or are variables or definite numerals. Definition scheme I is called iteration; definition by iteration

DEFINITION BY RECURSION

is a particular case of definition by scheme is

RECURSION

19

of which the

f(x, 0) = a(x) f(x, Sy) = b(x, y, f(x, y»

R

(The difference between iteration and recursion is that in the latter, function b depends upon y as well as upon x and F(x, y).) An example of a function which may be defined under the scheme R is the sum of the geometric series

<.

1 +x+x2+ ... +x"=G(x, n)

which satisfies the equations G(x, 0)= 1 G(x, Sn) = B(x,

n, G(x, n»

where B(x, y, z)=Sum(Exp(x, y+ 1), z). More specifically the scheme R is called recursion in y; the second variable x is called a parameter of the scheme. No limit is placed on the number of parameters, so that for instance the scheme F(O, y, Z, w) = o(y, Z, w) R* F(Sx, y, z, w)=b(x, y, z, w, F(x, y, Z, w» is also called recursion (in x).

The variables in a function sign, like x, y, z in F(x, y, z) are sometimes called the arguments of the function. If a (unique) numeral / can be found such that Eta, b, ... )=/ where a, b, ... denote certain definite numerals then / is called the value of the function F(x, y, ... ) for the values a, b, ... of the arguments x, y, ... of the function F(x, y, ... ). The characteristic property of a function F(x, y, ... ) defined by recursion is that the value of F(x, y, ... ) is determinate for any set of values of the arguments provided that the auxiliary functions in the definition scheme have this property, for F(O, y, ... ) is a known function and the value of F(Sx, y, ... ) is determined when the value of F(x, y, ) is known, since F(Sx, y, ... ) is expressed in terms of F(x, y, ) by means of a function with known values.

20

DEl!'INITION BY RECURSION

1.6 It will be seen, in a later chapter that the introduction of additional parameters, as in the scheme R *, is in fact superfluous and that it suffices to consider only the simplest recursive scheme F(O)=O F(Sx) = b(F(x)),

in which there are no parameters at all. We collect here for future reference three functions which have an important part to play in this reduction. These are Alt x, Hf X= [xJ2] and Rt X= [xl/ 2] defined by the recursions (i) AltO=O,AltSx=l...:..Altx so that Alt 1 = 1 , Alt 2 = 0 , Alt 3 = 1 and so on; (ii) HfO=O, HfSx=Hfx+Altx so that Hf 1 = 0 , Hf 2 = 1 , Hf 3 = 1 , and so on, that is Hf 2x=Hf (2x+ l)=x ; (iii) Rt 0= 0, Rt Sx=Rt x+ {I...:.. p(x, Rt x)} where p(x, y)=(Sy)2...:..SX ;

thus Rt Sx is obtained from Rt x by adding 1 if Sx is the next square after (Rt X)2, and adding 0 otherwise, i.e. Rt x is the greatest number whose square does not exceed x.

1.61 Oontracted Notation When a function is defined by the simplest recursion F(O)=O, F(Sx)=b(F(x))

its successive values F(l), F(2), F(3), ... , are given by b(O), bb(O), bbb(O), .. , , and it is natural to denote these by b1(0), b2(0), b3(0), ... , so that F(x) may be expressed in the form b"'(O), where bO(O) stands for O. Thus, since x+y is defined by x+O=O, x+Sy=S(x+y)

it follows that x+1=Sx,x+2=S2(x),x+3=S3x, ... , so that x + y may be expressed in the form SlI(X) where SO(x) stands for x.

DEFINITION BY RECURSION

1.7

21

SUBSTITUTION

The simplest operation by which a new function may be constructed from given functions is substitution. Given a function of any number of variables, say a function of four variables P(a, b, c, d), and any four functions, say A(x, y, z), B(x, y, z), C(x, y, z), D(x, y, z) then a new function F(x, y, z) may be defined by the equation Bubst

F(x, y, z)=P(A(x, y, z), B(x, y, z), C(x, y, z), D(x, y, z)).

The function F so defined is said to be defined from P, A, B, C, and D by substitution in R For instance in the definition of addition, x + By is defined from x + y and B by substitution in S». A definition of the kind Subst is also said to be an explicit definition of the function F, by contrast with a recursive definition. In an explicit definition every variable which occurs on the right hand side of the equation must occur also in the new function on the left, but there may be additional variables on the left-hand side. 1.8

THE INITIAL FUNCTIONS

It is often convenient to be able to express any variable, or definite numeral, in the form of a function. For this purpose we introduce a number of special functions. The first of these is the zero function Z(x) defined by the equation Z(x)=o.

This definition could be given recursively in the form Z(O) = 0 ,ZB(x) =Z(x)

but the explicit definition is indispensible in the sequel. In addition to the zero function we define also the identity function ..F(x) by the equation ..F(x)=x.

From the zero function we can obtain any constant function by substitution in the successor function Sx. For example the function C(x), whose value is 2 for any values of x, is defined by C(x) =S2Z(X).

22

1.9

DEFINITION BY RECURSION

SINGLY RECURSIVE FUNCTIONS

A function is said to be singly recursive (or primitive recursive) if it may be defined from the zero function, the successor function and the identity function by substitutions and recursions. We may also express this definition by saying that the zero function, the successor function and the identity function are singly recursive, and further that a function is singly recursive if it is defined by substituting singly recursive functions in a singly recursive function or is defined by a singly recursive scheme employing only singly recursive functions. Thus all the functions defined up to this point are singly recursive functions and any function formed from them by substitution will be singly recursive. In a sense which will be made sharper in the next chapter a recursive function is totally eliminable, that is to say, when all its argument places are filled by definite numerals the function may be transformed into a single numeral. For the initial functions are certainly eliminable, and substitution of eliminable functions in an eliminable function results in an eliminable function; finally a function F(x), which may contain parameters, defined in terms of eliminable functions by a singly recursive scheme, is eliminable since F(O) is eliminable and F(Sx) is eliminable if F(x) is eliminable. 1.91

Doubly Recursive Functions

The totality of arithmetical operations, addition, multiplication, exponentiation, tetration, pentationand so on, may be expressed by means of a single function. If we define a function H(p, n, a) by the equations H

H(O, n, a)=Sn, H(I, 0, a)=a, H(2, 0, a)=O, H(p+3, 0, a)= 1 (so that H(p + 1, 0, a) = a(l-=- p) + {1-'-(2 -'-pm and H(p+ 1, n+ 1, a)=H(p, H(p+ 1, n, a), a)

then the value of H(p, n, a) is determined for any values of p, n, and a; for H(O, n, a) is given, and if H(p, n, a) is given for some p and any n and a then H (p + 1, n, a) is defined by single recursion in n, The functions H(I, n, a), H(2, n, a), H(3, n, a),

DEFINITION BY RECURS!ON

H(4, n, a), tetration,

23

define in turn addition, multiplication, exponentiation, ; for

H(I, 0, a)=a, H(I, n+ 1, a)=SH(I, n, a) so that H(I, n, a)=a+n, and therefore H(2, 0, a)=O, H(2, n+ I, a)=H(2, n, a)+a

so that H(2,n,a)=na, and hence we find in turn H(3, 0, a)= 1, H(3, n+I, a)=H(3, n, a)·a,

so that H(3, n, a)=a" H(4, 0, a)= 1, H(4, n+ 1, a) = aH<4. fl . Gl ,

so that H(4, n, a)="a and so on. The equations H do not fall under the singly recursive scheme R since the successors of both nand p are involved. H is an example of definition by double recursion. The general form of a definition by double recursion is as follows: A function G(p, n), which may depend also upon one or more parameters, is said to be defined by double recursion from certain given functions if G(O, n) is one of the given functions G(p+ 1, O)=a(p, G(p, b(p))) G(p+ 1, n + 1) =c{p, n,G(p, d[p, n, G(p+ 1, n)]), G(p+ 1, n)},

where a(x, y, z) , b(x) , c(x, y, z, w) and d(x, y, z) are given functions. We observe that G(p+ 1,0) is obtained by substituting a given function for n in G(p, n) and substituting the result in a given function; G(p+ 1, n+ 1) is also obtained by substituting G(p+ 1, n) in given functions and substituting for n in G(p, n). We shall show in Example 1.9 that the first two equations may be replaced by G(O, n)=G(p+ 1,0)= 1. If the given functions are singly recursive then the equations R 2 show that for' any given value of p, the function G(p, n) is singly

24

DEFINITIPN BY RECURSION

recursive in n, for G(O, n) is given to be singly recursive, and if for some p, G(p, n) is singly resursive in n, then G(p+ 1, n+ 1) is expressed by singly recursive functions in terms of G(p+ 1, n). It does not necessarily follow, however, that G(p, n) is a singly recursive function of n with parameter p and we shall subsequently show that there are doubly recursive functions which are demonstrably different from any singly recursive function; in particular this is true of the doubly recursive function H(p, n, a) defined above (see Peter [1]) * By analogy with the equations R 2 one may readily define recursions in three or more variables, but we shall have no occasion to employ more than a double recursion. 1.92 From single and multiple recursions we derive the general notion of a recursive function. A function said to be recursive if it is singly recursive, or defined by a recursion in any number of variables with recursive given functions, or defined by substitution from recursive functions. We shall show that all the processes of classical arithmetic may be described by means of recursive functions.

* The

number in square brackets refers to the bibliography.

Examples I 1. Formulate the recursive definitions of the arithmetical processes of hexation and heptation.

1.1 Evaluate Tet (2,3) and Pent (2, 3) and express Hex (2, 3) as a repeated exponential. 1.2

Ifj(O,n)=I,j(p,O)=p+l and j(p+ 1, n+ 1) = j(p, pn) .j(p+ 1, n),

find the values of j(2, n), j(3, n) and j(4,4). 1.3

---

If
prove that j(x) is recursive. 1.31 If g(x) and k(x) are both recursive and if j(x)=g(x) for x<2. =k(x) for x> 2, prove that j(x) is recursive. 1.4

Prove that the following functions are recursive:

1.41 The remainder when x is divided by 2, and the quotient; the highest common factor of x and 2, and their least common multiple. 1.42 1.5

As in 1.41 with '3' in place of '2'. If g(x), h(x) are recursive and j(x) = g(x) if x is even, =k(x) if x is odd,

show that j(x) is recursive. 1.6 Show that the coefficient of x' in the expansion of (1 + x)" is a doubly recursive function. 25

26

DEFINITION BY

~ECURSION

1.7 Assuming that a+b=b+a and (a+b)+c=a+ (b+c) prove that the sum a + b + c + d is independent of the order of its factors. 1.8 Define primitive recursive functions i(x), j(x, y), k(x, y) such that i(O)= 1, i(x+ 1)=0; j(O, y)=O, j(x+ 1,0)= 1, j(x+ 1, y+ 1)=0; k(O,y)=O, k(x+l,O)=O, k(x+l,y+l)=1.

1.81

If a(x), b(x) and c(x, y) are primitive recursive and f(O, y) = a(y) f(x+ 1, O)=b(x) f(x-+ 1, y+ 1)=c(x, y)

prove that f(x, y) is primitive recursive. 1.9

If A(p, n, u, v)=i(p)·a(n)+i(p, n)·b(p---l, u) +k(p, n).c(p---l, n---l, u, v)

and /-l(p, n, w)

=

i(p) + j(p, n) .Sd(p -'-I) + k(p, n) ·Se(p --- 1, n --- 1, w)

and if 1p(p, n) is defined by the double recursion 1p(0, n)=1p(p+ 1,0)= 1 1p(p+ 1, n -+ 1) =A(p, n, ljJ(p, /-l[p, n, 1p(p+ 1, n)]), 1p(p+ 1, n»

prove that, if G(p, n)=1p(p+ 1, n+ 1), then G(O, n)=a(n)

G(p+ 1, O)=b(p, G(p, d(p»)

G(p+ 1, n+ 1) =c(p, n, G(p, e[p, n, G(p+ I, n)]), G(p+ 1, n»

1.91 Formulate the introductory equations of a function f(x, y, z) defined by triple recursion.

CHAPTER I1 THE EQUATION CALCULUS 2. We have described the two processes, substitution and recursion, by which new functions may be defined from functions previously introduced, and have shown how all recursive functions are built up from the zero and successor functions. We turn now from the construction of functions to the proofs of equations. The equations we consider express the fact that certain function signs may be substituted for others; for instance the equation x +y = y + x says that x + y may be substituted for y + x , and vice-versa. The general form of such equations is F = G where F and G are recursive functions.

DEFINITION OF A PROOF A proof is a table of equations each of which is either (part of) the definition of a function, or an equation of the form F = F , or is a proved equation. If F=G is one of the equations of a proof, then a proved equation is obtained by replacing the function F by the function G at one or more of the places at which F occurs in some equation of a proof. Furthermore, the equation formed by replacing a variable at all the points at which it occurs in some equation of a proof, by another variable, or by a definite numeral or function, is a proved equation. Finally, F=G is a proved equation if equations of a proof are obtained by substituting the function F for a function H , and by substituting the function G for H , in the equations which define the function H . If the results of substituting F and G in the defining equations of some function are equations of a proof, we shall say that F and G satisfy the same introductory equations. The rule that F=G 2.1

21

28

THE EQUATION CALCULUS

if F and G satisfy the same introductory equations is just another way of saying that the processes of recursion and substitution define unique functions. For instance, the equations x 0=x, x+Sy =S(x+ y), which introduce the sum function, define this function completely, so that any f(x,y) which satisfies the same equations, namely f (x,0) =x, f (x,Sy) =Sf (2,y), is just another notation for the same function.

+

To illustrate the proof technique, and in preparation for subsequent applications, we collect together here proofs of the principal properties of the functions we introduced in the previous chapter. 2.11

Properties of the #urn Function As the first example of a proof we consider the commutative property of addition which is expressed by the equation 2.2

x+y=y+x

To make it easier to follow we divide the proof into two parts, starting with the proof of the equation

8 x + y =S(x + y) The proof consists of the nine equations (2.21)

x+O=z

(2.22)

sx+0=sx

(2.23)

xx =sx

(2.24)

X(z+ 0) =sx

(2.25)

X +xy

=X(z

+ y)

(2.26)

xx +sy =X(Sx + y)

(2.27)

S(X +Xy) =S(x +sy)

(2.28)

S(x+Sy) =SS(x+y)

(2.29)

xx+ y=x@ + 9).

29

THE EQUATION CALCULUS

We observe that (2.21), (2.25) are defining equations of the sum function; (2.22) is derived from (2.21) by substituting Sx for x; (2.24) derives from (2.23) by substituting x+O for x (on the left hand side) as is permitted by (2.21); (2.26) derives from (2.25) by substituting Sx for s ;(2.28) is obtained from (2.25) and (2.27). Theequations (2.22), (2.24), (2.26) and (2.28) show that both Sx+y and S(x+ y) satisfy the introductory equations T(X, O)=Sx

V(G BY)=S d G Y)

so that (2.29) is a proved equation. Substituting x for y and y for x in (2.29) we obtain

+

+

2.291

sy x =S(y x)

The next step is the proof of the equation

o+x=x ;

2.3

the pair of equations 0

O+O=O

+sx =S(O +2)

which follow by substitution in (2.21) and (2.25), show that the function 0 +x satisfies the introductory equations 940)= 0

=&w

Y(S4

which are also satisfied by the identity function 4(x), proving 2.3. The four equations 2.21, 2.3, 2.25 and 2.291 show that x+y and y +x both satisfy the introductory equations 0)=5

Y(Z,

V(X9

fly) =&(x,

Y)

which completes the proof that s+ y =y+x.

We take as the next example the important equation 2.4

'

(x+y) +z=x+ (y + z )

which expresses the associative property of addition.

30

THE EQUATION CALCULUS

+y)+2) and x + (y +Sz)= x +S(y + 2) =six + (y + 2))

Since (x+y) +Sz =S((x

(where we are using an abbreviation of the form a = b = e to denote the pair of equations a=b, b = c ) it follows that both the functions (x+y)+z and x+(y+z) satisfy the equation y(x, y,Sz)=Sv(x, y, z ) ; furthermore (x+ y) + 0 = x

+y =x + (y + 0)

so that these functions also satisfy the equation

d x , Y, O)=x+y and so 2.4 is proved. 2.5

PRODUCTS

The commutative property of multiplication, which is expressed by the equation

xy =yx is readily proved with the help of equations 2.2 and 2.4. The defining equations of the product yx are 2.51

o * x = o ,(Sy)*x=(y-x)+x ;

;ts in the proof of 2.2 we start by establishing the companion equations

2.52

x.o=o, x.Sy=(x*y)+x.

The first of these is proved by the two equations 0 . 0= 0,Sx.0=x .0 (instances of the introductory equations of the product), which in conjunction with Z(O)=O,ZSx=Zx (where Z x is the zero function), show that x.o=zx=o.

For the second we observe that 0 .sy = 0 = 0. y

+0 ,

s x .sy = x .sy+sy

31

TEE EQUATION CALCULUS

sx .y +sx = (x.y +y)+Sx

and

=(X.Y+SX)+Y by =S(X*y +x) + y

+

= S ( (X*y+ X) y)

2.4

by 2.29

(x.y +x) +sy which shows that x.Sy and x.y+x both satisfy the introductory =

equations

do7 Y)=O

9tSX7Y)=9(%9)+BY7

proving 2.52. The equations 2-51, 2.52 show that the functions y - x and x y satisfy the same introductory equations, which completes the proof.

-

2.53 If for some f(x), f ( O ) = O f (x)= 0 is provable. For from

and f(Sz)=O are provable then

~(SX) = 0 and 0 .f (x)=0 we derive f(Sx)=O.f(x) and since ZSx=O-Zx therefore f(x) and Zz both satisfy the introductory equations

d o )= 0 , VtS4 = O-vtx) so that f (x)=Zx and f (x)= 0 is provable.

DIFFERENCE FUNCTION We commence by proving the important equation

2.6

PROPERTIES OF THE

x y =sx 'SY.

2.61

A

From the equations

x 2- 0=x ,sxAS0 = (sx 0) 1 =sx A 1 =x -L

and

X'Sy=

A

(xA y) '1 sx-ssy = (Sx+Sy)A I

(which are instances of the introductory equations of the difference function) it follows that X A y and Sx-Sy both satisfy the introductory equations 9@, 0) =x 80

that

Sy)= 9(z7 y)A 1 x-Ly=sx43y. Y

dX7

32

THE EQUATION CALCULUS

It follows that (z+y) y =z for (z+Sy) -By =S(y +z)-Sy = (z+y) Ly and therefore if Il(z, y) is defined explicitly by the equation Il(z, y) =z both (z+y) y and Il(z,y) satisfy 2.611

qJ(x, O ) = X

qJP(x,S Y ) = q J ( X , Y ) .

2

From the equations 0-10=0,0-Sz=(O~z)~l and ZO=O, 2 S z = Z z ~1 follows 0 - x = 0 and similarly using Z(z, y) defined explicitly by the equation Z(z, y) = 0 we may prove y (z+y) = 0, and thence y-y=O. We define the positive difference of z and y

-

)z,yl by the equation

It follows that

Iz, yJ=(z-+/)+(y--) $1, and using 2.61,

12, yI = 19,

1%

2.612

yl = pz, Syl

that

-

The only applications of the equalking rules for functions which we have made so far have been confined to single recursions. However, the following proof of the equation

+

2.62

z (y

-

2)=y

+ (z y) 2-

requires an application of the rule to a double recursion. We observe that z+ ( 0 Az)=z= o+ (zz0 ) 0

and

+ (Sy

2

0 ) =sy =Sy

+ (0

-+

Sy)

8%+ (Sy 4!3z)=Sz + (y A 2)=S(z (yA z))

+

Sy (Sx-1Sy) =Sy

+(x

A

y) =S(y

+ (z y))

so that z+ (y- 2 ) and y + (z-1y) both satisfy the doubly recursive equations Y(%J o ) = x

9

Y(O, sy)=Sy

dSzJ ='dz,

$)

33

THE EQUATION CALCULUS

which completes the proof. The common value of z+(y-z) and y+ (x-y) is the greater of x and y, denoted by max (z, y). In preparation for a theorem connecting equality with zero positive difference we prove next that for any recursive function

f (4 2.63

(1-1-15, !lI)f(4=(1'lx,

YMY)

The special case of 2.63 obtained by substituting x+y for x, namely 2.631

(lIx)f(IZ:+y)=(l'x)f(y)

is evident, since (1 2 x)f (x+ y) and (1 -x)f (y) both satisfy $40, Y)=f (Y)9 P ( f k

Y)=ZdX, Y) ;

-

substituting y'x

for x in 2.631 we find

(1 '(x2-? / ) ) f ( Y= ) (1 (x' Y ) ) f + (x'?I)) and after multiplying by 1-12, yI and using the provable equation { l ~ ( x - y ) ) { l ~ ~yI}= x , {l-lx, yl}, (see Example 2.241),

2.64

this becomes 2.65

(1

-

-

1% YIIf (?A= (1

1 x 9

(v

YI ) f ( Y+ (x'Y))

-

Substituting z for y and y for x in 2.65, since Ix,yI = Iy,21, we obtain 2.66

(1'1x7 YI )f (5)= (1

1% YI )f (X+ (Y

2))

and so 2.63 follows by 2.62. Taking f ( z )to be the identity function 9 ( x ) = z in 2.63 we have ?/I)x=(1-1x, Yl)Y and so if IF, GI = 0 is a proved equation for certain functions F , G then substituting F , G for x,y in 2.67 we see that F=G is a proved equation. Conversely from P = G we derive, by means of F-F=O, both F-G=O and G-F=O and so IF,Gl=O. 2.67

(1+,

Thus we have shown that euch of the equations IF,Gl=O, F=G may be deduced from the other.

34

THE EQUATION CALCULUS

The equation 2.63 may be expressed in another form which is often more convenient to use; if we multiply 2.63 by I ~ f ( x )and , use the equation (1

f ( x ) ) f ( x=) 0

which is obtained by substituting f ( x ) for x in the provable equation x(l-z)=O (see Example 2.1) we obtain 2.68

(1+

YI)(1+(z))f(Y)=O.*

2.7 As a first example of the use of the equivalence of the equations IF, GI = O and F=G we prove the equality of two functions f and g which are known to satisfy the equations

f (0)= d o ) > f (8x1=g(flx) q ( x )= I f ( x ), g(x)( so that

We define q(0)= 0 , pl(Sx)= 0 are proved equations, and therefore q(Sx)=Zq(x). Since Zx also satisfies the introductory equations

zo=o,

zsx=zzx

therefore q ( x )=Zx, and so ~ ( x=)0 , are proved equations, whence, by the previous theorem, f ( x ) = g ( x ) is a proved equation. A generahation of this result is given in example 2.73. To record the fact that a proof of some equation F =G may be obtained by combining the proofs of one or more equations f i =gl, f 2 = g2, ... , fk =gk, we write simply fi = 91 f2=92

.......

fk=&

F=G and call this a proof schema.

* Multiplying an equation by a factor is a shorthand way of describing the derivation of an equation ca =cb from the equation a = b ; this derivation is of course effected by substituting “b” for “a” on the right-hand side of the equation ca = m.

TRE EQUATION CALCULUS

35

Thus the foregoing theorem may be represented by the schema

f (0)= g ( O ) f (S4=g ( S 4

f (4=g ( 4 * 2.8 If for some function f ( z ) the equations f ( O ) = O , (1-f(z))f(fi4=0

are provable, then the equation f (x)= 0 is provable; in schematic form f(O)=O (l-f(4)f(W=O

f(x)=O. =0 By example 2.232, (1 a)b=b 2 ab and so from (12 f (x))f(Sx) we deduce

-

( 1 A f (x))( 1A f (Sx)) = ( 12- f (5))2. ( 1 f (z))f (Sx) = 1 f (2),

and so if g ( x ) is defined by the recursion g(0)= 1 7 g ( W =g(x)( 1

f (4)

and if h ( z )= g(Xz) then

h(Sx)=g(Sx)(1L f ( S 2 ) )= g(z)( 1 -= f (2))=g(Sx) ; but h ( O ) = g ( O ) (since f ( O ) = O ) and so h(x)=g(x) ; Thus g(O)=1, g ( S x ) = g ( s ) and so (by example 2.7304), g(z)=l, and therefore 1 f (2)= 1. Hence f(x)=f(~)(l-f(z))=O, using example 2.1. We shall show later that theorem 2.8 establishes the validity of the method of proof known as mathematical induction.

THE FUNCTIONS Z;, II, and p,. For any function f ( z ) we define

2.9

36

THE EQUATION CALCULUS

ao that for any numeral p

f (0)+f ( 1 )+ ... + f (PI m P ) =f (0).f ( 1 )....-f(P) ; z;(P)

and

=

for a function f (2, a ) depending also upon a parameter a we write

‘ q o , a )= f ( O 7

a)

z;(n+1, a)=Z;(n, a ) i - f ( n + 1, a ) with an analogous formulation for the ease of more than one parameter. We take for granted the companion definitions for the function 17. For any assigned numeral p we have

Zf(P, a)=f(O, a ) + f ( l ,a ) t ...+f ( P , a )

17,(p, a ) = / ( ( ) ,a)*f(l, a ) ..... f ( P , a ) .

and

I n using the functions Z;, ITt we take the first rariatble in f to be the operative variable; to study such a sum as

/ ( a , O ) + f ( % I ) + * . * +/(a7 P) we introduce ~ ( xa ), by the explicit definition ~ ( xa ),= / ( a , z) and express #(a,O ) + f ( a , I ) + ... + / ( a , p) as Xp(p,a). 2.91

If for some functions f ( x ) ,g(z) the equation ( 1 -g(n))(SnLa)f(a)=O. . . . . . . . . . . . . . . . . .

(i)

is provable, then (1

g(n))Z,(n)= 0 . .......................

(N

is provable. Denote (Sn-m)(l-..g(n))Z,(m)by A(m) ; then by (i) A(O)=O, and A (Sm)= (Sn Sm)(1 g ( n ) ){Z,(m) f (Ism)).

-

-

=(Sn-Sm)(l -q(n))Zt(m) =A ( m )

(1 g(n))C,(m)

+

37

THE EQUATION CALCULUS

80 that (1 : A ( m ) } A(Sm)= 0 , and therefore by 2.8, A(m)=O. In particular A(n)=O, that is, {1~g(n)}L',(n)=O.

2.92 If for some f(s), g(x) (Sn2- a)(1A f ( a ) )g(n)= 0

then {l-~-n,(n)> g(n)=O. Let A(m), B(m),G(m) denote respectively (Sn:m)(l I n , ( m ) ) g ( n )(Sn-Sm)(l , -n,(m))g(n)

and (Xn-Sm)(l f (Xrn)) g(n) . By hypothesis C(m)= 0 and so (1 IA ( m ) }C ( m )= 0 ; moreover { l - A ( m ) ) B(m)=(1 -I.A(m)){A(m)-(14f(m))g(n)}=0 and so (by example 2.47, with n , ( m ) for a, f(Sm) for b and (Xn-~-Xm)g(n) for c), (1 A(rn)}A(Srn)= 0,

and therefore A(m)=O , whence A(n)=O, that is (1 4 7 , ( n ) }g(n)= 0 , (1If (x)}IT,@) = 0

2.921

For and

.

{l-l.f(O))n~(o)={l-f(O)}f(O)=O

{ 1 2-f (XZ)}n,(Xx) = { 1If (Xx)} f (Sx) .IT,(x)= 0.

2.93 In the following section we require a number of the elementary properties of the function a(x)= 1A (12 x) which are proved in Examples 2.26. I n particular we shall use the equations

a(lLz)=l'a(X),

oI(Ly(z))=Ly.(z),

Z*a(Z)=X

and

a(xy) = .(z) * a(y) . The function a(%)obviously satisfies the equations Oc(O)=O, a(Sz)=l .

For some given f ( x ) we define e(z)=Oc(f(x)). It follows that a(&))

and

a( 1

-

= ..(a(f(z))) =

..(f(4) =e ( 4

e(z))= 1 2 &(@(Z))= 1 2 @(X) .

2.932

For 2.933

f ( 4 . l e ( z ) 1.1= O .

) 1)= 0 ; but For .(z) 1.1 = 0 and so e(x) 1 = 0 whence f ( x ) ( e ( z 2 f ( x ) ( l:e(s))=O and so, by addition, f(z).(e(z),11=O. 2.934

For

I f e ( 4= .(f(4)

then

r;r(4=4q4).

a n , ( s x )= anf($). af(8z)

@(sx)

= (xU,(S).

and

n e ( 8 x )= n e ( x ) .e('s)

whilst

&lIf(0)= W f (0)= e(0)= De(0)

so that I l e ( x ) and a n , (x)satisfy the same introductory equations. 2.935

For

If a(f(x)).g(x)=Othen f ( z ) . g ( s ) = O .

-

0 =f ( 4 * 4 f W )* 9 ( 4=f (4* g ( 4

39

THE EQUATION CALCULUS

-

We note the following immediate consequences of the definitions: (1 O,(n))Pr(Sn)= (1

e , w Pf(4

9

W n )P , P 4 =%(n)*fin

(since e,(n).B,(n)=ornf(n).(lLe(Sn))-aD,(n)-(l =-e(Sn))

-

.

= arlirf(n) (1 -L e(Sn))=e,(n))

If for some a, ne(u)=1 and e(flu)=O then O,(u)= 1 and so p,(Su)=Su; if either ITe@)= O or e(8u)= 1

-

then p,(Su)=pf(u).

2.941 I / R , ( ~=) 1 e,(z) f ( 0 ) (1-n,(Sz))nRI(z)=o.

then

- - - -ne( n,(flz))

Denote e(0) (1 -lIT,(Sx))nB,(z)by A(%),then:

A(0)=e(o) (1 2 Do(1)) (1 e,(O)) =e(0) (1 e(0) (1

1)) = 0, and

A ( S z )= e ( O ) (1 -ne(SSz))nq(z){l (IIe(Sz)-nB(SS2))} =e(0) = (1

(1

n,(Sflz))nx,(z) (1

-n,(Ssz))A(z)

.

so that (1 -A(x))A(Sz)= 0 and therefore A($)= O Since e(0) (1 +Z7,(Sz))=oIf(O)(1 ~aZI,(Sz))=a(f(O) (1 -ZI,(Sz))), 2.941 now follows by 2.935. 2.942 Since and

-e , w

p,(z) 2- 2 =0.

(1

e,(z)

cu,(fl4 -1u,(41= 0

{p,(sx) -xz} = o

therefore, by example 2.36, (lUi(fl42flz) ( c l l ( f l 4 ' P , ( 4 ) = 0 ; hence, by example 2.422, {Sz-p,(z)} {p,(IYz)432)= 0

.

Since (by example 2.41)

flx -p,(z)

+ {1

= {z' p , ( z ) }

A

(p,(2)-2))

41

42

THE EQUATION CALCULUS

Theorem 2.9471 follows from 2.947 by means of example 2.741. (PAS4 I.I

2.948

f (Pr(fiX + r ) )= 0

*

From 2.946 and 2.9471 we obtain (as in example 2.36) {pr(SZ)2 x} I,U,(XZ

+ r ) , pr(SZ)1= 0 . . . . . . . . . . . 0)

and so, by example 2.472, (1

IPct(~4> 1 I~d4S x + 4, P f ( S 4 = 0

and similarly from 2.947 {I IPf(S4,SZl} f (SZ) = 0 ;

but (1

I P f ( W ,SZl} f ( P f ( W )= { 1 IPr(S49XZl t f (84

and so I P , ( f w , 84 1f (PCf(S4)= 0

(1

which, by example 2.481, transforms into {pt(SZ)'z}f(Pu,(Sx))=O.. . . . . . . . . . . . . . .

By (i), (ii) and theorem 2.63 we obtain 2.948, which may also be written in the form (1 2.949

Ilu,(S4,W } f (P,(flZ + r))= 0 {1-1T,(x)}f(Pu,(4)=0

Since aOt(x)= an,(^) .a( 1 -L pS'x) =I7,(x) ( 1 therefore 1IR,(z)= O,(x) and O,(z).

esz)= Oj(z)

= O,(z)

and so (1 ~ R , ( z Ip,(Sx), )} Sxl= 0 ; hence by the last theorem

-

(1 I R , ( 4 1 f ( P r ( S Z + ~ ) ) = O

It follows from example 2.711 that (SZ Ia ) ( 1 2- R,(a))f(p,Sz) = 0

and therefore by theorem 2.92 (1

17n,(x))f(Pui(X4) = 0

;

43

THE EQUATION CALCULUS

hence by theorem 2.941 (and example 2.36)

. . . . . .. . . . .. . (9

f (0) (1 2- IIt(8z)) f(pr(8z)) =0. .

Since, by 2.63,

- - - - - - - - - (W

(1 -rU,(r))f(pt(r)) = (1 -I.t(r))f(O)

*

*

and by 2.946 (taking z=O),{ l i f ( 0 ) } p t ( ~ ) = O , we have, on multiplying (ii) by l - f ( O ) , (1 - f ( O ) )

f(FC,(T))=O

= 0, which added to (i) yields and so ( 1 A f (0))(1 IIt(Sz))f(p,(Sz))

Dj(84)f ( P f ( f i 4 )= 0 -

(1

Finallywe observe that (1 2 - I I f ( 0 ) ) f ( p f ( 0 ) ) = ( 1 - - f ( 0f)()0 ) = 0 which completes the proof. f ( 0 ) (1 - f ( 4 ) (1 LPr(z))=o

2.941

For (1 A f(z))II,(z) = 0 and so, by the previous theorem, (1-f(4)f(Pt(X))=O ;

however (1 ' P t ( 4 ) f(p,(z))=( 1 -PA")) f ( 0 )

and so 2.9491 follows on multiplication by l - f ( z ) . From 2.9491 and 2.944 we deduce, by example 2.462,

@ a i 4 ( l z f ( 4 ) f ( O ) (l-lu,(a))=O and so, by 3.92, we have f ( 0 )(l'flt(4)

2.9492

(1-p,(a))=O *

We prove next (1 -rc4> (rU,(44=0

2.9493

*

First we prove that (l-f(a)) {pr(a+n)-a}=O. Denote (1 ~ - f ( a ) ) p , ( a + nby ) g(n); then since, by example 2.74, { l ~ f ( a ) ) B , ( a - t - n ) = Owe , have g(Sn)=g(n) and so g ( n ) = g ( O ) that is (1

/(.)I

Pr("

+4= (1 f (4>P l ( 4 A

*

44

TIIE EQUATION CALCULUS

But p,(a) A a = 0 and therefore { 1 2 f ( a ) }{/.,(a + n)-a} may be expressed in the form ( S X L U ) ( l L f ( U ) ){pt(x)-a)=O..

=0

which

. . . . . . . . . . . 0)

However, by 2.944 and 2.942 (#a-2) ( p U f ( z ) - p , ( a ) ) = 0p,(a)-a=O , and so (&a-2) (p,(z)-a)=O

................... (ii) ;

multiplying (ii) by l - f ( a ) and adding (i) we find, since

(Sx a ) + (Sa2.z) = 1 + { 1 (a2 z)} + ( a

-+ z)

(z

#a)

(by examples 2.245 and 2.41), ( l - f ( a ) )(p,(z)-a)=O

*

We shall show in the next chapter that theorems 2.945, 2.949 and 2.9493 prove that when f ( z ) is zero for some x from 0 to n then p U f ( n ) is the least value of x, between 0 and n, for which f (x)is zero, but if none of f ( o ) , / ( I ) , ... , f(n) is zero then put(n)=O. 2.95 FURTHER PROPERTIES OF THE FUNCTION ,u, 2.951

and then

If I , ( O ) = O A,(Sn)=A,(n)tSn-{l-(A,(n)+f(Sn))]

A n )= e(0)' 4 ( n )

In the following proof, to simplify the typography, we shall drop the suffix and write simply A(n)and p ( n ) for A,(n)and p,(n). The proof turns on the following properties of A(x), obtained from the defining equations by multiplication by an appropiate factor. e(Sn).,l(Sn)=e(Sn).A(n)............................ (i) I ( n )-A(Sn)=A(n).A(n) .............................. (ii) {l-A(n)){1~e(8n)}A(~Sn)={l-A(n)) ( l - ~ ( S n ) } S n ... . .(iii)

It is readily seen that (apart from the introduction of the factor

THE EQUATION CALCULUS

45

e(0)) these equations still hold if we replace ;Iby p. The equations obtained from (i) and (ii) by replacing 4, by ,u are proved by means of the relations e(8n).O(n)= 0 and p ( n ).l7(n)= 0 previously established; instead of (iii) we have

e(O){l

e ( 8 4 ) (1 :l.(n)) p(8n) =@(o)(l:@(Sn))(1 1 p ( ? &n(n)-xn )) = p ( o ) { l ~ ~ ( X n {l.-p(n)).Xn, )} sincen(n) (1 ~ p ( n ) ) = I T ( n ) .

. . . .(iv)

. .(v).

2.96 Inequalities 2.961

We define the inequazity a>b to stand for the equation a = b + (a- b)

and the inequality a
a = b A ( b ~ a; ) The inequality a> b is read as ‘a is greater than or equal to b’ and a g b as ‘a is less than or equal to by. We introduce also two further inequalities

a< b ( a is less than b), a> b (a is greater than b ) which are defined to stand for Sa Q b, a >Sb respectively. It follows from example 2.5 that the inequality a g b may be derived from the equation a-b=O, and conversely, and that similarly the inequalities a g b , b>a (and therefore also the inequalities aa) may each be derived from the other. The following simple inequalities derive immediately from the definitions and the properties of the difference function :

a+x>a, a>a-x, a+Sx>a, S a > a - x .

T1XE EQUATION CALCULUS

2.962 2.9621

47

The inequality relations are transitive, that is to say a > c follows from a > b and b > c

2.9622 a > c follows from a> b and b > c

.

Since a < b is equivalent to b >a (that is, each is derivable from the other) the transitivity of the relations Q and < follows from 2.9621 and 2.9622. The proof of 2.9621 is contained in example 2.461 ; we observe next that since ( y ~ S x ) (yiz) = 2 1, therefore Sx>y follows from x>y and therefore 2.9622 follows from 2.9621. In fact 2.9622 is a consequence of 2.963

a > c follows from a>b and b>c

which follows from 2.9621 and 2.964

if b > c then Sb>Sc

which is proved by observing that from b=C-/-(b-I-C) we derive

Xb =Sc + ( b Ic )

and so

Sb =SC+ (Sb2SC)

The inequalities a > b, a -+ c >b + c may each be derived from the other, and so too may each of the inequalities a> b, a.Sc> b-Sc. For ( b -tc) I( a+ c) = b Ia, b .Sc 2 a .Sc = ( bA a )-Sc and ( b1a)Sc= 0 if and only if b ~ a - 0 , since ( b A a ) S c = ( b I a ) c + ( b . - a ) . 2.97

2.98 Recursive arithmetic as it has been constituted in 8 2.1 is demonstrably free from contradiction, in the sense that we can show that if p = q is a provable equation, where p and q are definite numerals, then p and q are the same numeral. In other words it is impossible to prove the equation O = 1 in recursive arithmetic. If we say that an equation F =G is verifiable only if F and G are the same numeral or the substitution of numerals for the variables in F and G always reduces F and G to the same numeral, then we may express freedom from contradiction by saying that only verifiable eqzmtions are provable. As we observed in 5 1.9, when the variables in its argument

48

THE EQUATION CALCULUS

places are replaced by numerals, the sign of a recursive function is totally eliminable, that is to say, for any recursive function f (xl, xz,... , xp) of p variables, and for any set of p numerals N,, N,, ... , N,, there is one and only one numeral V such that the equation

f Wl, NZ,* . Np)= V 4 ,

is provable. This is obviously true of the initial functions Z ( x ) , S(x),9 ( x )and it is a property which is preserved under substitution; for example if f ( u ,w), g(x,y), h(x, y) are eliminable then, for any given set of numerals M , N there are unique numerals, U , V, W such that the equations

q ( M , N ) = u , h ( M ,N ) = v

w

and f ( U ,V)= are provable, and therefore, for the function

dz,Y)= f ( g ( z , Y), h(x,Y)) defined from f, q and h by substitution, the equation

q ( N ,N ) =

w

is provable for one, and only one, W corresponding to the given

pair M , N . Next we remark that the property persists under recursion. If f (a, n), (with any number of parameters) is introduced by the single recursion

/ ( a ,O ) = d a )

/(a, n + l ) = B ( a , n, / ( a , n ) ) where ac and 9, are eliminable functions, then f is eliminable; for corresponding to any set of numerals A , N there are nunaerals Vl, V, ... , V, such that me can prove in turn &(A)= yo, @(A,0,VO)= Vl, @ ( A ,1, V,)

= V,,

...

3

B(A, N - 1, V,-J= v, and / ( A ,0 ) = V,,, / ( A , 1 ) = V,, ... , / ( A ,N ) = V , .

If f(a, m, n ) is doubly recursive then for any assigned set of

THE EQUATION CALCULUS

49

numerals A , M ,/ ( A ,M , n ) is single recursive and so to any chosen N corresponds a unique V so that f ( A , M , N ) = V is provable. A similar argument may be applied to show that a k-ply recursive function is eliminable (for each value of k). Since the substitution of a numeral for x in the equation x=x yields a verifiable equation, to complete the demonstration that only verifiable equations are provable, we have to show that the lawful transformations in a proof yield only verifiable equations from verifiable equations. There are just two cases to be considered. The first of these is the substitution of G for F in some equation containing F , when F=G is a verifiable equation. If F and G are just numerals then since by hypothesis F = G is verifiable, therefore F and G are the same numeral and the substitution leaves the equation unchanged; if F and G are functions then, by hypothesis, the result of substituting numerals for the variables in F and G is to transform both F and G to the same numeral v , say, and so if we substitute G for F in some verifiable equation and then substitute numerals for the variables in the equation, the places where G replaced F are filled by the numeral v just as if the substitution had not occurred. We come finally to the case of an equation F=G proved by showing that F and G satisfy the same introductory equations; let us denote by ip the function whose introductory equations both F and G satisfy. If ip is singly recursive and if its introductory equations are

then, by hypothesis, the equations

are all verifiable. Hence, if for some set of numerals A , N the values of F ( A , 0 ) , F(A, l), ... , F ( A , N ) and G ( A ,0 ) ,G ( A ,I), ... ,G(A,N ) are V,, V,, ... , V , and W,, W,, ... , W,

60

THE EQUATION CALCULUS

respectively, then V, and W , each equal &(A)whence

V,=B(A, 1, V,)=B(A, 1,Wo)= Wl V,=B(A, 2, V,)=B(A, 2, W J =

wz

and so on, up to V N = W , which shows that the equation F=G is verifiable. If tp is doubly recursive, and if its introductory equations are

~ ( a0,, n)=y(a, p + L 0 ) = 1 q(a, p+ 1 , n+ 1) = c (a, p , n, q7@, p , d(a, P, n, d a , P+ 1, n ) ) )tp@, , P+ 1, m ) ) , then P ( A , 0, N ) = G ( A , 0, N ) for any numeral N , and

P(A7 1 , 0 ) = G ( A , 1,o)=vio, say, d(A, 070,~1o)=dm,say, l)=c(A, O , O , vO> dm, wlo)=G(A,

P(A7

'>

1)=v117

say, d(A7 0,1, .,,)=do,,

P(A, 2 ) = c ( A , O, v07 dO1, v l l ) = G ( A 1 2, and so on, up to P ( A , l , N ) = G ( A , l,N)=v,, say, for any numeral N . Then P ( A , 2, 0) =G(A, 2, 0)=vzo, say, &(A,1, 0, vm)=dlo, say, 17

'7

' 7

P ( A , 2 , 1)=G(A, 2, 1)=c(A, 1, 0,V l , dl,, and so on up to P ( A , 2, N )= G (A , 2, N ) .

VZO)

Thus step by step we reach

P ( A , P,N ) =G(A, P,N ) for any set of numerals ( A , P , N ) and so F = G is verifiable. A similar argument applies if tp is recursive of higher order. This completes the proof that every provable equation is verifiable. We shall show subsequently that the converse does not hold, since there are verifiable equations which are not provable.

Examples II 2.

If !(O,y)=y and !(Sx,y)=!(x,y) prove that !(x,y)=y. Prove the equations 2.01- 2.251

2.01

a(b+c)=ab+ac

2.02

(ab)c=a(bc)

2.03

(ab)·(cd)=(ac)·(bd)

2.1

x(l-=-x)=O

2.2

1 -=- x = 0'"

2.201

(l-=-x) + {I-=-(l-=-x)}= 1

2.21

a-=-(b+c)=(a-=-b)-=-c

2.22

(a-=-b)-=-c=(a-=-c)-=-b

2.23

(a+x)-=-(b+x)=a-=-b

2.231

a(x-=-I)=ax-=-a

2.232

a(b .r: c) =ab -=- ac

2.233

x-=- x 2 = 0

2.234

(l-=-x)(I-=-x)=I-=-x

2.24

(1-=-la, bl) (b-=-a)=O

2.241

(1-=-la, bl) (1-=-(b-=-a»=I-=-la, bl

2.242

(b-=-a) (Sa-=-b) = 0

2.243

{I-=- (Sa-=-b)} {I-=-(b-=-a)} = 0

2.244

(Sa-=-b) (Sb-=-a)!a, bl =0

2.2441 {1-=-(Sa-=-b)}+ {1-=-la, bl}= 1--'-(a--'-b) 2.245

(b-=-a) + (Sa-=-b) = 1 + (a-=- b) + (b --'-Sa)

2.246

Sa-=-b= (a--'-b) + {1--'- (b-=-a)}

2.25

1-=-(p+q)=(I--'-p) (I--'-q)

2.251

{l--'-- (q+r)}+(I-=-q)r= (l-=-q)+ (l-=-q) (r-=-I)

2.26

If cx(x) = I-=- (I-=- x) prove that iX(X) -'- 1 = 0 , 51

52

THE EQUATION CALCULUS

IX(X) + {1-'-IX(x)}= 1, .x(x)...:..x=O, X'IX(X)=X, 1X(1"':" x) = 1"':".x(x) , 1X(.x(X)) = IX(X) , 1-'-IX(x) = 1...:.. x , IX(X) 'IX(X) = .x(x) , .x(xy) = .x(x) . lX(y) , and that 1X(f H1 = 0 follows from t- g = O.

2.261

If .x(x, y) = .x(lx, yi) prove that

{I...:.. .x(x, y)}x = {I...:...x(x, y)}y 2.27 2.271

and {b-'-b·.x(c, b)}+c . .x(c, b)=c. Prove the equations 2.271- 2.274 (I...:..c)-'- (l"':"ac)=O

2.272

(I...:..ac) -'-(1...:.. c) = (1...:.. a)lX(c)

2.273

(1 ...:.. (1 ...:.. a )b) (1 -'-ac)b = (1 ...:.. (I

2.274

(1"':"(I...:..ab)c) (l-'-a)c=O.

2.28

Prove the index laws 2.281- 2.283

.s:

a)b) (1 ...:.. c )b

2.281 2.282

(xm)n = x mn

2.283

(x·yt=xn.yn

2.29

Prove that 3.2=6, 4.2=8, 3.7=21,4.7=28 and 34.27=918

2.3

Establish the following proof schemata

2.31

l+g=O

2.32 2.33 2.331

1=0 {z+(I-'-z)}/=O

1=0 {1-'- (g -'- f)}h=O (SI-'-g)h=O (SI...:..g)h=O

{I"':" (g-'- f)}h=O

2.332

(f...:..g)h=O (f-'-Sg)h=O

2.34

{l--'- (SI .s: g)}h=O {I"':" II, gi}h=O {I"':" (I -'- g)}h=O I.or-(g) = 0

2.341

(1"':" 1)( 1-'-IX(g» =0 lX(g) = 1-'- f

THE EQUATION CALCULUS

2.35

fg=fh fig, hl=o

2.351

Pg=o jg=o

2.36

jg=o (l...:...g)h=O jh=O (1"':'" jHg+( 1...:... h) (1"':'" g).p}=O

2.37

h=O (1"':'" j).p=O

2.4

Prove that

2.41

(x+y) "':"'Z= (x...:...z) +

2.42

Sx-: (x...:... a) =S(x...:... (x...:... a»)

2.421

a...:... (a...:... b) = b...:... (b...:...a)

2.422

(a"':'" b) (a"':'" c) =0 (a"':'" b) (Sb...:...c)=O

2.423

(a"':'" b) (Sc...:...b)+(a...:...c)=O a...:...b=O

2.43

Ib, a+(b...:...a)1 =a...:...b

2.44

la, b...:... (b...:...a)1 = a-: b

2.45

b...:...a=O a...:...b=c b+c=a

2.451

plb,cl=O pia, bl =pla, c]

2.452

plb,cl=O {1...:...pla,bl}pla,cl-O

2.453

pla,bl +pla,cl=O plb,cl=O

2.46

a...:...b=O (c -'- b) (Sa"':'" c) =0

2.461

b...:...c=O {l...:...(a"':"'b)} (a...:...c)=O

2.462

p{(a...:... b) +(b"':'" c)}=O p(a...:...c)=O

{y...:... (z...:...x)}

53

54

THE EQUATION CALCULUS

HI-'- a) +(1-'- b)}c=O 2.47

(l-'-ab)c=O

(l-'-ab)c=O 2.4701 {(I-'-a)+(I-'-b)}c=O 2.471 2.472

(l-'-a)bc=O (l-=-ab)bc=O (Sa-=-b)c=O

{l-=-la,bj}C=O 2.473

1-=-(a-=-b)=O a=b+(a-=-b)

2.48

{1-=-(I-'-x)y} (l-'-xz)yz=O

2.4801 {I-=- (l-'-ab)c} {(l-'-a) + (l-'-b)}c=O

Dc

2.481

(I -=-Ia, b = 0 (a-'-b)c=O (Sa-=-b)c=O

2.49

{I -'-(1 -'- a )b} (1 -'-a) = (1 -'-a) (1 -'- b)

2.491

.,..,-----,-.,-,...,..,c..:(1:..--'-:..-d)c= 0

(1-'- a) (1-'- b)c =0

{I-=- (1-'- a)b}( 1-'- ad)c=O

2.5

Prove that each of the equations (x) b=a+(b-=-a), ({J) a=b-'-(b-=-a), (y) a-'-b=O follows from any other

2.6

Prove the equation {I -'- (a -'- b)} + {I -'- (Sb .z: a)} = 1

2.7

Establish the schemata 2.701-2.72

2.701

f(a,a+c)=O

{I-=- (a -'- b)} f(a,b)=O

2.71

f(a,a+c)=O (Sb-=-a)j(a,b)=O

2.711

j(b+Sc,b)=O (a-'- b)j(a,b)=O

2.72

j(a,a+c)=O j(a+Sc,a)=O f(a,b)=O

THE EQUATION CALCULUS

55

2.7201 Prove the equations a-'-(a-'-(a-'-b» =a-'-b {1-'-(b-'-a)} {b-'-(a-'-(a-'-b»}=O

2.73

Prove that, for any definite numeral p

f(p-'-r)=O 2.7301 f(p+Sr)=O f(x)=O 2.7302

f(O) +f(l)+ ... +f(p)=O f(p+Sr)=O f(x) =0

f(p -'- Sr)=O f(p)=O 2.7303 f(p+Sr)=O f(x)=o f(O)=p 2.7304 f(Sx)=f(x) f(x)=p

2.74

If tp(x, 0)=/(8x) and tp(x, Br)=tp(x, r)'/(x+8Br) prove that IIj(x+Br)=IIj(x)·tp(x, r)

and deduce that {1-'- IIj(x)} IIj(x+ r) = O.

2.741

Prove that {1-'- 1(x)} IIj(x) = 0

2.8

With the notation of § 1.6 prove

2.81

Alt x·Alt Bx=O , Alt x+Alt Bx= 1 Alt 2x = 0 , Alt (2x + 1) = 1 .x(Alt x)=Alt x

2.82

Hf(2x)=Hf(2x+ 1)=x

2.83

Bx -'-(BRt

2.9

If the function n! is defined by the equations O! = 1 , (Bn)! = (n!)Bn prove that l-'-n!=O and (Bn)!=IIs(n).

2.91

X)2 =

0 , (Rt

X)2 -'-

x=0

Prove the equation {1-'-(I-'-a) (l-'-b)) (l-'-ac) {1-'-(b+(I-'-c»}=O.

CHAPTER III THE LOGICAL CONSTANTS 3. If a and b are natural numbers and if [a, b] = 0 is a provable equation then we call the equation a = b a true proposition, and if the inequality Ia, b] > 0 is provable then the equation a = b is called a false proposition. Since the equations a = b, [a, b] = 0 may each be derived from the other it follows that a true proposition is a provable equation and conversely. Moreover, as lx, yl is a recursive function, for any natural numbers a, b there is a unique natural number c such that

[a, bl=c is provable; if this number c is zero then a = b is a true proposition, and if c is not zero then 1 -"-I a, b] = 0 is provable, so that a = b is a false proposition. Accordingly, any proposition is necessarily either true or false, and no proposition is both true and false. 3.1 We call tX(la, hi) the number of the proposition a=h, so that a true proposition has the number zero and the number of a false proposition is unity. Conversely, if tX(ja, bl) = 0 then [a, hi = 0 so that the proposition a = b is true, and if tX(la, hi) = 1 then [a, hi> 0 and a = b is false, and therefore the number of a proposition is zero if and only if the proposition is true and the number is unity if and only if the proposition is false. 3.11 The proposition l-"-la, hi = 0 is called the negation of the proposition a = h. Since tX(l-"-lx, YI)= l-"-lx,

yl

it follows that if the proposition a = h is true, so that [a, hi = 0, then the number of the negation of a = b is unity, and if a = b is false then the number of its negation is zero. Thus the negation 56

57

THE LOGICAL CONSTANTS

of a true proposition is a false one, and the negation of one that is false is a true proposition. 3.12 We denote propositions by single letters p, q, r with or without numerical subscripts. The negation of the proposition p is denoted by """ p, which is read 'not-p'. When the context renders the usage free from ambiguity we shall write simply 'p' for 'p is true' and <; p' for 'p is false'. 3.13 If p, q denote the propositions a = b, C = d respectively then the proposition

[a, bl+Ie, dl =0 is denoted by p & q which is read 'p and q', the proposition

[a, bl·le, d] =0 is denoted by p v q which is read 'p or q', and

(I

-'-I a, bl)·1 c, d] =

0

is denoted by p --+ q read 'p implies q', p --+ q is thus the same proposition as ,....., p v q. Finally, we denote the equation

(l-'-Ia, bj)·le, d] +(l-'-Ic, dj)·la, hi =0 by p ~ q which is read 'p is equivalent to q', so that p the same proposition as (p --+ q) & (q --+ pl. We observe that, since Ila, b], 01 = [a, b], therefore

~

q is

(a= b)~(la, b] =0)

and since and therefore

(l-'-x)(X(x)=x(I-'-(X(x))=O

(X((X(x))= (X(x) (x=O)~((X(x)=O)

(z »

O)~((X(x) =

I).

We shall justify the suggested readings of the signs and ~ in the next section.

&,

v, ,....." --+

58

THE LOGICAL CONSTANTS

3.14 From the equation [a, b] + [c, d] =0 it follows that [a, b] =0 and [c, dl =0, and conversely, and therefore p & q is true, if, and only if, p and q are both true. 3.141 The product [a, bj-]c, d] vanishes if either [a, b] = 0 or [e, dl =0; and if both [a, b] >0 and [c, d] >0 then [a, bl·le, d] >0, and so: p v q is true if and only if either p is true or q is true. 3.142 It follows that p -+ q is true if and only if either p is false or q is true, which we take to be the sense of the expression p implies q. 3.15 The definition of equivalence makes p and q equivalent if and only if p and q are both true or both false. From the proposition p +)- q and a proof of q we derive a proof of p, for if {1~la, blHe, dl+{I~le, dl}·la,

bl=O

[c, dl=O

and

are proved equations, then [a, b] =0 is a proved equation; accordingly, to prove some proposition p it suffices to prove a proposition equivalent to p. 3.16 The signs &, v, ,...." -+ and +)- are known as logical constants; their introduction effects a considerable economy in proof technique by revealing a variety of easily recognisable patterns of procedure. 3.2

PROPERTIES OF THE LOGICAL CONSTANTS

(1 ~x)Y= 0 (1 ~y)z=O

(1 ~x)z=O

and therefore, if p, q, r are any propositions ~ =~, bl = b2 , el =~, say, taking lal , ~I for x, IbI> b2 1for y and I~, <;1 for z respectively, we obtain the proof schema

p-+q q-+r ---_ .. p-+r

-

THE LOGICAL CONSTANTS

59

which is summed up by saying that '-)0' is a transitive relation. Similarly from the schema

x=o (l-"-x)y=O y=O

we derive the fundamental schema for implication (known as modus ponens)

3.22 Since addition and multiplication are commutative the following proof schemata are valid: p&q

pvq

»->«

q&p

qvp

q-
--, --, --. These show that '&', 'v' and '-
Since f=O and g=O follow from !+g=O, and conversely,

we have

p&q

--, P

and

p q p&q

It follows that '-
3.3 We consider next some of the important relations which hold between the logical constants. Since I .z: {I -"- (I -"- x)} = I .z: x it follows that 3.31 From the equation l-"-(x+y)=(l-"-x) (l-"-y) we deduce

60

THE LOGICAL CONSTANTS

3.32

"-' (p & q) ~ "-' P v "-' q

and from 1-'- xy = iX«1 -'- x) + (1-'- y)) follows 3.321 Furthermore, since x(y+z)=xy+xz, P v (q & r) is the same proposition as (p v q) 3.33

P v (q

& (p v

r) so that

& r) ~ (p v q) & (p v r)

and since {1-'-(x+yz)} (x+y) (x+z)={I-'-(x+y) (x+z)} (X+yz) = 0 therefore 3.331

P

&

(q v r)

~

(p

&

q) v (p

&

r)

For equivalent propositions we can prove interchangeability; this is given by following theorem: 3.34

If PI ~ P2 and ql ~ q2 then

We observe first that since the suffixes 1,2 are interchangeable in the given conditions PI ~ P2 , ql ~ q2 therefore it suffices to prove the foregoing propositions with implication in place of the equivalence relation. Since P -+ q is the same proposition as "-' P v q, the truth of the fourth proposition follows from that of the first and third, and the fifth then follows by means of the second. That "-' P2 -+ "-' PI follows from PI -+ P2 is shown by the equation

{I .s: (1 -'-x2 )} (1 -'-Xl) = (1 -'-Xl)-'-{(1 ..=... Xl) .z., x2 ( 1 .s: Xl)} by means of which {I -'-(1 -'-x 2 ) } (1 .z; Xl) = 0 follows from (I-'-X I )X 2=O.

Since "-' "-' r 3.341

+->-

r we deduce from 3.321 that Pvq

~

"-' ("-' P

& "-'

q)

so that the truth of the third proposition follows from that of the first and second.

THE LOGICAL CONSTANTS

61

It remains only to prove the schema PI -+ P2 gl -+ g2 PI & gl -+ P2 & g2

and this follows from the equation {I-'- (Xl + YI)}(X2+ Y2) = {(I-'-XI)X2 -'- YIX2}+ {(l-'- YI)Y2 -'-XIY2}

by means of which we derive {l-'-(xt +YI)}(X2+Y2)=O

from the equations (I-'-XI)X2=0, (I-'-Yl)Y2=0. 3.342 The importance of the results contained in theorem 3.34 lies in the fact that it allows us to replace any proposition, in an expression made up of logical constants, by an equivalent; the resulting expression will be equivalent to the original one and so by 3.15 a proof of the transformed expression suffices to prove the original. In particular, since the propositions Ia, hi = 0 and a = h are equivalent, we may without loss of generality suppose any proposition to have the form c=O. 3.343 As an example of the use of this principle we consider in detail a proof of the proposition

P:

(p -+ q) -+ {(1' V p) -+ (r v g)}

s', r' are equivalent to P, g, r respectively then p' -+ g' , r' v p' , r' v q' are equivalent to

If p',

p

-'>-

pi:

g ,r

v

P ,r

v

(pi

g respectively and so P is equivalent to -'>-

g/) -+ {(r' v pi)

-'>-

(r ' v q')}

Whatever the propositions p, g, r we may take pi, s'. r' to have the forms a = 0, b = 0, c = 0 respectively, and so P is equivalent to P*:

{I-'-(I-'-a)h} (l-'-ca)cb=O

where a, b, c are certain numerals, which may be derived from the provable equation (example 2.48)

G:

{I -'- (1 .s: x)y} (1 -'- zx)zy = 0

62

THE LOGICAL CONSTANTS

(where x, y, z are numeral variables) by substituting a, b, and c for z, y and z respectively. If we allow the letters p, q, r to have the dual roles of names for propositions and numeral variables then we may write G directly in the form {l--=-(l--=-p)q} (l--=-rp)rq=O; that is to say, we may formulate an equivalent of P simply by writing (l--=-p)q for p -7 q, rp for r vp and so on (though of course we change the significance of the letters when we make the transcription). 3.35

From the provable equations x(l--=-x)=O,l--=-{x+(l--=-x)}=O

we derive the truth of the propositions p v ~ p and ~ (p & ~ p)

which are known respectively as the principle of excluded middle (or tertium non datur) and the principle of noncontradiction. 3.4 PROPOSITIONAL FUNCTIONS If X is a variable and f(x), g(x) are two given recursive functions then the equation f(x) =;:g(x)

is called a propositionaljunction; if for some value a of x, f(a)=g(a) is provable then the propositional function is said to be true for the value a; if If(a), g(a)1 >0 is demonstrable then the propositional function is said to be false for the value a. More precisely, the equation f (x) = g(x) is called a one-variable propositional function, and f(x, y)=g(x, y) is a two-variable propositional function, and so on. Propositional functions are denoted by p(x), p(x, y) etc., according to the number of variables. 3.41 As in the case of propositions, if p(x), q(x) denote the propositional functions F(x)=f(x) , G(x)=g(x)

THE LOGICAL CONSTANTS

63

then we denote 1-'-IF(x), f(x)1 = 0 by '"'" p(x), IF(x), f(x)1

+ IG(y),

g(y)1 = 0 by p(x) .. q(y),

IF(x), f(x)j·IG(y), g(y)j = 0 by p(x) v q(y), r-..J

and

p(x)

V

q(y) by p(x) -+ q(y)

{p(x) -+ q(y)} .. {q(y) -+ p(x)} by p(x) ~ q(y).

3.42 The relations between the logical constants which we have established for propositions hold also for propositional functions, the proofs proceeding on the same lines; for instance we derive p(x) v'"'" p(x)

from y(l-'-y)=O by taking IF(x), !(x)1 for y. We use the term formula to cover both propositions and propositional functions. In the notation of this chapter the fundamental theorem 2.68 takes the form 3.43

(x=y) -)- {p(x) -+ p(y)}

3.5 The logical constants enable us to introduce the conditional equations of elementary algebra. Unlike the variable x which is characterised by the property that it may be replaced by zero or by S» wherever it occurs, the x in a conditional equation is a missing number sign which may only be replaced by some definite numeral. For instance, when we say that x = 3 is a solution of the equation x 2 = 9, we mean that a true equation results from replacing x by 3 in the second equation, but neither in x = 3 nor in x 2 = 9 may we replace x by zero or by Sx. In fact x = 3 and x 2 = 9 are not equations but propositional functions; and the fact that 3 is a value of x which satisfies x 2 = 9 is expressed by the implication F:

(x= 3) -+ (x 2 = 9) .

Formula F holds for any value of x as may readily be verified, for 1-'-13+Sr,31=0 and 1-'-13--'--Sr,31=0 (by example 2.42) and 32, 91 = 0, so that, by example 2.7303, {l-'-Ix, 31Hx 2 , 91=0

which completes the proof of formula F.

64

THE LOGICAL CONSTANTS

Similarly, the fact that the 'equation' x 2+6=5x has only the two solutions x = 2 and x = 3 is expressed by the implication (x 2+6=5x) --+ (x=2)

v

(x=3)

r.e. {1~lx2+6, 5xl}·lx, 21·jx, 31=0; denoting the left-hand side of this equation by f(x) we have f(3+r)= {1-=-r(r+ 1)}r(r+ 1)=0

and f(0)=f(I)=/(2)=0 which proves that f(x)=O. In addition to the logical constants we introduce the limited universal, existential and minimal operators A:, E: and L: as follows: A~U(x)=O) stands for the propositional function Lf(n)=O ; E:U(x)=O) for the propositional function Ilf(n)=O, and L:(!(x)=O) for the function f-lf(n). The operators 'A:', 'E:', 'L:' are read as 'for all x from 0 to n', 'for some x from 0 to n' and 'the least x from 0 to n'; we proceed to justify these suggested readings. (We use the term 'justify' to cover an informal discussion - there is no question of a formal proof, since only the sign itself and not the interpretation finds a place in the formal work.) 3.6

3.61 The logical constants and the operators may be regarded simply as abbreviations for the expressions by means of which they were introduced, in which case in any formal proof the logical constants and operators must be eliminated and replaced by the expressions which they denote. Alternatively we may regard these signs as an additional part of the formal system satisfying, by definition, the equivalences

{(a=b) v (e=d)} ** {la, bl·le, dl =O} {(a=b) & (e=d)} ** {la, bl + [c, dl =O} {(a=b) -)- (e=d)} ** {(l~la, b\)le, dl =O} {(a=b) ** (c=d)} ** {(l .z: la, b\)lc, dl + (1 ~ [c, d\)la, bl =O} A:U(x)=O) E:U(x)=O)

** (Lf(n) = 0) ** (Ilf(n) =0)

L:U(x) =0) = flJ(n)

THE LOGICAL CONSTANTS

65

which must be added to the list of permitted formulae in a proof; in this case the new signs cannot be totally eliminated but expressions containing them may be transformed into equivalent equations in which they do not appear. The sign x in the operators A~(/(x)=O), E:(/(x) = 0) and L:(f(x)=O) is not a true variable but an auxiliary sign known as a bound variable. We could readily reserve a special class of signs for bound variables but since there is very little risk of confusion it is customary to use the same signs as for variables. The fact that the x in A:(f(x)=O) is a bound variable may be expressed formally by the rule that A:(f(x)=O) may be replaced by A~(f(y)=O), or by the expression obtained by replacing x by any other variable, but substitution for the bound variable is not allowed, with a corresponding rule for the other operators. Alternatively we may formulate the rule as an equivalence

in which x and y may be replaced by any other variables, with corresponding equivalences for the other operators. 3.62

A~(f(x)=O)

is the proposition 1:'j(O)=O, that is 1(0)=0, and is equivalent to

A~(f(x)=O)

if for some P,

1(0)=0 & 1(1)=0 & ... &l(p)=O,

then since 1:'j(n+ 1)=1:'j(n)+/(n+ 1), so that A~+l(f(x)=O) is the propositionA~(f(x)=O)&/(p+I)=O, it follows that A:+l(f(x)=O) is equivalent to 1(0)=0 & 1(1)=0 & ... & I(p+ 1)=0,

so that for any assigned p, A:(f(x)=O) is equivalent to 1(0)=0 & 1(1)=0 & ... &l(p)=O.

Similarly, since IIj(n+ 1) = IIf(n)· I(n + 1), E:(f(x)=O) is equivalent to 1(0)=0 v 1(1)=0 v ... v l(p)=O.

66

THE LOGICAL CONSTANTS

L:

For the interpretation of the operator we recall the characteristic properties of the function pt(n) proved in the last chapter. From the proved equations (2.942 and 2.949) we have pt(n),;;;;;,n,

and whence we obtain the formula E~(f(x)=O) ~ {f(pt(n))=O &pt(n) ,;;;;;, n}

which says that if there is a value of x between 0 and n for which f(x) vanishes, then pt(n) is one of these values; and from the

equation (2.9493) we have j(n)=O ~ Pt(x) ,;;;;;, n

which is equivalent to n
which says that f(n) does not vanish for an n less than Pt(x), so that if f(x) vanishes for some value of x from 0 to n, then pt(n) is the least of these values. If f (x) > 0 for all x from 0 to n, it follows from the equation (2.945) IIj(n)pt(n) = 0

that 3.7

L~(f(x) = 0) = pj(n) = 0 MATHEMATICAL INDUCTION

It follows from theorem 2.8 that the proof schema p(x)

p(O) p(x+ 1)

~

p(x)

is valid. This schema is known as the schema of mathematical induction:

67

THE LOGICAL CONSTANTS

The companion formula 3.8

[p(O)

& A~{p(x) ~

p(x+

1m

~ p(n)

may be called the principle of mathematical induction. We derive formula 3.8 from 3.81

[p(O)

& A~{p(x) ~

p(x+ I)}] ~ p(n+ 1) ;

any p(x) has an equivalent propositional function, f (x) = 0, say; by examples 3.01, 3.02 and 3.322 we see that formula 3.81 is equivalent to r(n)=O where r(n) = {I--'- f(O)}l1o(n)f(n + 1) , O(x) = f(x) + {I--'- f(x+ I)}.

Since r(O)={I--'-f(O)} {/(O)+(I--'-f(I))}f(I)=O and r(n + 1) = {I--'- f(O)}l1o(n) {/(n + 1) + (1--'- f(n+ 2))} f(n + 2) =r(n)f(n+2)

so that {I--'-r(n)}r(n+I)=O, whence r(n)=O follows by the induction schema. If we denote the propositional function 3.8 by P(n) then P(O) is p(O) & {p(O) ~ p(I)} ~ p(O)

which holds by example 3.031, and P(n+ 1) is equivalent to p(O) & A:{p(x) ~ p(x+ I)} & {p(n+ 1) ~ p(n+2)} ~ p(n+ 1) which follows from 3.81 (and example 3.031). Since P(O) and Pin-i- 1) are proved then P(n) follows by 2.7. 3.9 We collect here for reference the principle properties of the operators A, E and L; the fractional part of the numbers of the following formulae are the same as in the numbers of the theorems in the previous chapter of which these formulae are transcriptions. 3.91 3.92

q(n)->-{a<:n~p(a)} q(n)~ A:p(a) {a<:n&p(a)}~q(n) E:p(a)~q(n)

~

3.921

p(n)

3.942

L~p(x)

E:p(a) ;

<: n ;

68

THE LOGICAL CONSTANTS

3.945 3.949 3.9493

r-o-IE".,p(x) -+ {L:p(x)=O} ; E".,p(x) -+ p(L:P(x» pea) -+ {L:p(x)
3.95 We consider next some further theorems on these operators which are needed in the sequel. In the following proofs we systematically employ the corresponding lower case letter as the representing function of a predicate expressed by a capital letter. The product IIj(n) and the sum Ej(n) will also be denoted by II I (x) and E I(x) respectively.

lt~n

:r:~"

H~{F(x)~G(x)}

3.951

H~{~F(x)-+~G(x)}

In terms of representing functions we have to derive the equation ~ h)

IIj(n»IIIl(n) = 0

(i)

(I

from

(I~h) (l~f(x»g(x)=O.

(I

.i:

For n=O the derivation is obvious. By 2.7, (l~p)..:...pq=I":"'p, and thence by 2.7 (over q) l..:...pq=(I..:...p)

{1":"'(I~q)}+(I..:...q).

It follows that [l~{(I~h} (I~IIj(n})IIIl(n)}](1~h) (I~IIj(n+I})IIIl(n+I)=

= [I ~ {(I ~h) (I ~IIj(n»IIIl(n)}] (I ~h)g(n+ I)IIIl(n) {(I":'" IIj(n}) (I..:... (I ~ f(n+ I») + (I ~ f(n + I»}=O,

which completes a proof of (i) by 2.7. 3.9511

For

H~{ F(x)-+G(x)}

H

-+{A~F(x)-+ A:G(x)}

[1~{(I~h) (I~Ej(n})EIl(n)}](I~h) {(1~Ej(n})~f(n+l)}

(EIl(n)+g(n+ I»

=

=[I":"'{(I~h) (I":'" Ej(n})EIl(n)}] (I~h) {(I~f(n+I»~Ej(n)}

g(n+ 1)=0,

using the hypothesis (1 ~h) (I

~f(x})g(x)=O.

THE LOGICAL CONSTANTS

69

A similar proof shows also that H~(F(x)~G)

3.952

H~(~F(x)~G)

is valid. In either of these schemata, by taking 0 = 0 for H we see that the schemata are valid also if H is suppressed. E';(x.;;;n" F(x» ~ E:F(x)

3.953

x.;;;n " F(x) ~ F(x)

For

whence the result follows by 3.951. 3.954

By example 3.831. E:F(x) "A:G(x) ~ E:{F(x) "G(x)} .

3.955

The result is evident for n = O. Now [l--'-(l--'-a) (l-=-b)c] (1--'- (a+q» (l-=-bp)c(p+q) =

[1--'-(I-=-a) (l-=-b)c] (l-=-(a+q» (l-=-b)cp, since (l--'-q)q=O and (l-=-bp)p= (l-'-b)p,

= [l--'-(I-=-a) (l-=-b)c] {(l-=-a) (l-=-b)cp--'-q(l-'-b)cp}=O,

and therefore [1-'-(1-=-

.2 g(x» (I-=- II j(x» II (f(x)+g(x»]

z~n

(I-=-

a:~n

L

x~n

g(x» (I-=-

x~n+l

II /(x» II x:S;;;n+l

x~n+l

(f(x)+g(x»=O,

whence 3.955 follows by 2.8. 3.956

For 3.957

E:F(x) A~(x .;;; n)

-7>

-7>

E: (x .;;; n " F(x» .

{E:F(x)

E:F(x)

-<8-

-7>

E: (x .;;; n " F(x»)} .

E: (x .;;; n " F(x» •

By 3.952 and 3.956. 3.958

F(n)

-7>

E:F(x)

For (1-'-/(0»/(0)=0 and {1-=-/(n+l)}

{II l(x)}/(n+l)=O. x~n

70

THE LOGICAL CONSTA1'i'l'S

3.959 If G does not contain the variable x and n is a variable not contained in F or G then F(x)~G

EJ7;,F{x)~G

The case n = 0 is obvious. Since (p -> q) & (r -+ q) -+ (p fore (E:F(x) -+ G)

Y

r -+ q), as is readily proved, there-

(F(n + I) -+ G) ->- (E~+l F(x) -+ G)

&

and so, from F(n+ I) -+ G follows (E:F(x) -+ G) -+ (E:+l F(x) -+ G)

and the proof is completed by 3.7. It follows that the schema x
is valid, and so in particular is the schema x
3.96

F---+ A~G(x)

The validity of the schema for n = 0 is obvious. Since {I-'- (1-'- f)

L

x~n

g(x)} (1-'- f) {

={l-'-(I-'-f) =

L

x~n

g(x)+g(n+ In

L g(x)}(I-'-f)g(n+l)

:z:':::;;n

0, using the hypothesis (1-'- f)g(x) = 0,

the proof is completed by an application of 3.7. 3.961

For G(O) -+ G(O) and

{I-'-

L

"',,;;,,+ 1

g(x)}

L

,,"',,+ 1

g(x)={(I-'-g(n+I»-'-

L g(x)}g(n+ 1) II g(x)=O .

x~n

x~n

71

THE LOGICAL CONSTANTS

3.962 By 3.96 and 3.961. F(x)~(1(x)

F(x)-G(x)

3.963

~F(x)~~(1(x)

By 3.951, 3.9511. 3.964 The case n=O is obvious. To prove the equation

",-,A~G(x) --0>-

E: ",-,G(x) we use

[1 ~ {I ~ (1 ~ a)} b] {1-'-(1-'- (a + c»} b(1-'- c) = 0 which is proved by applying 2.8 to the variable c; taking ~ g(x) for a, II (1 ~g(x» for band g(n+ 1) for c, the z~n

z~n

proof is completed by an application of schema 2.8. Similarly to prove E~

"'-' G(x)

--0>- "'-'

A: G(x)

we use the equation {1-'-(I~a) (l-'-b)}

{(I-'-b)-'-bc} {(I-'-a)-'-c}

(which is also proved by applying 2.8 to the variable c) with the same substitution for a, band c. 3.9641 have

Taking "'-' G(x) for F(x) A:(x <; n

--0>-

III

3.957 and using 3.964 we

G(x»**A:G(x) .

From the propositional schema p-+-q --q-+-r-p

and 3.964 we derive immediately 3.965

F(n)-+ A~(1(x) E~-.-G(x) -+-.-F(n)

72

THE LOGICAL CONSTANTS

x
3.966

Denote this implication by P(n). P(O) is evident, and {x
But

E: G(y) -+ E:+l G(y)

so that

{P(n) -+ E:G(y)} -+ {P(n) -+ E:+lG(y)}

whence, taking y=x" f(y)=O for G(y), {x
But {x=n+ 1 "f(x)=O} -+ (x=n+ 1) -+ E:+l{x

=

{f(n+ 1)=O}

&

y

f(y)=O} , by 3.958,

&

so that x=n+ 1 & f(x)=O -+ [P(n) -+ E:+l{y=x & f(y)=O}].

Since x
&

f(x)=O -+ {x
&

f(x) =O}

Y

{x=n+ 1

&

f(x) =O} ,

it follows that {x
&

f(x)=O} -+ [P(n) -+ E:+ 1{y=x

&

f(y)=O}]

whence P(n) -+ [{x
&

f(x)=O} -)- E:+l{y=x

&

i.e. P(n) -+ P(n+ 1), which completes the proof. If F does not depend upon x then: 3.9661

A~(F "G(x» ~ F " A:G(x)

3.9662

A:(F

3.9663

E:(F & G(x» ~ F & E:G(x)

3.9664

E~(F

Y

Y

G(x» ~ F

G(x»

~

F

Y

Y

A:G(x)

E:G(x) .

f(y)=O}]

73

THE LOGICAL CONSTANTS

F .. G(x) ~ G(x)

From

A~(F" G(x» ~ A~G(x)

by 3.9511 we derive

F .. G(x)

and from

by 3.959, 3.961 we derive

~

F

A~(F" G(x» ~

F

A=(F .. G(x» ~ F .. A:G(x) follows.

whence Conversely from

A=G(y) ~ {x
follows

F .. A=G(y) ~ F .. {x <; n ~ G(x)} ,

whence

F .. A:G(y) ~ {x,;;;;n ~ (F .. G(x»} ,

and so, by 3.96, 3.9641 F .. A:G(y) ~ A~(F .. G(x»

which completes the proof of 3.9661. Again, by 3.9641, A:G(y) ~ {x,;;;;n ~G(x)}

and

80

F v A:G(y) ~ F v {x
F v A:G(y)

and so

F v A:G(y) ~ A:(F v G(x»

Conversely

A:(F v G(y» ~ {x,;;;;n ~ (F v G(x»}

and so

A:(F

and therefore

A:(F v G(y» .. ,......, F ~ A:G(y) ,

v

{(x,;;;;n)

~

whence

(F v G(x»} ,aB

above.

G(y» .. ,......, F ~ {x
whence 3.9662 follows. Theorems 3.9663 and 3.9664 follow from 3.9662, 3.9661 by taking ,......, F for F and '" G(x) for G(x).

3.97 The Oounting Operator N:F(x) '1'0 express the number of roots of an equation f (x) = 0 in the

74

THE LOGICAL CONSTANTS

range O.;;;;x.;;;;n, or what is the same thing, the number of values of x in this range for which the propositional function f (x) = 0 {I(x) = O} defined by the is true we introduce the function recursion

N:

N~{t(x) = O}= 1..:.. 1(0)

N:+1{/(x)=O}=N:{t(x)= O}+ {I..:.. I(n+

I)} .

If I(x) is the representing function of the propositional function F(x) then we write N:F(x)=N:{t(x)=O}. The first theorem we prove for the counting operator is F(x)_G(x)

3.971

~F(x) = N~G(x)

If D(a, b) denotes the positive difference between II..:..a, I..:..bl and I..:..{1..:..!(l..:..a)b, (I":"b)al}

then

Dia, 0)=0, D(O, Sb)=D(Sa, Sb)=O,

so that D(a, Sb) = 0 and therefore D(a, b) = 0 and finally II..:..a, I..:..bl=l..:..{I..:..I(I..:..a)b, (I..:..b)al}.

Thus from {l..:.. f(x)}g(x) = {l..:..g(x)}f(x) = 0 we derive I..:..f(x)=1..:..g(x). N~F(x) = N~G(x)

It follows that and

IN~+1F(x), N:+1G(x)1 = IN:F(x), N:G(x)1

whence, by 2.8, 3.972

I

IN~F(x), N:G(x) =

{N:F(x) = O}

+r

0 , which completes the proof. A~ '"'"' F(x) .

For n=O the equivalence becomes simply (1..:../(0»=0 +r(I..:..f(O»=O.

Since N:+ 1(f(x) = 0) =N:(f(x) = 0) + (1..:.. I(n +-1» therefore {N;+ I F(x) = O}

+r

{N~F(x) =

O}

& '"'"'

F(n+ 1) ;

75

TH}, LOGICAl. CONSTANTS

moreover

A~+I

[{N:F(x)=O}

,...., F(x)

--3>-

** {A~

A:,...., F(x)]

,...., F'(x)} &: ,...., F(n+ I), and so

--3>-

{[N:+IF(x)=O}

--3>-

{A:+I,...., F(x)}]

whence, by 3.7, Conversely [A: ,...., F(x)

--3>-

{N:F(x) = O}] A~,....,

whence

F(x)

--3>-

[A:+ 1

--3>-

,.....,

F(x)

--3>-

{N:-H F(x) = On

{N:F(x)=O}.

The equivalence 3.972 may also be expressed in the form {N:F(x) > O} 3.973

** E:F(x)

.

N:F(x)=N:(F(x) &:G(x))+N~(F(x) &: ,.....,G(x))

Let 8(n) denote the positive difference between the two sides of the equation. The proved equation l-.:-p={I-.:-(p+q)}+[I-,:-{p-l-(I-.:-q)}] suffices both to provc 8(0)=0 and 8(n)=8(n-+-I). 3.974

N:{F(x) v G(x)}=N:F(x) +N:(,...., F(x) &: G(x)).

Proof as for above, using the equation, l-.:-pq=(I-.:-p)+[I-.:-{q+(l-'-p)}] N:(F(x) &: G(x)) <: N:F(x)

3.975

In the equation {1-.:-(a-.:-b)}([a+(I-'-(p+q))]-'-[b+(l-'-p)]}=O which is proved by applying schema 2.7 to the variable q, we take N:(.P(x) &: G(x)) for a, N:F(x) for b, F(n+ I) for p and G(n+ I) for q. z » n--3>- {N:(y=x)=O}

3.976

Let P(n) denote the implication; then P(O) is equivalent to {l-'-(I-'-x)} (l-'-x)=O.

Since x> n-tl--3>-x > n and x> n+I-*(I-.:-!x,n+l!)=O therefore and so P(n)

P(n)

--3>-

&: x>n+

{x > n-t-l

-+ [N~(y=x)=O]}

1 --3>- {N~(y=x)+(I-'-lx, n+ I\)}=O

76

THE LOGICAL CONSTANTS

P(n) -+ {x>n+ 1 -+ N;+l(y=x)= O}

whence

P(n) -+ P(n+ I) .

i.e.

N;(y=n) = 1

3.977

For n=O this is simply 1-'-(0-'-0)=1 ; and N~+l(y=n+ I) =N;(y=n+ I) + (I-'-In+ I, n + 11)= I, by 3.976.

N;+'(y=r) = 1

3.978

For n=O this reduces to the previous theorem; furthermore N~+'+l(y=r)=N;+'(y=r)

+ {I-'-In+r+ I, r\}= N;+'(y=r). N~(y=x) =

It follows that

n;>x -+

for

n;>x -+ n=x+(n-'-x)

1,

-+ N~(y=x)=N~+(n-'-")(y=x)=1 , by 3.978.

From

x>n -+ {N;(y=x)=O}

follows

x>n -+ {N;(y=x) < I}

and from

x
follows

xn)

which proves and since (x>n) conclude

v

v

(x< n) -+ N;(y=x),;;;;, 1

(x,;;;;,n) is provable by double recursion, we

3.979

3.980 Since and

N:(x
~

~

(x
v

(x=n+ I)

(x=n+ 1) & ,...".,

the result follows by 3.971 and 3.974.

3.981

N:(x
For n = 0 this is evident; and

(x
77

THE LOGICAL CONSTANTS

N~+l(x,,;;;n+1) =N:+

1(x,,;;;n)+N:+l(x=n+

1) , by 3.980,

= N:(x,,;;;n) + 1, whence the result follows.

3.982

by 3.977,

x<.nBrf(x)=O-'?- [NZ{y=x Brf(y)=O}=I]

We consider first the case n=O. Since

x,,;;; 0

{f(x)=O

therefore

x<.O Br f(x)=O

-'?-

/(0)=0

and so

x,,;;; 0 Br /(x)=O

-'?-

x+f(O)=O

whence

x
-'?-

-'?-

f(O)=O}

which proves the case n= O.

By 3.972 [N;{y=x

Br

and so E;{y=x Br f(y)

f(y)=O}=O] =

whence x,,;;;n Br /(x)=O But

N~{y=x Br

O} -'?-

-'?-

-'?-

A~

N~{y=x Br

N~{y=x Br

f'J

{y=x Br /(y)=O}

f(y)

=

O}>O, by 3.965,

f(y)=O}>O, by 3.966.

!(y)=O}";;; 1 , by 3.979 and 3.973

and therefore x,,;;;n Br f(x)=O

-'?-

N:{y=x Br f(y)=O}= 1.

3.983

{NZ(f(y) From

=

0) = k+ I}

-<.+

E:U(x) = 0 Br N:(f(y) = 0 Br yi'x) = k}.

N~(f(y)=O)=N:(f(y)=OBr

follows x
=

Yi'x)+N:(/(y)=O Br y=x)

0 -+ N:(f(y) =0) = N:(f(y) = 0 .. y~x) + 1

whence x,,;;;n" f(x)=O -+ {N:(f(y)=O Br yi'x)=k -'?- N:(f(y) = 0) = k + I} i.e, x,,;;;n Br f(x)=O Br {N:(f(y) = 0 Br y~x)=k} -+

whence

{N:(f(y)=O)=k+ I}

78

THE LOGICAL CONSTANTS

Oonversely from

follows x.c;;;n & !(x)=O ~ {N:(f(y)=O)=k+ 1 ~ N:(f(y) = 0 & y#x)=k}

whence {N:(f(y)=O)=k+ I}

~

{(x.c;;;n & !(x)=O) ~ N:(f(y)=O & y#x)=k}

and so

{N~(f{y)=O)=k+

I} ->- [x.c;;;n

&

!(x)=O ~ {x.c;;;n & !{x)=O

&

N:(f(y) = 0

Y?'ox)=k}]

&

whence, by 3.951, {N:(f{y)=O)=k+ I} ~ {E:{x .c;;;n & !(x)=O) ->- E~(x.c;;;n & !(x)=O & N:(f(y) = 0 & y#x)=k)}

and therefore, by 3.957, {N:(f(y) = 0) = k + I} ~ {E~(f(x) = 0) -;. E~(f(x) = 0 &

N:(f(y) = 0 & y#x)=k)}.

But N=(f(y)=O)=k+I->-E:(f(x)=O), by 3.972, and so N:(f(y)=O)=k+ 1 ~ E:{!(x)=O

lit

N:(f(y)=O & yolx)=k}

which completes the proof. It follows from 3.972 and 3.983 that {N~(f(t)=O)=I} +'7 E:{!(x)=O lit NW(t)=O & t#x)=O}
{N~(f(t) = 0) = 2} - E:{f(x)=O & N~(f(t)=O & t#x) +'7

E:{!{x) = 0

&

& N~(f(t)=O

=

I}

E:[f{y) =0 &yclx &

t#x & t#y)=O]}

+->-E:E:U(x)=O &/(y)=o &x#y 8< A7(f{t)=O ~ (t=x) v (t=y))};

{N~(f{t) = 0) = 3}

+'7

E~{!{x) = 0 & N~(f{t) = 0 & t?'ox) = 2}

-o(~ E~[f(x)

=0 & E:E=U(y) =0 &x¥y&x#z&y#z&

&. !(z) =

0

79

THE LOGICAL CONSTANTS

& AW(t)=O & t#x ~ (t=y) +-> E:E:E:{f(x)

=0

&

& A~(f(t) = 0

~

(t=x)

f(Y)=o &x#y&x#z&y#z y

&

y

(t=z»)}]

f(z)=o

(t=y)

y

(t=z»}

and so on. These equivalences show that the operator N~{f(x)=O} has the desired properties. Thus N:(f(x)=O) has the value unity if and only if there is a unique x from 0 to n for which f(x)=O. And N:U(x) =0) has the value 2 if and only if there is an x and y from o to n, x different from y, such that f (x) = 0 and f (y) = 0 and f (t) docs not vanish for any other value of t from 0 to n, and so on.

Examples

m

3.

Prove the formulae 3.01 - 3.05

3.01

P &. q

3.02

,...., (p -').- q)

3.03

pvp-').-p

3.031

p&'q-').-p

3.032

P &. (q &. r)

3.04

q -').- (p v q)

3.041

(p -').- q) -').- (p &. r -').- q)

3.05

(p v q) -').- (q v p)

~

r-..I (r-..I P

V

r-..I q)

~

P &. ,...., q

~

(p &. q) &. r

Establish the schemata 3.051- 3.093 3.051

p-'?-r (p &. q)-'?-r

3.052

p-'?-(q-'?-r) (p &. q)-'?-r

3.06

p-'?- (q-'?-r) (i-'?- (p-'?-r)

3.0601

p-'?-q&.r p-'?-q

3.061

p~(q~r)

q p~r

3.0611 3.062

p-'?-p* q-'?-q* p&'q-'?-p*&'q* p~(q-'?-r) ~(......".-'?-.-q)

80

THE LOGICAL CONSTANTS

3.063

p-'?q (pv1') -'? (qvr)

3.064

a-'?{p-'?(p-'?r) } a-'?(p-'?r)

3.07

P-'?( q-'?r) 1'-'?s p-'?( q-'?s)

3.071

p-'?(qvr) 1'-+8 p-'?(qvs)

3.08

p-'?q r-+(q-'?8) p-'?(r-'?s)

3.081

p-+qvr q-'?q* r-'?1'* p-'?q*vr*

3.082

q-'?{ r v (p-'?r) } q-'?(p-'?r)

3.083

a-'?b (p-'?a)-'?(p-'?b)

3.084

p-'? (p--""q) p-'?q q-'?8V t

3.085

8-'?r t-'?(p-H') (p-'?q)-'? (p--""r)

3.09

p-'?(q -'?r) p-+(q*-'?r*) p-'?{ (q&:q*)-'?(r&:r*)}

3.091

p-+ (q &:r-'?s ) q-'?r (p&:q)-'?8

3.0911

p&:q-'?r q-'?(1'-'?S) p&:q-'?s

81

82

THE LOGICAL CONSTANTS t~(1'~8)

3.092

t~r ---..:._t-s-s (p&:q)~(1'~ 8)

3.093

p--.+r p&:q--'+8

Prove the formulae 3.1 - 3.194 3.1

,....., (a 0;;;; b)

-?

3.11

(a 0;;;; b)

(a
3.12

(ab) v (a=b)

3.121

(ao;;;;b) &: (b
3.13

(a;;;.b) &: "" (a=b)

-?

3.131

(a=b) &: (b=c)

(a=c)

3.132

a=b

-?

la,

xl =Ib, xl

3.14

(ab=O)

-?

(a=O) v (b=O)

3.15

ab>O

3.16

x 2+1;;;.2x

3.17

a-cb

3.171

(a=A) &: (b=B)

3.18

ko;;;;b &: i-;«

3.19

b z:c

3.191

lo;;;;n &: mo;;;;n Be n-'-l
3.192

p;;;.q &:1';;;'8

3.193

xo;;;;n

-?

-?

-7

-?

-?

(a>b)

-?

-?

(a
(a>O) &: (b>O)

{b=a+ (b-'-a)}

-?

-?

{f(a, b)=/(A, B)}

{ka-'-lb= (a-=-l)b-'- (b-=-k)a}

{(a+b)-'-c=a+(b-'-c)}

-?

-?

m «l

{(p-'-q)-'-(1'-'-s)=(p+s)-'-(q+1')}

{/(x)o;;;;.Et(n)}

THE LOGICAL CONSTANrS

3.194

s>« -+ {(p-'-r)-'-(q-'-r)=p-'-q}

3.2

Prove the schemata 3.31- 3.38

p(x,a) A;'p(m,a)

p(x) ,

E;p(n)

,

p(x,a) A:p(x,a)

3.3

Prove the formulae

3.31

(1-'- n)L:{qJ(n, x) = O} = 0

3.32

r-.J

3.321

E:p(x) -+

r-.J

A:

3.222

,..." A~p(x)

+»-

E: ,..." p(x)

3.33

A:p(x) --+ p(n)

3.35

E~p(x) -> E~p(n-'-r)

3.36

A:+C{g(t, y)=O} --+ g(x, y)=O

3.37

A~{/(t, a)=O} & A~{/(x, u)=O} -+ {f(x, a)=O}

3.38

E';(f(x)=a) & p(a) -+ E':p(f(x»

A:p(x) --+ E';

r-.J

p(x)

r-.J

p(x)

(use example 3.42 below). Prove the schemata 3.39-3.392 3.39

(a=a)-+p(a) p(a)

3.391

p(a)-+--(a=a) --p(a)

3.392

p(b)~q(b .s: r)

3.4

Prove the formulae 3.41- 3.42

3.41

E~p(x) -+ E~"p(x)

p(a+r)-+q(a)

83

84

THE LOGICAL CONSTL."ITS

3.42

a-cb "p(a) -+ E~p(x)

3.5

Establish the schema A~q(x) q(p +Sr) q(n)

where p is any definite numeral. 3.6

If (i) (ii)

l=L~p(n..:.-x) E~p(x) -+

and g=n..:.-l prove

p(g)

g
r-...J

p(a)

so that g is the greatest x not exceeding n for which p(x) holds.

3.7

If ab » 0 prove that there is a greatest value of n for which nb c;« &na <. y.

3.8

Prove the schemata :i.81- 3.82

3.81 3.82

p-+q(x) p-+A~q(x)

q(x)-+p E~q(x)-+p

where x does not occur in p. 3.83

Prove the formulae 3.831- 3.834

3.831

A~(x<.n)

3.832

A~p(x) -+ (y<.n -+ p(y))

3.833

A~(p v

3.834

A~(p -+ q(x)) -+ (p -+ A:q(x))

3.84

Prove the schema

q(x)) -+ (p v A~q(x))

p-+(q-+r(x)) p-+(q-+ A~T(X))

85

THE LOGICAL CONSTANTS

3.9

Prove the following formulae

3.91

p

3.92

A~(p & q(x))

3.93

A~(p(x) & q(x))

** A:p(x)

3.931

E~(p(x) v

** .w;p(x)

3.94

A~(p(x) --+ q(x)) --+ (A~p(x) --+ A:q(x))

3.95

A~(;:r p(x))

3.96

E:(x
&

p(x))

** E:p(x)

3.97

A~(p(x) --+

q(x))

-;>

3.98

A:(p(x)

3.981

A:(p(x) -+ q)

3.982

A~A=p(x, y)

** A:A:p(x, y)

3.983

E~E=p(x, y)

** E:E:p(x, y)

3.984

E~A:p(x, y) --+ A=.w;p(x, y)

3.99

E~E:(p(x) & p(y) & x>y)

-;>

A:(p v q(x))

** P &

q(x))

** q(x))

A:q(x) &

A:q(x)

v E~q(x)

** A:p(x)

(E:p(x) --+ E:q(x))

--+ (E~p(x)

** (E~p(x)

** E~q(x))

-+ q)

** E:E:(p(x) ** E:E:(p(x)

&

p(y)

&

y>x)

&

p(y)

&

xcFy).

CHAPTER IV

THE FUNDAMENTAL THEOREMS OF ARITHMETIC 4. The notions of quotient and remainder are introduced into recursive number theory by means of the recursive functions Q(a, b) and R(a, b) which we define as follows: To simplify the formulae we write tX(c, d) for tX(lc, dl) so that tX(c, d)=O,l according as c, d are equal or unequal, and define R(O,b)=O R(Sa, b)=SR(a, b)·!X(SR(a, b), b)

and Q(O,b)=O Q(Sa, b)=Q(a, b)+{l-'-!X(SR(a, b), b)}

That these functions have in fact the required properties is shown by the following formulae: 4.01

a=b·Q(a, b)+R(a, b)

4.02

b>O

4.03

{(a=bc+r) .. (r
-?

R(a, b)
-?

{c=Q(a, b) .. r=R(a, b)} •

Proof of 4.01

If then and

f(a)=bQ(a, b)+R(a, b) f(O)=bQ(O, b)+R(O, b)=O f (Sa) = bQ(Sa, b) + R(Sa, b) =bQ(a, b)+b(l-'-tX(SR(a, b), b»+SR(a, b)'IX(SR(a, b), b) =bQ(a" b)+SR(a, b), by examples 2.27, 2.08, =Sf(a),

so that f(a)=a. 86

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

87

Proof of 4.02

From the introductory equations of R(a, b) it follows that IX(SR(a, b), b) = 0 --+ R(Sa, b) = 0

and so, by theorem 3.43 4.021

IX(SR(a, b), b)=O --+ {O
Since X'IX(X)=X therefore x>O --+ IX(X) = 1 and since x < y --+ !x, yl > 0 therefore SR(a, b)
By theorem 3.43 {1X(SR(a, b), b)=I} --+ ([R(Sa, b)=SR(a, b)'IX(SR(a, b), b)] --+ [R(Sa, b) =SR(a, b)]}

whence IX(SR(a, b), b) = 1 --+ R(Sa, b) =SR(a, b)

and so SR(a, b)
But

R(Sa, b)=SR(a, b) --+ {SR(a, b)
so that 4.022

SR(a, b)
Since x = y --+ IX(X, y) = 0 therefore 4.023

Ri«, b) < b --+ {SR(a, b) < b v IX(SR(a, b), b) = O} .

Denoting the formula O
Since P(O) holds by definition of Ria; b), 4.02 is proved. Proof of 4.03

Writing Q, R, for Q(a, b), R(a" b) respectively we have, by theorem 3.43,

88

THE FUNDAMENTAL '.rHEOREMS OF ARITHMETIC

a=bQ+R

---?

{(a=bc+r)

(bc+r=bQ+R)}

---?

whence, by 4.01 4.031

a=bc+r

---?

(bc+r=bQ+r) ;

but {Q=Sc+(Q--'-Sc)}

and

80

---?

[{be+r=bQ+R}

{r=b+b(Q--'-Sc) + R}]

---?

(using example 3.061) (a=bc+r)

---?

(Q>c -+ r;;.b)

whence (by example 3.062) (a=bc+r)

Similarly

---?

ir-cb -+ Q
(a=bc+r) -+ (R
and so (using examples 2.043, 3.09) (a=be+r) -+ {(r
Since r-eb -+ O
{(a=bc+r) Be (r
but 4.033

and

80

4.034

Since and

(Q=e) -+ {(be+r=bQ+R)

---?

(r=R)}

(by 4.031,2,3 and example 3.093) {(a=bc+r) Be (r
---?

---?

(r=R) .

{(be+r=bQ+R) -+ (be = bQ)} ,

(a=bc+r)

---?

(bc+r=bQ+R)

it follows from 4.034 that (a=bc+r) Be (r
but

r-eb

---?

---?

O
(be=bQ) ;

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

89

and r-eb -+ (bc=bQ -+ Q=c), using example 3.095, and so a=bc+r & r-eb -+ Q=c using example 3.0911, which completes the proof of 4.03. 4.1 It follows from formula 4.03 that if a=bc thenR(a, b)=O, and conversely if R{a, b)=O then a=bQ(a, b), so that the truth of the equation R(a, b) = 0 is both necessary and sufficient for a to be divisible by b. 4.2 PRIME NUMBERS If p{n) = 0 denotes the propositional function

n>1 &A:{a<1 va=nvR(n,a»O} then the equation p{n)=O says that n exceeds unity, and every number from 0 to n which exceeds unity but is not n itself, does not divide n, i.e, that n has no divisors apart from itself and unity, so that n is a prime number. Thus p{n)=O asserts that n is a prime number. 4.21 If q{n) is the least divisor of n, after unity, that is, if q(n)=L~{R(n, x)=O &

x> I}

then, using examples 4, 4.1, R{n, q{n»=O,

q(n)~n,

n>l-+q{n»I,

and

a
-+

{R(n, a»O v a «; I}

so that (using theorem 2.91 and example 3.063) n> 1 -+ {q(n) > 1 & A~(n)(a< 1 v a=q(n) v R(n, a»O)} , that is

n> 1 -+ p(q{n)) = 0

which says that q{n) is prime if n> 1. From the formula {q{n)=n}

-+

{p{q(n»=O

-+

p(n)=O}

90

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

we deduce that n> 1 8< q(n)=n

-'>-

p(n)=O.

From R(n, n)=O and (p(n) = 0)

(n> 1) 8< {a 1 -'>- R(n, a»O}

-'>-

we derive, firstly, (p(n) = 0)

-'>-

[{q(n) > I}

whence, since R(n, q(n»

=

&

{R(n, a) = 0

-'>-

a.,;;; 1

Y

a;» n}]

0, using example 2.37

p(n)=O

-'>-

{q(n);;;.n}.

By means of the proved formula q(n).,;;;n we derive p(n) = 0 4.22 Thus q(n)=n be a prime number.

-'>-

q(n) = n .

is a necessary and sufficient condition for n to

4.3 Following the idea of Euclid's famous proof of the infinitude of primes we can readily construct a formula giving prime numbers as great as we please. We prove in fact that for any n, there is a prime between nand n! + 1. By the characteristic properties of q(n), since n! + 1> 1 (by example 2.82) p(q(n!+I»=O and q(n!+I).,;;;n!+I.

Next, from example

4.44

we have

{R(n!+I,a)=O 8< a>l}-'>-a>n

and so {R(n! + 1, q(nl + 1» = 08< q(n!

whence q(n! + 1) > n, greater than n and 4.4

If

+ 1» I} -'>- (q(n! + 1»n) which proves that q(n! + 1) is a prime number not greater than n! + 1.

THE SEQUENCE OF PRIMES

.j)(0) = 2, .j)(n+ 1)=L~(")!+l{x>~(n)8< p(x)=O}

then (for n>O),

~(n) is the nth odd prime number.

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

91

We show first that p(n) is prime for any n. Since q(p(k)! + 1) >P(k) p(q(p(k)! + 1)) = 0

and therefore

p(k+l»p(k) , p(p(k+l))=O p(k+ 1) .,;;;q(p(k)! + 1) ,

and

so that P(k+l) is prime; but, 1::Jy example 4.45, P(O) is prime, and so p(n) is prime for any n. It follows by examples 4.5 and 4.61 that p(n+ 1) is an odd prime. To show that (for n>O) p(n) is in fact the nth odd prime, we shall show that every prime is a p(n) for some n, that is p(m)=O -+ E;;'{m=~(k)}.

By the properties of the function #- we have m
i.e.

Let 9?(m) = L:,{m m, we have ~(9?(m») > m and 9?(m).,;;; m; but

{m>1 "p(x»m}-+p(x»P(O)

and so

{m>1 "~(x»m}-+x>O

whence

m> 1 -+ 9?(m) .> 1

that is

m:» 1 -+ 9?(m) = 1 + (9?(m) -'--1).

Writing 1p(m)=9?(m)-'--I, so that m>I->1p(m)<9?(m), we have (since 9?(m) is the first x for which P(x) exceeds m) m » 1 -+ {~(1p(m».,;;;m "p(1p(m) + 1»m}

that is m> 1 -+ ([p(1p(m») « m. "~('lJ'(m}

+ 1) >m]

Y

p(1p(m»=m} .

92

Since p( m)

THE FUNDAMENTAl, THEOREMS

=

0

--+ m

»

O~'

ARITHMETIC

and

1

p(m)=O --+,......" {~(k)<m

&

m<.\J(k+

In

it follows that p(m)=O --+ .\J(1p(m»=m.

Accordingly, by example 3.42 p(m)=O --+ E~(m=.\J(k»

4.5 In preparation for theorem 4.51 below we prove next: if bd>O and R(ab,d)=O, and if l=L~{x>O &R(ax, d)=O}

then

R(b, 1)=0.

We observe first that b>O & R(ab, d)=O --+ 1>0 & R(a1, d)=O

and

d>O & R(ad, d)=O

-?

l~d.

Let (J, e denote Q(b, 1), R(b, 1) respectively so that b={J1+e and e
since R(ab, d) = 0 therefore R(a1{J+ ae, d) = 0 ; whence, since R(a1, d)=O, we have R(ae, d)=O; but

{x
&

R(ax, d)=O} --+ x=O

and so, from the inequality pletes the proof. 4.51 If p is prime then R(ab, p)=O

&

e< l

it follows that

e=

0, which com-

R(a, p»O --+ R(b, p)=O.

For by the previous result, since p> 0 & R(ap, p) = 0, it follows that R(p, l) = 0 and 1< p, and therefore, since p is prime, either 1=1 or l=p. But R(a1,p)=0 and R(a,p»O so that ,......,,(1=1), and therefore l=p. By 4.5, Rib, l)=O and so R(b, p)=O.

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

4.6

93

RESOLUTION INTO PRIME FACTORS

In this and the following section it is desirable to re-introduce the conventional sign by which to express the product of a number of factors. For any function j (x) we define (as in § 3.95)

II j(x)=II/n). x~"

Our principal result is that any number may be expressed in one way only as a product of prime factors. vVe start by considering the highest power of a given prime contained in a given number. We define '11(0, k)=O

v(Sn, k) = L:R(Sn, {1)(k)}x+!) > 0

Since tJ(k) > I, therefore by example 4.701, {tJ(k)}">n and so R(Sn, {tJ(k)}"+l) > 0

and therefore (by theorem 3.949) R(Sn, {+J(k) y<S".kl+!) > 0

i.c, 4.61

n>O ~> R(n, {tJ(k)}'·("·kl+!»O

and (by 3.\)493), v(n,k)<.n, and x < v(n, k) -I- 1 -+ R(n, {tJ(k)}X) = 0

so that, in particular, 4.62

R(n, {tJ(k)y
=

0.

Formulae 4.61 and 4.62 show that v(n, k) is the exponent of the greatest power of the kth odd prime contained in n. 4.63

m> 0 -+ m = II {tJ(x)Y<m.xl z':;;m

By example 4.86, m is divisible by

II {tJ(X)}v(m. Xl, and if Q is x~m

the quotient, it follows from formula 4.61 and example 4.87 that

m > 0 -+ R(Q, tJ(x»> 0 and hence (by example 4.84) that m> 0 -+ Q= 1.

94

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

4.64 To show that factorisation is unique we isolate an individual factor in a product by means of the following theorems: if cp (m, r, 0)= 1

and !p(m, r, Sk)=<x(r-'-k) II {.\J(x)}·(m,:rl+ x~k

+ {1-'- <x(r -'-k)} {.\J(Sk)}·(m.k). !p(m, r, k)

then 4.641

r <, k -;.. II {.\J(x)}v(m,x) = {.\J(r)}V(m,1). !p(m, r, k) "'';;k

and 4.642

r <, k -;.. R(cp(m, r, k), .\J(r)) >

°.

Denote formulae 4.641 and 4.642 by "r <, k -;.. H(k)" and "r<,k-;..J(k)" respectively; then r<,O-;..H(O) and r<,O-;..J(O) are clearly true. If r =Sk then II {.\J(x)}"(m,x) = {.\J(Sk)},'(m,Sk) II {.\J(x)}"(m,x) x';;Sk x';;k = {.\J(Sk)y(m,Skl . cp(m, r,Sk) so that (r=Sk) -;.. H(Sk) ; (i) if r <, k then cp(m, r, Sk) = {.\J(Sk)}·(m.sk). !p(m, r, k) and so {r <,k} -;.. [H(k) -+ { II {.\J(x)y(m,x) = {.\J(r)y(m,r)cp(m, r, Sk)}] "'';;Sk i.e. (r<,k) -;.. {H(k) -;.. H(Sk)} , (ii) It follows from (i) and (ii), using example 2.491, that

((r<,k) -;.. H(k)} -;.. {(r<,Sk) -;.. H(Sk)} which completes a proof by induction of 4,641. The proof of 4.642 proceeds on the same lines; if r=Sk then ipim, r, Sk) = II {.\J(x)}v(m,x) and so (by examples :e~k

4.82, 4.85)

(r =Sk) -;.. R(cp(m, r, Sk), .\J(Sk)) > and so

(r=Sk) -;.. J(Sk) .

If r < k then cp(m, r,Sk) = {.\J(Sk)}v(m,Sk). cp(m, r, k)

°

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

95

and so, by theorem 4.51 (and example 4.82 again),

r-ck. ~ {J(k)

~

J(Sk)}

and the proof is completed as before. 4.7 THE UNIQUENESS OF THE RESOLUTION INTO PRIME FACTORS We have to prove that if m admits two resolutions into prime factors, II {p(x)ylm",) and II {p(x)}
~(m,

~~m

x)=v(m, x) for x-cm,

We have

x<,m ~ R(m, {p(xW(m."'l)=0 x<,m ~ m={p(x)}'lm'''''qJ(m, x, m)

and

x<,m

~

R(qJ(m, x, m), p(x)) > 0

so that, (by example 4.801) x <, m ~ R( {P(x) r(m,.,), {p(x)y(m,,,') = 0

and so (by example 4.711), v(m, x»~(m, x) for all x<m. Similarly ~(m,x»v(m,x), so that ~(m,x)=v(m,x), for all x<m. 4.71

{m> 0 & R(m, p(x)) = O} ~ v(m, z)

0

Since {v(m, x)=O} ~ {[m>O ~ R(m, {p(x)ylm,"l+!»O] ~ [m>O ~ R(m, p(x)) > OJ}

and

m>O ~ R(m, {p(x)y(m,.,l+!»O

therefore, by example 3.064, v(m, x)=O ~ {m>O ~ R(m,p(x))>O}

which proves 4.71. It follows that {m>O & Rim, P(x))=O} +l>v(m, x»O

so that, if m> 0, v(m, x) > 0 is a necessary and sufficient condition for m to be divisible by p(x).

96

THE FUNDAMENTAL THEOREMS OF AIHTHMETIC

4.72 THE GREATEST PRIME FACTOR If l(n)=L:{v(n, n...:...x»O} and g(n)=n...:...l(n) we shall show that .\)(g(n)) is the greatest prime factor of n. For, by example 4.83, n » 1 ~ E:{R(n, .\)(x))=o}

whence n> 1 ~ E:{v(n, x»O}.

It follows, by example 3.36, that 4.721

n> 1 ~ v(n, g(n)) >

°

and 4.722

g(n)
a c n. ~ v(n, a)=O .

Since a z-n. ~ R(n, .\)(a))=n, therefore a>n may be rewritten in the form 4.723

~ v(n, a)=O

and 4.722

g(n)
Formulae 4.721 and 4.723 show that 1J(g(n)) is the greatest prime factor of n. 4.8 The function v(m, x) makes it possible to assign a unique number to any given collection of numbers. For given a collection ao' all ... , a", with a,,> 0, we assign to this collection the number N given by

N = II .\)(x)a",. X~D

The number N is completely determined by the given set of numbers ao' aI' ... an' with aD>0, and conversely the set itself is uniquely determined by the single number N, for N = II 1J(X)"(N."l by theorem 4.63, x!{"N and so, since factorisation is unique, x<.g(N) ~ v(N, x)

=

a",

i.e. the numbers v(N, 0), v(N, 1), v(N, 2), ... , 11(N , g(N)) reproduce the given collection (where g(N) is defined as in 4.72).

97

THE FUNDAMENTAL THEOltEMS OF ARITHMETIC

4.9

THE GREATEST COMMON FACTOR

Following example 3.36 we can readily define the greatest value of x which divides each of two non-zero numbers a and b. We prefer however to proceed differently and to establish the greatest common factor as the least non-zero value of the function ax-'-by. We shall determine functions k(a, b) and l(a, b) such that h(a, b)=ak(a, b)-'-bl(a, b) has the least possible value, not less than unity. From the equation boa -'-

° 0

b = ab

we derive Ef(b·a-'-l. b=ab)

and from this in turn we derive E~{Ef(ka-'-lb= ab)} •

From this, and the inequality ob:» 0, follows E':'{c;;. 1

&

EZ(Ef(ka-'-lb=c))}

and, therefore, if h(a, b)=L':'{c;;. 1

& E~(Er(ka-'-lb=c))}

we have, for a, b satisfying ab > 0, E~(Er(ka-'-lb=h(a, b)))

and h(a, b);;. 1.

Hence if then E~(a·k(a, b)-'-lb=h(a, b))

and k(a, b)
and so if l(a, b)=L:(aok(a, b)-'-yb=h(a, b))

we have ab;» 1

~ {a k(a, b)-'-b·Z(a, b)=h(a, b) & k(a, b)
&

h(a, b);;. I}.

98

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

4.91 When one of a or b is zero we define h(a, b)=a+b, so that a>O

and 4.92

&: b=O -+

l·a-'-O·b=h(a, b)

a=O &: b>O -+ l·b-'-O·a=h(a, b). k cb &: l c a &: c> 1 &: ka-i-lb »:» -+ h(a, b)<,c.

For k-cb

&: l e;o. &: ka-s-lb»:« -+ E~(Ef(ka-'-lb=c)) and

E:(Ef(ka-'-lb=c)) &:c>l-+h(a,b)<,c (by theorem 2.9493).

4.93

By example 3.18 k <, b

&: l <, a -+

{ka -'-lb = (a -'-l)b -'- (b .z: k)a}

whence, since a-s-lia, b)<,a, b-'-k(a, b)<,b, h(b, a) <,h(a, b) .

Similarly

h(a, b) <,h(b, a)

so that

h(a, b) = h(b, a) .

4.94

R(a, d)=O &: R(b, d)=O -+ R(h(a, b), d)=O.

For if R(a, d) = 0 and R(b, d) = 0 then, denoting Q(a, d) and Q(b, d) by IX and fJ respectively, h(a, b) = {IXk(a, b) .s: fJl(a, b)}d . 4.95

If ra-'-sb is not zero then it is a multiple of h(a, b). We may without loss of generality suppose that ab > O. Denoting k(b, a), l(b, a) and h(a, b) by k, land h for short we have kb-'-la=h.

Let q be the quotient and e the remainder when ra .z; sb > 0 is divided by h, so that e< h, and ra-'-sb=hq+e ;

but

kqb-'-lqa=hq

and so (r+ql)a= (s+qk)b+e

THE FUNDAMENTAl. THEOREMS OF ARITHMETIC

99

which we shall write for short in the form xa -"- yb =

e.

By example 3.193, if nb «;«, na,y and yb,ax then a(x -"- nb) -"- b(y -"- na) = e ;

hence if N is the greatest n satisfying nb «; x .. na <, y (given by example 3.7) and if we denote »
If Y>a, X>b, then by example 2.41, from the inequalities Nb <,X, Na,y we derive x> (N + l)b, ll> (N + l)a contradicting the definition of N, and so either Y b and Y ab-"-bY =b(a.:... Y»O and so X>b .. Y O" c
whence, from E':,E:(ax.:...by=c)" c>O ~ c>h (theorem 4.92) we conclude

,...., (X>b .. Y
If X «b, since Yb c Xa we have Y O); but e
4.96 It follows from the last result that h(a, b) divides both a and b, for l·a.:...O·b=a is a multiple of h(a, b) and l·b.:...O·a=b is a multiple of htb, a)=h(a, b). Thus h divides both a and b. Since every common factor of a, b is a factor of h, it follows that h is the greatest common factor of a and b. In particular if h= 1, a and b are relatively prime. 4.97 From the equation ka-"-lb=h, where k, land h denote k(a, b), l(a, b) and h(a, b), and ab>O it follows (by example 4.31) that ka=h (mod b) and from (a .:...l)b .z: (b -"- k)a = h we derive (a-"-l)b

=

h (mod a) .

100

THE FUNDAMENTAL THEOREMS OJ;' ARITHMETIC

Since ka=O (mod a) and (a-'--l)b=O (mod b) we find, for any r and s, (using examples 4.322, 4.323) (a-'--l)br+kas=rh (mod a) (a-'--l)br+kas=sh (mod b).

Whence, by example 4.91, if n denotes R((a-'--l)br+kas, ab), n=rh (mod a) n=sh (mod b)

Accordingly, if a and b are relatively prime, so that h= 1, we have proved: II a and b are relatively prime and il r, s are any two numbers, there is a number n < ab such that n=r (mod a) ,n=s (mod b) ,

Examples IV Prove the formulae 4-4.:> 4.

R(n, n)=O

4.01

R(a, 0) = a , Q(a, 0) = 0

4.0ll

R(x,y)=O~(y>O)v(x=O)

4.012

R(a, b)=O

&

4.013

R(a, x) = 0

~

4.014

x> 0 ~ {R(ax, bx) = 0 ~ R(a, b) = O}

4.02

R(p, a)=O

4.03

R(x, ])=0

4.04

Ria, be) = 0 ;. R(a, b) = 0

4.1

Ria s-b, e)=O

4.2

R(ab, b)=O

4.21

R(a,b)=O&b>I~R(a+l,b)=1

4.3

b» 1

4.31

If a-"--cd=b and b>O prove that

~

R(b, c)=O

&

~

R(a, c)=O

R(ab, x) = 0

R(q, b)=O

&

~

R(pq, ab)=O

R(a, e)=O -,.. R(b, c)=O

R(ab+ 1, b)= 1

R(a, d) = R(b, d) 4.32

Show that a=b+dx

~

a=b (mod d)

Prove the schemata 4.321- 4.323 4.321

4.322

a=b (mod b=c (mod

d) d) a=c(mod d)

a= b (mod d)

ar=br (mod

d)

101

102

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

=b l (mod d) a 2 = b2 (mod d) at +a2 =b1 +b 2 (mod d) al

4.323

Prove that is divisible by /(1 )

4.4

Ilf(11,)

4.41

Ilf(r+11,)

4.42

If a<11, then Ilf(11,) is divisible by llf(a).

is divisible by Ilf(r)

Prove the formulae 4.43 - 4.44 4.43

O
4.44

1
4.45

Prove that 2 is a prime number

4.46

Prove that 3 is a prime number

& a<11,

--+ R(11,! + 1, a)= 1

Prove the formulae 4.5 - 4.711 4.5

.\:)(11,+1»2

4.51

.\:)(11,»11,

4.6

11,> 2 & R(11" 2)=0 -> p(11,»O

4.61

p(11,)=O -+ (11,=2) v {R(11" 2)= I}

4.7

(l+x)rn>l+mx

4.702

a>1 &p>q-+a'P>a

4.71

a>O & b i- a -+ R(a, b»O

4.711

x>

q

1 & R(xa , xb)=O --+ b «;a

If p is a prime number prove that 8 4.801

R(af<+I, p)=O -+ R(x, p)=O R(a, p)

>0

& R(ab, pk) =

0 -+ R(b, pk) = 0

THE FUNDAMENTAL THEOREMS OF ARITHMETIC

4.802

103

R(m, a)=O Br R(m, pk)=O Br R(a, p»O -+ R(m, apk) =0

Prove the formulae 4.81

R(~1c' ~,)=O -+

4.82

,...., (k=l) -+

4.83

m> 1 -+ E':R(m, V",) =0

4.84

Show that·

k=l

R(~:i:, ~,»O

x<m~R(m,p",»O

m=1

Prove the formulae (where p is a prime number) 4.85

A~(f(x), p»O -+ R ( II f(x), p»O ",';;;k

4.86

R(m, II

~;(m.",) =

0

z~k

Establish the schema ~{R(b,pJ=O}

4.87

x~m-+R(a,p:') =

0

E:'{R(ab,p:.,+l) = O}

4.9

If a(l) > 0 and N

=

II V(x)lJ("') prove that, in the notation z';;;l

of theorem 4.7, g(N) = l. 4.91

If b » 0 prove that R(a, bc)=a (mod b)

4.92

If h is the H.C.F. of a and b, and H is the H.C.F. of h and C, prove that H is the H.C.F. of a, band c.

4.93

Prove that the L.C.M. of x and y is a primitive recursive function.

CHAPTER V

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC In this chapter we consider a codification of primitive recursive arithmetic in which the only axioms are explicit and recursive function definitions, and the only inference rules are the substitution schemata

se,

F(x)=G(x) F(A)=G(A)

Sb 2

A=B F(A)=F(B)

T

A=B, A=O B=O

where F(x), G(x) are recursive functions and A, B, C are recursive terms, and the primitive recursive uniqueness rule

u

F(Sx) =H(x, F(x» F(x)=H"'F(O)

where the iterative function H't is defined by the primitive recursion HOt=t, HSXt=H(x, H"'t); in the schema U, F may contain additional parameters but H is a function of not more than two variables. In Sbl , the function G(x) may be replaced by a term G independent of x, provided that G(A) is also replaced by G. The novelty in this codification lies in the derivation of the key equation a + (b-'-a) = b + (a -'- b) by means of the primitive recursive uniqueness rule, instead of requiring a doubly recursive equalising rule as before. We start by proving a few auxiliary schemata. From the defining equations x+O=x, x+O=x follows x=x by T, and thence by 104

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

105

S~ we reach A=A. Since B=A follows from A=B, A=A by T, we have proved

K

A schema equivalent to V is j(O}=g(O) f(Sx}=H(x,f(x» g(Sx} =H(x,g(x}) f(x)=g(x}

The passage from VI to U is obvious; for the converse we derive j(x) = H'] (0) ,g(x)=H"'g(O)

from the stated hypotheses, by V, and H"'t(O) = H"'g(O) from j(O)=g(O) by Sb 2 , whence t(x)=g(x) follows by T and K. As an illustration of the use of Sb 2 we derive F( a, b) = F(A, B) from the pair of equations a = A , b = B. First we derive F(a, b) = F(a, B) from b = B by Sb2 , and similarly F(a, B) = F(A, B) from a=A; hence by using K and T we derive F(a, b)=F(A, B) from a=A, b=B. Two further schemata of importance are F(Sx}=F(x} F(x}=F(O)

•

F(O}=O, F(Sx}=O F(x}=O .

To prove E I , we define HI(x, t), C(t) explicitly by the axioms HI(x, t) = t , C(t)

=

F(O) ,

whence we readily derive 0(0) = F(O), C(Sx) = HI(x, O(x», F(Sx) = HI(x, F(x» which, by VI' yields F(x) =C(x), and from this we reach F(x)=F(O) by Sb., T. For E 2 we define Z(t)=O, so that Z(F(x»=O whence from F(Sx)=O follows F(Sx)=Z(F(x». This equation together with Z(Sx)=Z(Z(x» and F(O)=Z(O) yields F(x)=Z(x), by VI> and E 2 follows.

106

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

We establish next some results for addition, subtraction and multiplication, taking the defining equations for these operations to be: a+O=a, a+Sb=S(a+b) ; O~ 1 =0 .Ba-» 1 =a ,a~O=a ,a-=-Sb = (a-=-b) -=-1; a·O=O, a-Sb-s a-b-s- a , (a~b)~I=(a~I)-=-b.

5.01

For

(a~O) ~

(a-'-I)-'-Sb=

1=

(a~

Sa-'-Sb=(Sa-=-I)-'-b=a~b, using a~a=O

5.03

For Sa-'-Sa=a-'-a and so

{(a~b)

1=

-=-1 }-=-l,

and the result follows by U1 .

Sa~Sb=a-'-b

5.02

For

(a~Sb) ~

1) -=-0 ,

{(a-=-l)-'-b}~1,

. (5.01).

.

a-=-a=O~O=O. O~a=O .

5.04

Proof by El' using O-=-Sa= (O~ 1) -=- a= 0 -'- a. (a+b)~b=a.

5.05

For (a+Sb) ~Sb =S(a+ b) -'-Sb = (a+ b) -=-b so that (a+b) -=-b=a, by ~. 5.051

(a+n) -'-(b+n) =a-'-b .

For (a+Sn)-'- (b+Sn)=S(a+n)-'-S(b+n)= (a+n) -=- (b+n), and (a+O)~(b+O)=a~b.

5.052

n~(b+n)=O .

By 5.051 and 5.04. 5.06

O+a=a.

For 0+0=0, O+Sa=S(O+a), Sa=Sa, and the result follows by U1 . 5.07

a-t-Sb=Sa+b.

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

107

We use a+SO=Sa=Sa+O, a+SSb=S(a+Sb) , Sa+Sb=S(Sa+b) and U 1 • 5.08

a+b=b+a.

From a+O=O+a, a+Sb=S(a+b), using 5.07, follows Sb+a=b+Sa=S(b+a). Then 5.08 follows by U 1 .

(a+b)-'--a=b.

5.09 By 5.08 and 5.05. 5.10

(a+b)+c=a+(b+c) .

With c as variable, apply U1 . 5.1I

Sa-b-s a-b e-b ,

For Sa·O=a·O+O .Ba-Sb v Sa-b s-Ba, a·Sb+Sb=(a·b+a)+Sb= =S{(a.b+a)+b}=S{(a.b+b)+a}, by 5.08, 5.10, and so a·Sb+Sb= (a·b+b) -s-Sa, whence 5.1I follows by U1 . 5.12

O·a=O.

For O·Sa=O·a so that O·a=O·O=O. 5.13

a(l-'--a)=O.

For 0(1-'--0)=0 and Sa(I-=-Sa)=Sa(O-'--a)=Sa.O=O. 5.14

a·b=b·a·

For a·O=O·a and a-Sb-e ab-s-a , Sb·a=b·a+a. 5.15

a(b+c) =a·b+a·c .

This is a consequence by U 1 of the provable equations

a(b+O)=a·b=a·b+a·O, a(b+Sc) =a·S(b +0) =a(b+c) +a, a-b-s-a-Sc-s ab s: (ac+a)= (ab+ac) +a . 5.151

a(bc) = (ab)c .

108

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

For a(b·O)=O=(a·b)·O and a(b·Sc)=a(bc+b)=a(bc)+ab, (ab)·Be = (ab)c+ab. We prove now an ·extension of schema E 2 • E

f(O) =g(O), f(Sx) =g(Sx) f(x)=g(x)

3

Define H 2(x, t)=O·t+g(Sx)=O·t+f(Sx), then f(Sx)=O. f(x) + f(Sx)=H 2(x, f(x», g(Sx) =

o· g(x) + g(Sx) =

H 2(x, g(x»,

whence E 3 follows by U 1 • (I.:....a)b=b-=--ab.

5.16

For (l-=--O)b=b=b.:....O·b, (I-=--Sa)b=(O.:....a)b=O, and b-s-Sa-b=b-=-- (b +a·b) = O. Next we prove the key equation a+ (b-=--a) =b+ (a-=--b) .

5.17

Define f(a,b)=a+(b-=--a), so that f(a,O)=a, f(O,b)=b, f(Sa,Sb)=Sf(a, b); and define g(a, b)=b+(a.:....b) so that g(a, O)=a, g(O, b)=b, g(Sa, Sb)=Sg(o" b). We start by proving 5.171

f(a, b) = f(o,-=--l, b-=--l)+ {I-=-- (I-=-- (a+b»)

By E 3 , a=(a-=--I)+{I-=--(I-=--a)}, whence f(a, O)=f(a-=--I, 0)+ {l-=--(I-=--a)} which establishes 5.171 with for b. With Sb for b, 5.171 becomes

°

f(a, Sb)=Sf(a-=--l, b)

which is a consequence of the equations f(O, Sb)=Sb=Sf(O, b), f(Sa, Sb)=Sf(a, b), completing the proof of 5.171. Next we define lp(O, a, b) = 0, lp(Sn, a, b) = lp(n, a, b) + {I-=-- [1':"" ((a-=-- n) + (b-=--n»]}

and prove f(a.:....n, b.:....n)+lp(n, a, b)=f(a-=--Sn, b.:....Sn)-+-lp(Sn, a, b);

in fact, by 5.171

]i'Oltl\IALISATIONS OF PRUIITIVE RECURSIVE ARITHMETIC

109

[ta-i--n, b-'-n)+q>(n, a, b)

= j(a-'-Sn, b -'-Sn) + q>(n, a, b) + {1-'- [1-'- «a -'-n)

+ (b .s: n»)]}

= j(a -'-Sn, b -'-Sn) + q>(Sn, a, b) sothatj(a-'-n, b-'-n)+p(n, a, b)=j(a, b)+q>(O, a, b)=j(a, b), whence j(a, b)=j(a-'-b, O)+q>(b, a, b)=(a-'-b)+p(b, a, b). Similarly g(a, b)= (a-'-b)+q>(b, a, b) whence equation 5.17 follows. 'Ve derive from 5.17 the schema jA,B:!=O. A=B '

for from lA, BI =0 follows lA, BI-'-(B-'-A)=O by 5.04, whence by 5,05, A -'-B = 0, and similarly, B -'- A = 0; from these we reach A+(B-'-A)=A, B+(A-'-B)=B

and thence A=B, by 5.17. The derivation of

lA, BI=O

from

A = B is of course trivial. We come now to some induction schemata. Let P(x) denote the equation j(x)=g(x). The familiar induction schema is P(O) ,P(x)-+P(Sx) P(x)

or writing p(x)

=

If(x), g(x)1 p(O) =0, (l-'-p(x»p(Sx)=o

p(x)=O

As in the proof of 2.8 we define q(O) = 1, q(Sn)=q(n) (l-'-p(n»; then q(SSn)=q(Sn) (l-'-p(Sn»=q(n) (l-'-p(n» (l-'-p(Sn» = q(n){(1 -'-p(n» -'-- (1-'--p(n»p(Sn)} =q(Sn)

where the last equality sign holds according to hypothesis since (1-'-p(n»p(Sn)=O;

hence q(Sn)=q(SO) = 1, that is q(n) (l-'--p(n»= 1, and multiplying by p(n), p(n)=O, by 5.13.

110

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

f(a,O) =0 ,f(O,Sb) =O,{f(a,b) =O}---,>-{f(Sa,Sb) =O} f(a,b) =0

We observe first that from f(O, 0) =0, f(O, Sb)=O follows f(O, b) = The implication hypothesis stands for the equation

o.

{l-'--f(a, b)} f(Sa, Sb)=O.

Now {l-'-- f(O, b-'-l)}f(O, b)=O and from

{l-'-- f(a, O)} f(Sa, 0)= 0, {l-'-/(a, b)}/(Sa, Sb) =0 follows

{I -'-I (a, b -'- I)} 1(Sa, b) = 0, therefore {I-'-/(a-'-I, b-'--l)} I(a, b)=O

and so

Next we show that (j)

f(a, b){I-'-/(a-'-n, b-'-n)}=O.

To this end we prove [1-'-/ (a, b){l-'-I (a -'-- n, b .z: n)}] 1(a, b){l-'- f(a --Bn; b -'-Sn)} = 0;

with p, q, r standing for [ta, b), f(a-'-n, b-'-n) and I(a-'--Sn, b-'--Sn), respectively, the left hand side of this equation has the form {I -'-- p(l -'-q)} p( 1 -'-r) = p {( 1 -'-r) -'- p(l -'-q) (I -'-- r)} =p{(l-'--r)-'-p(l-'-r)} , since q(l-'--r)=O, =p(l-'--r) (l-'--p)=O

which completes the proof of (j) by II (the validity of (j) with 0 for n being evident). By writing IIp(a, b), 1p(a, b)1 = I(a, b) it follows from 12 that the schema lp(a, O)=1p(a, 0), lp(O, Sb)=1p(O, Sb), {rp(a, b) =Ip(a, b)} - { rp(Sa,Sb) =Ip(Sa,Sb)} rp(a,b) =Ip(a,b)

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

III

is valid. As particular cases of 12 , 13 we note that from I(a, 0)=0, 1(0, Sb)=O and I(Sa, Sb)=O follows [ia, b)=O; from I (a, 0)=0, 1(0, Sb)=O, I(a, b)=/(Sa, Sb) follows I(a, b)=O; and from qJ(a,O)=1fJ(a,O), qJ{O,Sb)=1fJ{O,Sb), qJ(a,b)=qJ(Sa,Sb) and 1fJ(a, b)='IjJ(Sa,Sb) follows ep(a, b)='IjJ(a, b); for if we denote IqJ(a,b),'IjJ(a,b)1 by I(a,b) then [ta, 0)=0, I(O,Sb)=O; and from qJ(a, b) =qJ (Sa, Sb), 1fJ(a, b) ='IjJ(Sa, Sb) follows [t«, b) = I(Sa, Sb); whence [ia, b)=O and so ep(a, b)='IjJ(a, b). As instances of this last schema we mention 5.18

c(a-'-b)=ca-'-cb, a-'-(b+c)=(a-'-b)-'-c.

To complete the construction of recursive arithmetic there remains only to prove the substitution theorem (x=y) ~ {F{x) = F{y)} .

This is readily derived from the equation

(l-'-Ix, yi)F(x) = (l-'-Ix, y\)F{y) . Exactly as in 2.63, we start from the equation (l--"--z)F(y+z)= (l-'-z)F(y) ,

which is proved by applying E z with z as variable, and derive

{I -'- (x y) }F (y + (x -'-y) ) = {I -'- (x -'-y) }F (y) .z.,

and multiplying by l-'-Ix,

(l-'-Ix,

yl,

we reach

yl )F(y + (x -'-y»

=

(l-'-Ix, yi)F(y)

since

{l--"-- [(x--"--y) + (y-'-x)]} {l--'--(x-'-y)} =

(l-'-Ix, yi) .s: {x -'- y){[l--"-- (x -'-y)] --"-- (y .z: x)}

= l-'-Ix,

yl ;

similarly {l--"--Ix, yi)F(x + (y .z: x» = (1 -'-Ix, y\)F(x) and since x+(y-'-x)=y+(x-'-y) ,

the required result follows by T.

112

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

We call the foregoing formalisation of recursive arithmetic system &/. The Deduction Theorem

If the equation A = B is derivable in &/ from an hypothesis F =G, (i.e. an unproved equation) and if the derivation does not involve substitution for the variables in the hypothesis, then

(F=G) -+ (A=B) is provable in i~. We multiply each equation of the derivation by 1--"-1 F,GI. The hypothesis becomes the proved equation {l--"-IF, GI}F= {l--"-IF, G/}G and the final equation becomes {l--"-IF, GI}A = {l--"-IF, G/}B from which we may derive {l-'-IF, GI}!A, BI = 0,

i.e. (F=G) -> (A=B). If P = Q is a proved equation then for any function R,

RP=RQ is a proved equation, and so multiplication by 1--"-1 F, GI turns a proved equation into a proved equation. We show next that multiplication by a factor does not invalidate an application of any of the schemata 8br, 8b2 , T and U. For 8b1 we have to prove that R. F(x) =R.G(x) R.F(A)=R.G(A)

is valid when the factor R does not contain the variable z, and

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITIIMETIC

113

this of course is a consequence of Sb, itself. For Sb 2 we have to prove the validity of the derivation R.A=R.B R.F(A) =R.F(B)

To this end we remark that, since RIA, BI 5.15 and 5.18, therefore

=

IRA, RBI, by equations

(RA=RB)

~

(A=B) v (R=O),

(F(A)=F(B» (R=O)

~

(R.F(A)=R.F(B», (R.F(A)=R.F(B» ,

~

and by the substitution theorem (A=B)

~

{F(A) = F(B)}

whence, using the schemata

which follow from the provable equations (I-=-PI) + (l-=-p2)= (I-=-PIP2)+ {I-=- (PI +P2)}'

k+(I...:..k)=I+(k-=-l) ,

we prove (R.A=R.B) ~ {R.F(A)=R.F(B)}, and so, (taking 0=0 for H in the second of the above schemata] we see that R.F(A)=R.F(B) follows from R.A =R.B. For T we have only to prove the schema RA=RB RA=RO RB=RO

and this follows by T itself. It remains to prove that an application of U 1 remains valid under mutiplication by R, i,e. that the derivation R.F(O)=R.G(O) R.F(Sx)=R.H(x,F(x)) R.G(Sx) =R.H(x,G(x)) R.F(x)=R.G(x)

114

FORMA-LISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

is valid, when R does not contain the variable x. We start by proving the schema

(P=R)--+(Q =S)

By 8~, P=Q

R=S

IP,RI=IQ,RI

IQ,RI=IQ,SI

whence, by T, P=Q, R=S

IP,RI=IQ,SI and the desired derivation follows by the schema a=b (l-'-a)b=O

which is also proved by 8bz. From the formula (ka=kb) -+ {kJ(a)=kJ(b)}, which is proved above, follows {R.F(x)=R.G(x)} -+ {R.H(x, F(x))=R.H(x, G(x))}

whence, by the given hypotheses and the above schemata, {R.F(x) =R.G(x)} -+ {R.F(Sx)=R.G(Sx)}

and this, with the first hypothesis, proves R.F(x) = R.G(x), by induction schema 11' and the deduction theorem is proved. The deduction theorem holds for any number of hypotheses. For instance given a derivation of A = B from two hypotheses F l =Gl , Fz=Gz we obtain a proof of the implication (Fl =Gl )

-+

{(F2=GZ)

-+

(A

=

B)}

by multiplying each equation in the derivation by the factor (l-'-IFv GlD (l-'-IFz, GzD .

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

115

by multiplying each equation in thc derivation from these hypotheses by and so on.

Reduction of Schema U We consider first a system

(lJl*

with Sbl , Sb2 , T, the schema

j(O)=O,j(n)=j(Sn) j(n)=O

E

and the axiom A

a+ (b -'-a) = b + (a-'- b)

and, in place of the familiar introductory equations of the predecessor function, the axiom p The axiom a+(b-'-a)=b+{a-'-b) enables us to deduce a=b from a---b=O and b-'-a=O. For by Sb 2 , a-'-b=O b+(a-'-b)=b+O=b

a+(b-'-a)=a+O=a

and from a+{b---a)=a, b+(a-'-b)=b and a+(b-'-a)=b+(a-'-b) follows a=b. Derivation of a=b from a-'-b=O, b-'-a=O we call schema A. To prove schema E I , namely F(Sx)=F(x) F(x)=F(O)

we define (0)=0 and
whence
which completes the proof of schema

~.

116

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

We turn now to a reconsideration of the equations and schemata proved in fJi. Equation 5.01 we leave to the end. Equation 5.02 is now an axiom. The proofs of 5.03 and 5.05 remain unchanged. Proof in fJi* of 5.07 (a+SO) -'-- (Sa+O)=Sa-'--Sa=O, (a+SSb) -'-- (Sa+Sb) =S(a+Sb) -'--S(Sa+b) = (a+Sb) -'-- (Sa+b),

proving (a+Sb) .z, (Sa+b) =0. Similarly (Sa+b) -'-- (a+Sb)=O whence 5.07 follows by schema A. Proof of 5.06 (O+Sa) -'--Sa=S(O+a) -'--Sa = (O+a)-'--a

so that (O+a) -'--a= (0+ 0) -'--0= O. Similarly Sa-'-- (0 +Sa) =a-'--(O+a) whence 5.06 follows by schema A. Proof of 5.08 (a+O)-'--(O+a)=a-'--a=O, (a+Sb) -'-- (Sb+a) =S(a+b) -'--S(b+a) = (a+b) -'-- (b+a) ,

so that (a+b) -'-- (b+a)= 0, whence (b+a) -'-- (a+b) = 0 and a+b=b+a follows by A. The proof of equation 5.09 remains unchanged. Proof of 5.04 a+ (O-'--a)=O+ (a-'-- 0) =a, therefore {a+ (O-'--a)}-'--a=a-'--a=O, whence, by 5.09, O-'--a=O.

The schema f(O)=g(O) f(Sa)=f(a) g(Sa)=g(a) f(a)=g(a)

follows by two applications of schema E and two of schema T.

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

117

The proofs of 5.051, 5.052 remain unchanged, and from 5.051 it follows that !a+n, b+nl =Ia,

bl·

Proof of 5.10 la+(b+O), (a+b)+Oj=O, la+ (b+Sn), (a+b) +Snl =Ia+ (b+n), (a+b)+nl ,etc. Proof

0/ 5.11

Since Sa·Sb=Sa·b+Sa and a ·Sb +Sb = (ab -+ a) -+Sb =S{(ab -+ a) + b} =S{(ab -+b) +a}= (ab+b) -s-S«

therefore ISa. Sb, a- Sb -+ Sbl

=

ISa. b, a- b + bl

, etc.

The proof of 5.12 remains unchanged. For 5.14, 5.15 and 5.151 we note that la.Sb, (Sb)al=lab+a, ba-+al=lab,

bal,

la(b+Sc), ab+a.ScI=la(b+c)+a, (ab+ac)+al=la(b+c), ab+acj, la(b·Se), ab·SeI =Ia(bc)+ab, (ab)c+abl =Ia(bc), (ab)cl ,etc.

The schema F(x) +G(x) =0

F(x)=O

is proved by means of the equations 5.09, 5.04 which yield {F(x) +G(x)}~G(x)= F(x)

and

O~G(x)=O .

To prove schema E 2 F(O) =0 , F(Sx) =0

F(x)-O

118

FORMALISATIONS OF PRIMITIVE RECURSIVE ARITHMETIC

we define ep(O) = 0, ep(Sx) = ep(x) + F(x) so that ep(SSx) = ep(Sx) whence ep(Sx)=ep(SO)=O and so ep(x)+F(x)=O whence F(x)=O. The proof of 5.13 follows by E 2 in 5.16 remains unchanged.

~*

as in

~,

and the proof of

80, too, the proofs of the induction schemata 11 , 12 , 13 and the substitution schema carry straight through from £Jt to ~*. It remains only to prove schema U 1, F(O)=G(O), F(Sx) =H(x,F(x», G(Sx}=H(x,G(x}} F(x)=G(x).

Define ep(x) = IF(x), G(x)l, then ep(O) = 0 and by the substitution schema, {I-=-ep(x)}IH( t, !(x», H( t, g(x»)1

=

whence {I-=-ep(x)}IH(x, !(x)), H(x, g(x»)1 =

0

°

and so {I -'- ep(x) }ep(Sx) = 0

whence ep(x) = 0 and therefore

! (x) =

g(x).

The proof of equation 5.01 in !J2 is therefore valid also in ~*. We may obviously take the equation (a-'-b) -'-I = (a-'-l) -'-b as an axiom in ~* in place of Sa-'-Sb=a-'-b. The system ~* may be further modified by taking the schema

8

a-=-b=O a+(b-'-a)=b

in place of axiom A, provided that we add the axiom 0-'-1=0.

For by 8 and 8bz

a-'-b=O, b-'-a=O a=b

To prove O-=-a=O we use O-'-Sa=(O-=-a)-=-l = (O-'-l)-'-a=O-=-a, and 0-=-0=0 (in place of axiom A). Finally the equation a+(b-'-a)=b-f-(a-=-b) is proved exactly as in ~.

CHAPTER VI

REDUCTIONS TO PRIMITIVE RECURSION 6.

Course-of-values Recursion There are many kinds of recursive definition which, though seemingly very different from primitive recursion, may in fact be transformed into a primitive recursion. Consider, for instance, the sequence f (0), f (1), f (2), ... in which f(O)=a, f(I)=b, f(2)=f(I)+f(O)=a+b, f(3)=f(2)+f(I)=a+2b, and so on. The general law ofthe sequence is f(n+ 1) = f(n) + f(n~ 1) so that f(n+ 1) depends, not just on f(n) but also on f(n~ 1). To show that this sequence f(n) may also be obtained by primitive

recursions alone, we introduce the function g(n) = II ~~(r) r
so that f(n)=v(g(n), n), the exponent of the greatest power of the prime ~n which divides g(n). Let y(n, b)=v(b, n)+v(b, n-"-l), so that y(n, g(n))=f(n)+f(n----l)=f(n+ 1).

It follows that g(n+ 1) =t.l~~~ll.g(n) =t.l~~ig(n)) ·g(n)

so that g(n) is primitive recursive and therefore f(n) is primitive recursive. The function II t.l~(r) is called the course-of-values function of r~n

f(n), and a recursion in which f(n+ I} depends not just on f(n) but also on the values of f(x) for values of x
course-of-values recursion. The foregoing method of transforming a course-of-values recursion into a primitive recursion is of general applicability. Thus if fJ(n,~, b2 , •.• , bk ) is a primitive recursive 119

120

REDUCTIONS TO PRIMITIVE RECURSION

function of k+ I variables and if Ar(n) is primitive recursive and satisfies Ar(n)
is transformed into a primitive recursion by means of the courseof-values function g(n) =.\:>~(O)' .\:>im .... ,.\:>~,,) , g(O) =.\:>~(O), g(n+ I)

i.e.

= g(n) ·.\:>~~il)

For if y(n, b) = {3(n, v(b, A1 (n » , v(b, ~(n», ... ,v(b, Ak(n»), so that y is primitive recursive then y(n, g(n))

=

{3(n, !(A 1(n)), f(A 2(n)), ... , f(Ak(n») = f(n + I)

and g(n + I)

= g(n) ·.\:>~~il) =g(n) ·.\:>~~ig(,,)),

so that g(n) is defined by primitive recursion and !(n+ 1) is obtained by substitution from y and g. Similarly, if A(n) is primitive recursive and f(0)=1,f(n+1)= II (f(x) + A(n)) , ~~n

then we take g(n) as above and y(n, b)= II {v(b, x)+A(n)} X~'1"

so that y(n, g(n))

=

f(n+ 1) and g(n + I) =g(n) ·.\:>~~ig(,,))

as before. 6.1

RECURSION WITH PARAMETER SUBSTITUTION

Another recursion transformable to primitive recursion is recursion with parameter substitution. As an example of such a recursion we consider f(O, a)=a, f(n+ 1, a)=f(n, y(n, a)) .

REDUCTIONS TO PRIMITIVE RECURSION

121

To determine t(n+ 1, a) from the second of these equations we need to know the value of t(n, x) not just for x=a, but for the value y(n, a) of x, which of course varies with n. The method of transforming this recursion (due to R. Peter) is of considerable interest apart from the present application. If we calculate in turn the values of t(n, a) for n=O, 1, 2, 3, and so on, we determine the sequence of terms a, 1'(0, a), 1'(0, 1'(1, a)), 1'(0, y(I, 1'(2, a))),

and so on, which are formed by repeated substitution for the parameter. The essential idea of the transformation is to disentangle these substitutions by means of a function "P(n, a) with the following property: for any n we can find p, q and for any p, q we can find n so that "P(n+ 1, a) = y(p, "P(q, a)) .

With a suitable initial condition, like "P(O, a)=a, this function transforms any term like 1'(0, 1'(1, 1'(2, a))) successively into 1'(0,1'(1,1'(2, "P(O, a)))), 1'(0, 1'(1, "P(~, a))), 1'(0, "P(~, a)), and finally into "P("ka, a), for appropiate hI> h2 , ha, so that if "P can be defined by primitive recursion, and also the auxiliary function h r , then so can t(n, a). To construct such a function "P(n, a) we observe that, for any n, n+ 1 is expressible in only one way in the form 211(2q + 1), where in fact p='JI(n+ 1,0) and q= [(n+ 1)/211+1 ], so that p and q are primitive recursive functions of n (and of course n is primitive recursive in p and q). We define "P(O, a) = a, "P(n + 1, a) = y(p, "P(q, a))

(where p, q are the functions of n just introduced); since q < n + 1 this is a course-of-value recursion so that"P is primitive recursive. To complete the transformation it remains to show that there is a primitive recursive k(n) such that t(n, a) ="P(k(n), a); first however we introduce g(n, a) such that "P(n, tp(8, a))=1jJ(g(n, 8), a) .

122

REDUCTIONS TO PRIMITIVE ItECURSION

To determine g(n, a) we consider the relation 'IjJ(g(n + 1, s), a) ='IjJ(n + 1, 'IjJ(s, a)) =y(p, 'IjJ(q, 'IjJ(s, a))) =y(p, 'IjJ(g(q, s), a» ='IjJ(2P(2g(q, s)+ 1), a)

which reveals the definition by course-of-values recursion g(n+ 1, s)=2 P + 1.g(q, s)+2 P

,

the definition of g being completed by taking g(O, s)=s. Similarly,

to determine k( n) we consider 'IjJ(k(n+l), a)=f(n+l, a)=f(n, y(n, a)) = 'IjJ(k(n) , y(n, a)) ='IjJ(k(n), 'IjJ(2 n , a))

=

'IjJ(g(k(n), 2n ) , a)

whence k(n+ 1)=g(k(n), 2n ) , which together with the initial condition k(O) = 0 is a primitive recursive definition of the function k(n). 6.2

SIMULTANEOUS RECURSIONS

As a final illustration of an indirect definition of a primitive recursive function we consider the simultaneous recursions f(O)=g(O)=O f(n+ l)=P(f(n), g(n)) g(n+ 1)=Q(f(n), g(n))

where P, Q are primitive recursive. Here again we introduce an auxiliary function

so that

f(n)=v(h(n), 0) ; g(n)=v(h(n), 1).

It remains to show that h(n) is primitive recursive. Writing p(x)=P(v(x, 0), v(x, 1)), q(x)=Q(v(x, 0), v(x, 1)) and

RED{;CTIO~S

TO PRIMITIVE RECURSION

123

we have f(n+ I)=P(f(n), g(n»=p(h(n» g(n+ 1) =Q(f(n), g(n» =q(h(n»

and so h(n -I-I)

=

2/(" -H

).

3°("+11 = 2P(I« n l l . 3q (h(n l l =p(h(n»

which, together with the initial condition h(O) ~ I, completes the primitive recursive definition of h(n), in terms of the given functions Piu, v) , Q(1l-, v).

6.3 GENERALISED INDUCTION SCHEMATA 6.31 We consider now some generalisations of induction which are provable in &t. The first of these is the schema (j)

P(a, 0) P(f(a, n), n) --+ P(a, Sn)

(k)

P(a, n)

(i)

Ig

This schema has an obvious affinity with recursion with parameter substitution, and the proof of the schema depends in essense upon the reduction to primitive recursion of a recursion with parameter substitution. In the following proof however this connection is not brought out, the reduction having been made in advance. In preparation for the proof we shall establish the equations (11)

Sn<;b--+n=b--'-S(b--'-Sn),

(12) Sn<;b--+S(b--'-Sn)=b-'-n,

and the proof schema (m)

p-+q p-+ (r-+8) (q-+r) -+(P-+8)

The equation (11) is evident with 0 for b; with Sb for b the equation becomes n <; b --+ n=b--'-(b--'-n)

which follows by the substitution theorem (3.43) from

124

REDUCTIONS TO PRIMITIVE RECURSION

b-=-(b-=-n)=n-=-(n-=-b), (example 2.421).

For

(~)

we have

Sn <; b -'7 b =Sn + (b-=-Sn) -'7 b=n+S(b-=-Sn) -'7 b-=-n=S(b-=-Sn) .

For the schema we note that the hypotheses are equivalent to ,....., p v q and e-, p v ,....., r v 8 and so imply e-, P v 8 v (-.. r & q) which in turn is equivalent to (q -'7 r) -'7 (p -> 8) as required. To prove schema I g we define g(n, a, b) by the primitive recursion g(O, a, b)=a g(Sn, a, b)=f(g(n, a, b), b-=-Sn).

From hypothesis (i) follows P(g(b, a, b), 0) and thence 0 <; b -'7 P(g(b-=-O, a, b), 0)

(n)

follows. From (II) and the substitution theorem (3.43) we find Sn ~ b -'7 {f(g(b-=-Sn, a, b), n)=f(g(b-=-Sn, a, b), b-=-S(b-=-Sn))}

and so, by (12) , Sn <; b -'7 {f(g(b-=-Sn, a, b), n)=g(b-=-n, a, b)}.

But by hypothesis (j) P(f(g(b-=-Sn, a, b), n), n)

-'7

P(g(b-=-Sn, a, b), Sn)

whence we derive Sn

< b -'7 {P(g(b-=-n, a, b), n)

and since Sn < b -'7 n

-'7

P(g(b-=-Sn, a, b), Snn

b, it follows by schema (m) that

<;

{(n <; b)-'7P(g(b-=-n,a, b),

nn

-'7

{(Sn <; b) -'7 P(g(b-=-Sn, a, b),Sn)}

which in conjunction with (n) completes an inductive proof of n <; b -'7 P(g(b-=-n, a, b), n)

and substituting b for n we derive n-c n. -'7 Pta; n)

whence (k) follows, since n<;n is provable.

REDUCTIONS TO PRIMITIVE RECURSION

125

6.32 The single premise in hypothesis (j) in schema I g may be replaced by a sequence of premisses, in fact a sequence of variable length, leading to the schema (i')

I;

A~(1J·f1}PU(x,

(j')

P(a, 0) a, n}, n) _ P(a, Sn)

(k')

P(a, n)

(The hypothesis (j') is the conjunction of k(a, n) terms PUI' n) & PU2' n) & ... & P(ik, n) where

To prove schema

I; we

t. stands for f(r, a, n».

introduce first the function

F(a,n)=

E

f(x,a,n)

z~k(a,lI)

which by example 3.193, satisfies »-cki«, n) _ {f(x, a, n)<.F(a, n)}.

Next we define G(O, n) = F(O, n) G(Sa, n)=G(a, n)+F(Sa, n)

from which it readily follows that F(a, n) <.G(a, n) and G(a, n) <.G(Sa, n) .

From the second inequality, we derive, by induction over m, G(a, n) <.G(a+m, n)

and hence, taking b -'-a for m and using the implication a-cb _ (a+(b-'-a)=b)

we derive

(1')

a-cb _ {G(a, n)<.G(b, n)}.

Let p(a, n) be the representing function of P(a, n) and let Q(a, n) denote the equation E p(x, n)=O so that the equivalences Q(a, n) ~ A:P(x, n) Q(O, n) ~ P(O, n) , Q(Sa, n) ~ Q(a, n) & P(Sa, n) hold.

126

REDUCTIONS TO PRIMITIVE RECURSION

It follows that Q(a, n) -+ Pia, n) and Q(Sa, n) -+ Q(a, n). By considering in turn the cases with 0 for a and Sa for a we readily prove Q(a, n)

-~

Q(a-=-l, n)

and hence, by induction over m Q(a, n) -+ Q(a-'--m, n).

Taking a-'--b for m and using the implication b «;a -+ {a-'-- (a-'--b)=b}, we derive

b «;a -+ {Q(a, n) -+ Q(b, n)} and hence, from (I') (m')

b <;a -+ {Q(G(a, n), n) -+ Q(G(b, n), n)}

and since x<;k(b, n) -+ {!(x, b, n)<;G(b, n)} ,

therefore x<;k(b, n) -+ {Q(G(b, n), n) -+ Q(f(x, b, n), n)}

from which we derive in turn x
and then, by hypothesis (j'), Q(G(b, n), n) -+ P(b, Sn)

whence, by (m'), b «;a -+ {Q(G(a, n), n) -> Pib, Sn)}

from which follows in turn Q(G(a, n), n) -+ A~(x, Sn)

and (n')

Q(G(a, n), n) -+ Q(a, Sn) .

REDUCTIONS TO PRIMITIVE RECURSION

127

Finally, from the equivalences Q(O, 0) ~ P(O, 0) , Q(Sa, 0) +-> Q(a, 0)

&

P(Sa, 0)

and hypothesis (i') we derive Q(O, 0) and Q(a, 0) ~ Q(Sa, 0)

which proves Q(a, 0). From Q(a, 0) and (n') we derive (k') by I g , proving the schema I~. 6.34 The last generalisation of induction which we consider is the schema (h")

I;

P(a, 0)

(i")

P(f(O, n), n) ~ P(O, Sn)

(j") (k")

P(f(Sa, n), n) ~ {P(a, Sn) ~ P(Sa, Sn)}

Pia, n)

We start by defining g(O, n) =

f (0, n)

g(Sa, n)=g(a, n)+t(Sa, n)

from which it follows that and

g(a, n)
and hence (I")

b
We define Q(CL, 11,) as above, so that Q(a, 0) is provable, and so too Q(a, 11,)

~

P(a, n) •

The next step in the proof is to establish the implication (m")

Q(g(a, 11,), n) -). Q(a, Sn) .

The case a=O is hypothesis (i"). To complete an inductive proof of (m") we use the schema A->-B,A.-+D {A & (B.-+O)}.-+(O&D)

128

REDUCTIONS TO PRIMITIVE RECURSION

which is proved by observing that {A & (B ~ O)} ~ (0 & D) is equivalent to "-' A v (B & "-' 0) v (0 & D), and the equation (l..:...a) (b+(l..:...c» (c+d)=O

follows from the equations (1..:... a)b = 0 , (I..:... a)d = O. We take Q(g(Sa, n), n) , Q(g(a, n), n) for A, B and Q(a, Sn), P(f(Sa, n), n) for 0, D. The implication A ~ B follows from (I") since g(Sa, n):>g(a, n) and similarly from t(Sa, n) <.g(Sa, n) follows Q(g(Sa, n), n) ~ Q(f(Sa, n), n) ; ~

but

Q(f(Sa, n), n)

and so

Q(g(Sa, n), n) ~ P(f(Sa, n), n) A~D.

i.e. From A i.e.

~

B ,A

~

D we derive {A & (B -+ O)}

~

(0 & D)

Q(g(Sa, n), n) & {Q(g(a, n), n) ~ Q(a, Sn)} ~

But

P(f(Sa, n), n)

Q(a, Sn) & P(f(Sa, n), n) .

Q(a, Sn) & P(f(Sa, n), n) ~

Q(a, Sn) & Pta, Sn) & P(f(Sa, n), n)

~

Pia, Sn)

& P(f(Sa, n), n) & Q(a, Sn)

~

P(Sa, Sn) & Q(a, Sn) , by hypothesis (j")

~

Q(Sa,Sn).

Thus we have proved [{Q(g(a, n), n) ~ Q(a, Sn)} & Q(g(Sa, n), n)] ~ Q(Sa, Sn)

which is equivalent to {Q(g(a, n), n) ~ Q(a, Sn)} ~ {Q(g(Sa, n), n) ~ Q(Sa, Sn)}

and this completes an inductive proof of implication (m"), From Q(a, 0) and (m") we derive Q(a, n) by I g and from Q(a, n) we derive P(a, n), proving the schema

I;.

REDUCTIONS TO PRIMITIVE RECURSION

6.4

129

PERMUTATION

In this section we define by recursion the processes of transposition and permutation, and apply the results obtained to prove that the sum of a variable number of terms a(I)+a(2)+ ... +a(p+l)

is independent of the order of the terms. We commence by defining in terms of a(k) the function l'(k, n) by the recursion -r(k, O)=a(k) , -r(k, n+ 1)=-r(k, n)+a(n+k+ 1);

it follows by induction over r that 6.41

xip, q)+-r(p+q+ 1, r)=-r(p, q+r+ 1).

Denoting 1-"- (l-"-x) by .x(x) we define a(p, q)=-r(p, q-'-p)·.x(Sq-'-p) so that a(p, p+n)=-r(p, n) , a(q+n+ 1, q)=O,

and

a(p, p+q+ 1) = a(p, p+q)+a(p+q+ 1).

We observe now that a(p, p+q)+a(p+q+ 1, p+q+r)=a(p, p+q+r);

for with 0 for r the equation is evident, and with r + 1 for r the equation becomes 6.4I. Next we prove 6.42

a(l, p+q+l)=a(l, p)+a(p+l)+a(p+2, p+q+l).

Consider first the equation with 0 for q, i.e. 6.421

a( 1, P + 1) = a( 1, p) + a(p + 1) ;

this equation is evident with p = 0 and with p + 1 for p it becomes a(l, p+2)=a(l, p+ 1)+a(p+2)

130

REDUCTIONS '.1'0 PRIMITIVE RECURSION

which is equivalent to 1'(1, p+ 1)=1'(1, p)+a(p+2) which follows from the definition of 1'. With q + 1 instead of q, (and using 6.421), equation 6.42 is equivalent to 1'(1, p+q+ 1)=1'(1, P)+1'(P+ 2, q)

which follows from 6.41, and this completes the proof of 6.42. 6.422

It follows that

a(l, r) +a(r+ 1) + a(r+ 2, q+r+ 1) +a(q+r+ 2) +a(q+r+ 3, p+q+r+ 2)

=a(l, q+r+ 1)+a(q+r+ 2)+a(q+r+3, p+q+r+2) =a(l, p+q+r+2).

6.43 To define an interchange of a(r) with a(r+8) we introduce the function (}(r,

8,

k)= [k+ {(r+8)--'-k}lk, rl*]--'-(k--'-r)lr+8, kl*

where [z, yl* = l--'-lX(lx, yD, so that if kter, kter+8 =r+8 if k=r =r if k=r+8.

(}(r, s, k) =k

Hence if b(r,

8,

k) = a((}(r, 8, k)) then

b(r,

8,

if k=/=r, k=/=r+8, =a(r+8) if k=r, =a(r) if k=r+8.

k)=a(k)

To form the sum a(l) +a(2) + ... +a(n) with a(r) and a(r+8) transposed we introduce 1'*(r, 8, k, n) and a*(r, 8, p, q) by the equations .*(r, e.k, O)=b(r,

8,

k)

1'*(r, 8, k, n+ l)=1'*(r, a*(r,

8,

8,

k, n)+b(r,

p, q)=lX(Sq--'-p) .•*(r,

8,

8,

n+k+ 1)

p, q--'-p).

REDUCTIONS TO PRIMITIVE RECURSION

131

It follows exactly as in 6.422 above that 6.431

a*(r+l,q+I,I,r)+b(r+l,q+l,r+I)+a*(r+l,q+l,r+2, r+q+ 1) +b(r+ I, q+ I, r+q+2)+a*(r+ 1, q+ I, r+q+3, r+q+p+2) =a*(r+l, q+l, I, r+q+p+2).

We prove next the three equations 6.45

a*(r-t-I, q+l, I, r)=a(l, r)

6.46

a*(r+ I, q+ I, r+2, r-t-q+ 1)=a(r+2, r+q+ I)

6.47

a*(r+ I, q+ I, r-t-q+3, r+q+p+2)=a(r+q+3, r+q+p+2)

Consider first 6.45. The equation with 0 for r is evident; the equation with r -l- I for r is equivalent to 6.451

T*(r-t-2,q-t-l,l,r)=T(I,r).

To prove 6.452 we consider the equation 6.452

T*(r-t-k+ 2, q+ 1, 1, r)=T(I, r)

which we denote by P(k, r). Then P(k, 0) is evident since T*(r+k+ 2, q+ I, I, O)=b(r+ k+ 2, q+ I, 1)=a(I)=T(I, 0). Further, since T*(r+k+3, q+ I, I, r+ 1)=T*(r+k+3, q+ I, I, r)+a(r+2) and T(I, r+I)=T(I, r)+a(r+2)

therefore P(k+ I, r) -+ P(k, r+ I) and P(k, r) follows by the generalised induction schema I g • Hence P(O, r) follows, and this is equation 6.451, which completes the proof of 6.45. The proof of 6.46 proceeds along similar lines; with q = 0 the

132

REDUCTIONS TO PRIMITIVE RECURSION

equation is evident. With q + 1 in place of q the equation is equivalent to 6.461

r*(r+ 1, q+2, r+2, q)=r(r-i-2, q)

which follows from

6.462

r*(r+ 1, q+k+2, r+2, q)=r(r+2, q).

We denote 6.462 by R(k, q). We readily establish

R(k, 0) and R(k+ 1, q) ---+ R(k, q+ 1) from which R(k, q) follows by I g , completing the proof of 6.46. The proof of 6.47 is rather simpler. With p = 0 the equation is evident; with p + 1 for p the equation is equivalent to

r*(r+ 1, q+ 1, r+q+ 3, p)=r(r+q+ 3, p) and this is proved by induction over p. From 6.422, 6.431, 6.45, 6.46 and 6.47 follows immediately

6.48

a*(r+ 1, q+ 1,1, r+q+p+2)=a(l, r+q+p+2)

since

b(r+ 1, q+ 1, r+ 1)=a(r+q+ 2) , b(r+ 1, q+ 1, r+q+ 2) =a(r+ 1) and for any j, k, 1, m, n

j+k+1+m+n=j+m+1+k+n. Equation 6.48 is equivalent to

6.481

r*(r+ 1, q+ 1,1, r+q+p+ 1)=i(l, r+q+p+ 1)

and, as we shall now prove, equivalent also to

6.482

{r;;;d &s;>1 &n;>(r+s)-'-I}---+{r*(r,s, l,n)=r(l,n)}.

Denote the equation r*(r,

8,

1, n)=r(l, n) by R(r,

8,

n).

Then, by 6.481, R(r+ 1, q+ 1, r+q+p+ 1) holds, and therefore

R(r+l,q+l,r+l +q+l+p) holds;

REDUCTIONS TO

PRIMITIVl~

133

l'tECURSION

substituting r-"-I for r we derive R(I-l-{r-"-I),q+I,I+(r-"-I)+q+l+p)

and hence, by the substitution theorem, using the equation 1 +(r-"-I)=r+(I-"-r), (l-"-r=O)

Substituting

8 -"-

-?

R(r, q+ I, r-j-l +q+p).

I for q it follows that

{(I--'- 8) = O} -? {(I--'- r) = 0

-?

R(r, 8, r+ (8 --'-I) + p)} .

Finally substituting n--'-{r+(8--'-I)} for p we derive {(r+ (s--'-I)) --'-n= O} - ? [(1-"-8) = 0

-?

{I--'-r= 0

-?

R(r, 8, n)}]

which is equivalent to 6.482. Equation 6.482 proves that a sum of n terms is unchanged by interchanging any two of them. Since a permutation may be regarded as a result of repeated transpositions, it remains only to define recursively a sequence of transpositions. Given three functions r(n), 8(n), and a(n) we define B(O, Ie) = a(k) B(n+-I, k) = B(n, (j(r(n), 8(n), k))

so that by 6.1, B is primitive recursive. For each n, B(n+ I, k) is a transposition of B(n, k) such that B(n+I, k)=B(n, k) if kc/=r(n), kc/= r(n) +8(n), =B(n, r(n)+8(n)) if k=r(n) =B(n, r(n)) if k=r(n)+8(n).

Define the sum function T(n, k, p) by the equations T(n, k, 0) = B(n, k) T(n, k, p+ I)=T(n, k, p) + B(n, k+p+ I),

then taking B(n, k) for a(k) and B(n+ I, k) for b(r, r=r(n), s=s(n) in the proof of 6.482, we find and so

8,

k) with

T(n+ I, I, p) =T(n, I, p) T(n, I, p) =T(O, I, p) =T(I, p)

which shows that the sum a( I) + a( 2) -I- ... + a(p + I) is unchanged by any rearrangement of its terms.

CHAPTER VII

ELIMINATION OF PARAMETERS 7. In this chapter we undertake the elimination of the parameters in the schema of definition by recursion to which we referred in § 1.6. This elimination is effected by establishing an appropriate numbering of the totality of pairs of natural numbers. A very simple numbering, sufficient for the first stages in the elimination is given by the function J(u, v)=(U+V)2+ U. To show that J(u, v) assigns a unique number to each ordered pair (u, v) we introduce in turn the functions

Ex = x .s: (Rt

X)2

where Rt x is the function defined in § 1.6 as the greatest integer whose square does not exceed x,

Ux=Ex, Vx= Rt x-'-Ex ; Ex is the amount by which x exceeds the nearest square contained in it. Since (U+V)2<J(U, v) «u+v+ 1)2 therefore and so

Rt J(u, v) =u-j-v UJ(u, v)=u, VJ(u, v)=v.

These equations prove that J(u, v) assigns each ordered pair a unique number, for if J(u, v) =J(u', v') then u = UJ(u, v) = UJ(u', v') = u'

and

V= VJ(u, v)= VJ(u', v')=v'. 134

ELIMINATION OF PARAMETERS

135

7.1 We start by showing how a single parameter may be eliminated. Consider the recursion R

f(u, v, O)=a(u, v), f(u, v, Sx)=b(u, v, x, f(u, v, x))

in which there may be parameters additional to the u and v indicated. We proceed to derive this function f from functions defined by recursion in fewer parameters. Let the function F(w, x) be defined by the recursion R'

F(w, O)=A(w) , F(w,Sx)=B(w, x, F(w, x))

where A(w)=a(Uw, Vw), B(w, x, y)=b(Uw, Vw, x, y) so that F(w, O)=f(Uw, Vw,O) and {F(w, x)=f(Uw, Vw, x)}

--+

{F(w, Sx)=f(Uw, Vw, Sx)}

which proves that F(w, x)=f(Uw, Vw, x) ;

whence, substituting J(u, v) for w, we find f(u, v, x) = F(J(u, v), x) ,

so that f is obtained by substitution from F, and F is defined by the recursion R' with one fewer parameter than there is in the initial recursion R. By repeated application of this reduction process we may derive t by substitution from a function defined by recursion in a single parameter. We may suppose that this function is in fact F defined by R' above, where the functions A and B contain no other variables than those indicated. 7.2 We show next that the variables wand x may be eliminated from B. Define qJ(w, x) by the recursion R"

where

qJ(w, 0) = lX(W) ,qJ(w, Sx)=P(x, qJ(w, x)) lX(W)=.J(w, Aw), P(x, y)=J(Uy, B(Uy, x, Vy));

it readily follows that qJ(w, x) =J(w, F(w, x)),

136

ELIMINATION OF PARAMETERS

for the equation holds for x = 0, and the equation in x implies that with Sx in place of x. F(w, x) is derivable from rp(w, x) by substitution only since F(w, x)= Vrp(w, x).

Next define rp'(w, x) by the iteration I

rp'(w, 0) =iX'(W) , rp'(w, Sx) =f3'rp'(w, x)

where

iX'(W)=J(O, iX(W)) , f3'y=J(SUy, f3(Uy, Vy));

as before we prove rp'(w, x)=J(x, rp(w, x))

by observing that this equation holds for the value 0 of x, and that J(SUJ(x, rp(w, x)), f3(UJ(x, rp(w, x)), VJ(x, rp(w, x)))) =J(Sx, f3(x, rp(w, x))) =J(Sx, rp(w, Sx)) ,

so that

rp'(w, x) =J(x, rp(w, x))

implies

rp'(w, Sx)=J(Sx, rp(w, Sx)).

The function

tp'

is defined by iteration without parameters and rp(w, x)= Vrp'(w, x).

This completes the reduction of the schema of primitive recursion to the simple form j(w, O)=a(w) , j(w, Sx) =b (f(w, x)) .

7.3

A further simplification may be effected by taking

R*

so that

F(w, O)=w, F(w, Sx)=b (F(w, x)) j(w, x)

=

F(a(w), x).

To eliminate the parameter w from the recursion R * we introduce new pairing functions J(u, v) = ((u+ V)2 + U)2+ v Ux=E Rt x, Vx=Ex.

ELIMINATION OF PARAMETERS

Since therefore

137

((U+V)2+ U)2<;J(U, v) < ((U+V)2+ U+ 1)2 RtJ(u, V)=(U+V)2+ U, EJ(u, V)=V

and

E Rt J(U, v) = u ,

so that

UJ(U, V)=U, VJ(U, V)=V.

This second set of pairing functions has the additional property: if E Sx>O then U SX= Ux and V Sx=S Vx; for if E So:» 0 then Sx is not a square and so RtSx=Rtx

whence and

U Sx=E Rt Sx=E Rt x= Ux V Sx=Sx-'- (Rt SX)2= (I +x) -'- (Rt X)2 = {x-'- (Rt X)2}+ (1-'- {(Rt X)2-'-X}) = {x-'- (Rt X)2}+ 1 =S Vx

since E Sx> 0 entails (Rt SX)2 « S» and therefore (Rt X)2 <; x so that (Rt X)2-,-X=O (which by example 2.83 holds also when E Sx=O). If we now define G(x) by the recursion without parameter II

where so that

GO=O,GSx=c(x,Gx) c(x, y) = (1-'- E Sx). U Sx + {I-'- (1-'- E Sx)}· b(y), c(x, y)= U Sx if E Sx=O =b(y)

if E Sx>O,

then we can show that G(x)

=

F(Ux, Vx);

the proof of the case x = 0 is obvious. If E Sx = 0, so that V Sx = 0 then

F(U Sx, V Sx)= U Sx =c(x, Gx) =GSx,

138

ELIMINATION OF PARAMETERS

and if E Sx » 0 then F(U Sx, V Sx)=F(Ux, S Vx)=b (F(Ux, Vx» =c(x, F(Ux, Vx»

and so from the hypothesis G(x) = F(Ux, Vx) we derive G(Sx)=F(U Sx, V Sx)

which completes the proof. F is obtained from G by the substitution F(u, v) =GJ(u, v) .

The variable x may be eliminated from c(x, y) exactly as from the function B above. We define HO=O,HSx=dHx

where dy=J(S Uy, c(Uy, Vy» , Hx-e Jt», Gx).

The proof of the case x = 0 is evident, and from the hypothesis Hx-e Jt», Gx) follows H Sx=d Hx-ed J(x, Gx) =J(Sx, c(x, Gx» =J(Sx, GSx)

as required. Thus Gx is derived from Hx by the substitution Gx=V Hx

and H x is defined by the schema H 0=0, H Sx=d Hx,

so that

Hx=d"'o.

To carry out this reduction we have introduced a number of

ELIMINATION OF PARAMETERS

139

auxiliary functions all of which are definable by substitution in terms of the functions u+v, U-'--v, u·v, Rt u,

which we shall call the initial functions. We have therefore proved that if these four functions are available then all primitive recursive functions are obtainable by substitution and application of the single schema Fx = A "'0 for generating functions of one variable. 7.31

It can be shown that in fact just the two initial functions u+v,Eu

are sufficient but this reduction in the initial functions though of great interest in showing that a single function of two variables enables us to reach all primitive recursive functions of any number of variables, is not essential to our purpose and will be omitted. 7.4 In the construction of primitive recursive functions of one variable it is never necessary to define by substitution a function of more than one variable. For instance if we can obtain a function f(n) from a function F(x, y) by taking, say, f(n) = F(g(n), h(n»

with F(x, y) defined in terms of given functions a(u, v), b(x, y), c(x, y) by the substitution F(x, y) = a(b(x, y), c(x, y»,

then since f(n)=a(b(g(n), h(n», c(g(n), h(n)))

instead of defining the two variable function F(x, y) we could define the one variable functions {3(n)=b(g(n), h(n» , y(n)=c(g(n), h(n»

and so define f(n) directly by f(n)=a({J(n), y(n».

140

ELIMINATION 0]' PARAMETERS

It follows that all one variable primitive recursive functions may be obtained using only the schemata t(n)=g(n}+h(n) , t(n)=g(n)-'-h(n) , t(n)=g(n)·h(n) , t(n)=Rt n, t(n)=g(h(n)) , t(n)=gno.

7.5 This last result enables us to enumerate all one-variable primitive recursive functions. The enumerating function is 91m(n) defined as follows: for the first four functions in the enumeration we take the functions 0, n, Sn, Rt n, so that !po(n) =0 ,91l(n)=n, CfJ2(n)=Sn, CfJ3(n)=Rt n.

Next we assign five ranges of values of m to the operations +, -'-, , substitution and recursion. Denoting as in § 4.61 the exponent of the greatest power of the kt h odd prime in n by v(n, k), we associate each operation with the cases v(m, 0)=0,1,2,3 and v(m, 0»3, for values of m>3. Consider now the function 91m(n) whose values are listed in the following table. Condition on m m=O m=1 m=2 m=3 m>3 m>3 m>3 m>3 m>3

I

Value of 91",(n)

o n Sn

Rt n &v(m, &v(m, &v(m, &v(m, &v(m,

0)=0 0)=1 0}=2 0)=3 0»3

91"(m,l) (n) + 91.(m.2) (n) CfJ'lm.l) (n) -'- CfJ'lm.2) (n) CfJ.(m.l) (n)· 91.(m.2) (n) CfJ.(m.l) (91.(",.2) (n)) {(1-'- (1-'- n)}CfJ"lnI, 11(CfJ",(n -'- I)}

We shall show that the sequence CfJo(n) , 91l(n) ,
is an enumeration of all one variable primitive recursive functions, (with repetitions).

ELIMINATION OF PARAMETERS

141

The sequence contains the initial functions 0, n, Sn, and Rt n. If it contains the functions g(n) and h(n) then there are integers p, q such that g(n)=!p1'(n) , h(n)=!p,An) .

If we choose a value of m. for which v(m, 1)=p, v(m, 2)=q then we shall have !Pv(m,l)(n)=g(n) , !Pv(m,21(n)=h(n);

such a value of m is for instance

where the factor 7 has been included to ensure that m:» 3. Denote 2/c· 31'· 5q • 7 by m/c so that v(m k, 0) = Ic , v(m k, 1) = P , v(m k, 2) =q and m k> 0

and therefore for the values k=O, 1,2,3 the functions tpmk(n) are respectively y(n) +h(n) ,g(n) -'-h(n) , g(n) ·h(n) ,g(h(n))

and for k> 3

whence it follows that the sequence !Pm(n) , m=O, 1,2, ... contains all one-variable primitive recursive functions. To represent !Pm(n) conveniently in a single expression we write o,,(n) for 1-'-llc, nl and ek(n) for 1-'-(k-'-n), so that °k(n) = 1 ifn=k and is zero otherwise, and ek(n) = 1 if nr»]: and is zero otherwise. Then !Pm(n) may be summarised in the expression 0r(m) ·n+o2(m) ·Sn+o3(m), Rt n + e",(m ){oo(v(m, 0))· (!Pv(m,!) (n) + !Pvlm, 21(n)) + or(v(m, 0))· (!Pvlm,!1 (n) -'-!Pv(m,2) (n)) + 02(v(m, 0))· (!Pvlm,!) (n). !Pv(m,2) (n)) + 03(v(m, 0))· (!Pvlm,!) (!P.(m,2) (n))) +e4(v(m, O))·er(n) !Pv(m,2)(!Pm(n-'-1))}.

142

ELIMINATION OF PARAMETERS

As a function of the two variables m, n the function Pm(n) is not primitive recursive; for if it were then Pn(n) + 1 would be primitive recursive. But Pm(m) + 1 > Pm(m) so that p,,(n) + 1 is not the function Pm(n) for any value of m and so Pn(n) + 1 is not contained in the enumeration of all one-variable primitive recursive functions. The definition of Pm(n) shows that the value of Pm(n+ 1) is given in terms of Pm(n) and Pm,( n), Pm, (n) where m I , m 2 are both less than m; thus Pm(n) is defined by a course-ofvalues double recursion which may be transformed into a standard double recursion by the method of the previous chapter. Accordingly !JJm(n) is an example oj a junction defined by double recursion which cannot be defined by primitive recursion and s1tbstitution alone.

CHAPTER VIII

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC 8. The principal aim of this final chapter is to show that there is a primitive recursive function I(n) such that each of the equations 1(0)=0,/(1)=0,/(2)=0, ... is provable in f!/t* but the equation l(n)=O, with variable n is not provable in f!/t*. The existence of such an 1 shows that f!/t* is incomplete, that is to say that there is an equation l(n)=O

which is verifiable but not provable in f!/t*. . This remarkable result, which was discovered by Kurt Godel in 1931, suggests that the natural numbers may not be the only class of objects for which the provable formulae of f!/t* are verifiable and that there may exist a class of entities which includes the natural numbers and other entities besides for which the provable formulae of f!/t* are verifiable. That such a class in fact exists was actually established three years after Godel's result, by Thoralf Skolem, the inventor of resursive arithmetic, who showed (independently of Gadel's methods and results) that not only systems like f!/t or f!/t*, but every [ormalieation. 01arithmetic fails to characterise the number concept completely and admits as values of the number variables a class of entities of which the natural numbers form only the initial segment.

8.1

The construction of the unprovable equation f(n)=O

is effected by means of a code in which the terms of f!/t* are expressed by numbers, and the syntax of f!/t* by primitive recursive relations. This code is called a Godel numbering of f!/t*. 143

144

GODEL NUMBERING AND THE INCOMPLETENESS 01' ARITHMETIC

We shall take the variables of [jt* to be m, n, Xl' X 2, X 3 ••• ; the functions to be 'Po(n), 9?1(n), 'P2(n), ... together with m-sn, m-i- n. and m- n; and the proof schemata to be

E

F{O)=O, F{n)=F{Sn) F{n)=O

T

A=B A=C B=C F{n)=G(n) F{A)=G{A)

A=B F'(A) = li'{B)

For axioms we shall take A

m+(n-"--m)=n+(m-"--n) ,

P

Sm-"--Sn=m..:-n,

the defining equations of the functions m. + n, m -"-- nand m· n, and the defining equation of lpm(n) only. Thus the various primitive recursive functions of one variable are all defined by the single axiom, the defining equation of 'Pm(n).

8.2 The several elements of [jt* are numbered as follows: to the functions 'Pk we assign the numbers 4k+ 17 , k= 0, 1, 2, ... and to the variables Xk the numbers 4k + 19 , k = 1, 2, .... The numbers of the remaining signs are as indicated in the table

OII(I)I+I-'-llmln 1

I

3

I

5

I

7

I

9

I 11 I 13 I 15 I 19

Now any formula of [jt* is simply a sequence of signs of [Ji*; if are the numbers of the component signs in a formula (in the correct order) then we assign to the formula the number 2"0. 31/,1 . 5"'2 ...• 'lJ~k

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC

145

where t:>k is the kth odd prime. From the number of the formula alone we can write down the full formula, simply by factorising the formula-number. For instance the number of the formula

m+O=m is 2I S • 39 . 5 . 73 . 11 15 and the number of the formula

m·Sn=m·n+m is

215.313.525. 719 · 113 .1315.1713.1919.239.2915

since'S' is standing for 'fP2'. Single signs all have odd numbers, but the number of a sequence is necessarily even. Of course 'nonsense' sequences of symbols like )+fP3= (

also have numbers but this does not give rise to any complications. 8.3 Next we observe that since a proof in f!,f* is a sequence of formulae then proofs too may be numbered. Thus a proof which consists of a sequence of formulae with numbers

10' II' 12' ... , Ire , is assigned the number

so that if N is the number of a proof and k is the greatest value of r for which p(N, r) > 0 (and therefore N is divisible by a non-zero power of t:>k' and t:>k is the largest prime for which this is true) then v(N, k) is the number of the formula proved by proof number N. We shall see how to single out the values of N which are in fact numbers of proofs. 8.31 Like all formulae the axioms of f!,f* may be numbered, and we shall denote by AI' A 2 , A 3 , A 4 , As and A 6 the numbers of the 6 axioms. We shall have no occasion to require the actual values of the numbers denoted by these letters. We denote by Ax(n) the

146

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC

disjunction n=A 1 v n=A 2 v n=As v n=A, says that n is the number of an axiom.

v

n=A 5

v

n=A 6 which

8.4 The first step in arithmetising the syntax of fJ.i* is to find the relation which expresses the property of being a variable. Since the numbers of the variables are 15 + 4k where k runs through the numbers 0, 1, 2, ... therefore the property 'n is the number of a variable' is expressed by the primitive recursive relation E:'(n=4m+ 15)

(there is an m between 0 and n such that n=4m+ 15). We denote this relation by V(n). 8.41 The relation 'n is the number of a one-variable primitive recursive function' is similarly E:'(n=4m+ 17)

which we denote by t(n). 8.42 We consider next the relation 'n is the number of a variable in formula number f'. This is expressed by the primitive recursive relation E~{p(f) m)=n &: V(n)}

which we denote by Vj(n). 8.43

The number of terms in sequence number f is If + 1 defined by

If= (L~{R(f, ~m) = 0

&: A~(n>m -+

R(f, ~m) >O)}). {R(f+ 1, 2)}

where R(a, b) is the remainder when a is divided by b. (Thus lj = 0 if f is odd, and if f is even l, is the least integer m such that f is divisible by the m t h odd prime but not by any greater prime). The predicate F(n) which says that n is the number of a formula may now be expressed by R(n, 2)=0 &:A;,,+1{R(v(n,r),2)=I}. 8.5 Before we consider the fundamental syntactical operation of substitution we introduce the primitive recursive function mAn which determines the number of the expression formed by writing expression number n after expression number m, For

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC

147

instance, the number of '9?4(rn)' is 233.35.515.77 and the number of '+n' is 29.319 then the number of '9?4(rn)+n' is 233 . 35 • 515 • 77 A 29 . 319 = 233 • 35 • 515 • 77 . 119. 13 19

•

The function rn A n is defined by the equation m. A n = m- II i~ln

".»

h VI

1"1",+0+1

provided that neither m. nor n is the number of a single term, i.e, provided that rn and n are both even. If m. is odd and n even n -~ ')m. II ~

""" A

h

V (" ' ; )

,,,,I,, 1"'+1

""

and if rn is even but n is odd then rn

A

h" n-rn'1" lm+1 ,

and if m. and n are both odd m. A n=2 m • 3" 8.51

The product

j(O)

A ... A

j(n)

is

defined to be II" j(i) i~n

where II" j(i)=j(O) , II" i~O

j(i)={ Il" I(i)}"f(n+l).

i~w.+l

i~n

8.6 We are now ready to express the relationship which holds between three numbers f, v, n and the number of the expression which is obtained by substituting object number n for the variable number v in formula number j. Now i> Il ~~(t,') and for each value of i for which v(f, i)=v, i",II+1

the variable whose number is v is to be replaced by object number n. Let us then define qi= (1 ~ Iv(f, i), v/)n+ {I ~ (l .s: Iv(f, i), v!}.v(f, i) ,

and Subl(vjn) =

Il"

q, ,

,,,,11+ 1

then Sub/.(vjn) is the desired number of the expression obtained by substituting object number n for variable number v in formula

148

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC

number

f. For instance, the number of the formula

IS

and the number of the formula

is the result of substituting

Xl +X2

for m in 1Jl",(m) is

<JI",(x1 +x2 )

with number which is equal to

33 A 5 A 223 • 39 • 527 A 7 .

8.7 Next we seek to characterise the concept 'recursive terms' or 'recursive function'. Now t is the number of a recursive term if t is the number of a variable or the number of one of the functions m+n, m-i- n, m·n, IJlk(n) or is the number of the expression obtained by substituting a recursive term for a variable in a recursive term. The numbers of the functions m+n, m.:-n, m·n and <JIk(n) are 215 • 39 • 519 ; 215 . 311 • 519 ; 215 • 313 • 519 ; and 2"'k+ 17. 35 . 519 • 77 ; and we denote these by aI' a2 , a3 and a(k) respectively. Thus the predicate T(n), which says that n is the number of a recursive term may be defined as (n=a1 ) v (n=a2 ) v (n=a3 ) v

E~(n=a(k))

v E~E~E={x
Vz(z) "n=Subx(z/y)}.

This is a course-of-values recursion so that T(n) is a primitive recursive relation. 8.71 It is now easy to formulate the condition E(n) that n is the number of an equation between recursive terms. E(n) may be expressed in the form E:E~(T(x) "T(y) "n=x

A

3

A

y)

8.8 The most important part in the arithmetising of syntax is

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC

149

the formulation of the predicate 'n is the number of a proof'. To this end we must consider the relations between the numbers of premisses and conclusions in the derivation schemata. 8.81 For Sb v we take the primitive recursive relation Sl(m, n) which expresses the relation: n is the number of an equation derived from equation number m by S~. Sl(m, n) takes the form E':E:E':E~{T(x) "T(y) 8< Vm(z) "m=x

A

3

A

Y "T(w) " n = Subm(zlw)}.

Sl(m, n) says that there exist x, y, z between 0 and m and w between o and n, such that x, y and ware numbers of terms, and z is the

number of a variable in equation number m; m is the number of the equation between the terms numbered x and y, and n is the number of the equation which results from it by substituting term number w for variable number z. 8.82

For Sb 2 we take S2(m, n) in the form:

E':E'::E:E:{T(x) "T(y) 8< m ».x A 3 A Y "T(u) " Vu(v) 8< n=Subu(vlx)

A

3

A

Subu(vly)}

(x and yare term numbers less than m, n is a term number and v

the number of a variable in term number u, m is the number of the equation between terms numbered x and y, and n is the number of the equation between the terms obtained by substituting first term number x, then term number y, for variable number v in term number u). 8.83

For schema T we take T(m, n, p) in the form:

E:E'::E~(T(x) "'J1(y)

"1'(z)

8<

m=x A 3 A Y & n=x A 3 A

Z

8
150

GODEL NUMBERING AND THE INCOMPLETENESS OF AltTI'HMETIC

8.84

Finally for schema E we take E(m, n, p) defined as:

E:E~(T(x) & V.,(y) & m=Sub.,(yj1) A 3A

1

&n=XA 3A Sub",(yj25A y) &p=XA 3A 1)

(x is the number of a term and y the number of a variable in this term; m is the number of the equation formed by equating to zero the result of substituting zero for the variable in term number x, n is the number of the expression formed by equating term number x to the result of prefixing'S' to the variable in this term and p is the number of the expression obtained by equating to zero term number x).

8.85 We are now ready to define the predicate 'n is the number of a proof', denoted by Pf'(n). We take Pf(n) to be R(n, 2) =

°

&

A;+l{Ax(v(n, x)) v E~(Sl(v(n, y), v(n, x)) v S2(v(n, y), v(n, x))) v E:E~(T(v(n, u), v(n, v), v(n, x)) v v E(v(n, u), v(n, v), v(n, x)))}

(n is the number of a sequence of equations each of which is either an axiom or is derived from one or two previous equations by one of the derivation schemata).

8.86 As we have already observed the relation Pr(m, n) which says that m is the number of a proof of formula number n, is expressed by Pf(rn) & virn, lm+ l)=n.

8.87 The numerals 0, SO, SSO, SSSO, ... are represented in /f4!* by 0, f{J20, f{J2f{J20, f{J2f{J2f{J20, ... and so their numbers are 1, 225.3, 225 . 3 25 . 5 , 225.325.525.7, ... ; if we denote these numbers by No, N v N 2, N 3 , ... then N. is primitive recursive and N o= 1, N 1 = 225.3, N.+ 1 =N r .~;4 ·~r+l. r;;?: 1 .

Write Stj('Oln) as an abbreviation for Subj('OjNn ) so that Stj('Ojn) is the number of the expression which is obtained when the numeral f{J2 f{J2 ••. rp20, with rp2 occuring n times, is substituted for variable number v in expression number [,

GODEL NUMBERING AND THE INCOMPL.ETENESS OF ARITHMETIC

151

Since "-' Pr(m, Stn (l 9jn )) is a primitive recursive relation it may be written in terms of the symbols of &f* as an equation, and we may speak of ',,-, Pr(m, St n (l 9jn ))' as an abbreviation for this equation in &f*. Let the equation in &f* for which Pr(m, St n (l 9jn )) is an abbreviation have number p. In formula number p substitute the numeral number N p (i.e. f{J2f{J2 ••• f{J20 with f{J2 occurring p times) for the variable number 19 (i.e, the variable n) ; the number of the resulting expression is Stp (19jp ) and the expression itself may be abbreviated to 8.88

e-,

"-' Pr(m, Stp (19jp )) and denoted by G(m), say. We shall show the expression thus denoted by G(m) is not provable in &f*. For if the expression which G(m) denotes were provable in &f*, and if k were the number of the proof then k is the number of the proof of equation number Stp (19jp ) and the relation 8.89

Pr(k, Stp (19jp )) holds; but if G(m) is provable then so is G(k) and therefore the relation ,...., Pr(k, Stp (19jp)) holds. As ,...., Pr(m, Stp (19jp)) is a primitive recursive relation, its representing function t(m), say, is primitive recursive and the value of t(k) may be obtained from the defining equations of t(m) by not more than k substitutions. This value will be either 0 or 1; if the former then e-, Pr(k, Stp (19jp )) holds and if the latter then Pr(k, Stp (19jp)) holds, but the two relations cannot hold simultaneously. Thus the expression denoted by G(m) is not provable, which is equivalent to saying that the equation f(m)=O is not provable in &f*, where f(m) is the term in f!£* by which the

152

GODEL NUll1BERING AND THE INCOMPLETENESS OF ARITHMETIC

representing function of the equation which G(m) denotes is expressed. However the assertion that this equation is not provable is equivalent to saying that none of the numbers 0, 1, 2, 3, ... is the number of a proof of equation number Stp(19/p), i.e, that each of the equations f(0)=0, f(1)=0, f(2)=0, ... holds and is therefore derivable in !Jl*. Thus there is a primitive recursive function the equations

f such that each of

f(0)=0, f(1)=0, f(2)=0, ... is provable in !Jl*, but the equation

f(m)=O with a free variable m is not provable in !Jl*. 8.9 The presence in !Jl* of an unprovable equation is not a mere accident arising from the particular axioms of !Jl*; we may add as many more (verifiable) axioms as we please to!Jl* and still carry through the foregoing construction. In particular we may add the equation f(m) = 0 itself as an axiom and still determine an unprovable equation in the extended system!Jl+ say. This result suggests that the natural numbers 0,1,2, .. , do not exhaust the values of the variable. We can in fact readily show directly that the natural numbers do not constitute the largest class of objects which satisfy the axioms of recursive arithmetic. 8.91 To this end we prove that, given any sequence of functions lot, lIt, Izt, lat, and so on, (recursive or not) there is a monotonic increasing function y(t) and a function v(i, j) so that, for all i, j, one of the relations I.y(t)
holds for all t » v( i, i). We start by arranging all pairs of functions I" I., with r «:s,

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC

153

in a simple sequence, starting with the pair 10' 11 followed in turn by 10' 12; 10' la; 11' 12; 10' 14; 11> t-: 10' 15; 11> 14; 12' la and so on, the pair hi; coming (e2+i+l)th or (e2+e+i+l)th in the sequence according as i + j is even or odd, where 2e is the greatest even number contained in i+j. For any two functions q;(t), 'IjJ(t) we can find a monotonic increasing function v(t) such that one of the relations cpv(t) <'ljJv(t) , q;v(t) ='IjJV(t) , epv(t) >'IjJV(t)

holds for all values of t; to determine v we divide the natural numbers t into 3 classes 0<, 0 ~ or 0> according as the relation <, = or > holds between cpt and 'ljJt. At least one of the three classes is infinite and we take the members of that class in increasing order of magnitude as the values of v(t) for the values 1,2,3, ... of t, and we write V for the relation «, = or » which holds between epv(t) and 'IjJV(t) for all values of t. Next we define in turn v1(t), v2(t) ... with associated relations VI' V 2, ... as follows; v1(t) and VI are the function and relation determined as above so that lov1(t) V1/1v1(t)

for all values of t; then we take lov1 and 12v1 for cp and determine v2 , V2 so that

'IjJ

and

lov1v2(t) V 2/2vl v2(t)

for all t. In this way we determine step by step Va' Va; v4' V4 ; V.. , V.. so that (E)

for all t, where n is the ordinal of the pair prescribed above. Write

g(n)=v1v2

•••

v"_lv..(n)

so that (taking t = n in equation E) l,g(n) Vnl i g(n) .

t.. i, in

the ordering

154

GODEL NUMBERING AND THE INCOMPLETgNESS OF ARITHMETIC

Similarly substituting v n+1V"+2 ... v,,+p(t) for t in equation E we find that l,v1v2 ... v,,+p(t) V"livlv2 ... vp+,,(t) for all t, and therefore, taking t = n + p,

which proves that

holds for all t ;»n. Since vn+1(t) is monotonic increasing (and its values are all different whole numbers) therefore vn+1(n+ l»n so that g(n+ 1)=V1V2 ... vnvn+1(n+ 1»V1V2 .,. vn(n)=g(n)

which proves that g(n) is monotonic increasing. 8.92 By means of this theorem we may introduce a linear ordering between the functions 10,11' 12' 13' .... We write

1,
1

or

1.=/1 or

li>11

according as V n has the value <, = or >, i.e. according as

•

for all t:» n. It is readily verified that

and t, < II ' t, < t, implies t,
GOOEL NUMBERING AND THE INCOMPLE'I'ENESS OE' AHITHMETIC

for if there were a function h such that should have

155

1
Ig(t) < hg(t) < Ig(t) + 1

for all sufficiently great t, which is of course impossible. Thus the terms of the sequence 10' 11' 12' la, ... share with the natural numbers the property of having a unique successor but the sequence has members greater than every natural number, since for instance the identity function I(t) =t exceeds every constant function, because Ig(t) has the value get) which grows with t beyond any assigned amount. 8.93 Let us now take for to' t.. 12' la, ... the sequence of all one variable primitive recursive functions with to as the zero function. We shall show that all the true statements of recursive arithmetic remain valid if we take to, tv t2' 13' ... in place of the natural numbers. Given any recursive function F(x l , x 2'

x,,)

... ,

of natural numbers, we introduce function variables '1'l(t), '1'2(t), ... whose 'values' are the functions t.. t2' .... For each assignment of values of the variables F( '1'I> '1'2' ... , '1',,)

becomes a one variable function

say, which we take to be the value of the function

for the values

1,,1'

tk2'

... ,

t.; of the

arguments

({iI'

({i2' ... , cp".

In particular when we give for values of CPl. ({i2' .,. , cP", n constant functions aI' a 2 , ••. , an say, F takes as value the constant function F(a t , a2 ,

... ,

a,,)

156

GODEL NUMBERING AND THE INCOMPLETENESS OF ARITHMETIC

which is of course exactly the same as the value of F(xl , x 2' .•. , x,,) for the values a v a 2, ••• , an of the arguments Xv X 2, ••• , X n• Thus a function of natural number variables may be interpreted as a function in the space of the functions 10' 11' 12' ... without changing its significance. 8.94 Under this interpretation a true equation (whether an axiom or derived from the axioms)

F(xl , x 2 ,

••• ,

x n ) =G(xl , x 2 '

••• ,

xn )

becomes a true equation in the function space; for if this equation holds for all Xv X 2' ... 'Xn and if i, j are the suffixes for which F(fkl(t), Ik2(t),

, Ikn(t)) = I.(t)

G(hl(t), Ik2(t),

, fx.,,(t)) = I;(t)

then

1;(t)=I;(t)

and so which proves

I g(t) = I;g(t)

I; = I

for all arguments

j

for all t for all t

and therefore

i.:r r= 1,2, ... , n.

8.95 Since we have defined equality and inequality independently it remains to be verified that the equations X.;;; y and X -'- Y = 0 are equivalent under the functional interpretation.

For the truth of li.;;;l; we require 1;fJ(t).;;;I,{J(t) for t>v(i, j); for the truth of 1;-'-1;=0, writing 1;-'-I;=lk' we require I~(t) = log(t)

for t> v(k, 0) .

Since I~(t) = hJ(t) -'-I;g(t) and I~(t) = 0, therefore the conditions I~(t) = I~(t)

and lo(t).;;; I;g(t)

are equivalent for any t, which completes the proof.

Solutions to Examples I 1.

We define in turn Pent(x, 0) = 1 , Pent(x, Sy) = Tet(x, Pent(x, y)) Hex(x, 0) = 1, Hex(x, Sy) = Pent(x, Hex(x, y)) Hept(x, 0) = 1 , Hept(x, Sy) = Hex(x, Hept (x, y)).

1.1

Tet(2, 3)=16, Pent(2, 3) = Tet(2, 4)=216=65536 2

Hex(2, 3)=Pent(2, 4) = Tet(2,65536) = 22 .' with 65536 factors. 1.2

1(1, n)=2, 1(2, n) = 3-2", 1(3, n)=3 n.2"2-..nu, 1(4,4)=2 116.3 18.5

1.3

1.31

I (x) = {I -'-(I -'- x)}g(x)

which is formed by substituting the recursive functions 1-'- (1-'- x) for u and g(x) for v in the recursive function uv. If k(x) = 1-'- (x -'- 2) then

I (x) = 1.4

k(x)g(x) + {I-'- k(x) }h(x).

Let Q(x, y), R(x, y) denote quotient and remainder when x is divided by y and Hf(x, y), Lm(x, y) the highest common factor and least common multiple of z, y. Then R(O, 2) = 0, R(x+ 1,2) = I-'-R(x, 2) , Q(O, 2)=0, Q(x+l, 2)=Q(x, 2)+R(x, 2), Hf(x, 2) = 2 --'-- R(x, 2) , Lm(x, 2) = x{I + R(x, 2)} . Similarly R(O, 3)=0, R(x+l, 3)={I-,-(R(x, 3)--'--I)}SR(x, 3), Q(O, 3) = 0, Q(x + 1, 3) =Q(x, 3) + {R(x, 3) -'-I)} Hf(x, 3) = I + 2(1-'- R(x, 3)) , Lm(x, 3) = x{3 -'- 2(1-'- R(x, 3))} .

1.5

I (x) =

{I-'- R(x, 2) }g(x) + R(x, 2)h(x) . 157

158

SOLUTIONS TO EXAMPLES

If C(n, r) is the coefficient of x' in the expansion of (I +x)" then C(n, 0)=1, C(O, r+l)=O C(n+ 1, r+ 1)=C(n, r)-+-C(n, r+ 1); the function C(n, r) is nevertheless primitive recursive, as n! is primitive recursive and C(n, r)=n!jr! (n-r)! 1.6

1.7

a+b+c+d=((a+b)+c)+d =

(a, + (b+c»+d

=((b+c)+a)+d =((c+b)+a)+d=c+b+a +d = (b+c)+ (a+d) =(b+c)+(d+a) = ((b+c) +d) +a =(d+(b+c))+a = ((d+b)+c)+a=d+b+c+a, etc. 1.8

i(x) = I--'-x, j(x, y)={I--'-(I--'-x)}(I-'--y) , k(x, y)={I--'-(I--'-x)} {1--'-(l--'-y)}.

1.81

j(x, y)=i(x)·a(y)+j(x, y)·b(x-'-I)+k(x, y).c(x-'-I, y--'-l)

1.9

G(O, n)=tp(l, n+ I) =A(O, n, 'lp(O, p(O, n, tp(l, n»), tp(l, n» =a(n) , G(p+ I, 0) =tp(p+ 2, I)

=

b(p, tp(p + 1, p(p + 1, 0, tp(p+ 2, 0»), tp(p+2,0»

=b(p, tp(p+ I, Sd(p))) =b(p, G(p, d(p»)

and G(p+ 1, n+ I)=tp(p+ 2, n+ 2) =I,(P+ I, n+ I, tp(p+ I, p(p+ 1, n+ 1, tp(p+ 2, n+ 1»), tp(p+ 2, nH» =c(p, n, G(p, e(p, n, G(p+ I, n))), G(p+ I, n».

SOLUTIONS TO EXAMPLES

159

This example shows that a function G(p, n) defined by doubly recursive equations R 2 may be obtained by the substitution G(p, n)=lp(p+ I, n+ I) from the doubly recursive function lp(p, n) which satisfies the simpler introductory equations

R;

11'(0, n)=lp(p+ 1,0)= I '!f'(P+ I, n+ 1) =A(p, n, lp(p, p(p, n, lp(p+ I, n))), ( "P(P+ I, n)).

It follows that we may take the equations R; as the standard introductory equations of a function defined by double recursion. 1.91

1(0, n 2 , n 3)=/(nl

+ 1,0, n 3)=/(nl + I, n 2+ 1,0)= I

l(n1+ I, n 2+ I, n 3+ I) =a(nl> n 2,

na, 11(nr, n 2, na), 12(n1, n 2, n 3,), 13(n1' n 2, n 3))

where

11(nl> n 2 , n 3) = 1(nl> b(n1, n 2, n 3, 1(n 1 + I, n 2+ I, na)), c(nl> n 2, n 3, l(n1+ I, n 2+ I, n 3))) Mn 1' n 2, n 3) = l(n 1+ 1, n 2, d(nl> n 2 , n 3, fa(n1, n 2, n 3) = l(n1+ 1,11 2 -i-I, na).

ti». + I, n 2 + 1, n 3)))

Solutions to Examples II 2.

Let
2.01

Write f(c), g(c) for a(b+c), ab+ac respectively; then I(O)=ab=g(O) and f(Sc)=a(b+Sc)=aS(b+c)=f(c)+a, g(Sc)=ab+aSc=ab+ (ac+a) =g(c) +a. Thus f(x) and g(x) satisfy the same introductory equations, so that f(c)=g(c).

2.02

(ab). 0 = O=a(b· 0); (ab)Sc = (ab)c+ (ab) and a(b·Sc) =a(bc+b)=a(bc) + (ab), whence the proof is com-

pleted as in 2.01. 2.03

(ab)· (cd) = aib- (cd)) = a«(cd)·b) = (a(cd))b =

«ac)d)b= (ac)·(db) .

2.1

If f(x)=x(l-'-x) , 1(0)=0 and I(Sx)=O.

2.2

1-'-0=0°, I-'-Sx=O=Os".

2.201

(1-'-0)+ {1-'-(I-'-O)}=SZO, (I--=--Sx)+{I--=--(I--=--Sx)}=SZSx so that (l--=--x) + {I--=-- (l-'-x)}=SZx= 1 .

2.21

a--=--(b+O)=a-'-b=(a--=--b)-'-O; a-s- (b+Sc) =a-'-S(b+c) = {a--=-- (b+ c)} -=-1 , and (a .z: b) --=-- Sc = {(a--=-- b) -'-c}--=-- 1 .

2.22

Use example 2.21.

2.23

(a+O)--=--(b+O)=a--=--b; (a+Sx) --=-- (b+Sx) =S(a+x) --=--S(b +x) = (a+x) -'-(b+x) ,

whence the result follows as in example 2. 2.231

a(O--=--I)=O=a·O--=--a; a(Sx--=--I)=a(Sx--=--SO)=ax and a·Sx--=--a=(ax+a)--=--a=ax, by 2.23. 160

SOLUTIONS TO EXAMPLES

161

2.232

a(b -=- 0) = ab = ab -=- a- 0 ; a(b -=-Be) = a{(b .z, e) -=- I} = a(b -=- c) -=- a by 2.231, and ab -=-a·Se =ab -=- (ac +a) = (ab-=-ae) -=-a, by 2.21.

2.233

x-=-x2=x(l-=-x)=0, by 2.232 and 2.1.

2.234

(l-=-x) (l-=-x)=(I-=-x)-=-x(l-=-x)=I-=-x, by 2.232 and 2.1.

2.24

By theorem 2.63, {1-=-la,bl}(b-=-a)={I-=-la,bl}(b-=-b)=O.

2.241

Use 2.232 and 2.24.

2.242

If !(a, b)=(b-=-a) (Sa-=-b), then f(a, 0)=0, !(O,b)=b(l-=-b)=O and !(Sa,Sb)=!(a,b) so that !(a,b) satisfies the same doubly recursive introductory equations as the zero function Z(a, b)=O.

2.243

Proof as for 2.242.

2.244

as above.

2.2441 as above (using 2.201). 2.245

If f(a, b)=(b-=-a) + (Sa-=-b) and g(a, b)=I+(a-=-b)+(b-=-Sa) then f(a, O)=Sa=g(a, 0), !(O, Sb)=Sb=g(O, Sb) and !(Sa, Sb) = !(a, b) ,g(Sa, Sb) =g(a, b).

2.246

Proof as in 2.245.

2.25

Use double recursion as in 2.245.

2.251

Substitute 0 and Sr in turn for r.

2.26

iX(X) -=-1 = (1-=-1) .s: (l-=-x), by 2.22, and so iX(X) -=-1 = 0 . iX(O)+{I-=-iX(O)}=1 ,iX(Sx)+{I-=-iX(Sx)}=1 so that iX(X) + {I-=- iX(X)}= 1. iX(O)-=-O=iX(SX)-=-Sx=O, proving iX(X)-=-X=O. X·iX(X) = x -=-x(l-=-x) = x; iX( I-=- x) = I-=- (I-=- (I-=- x» = I-=- iX(X). iX(iX(O»=iX(O)=O, iX(iX(Sx»=iX(I)=I=iX(Sx), so that iX(iX(X» = iX(X). Similarly I-=- iX(X) = I-=- x. iX(X) ·iX(X)= iX(X) -=- (I-=- X)iX(X) = iX(X) -=- {I-=- iX(X) }iX(X) = iX(X), by

2.1 Furthermore iX(O·y)=O=iX(O)·iX(y), iX(X, O)=O=iX(X)·iX(O) and

162

SOLUTIONS TO EXAMPLES

IX(Sx,Sy) =1X(S(x+y+xy» = 1 =IX(SX)·1X(Sy). Finally from j-g=O follows 1X(/·g)=O and so 1X(f)'IX(g) =0, IX(/) .g. lX(g) = 0 , IX(/)' g = O. 2.261

If f(b, c)= {b-'-b·IX(C, b)}+C'IX(C, b), g(b, c)=c, then f(O, c)=c=g(O, c); f(b, O)=O=g(b, 0) ; f(Sb, Sc) = (b + I) (I-'-IX(C, b» + (c + I)IX(C, b) =t(b, c) + {I-'-IX(C, b)} + IX(C, b)=St(b, c)

and g(Sb, Sc)=Sc=Sg(b, c), so that t(b, c)=g(b, c). Furthermore {l-'-lX(x, y)}x= {I-'-Ix, yl}x= {I-'-Ix, yl}y= = {I -'-IX(X, y) }y, by 2.26 2.271

Substitute 0 and Sc in turn for c.

2.272

Use double recursion as in 2.245.

2.273

Write f(a,b, c) for (1-'-(I-'-a)b)(l-'-ac)b and g(a, b, c) for (1-'- (I-'-a)b) (I-'-c)b then !(a, b, c)...:...g(a, b, c) =

(1-'- (l-'-a)b)b«I-'-ac) -'- (I-'-c»

=(I-'-(I-'-a)b) (l-'-a)b·cx(c)=O

and

g(a, b, c) -'-f(a, b, c) =

(1-'- (l-'-a)b)b«I-'-c) .z, (I-'-ac»

whence it follows that If,

gl =0

=0

and therefore f=g.

2.274

Substitute 0 and Sa in turn for a.

2.28

xm.xo=x'm=x'm+o ,xm·xS1>=(xm·x")x and xm+sn=xS(m+n)=xm+,,·x, whence 2.281 follows. (xm)O = 1 = x m.O , (xm)sn = (xm)n. xm and x m.Sn= x."",+m = x'm".xm , by 2.281. (xy)o=l=xO.yO, (xy)S" = (XY)"'xy andxS".yS"=(x".x)·(y".y)=(x".y").xy, by 2.03.

SOLUTIONS TO EXAMPLES

2.29

163

By repeated use of the formulae a+8b=8a+b, and 1 =80, 2=81, .,. , 9=88, 10=89, we have 3.2=3·81=3·1+3=3+3=4+2=5+1=6, and similarly 4.2 = 8. Next we proceed thus: 3·7=7.3=7·82=7.2+7={7+7)+7, and 7 +7=7 +{3+4)=(7 +3)+4= 10+4, (10+4)+ 7 = 10+ (7 +4)= 10+(7 + 3)+ 1 = 10+ 10+ 1 =2·10+ 1 so that 3·7 =2·10+ 1. Similarly 4.7=2.10+8. To evaluate 34.27 we take (3.10+4) (2.1O+7)=(3.10+4)2.10+{3.1O+4)7, by 2.01, =6.10 2+ 8.10+{2.1O+ 1)10+2.10+8 = (6+ 2)10 2+ (8+2)10+ 1.10+ 8=9.102+ 1.10+8, by 2.01 again, and 2.03.

2.31

1=(f+g)-=-g=O-=-g=O.

2.32

{x+{1-=-x)}/={1 +{x-=-1)}/=1+(x-=-1)1 whence the result follows by 2.31.

2.33

From {1-=- (g -=-/)}h = 0 we derive (81-=- g) {1-=- (g-=-/)}h = 0 and so (81-=-g)h=0, since, by 2.242, (81-=-g) (g-=- f) =0.

2.331

Use 2.246 and 2.31.

2.332

(f-=-8g)h=(f-=-g)h-=-h=0.

2.34

Use 2.2441.

2.341

0=(1-=-f) (1-=-<X(g))=(1-=-f).:-.<X(g) (1-'-f)=(1-=-f)-=-<x(g), since I· <x(g) = 0; but <x(g) -=- 1 = 0 and so (1-=- f)<x(g) -=- (1-'- f) = 0 whence <x(g) -'- (1 .s: f) = O. Accordingly I<X(g), 1-=-11 = 0 and so <x(g) = 1-=-1.

2.35

l(g-=-h)=lg-=-lh=O; l(h-=-g)=lh-=-lg=O.

2.351

To 12g=O add (1-=-l>lg=0 and use 2.32.

2.36

From Ig=O and (1-=-g)h=0 we derive Igk=O and (1-=-g)/h = 0, and the result follows by addition, using 2.32.

164

2.37

SOLUTIONS TO EXAMPLES

From the given equations follow (1-'-f) (g+(I-'-g)tp)=O, (1-'- f)gp=O , (1-'- f) (l-'-g)p= 0 and so (1-'-f)p(g+(I-'-g))=O, whence (1-'-f)p=O, since g+(l-'-g)= l+(g-'-l).

2.41

Write f(x, z)=(x+y)-,-z, g(x, z)=(x-'-z)+{y-'-(z-'-x)} then f(O, z)=y-'-z=g(O, z)!, f(x, O)=x+y=g(x, 0) and f(Sx, Sz)=f(x, z), g(Sx, Sz)=g(x, z).

2.42 2.421

Substitute 1 for y and x -'-a for z in 2.41. Double recursion, using 2.42. The common value of a-'-(a-'-b) and b-'-(b-'-a) is the lesser of a, b, denoted by min (a, b). Or we may use equations 2.62, 2.611 and

example 2.22 as follows: a -'- (a -'-b) = {(a + (b -'-a» -'- (b -'-a)} -'- (a -'-b) = {(b + (a .z, b» -'- (a -'- b»} -'- (b -'-a) =b-'- (b-'-a).

2.422

(a-'-b) (Sb-'-c) = (a-'-b) {((a+Sb)-'-a)-'-c} = (a-'- b){«a +Sb) -'-c) -'- a} = (a -'- b){[(a -'- c)

+ {Sb -'-- (c -'- a) } ] -'- a}

= (a-'- b){(Sb-'-a) .z: (e -'-a)} , since (a-'-b) (a -'-c) = 0, =

0 , by 2.242.

It follows of course that (a-'--b) (b-'--c)=O, since Sb-'--c= (b-'--c) + {l-'-- (c-'-b)} .

2.423

By 2.41 and 2.32, {a-'-b)(c-'--b)=O follows from (a-'-b) (Sc-'--b)==rO ,and (a-'--b) (c-'-b) = {a-'--b){«c+a),-'-a)-'-b} = (a-'- b){«c-'-a) + a) -'- b}= (a-'- b){a-'- b+ «c -'-a)

.z: (b -'-a))}

whence (a-'--b)2=0 and so (by 2.233) a-'--b=O. 2.43

Ib, a+ (b-'--a)1 = Ib, b+ (a-'-b)1 =a-'-b.

2.44

Double recursion, using 2.42.

2.45

a= {a+b)-'--b= (a-'-b) + {b-'- (b-'-a)} =;:c+ b.

SOLUTIONS TO EXAMPLES

2.451

165

From plb, e] =Owe derive both p(b-'-c) = 0 and p(c-'-b)=O; then p(a-'-c)=",p{«a+b)-'-b)-'-c}

= p{«b +a) .s: c) -'- b} = p(a -'-b), using 2.41. Similarly p(c-'-a)=p(b-'-a). Or from p!b, c] =0 we derive in turn Ipb, pcl =0, pb=pc and hence from pa-'-pb=pa-'-pb follows pa -'-pb = pa .s: pc, etc. 2.452

This follows from 2.451 and 2.1.

2.453

This follows from 2.451.

2.46

(c-'-b) (Sa-'-c) = ({(c+a)-'-a}-'-b) (Sa-'-c) = {«c+ a) -'-b) .s: a}(Sa .z; c) = {(c-'-(b-'-a»-'-a}(Sa-'-c) , since a s-b».«, = {(c-'-a)-'-(b-'-a)} (Sa-'-c)=O, by 2.242.

2.461

(a-'-c){l.:.. (a-'-b)) = (((a+b) -'-c}-'-b){I-'-(a.:..b)} = ({a-'-(c-'-b)}-'-b) {I-'-(a-'-b)} , since b-'-c=O, = {(a-'-b)-'-(c-'-b)} {l-'-(a-'-b)}=O.

2.462

p(a-'- c)=p{«a+b)-'-c)-'-b} =p{[(b-'-c) + {a-'- (c-'-b)}] .:..b} =p{(a-'-b)--'-(c-'-b)}=O, since p(b-'-c)=p(a-'-b)=O.

2.47

From (l-'-a)c=O and (l-'-b)c=O we derive c-'-bc=O and bC-'-abc=O whence, by 2.462, c.:..abc=O, that is (l-'-ab)e=O.

2.4701 From (l-'-ab)c=O follows (I-'-ab) (l-'-a)c=O which added to ab(l-'- a)e=O yields (I-'-a)c= 0; similarly (1-'- b)c= O. 2.471

From (l-'-a)bc=O and abc(l-'-b)=O we derive, by 2.462, bc(1 -'- ab) = O.

2.472

{I -'-la, bUe = (1-'- {(a -'-b) + (b .:..a)})e = [{I.:.. (b-'-a)}.:.. (a--'-b)]c=O since {1.:..(b-'-a)}e=O follows from (Sa-'-b)c=O.

2.473

By 2.242, (a--'-b) (b--'-a)=O and so {1-'-(I-'-(a.:..b»}(b.:..a)=O whence b »- a = 0 and thc result follows from 2.45.

166

SOLUTIONS TO EXAMI'LES

2.48

Substitute {l-'- (1 -'-x)y}y for c, x for a and z for b in 2.471; then from {l-'-(I-'-x)y}y(l-'-x)z=O we derive 2.48.

2.481

{1-'-la, bl}C= [{l-'-(b-'-a)}-'-(a-'-b)]c

= [{1-'- (b -'-a)} + (a -'- b)]c,since (a -'- b)c = 0, whence the result follows. 2.49

{1-'-(I-'-a)b}(l-'-a)=(I-'-a)-'-(I-'-a)b, by 2.234, =(I-'-a) (l-'-b)

2.491

We observe that since (l-'-d)c=O, we have

{1-'- (l-'-a)b} {(l-'-a)+ (l-'-d)}c = {l-'- (l-'-a)b} (l-'-a)c= (l-'-a) (1-'- b)c= 0, and so 2.491 follows by 2.47.

2.5

From (IX) we obtain b-'-(b-'-a)={a+(b-'-a)}-'-(b-'-a)=a and a-'- b =a-'- (a+ (b -'-a) = (a-'-a) -'- (b -'-a) = 0; from ({3), (b-'-a)+a=(b-'-a)+{b-'-(b-'-a)} =b+ {(b -'-a)-'-b}=b,

and a-s-b = {b -'- (b -'-a)}-'- b = (b -'-6) -'-(b -'-a) = 0; from (y), a+(b-'-a)=b+(a-'-b)=b and b -'- (b -'- a) = a-'- (a -'-b) = a.

2.6

Double recursion.

2.701

By theorem 2.68, since 1-'-j(a, a+(b-'-a)=I, {1-'-la+(b-'-a) , bl}j(a, b)=O

and the result follows by 2.43. 2.71

This follows from 2.701 by 2.33.

2.711

As in 2.701, {l-'-ISb+(a-'-Sb) , al}j(a, b)=O, and so {1-'-(Sb-'-a)} j(a, b)=O, whence multiplying by (a-'-b) the result follows from 2.242.

2.72

By 2.701 and 2.71 [{1-'- (a-'-b)}+ (a-:-b)]j(a, b) =0

and the result follows by 2.32.

SOLUTIONS TO EXAMPLES

2.7201

167

a~(a~ (a~b»= (a~b)~ {(a~b) ~a}= (a~b)~ {(a~a)~b} =a~b.

{I-'-(b-'-a)} {b-'- (a-'- (a~b»)= {1-'- (b-'-a)} {b~(b-'-(b~a»} = {1-'- (b -'-a)}(b ~a) = O.

2.7301 By theorem 2.68 {1~lx, p~(p~x)l}/(x)=O,

since

l(p~(p-'-x»=O,

and so {1-'-(x~p)}/(x)=O. Similarly {I-'- (Sp-'-x)} I(x) = 0, and the result follows on addition, using 2.6. 2.7302 For instance, consider the case p=2; we have 1(0)=0, 1(1)=0, 1(2)=0, 1(2+Sr)=0. Let q;(r)=/(2+r); from q;(O) =0, q;(Sr)=O follows q;(r)=O. Let "P(r)=/(1+r); from "P(O) = 0, "P(Sr) = q;(r) = 0, follows "P(r) = O. Finally, from 1(0)=0, I(Sr)="P(r)=O, follows l(r)=O. 2.7303 Let q;(r)=/(p-'-r) then q;(0)=0, q;(Sr)=O and so l(p-'-r)=q;(r)=O, and the result follows by 2.7301. 2.7304 H q;(x)=I/(x), pi then q;(0)=0, q;(Sx)=q;(x) and so q;(x)=Z(x) = O,whence f(x) = p. 2.74

IIf(x+SO)=IIf(x)'/(Sx)=IIf(x)'q;(x, O) and IIf(x+SSr) =IIf(x+Sr)· l(x+SSr) IIf(x)· q;(x, Sr) =IIf(x)· q;(x, r). f{x +SSr) ,

which proves the first equation; it follows that the second equation holds with r = 0 and with Sr for r, and so for any r. 2.741

H q;(x) = {1-'-/{x)}II f(x) then q;(0)=0 and q;(Sx) = {I-'- f{Sx)}/(Sx)IIf(x) = O.

2.81

Alt Sx·Alt X= (I-'-Alt x) Alt x=O; Alt x+Alt Sx=Alt x+ (I-'-Alt x) = 1; (1-'- Alt 2x) (1-'- Alt (2x + 1» = (I-'- Alt 2x) {1-'- (1-'- Alt 2x» and similarly {I-'- (1-'- Alt (2x + In Alt(2x + 2) = 0 whence by 2.36 (I ~Alt 2x) Alt 2 ·Sx= 0

=

0

168

SOLUTIONS TO EXAMPLES

which combined with Alt 2.0=0 proves Alt 2x=0; it follows that Alt (2x + I) = 1. Finally, IX(Alt O)=IX(O)=O=Alt 0 IX(Alt Sx) = x(I-'-Alt x) = 1-'- Alt x= Alt Sx, proving x(Alt x) = Alt x. 2.82

We observe first that Hf SSx = Hf x + Alt x + Alt Sx = Hf x + 1 whence Hf(2.Sx) =S ill (2x), which, with Hf (2.0) = 0, proves Hf (2x) =x. Hence ill (2x+ 1)=Hf (2x)+Alt 2x=x.

2.83

By example 2.243 and the introductory equation of Rt x, and denoting Rt x by Rx for brevity, (1-'- {SSx-'- (SRX)2})RSx= (1-'- {SSx-'- (SRX)2})Rx and so by 2.35. (1--'-- {SSx--'-- (SRX)2}) ·IRx, RSxl =0 (i) By theorem 2.68 (1-'-IRx, RSxl) (1-'- {SSx-'- (SRX)2}) {SSx-'- (SRSX)2} = 0 .. (ii)

Multiplying (i) by SSx-'-(SRSX)2 and adding to (ii) we find (1-'-{SSx-'-(SRx)2}) {SSx-'-(SRSX)2}=0

(A)

By example 2.241, multiplying the introductory equation by 1-'-I(SRx)2, Sxl we obtain {1-'-I(SRx)2, Sxl}.RSx= 1--'-- {I (SRX)2, Sxl}·SRx

and so {1--'--I(SRx)2, Sxl}·IRSx, SRxl =0

(iii)

Again by theorem 2.68 {1-'-IRSx, SRxl} {1-'-I(SRx)2, SxIH(RSx)2, Sxl =0 ..... (iv) and so adding I(RSX)2, Sxl times (iii) to (iv) we find

{I-'-I(SRX)2, Sxl }-I(RSX)2, Sxl = 0

(v)

It follows that {1-'-I(SRx)2, Sxl} {Sx--'-- (RSX)2} = 0

and therefore {1--'-- j(SRX)2, Sxl }{SSx --'-- (SRSX)2} = 0 . . . . . . . . . . . . . . .. (B) From (A) and (B) by example 2.34 (1--'-- {Sx--'-- (SRX)2}) (SSx-'- (SRSX)2) = 0;

but SO-'-(SRO)2=0 and therefore Sx--'--(SRX)2=0

(vi)

169

SOLlY.rIONS TO EXAMPLES

From (v) we obtain {1~I{SRx)2,SXI}{(RSX)2~SX}=0

(C)

Again from the introductory equation we find {{SRX)2~SX}'IRx, RSxl =0.

By theorem 2.68 (1 ~ IRx, RSxi) (I ~ {(RX)2~X}) «RSX)2~X)=0

and so by example 2.36 «SRX)2~SX) (1~{(Rx)2~x}) «RSX)2~X)=0

whence by 2.332 «SRX)2-"--SX) (1-"-- {(RX)2~X}) «RSX)2~SX)=0

(D) From (C), (D) (since l-"--la,bl={I~(a~b)}~(b~a) and pre = 0 follows from qrs = 0 and (p ~ q)r = 0) (1~{Sx~(SRx)2}) (1-"-- {(Rx)2~X}) «RSX)2-"--SX) = 0 and therefore by (vi) (1~{{Rx)2~X}) «RSX)2~SX)=0

which together with the equation

(RO)2~0=0

proves

(RX)2~X=0.

2.9

From

{I~(l-"--n!)}(l-'-n!)=O

{l-"--(l-"--n!)} (l-"--Sn)=O follows, by example 2.47 {l-'-(l-'-n!)} {l-'-(n!)Sn}=O i.e. {l-"--(l~n!)} {l-'-(Sn)!}=O which, with 1-"-- O! = 0, proves 1-'- n! = O. Further, (SO)!=l=11s(O); (SSn)!={(Sn)!}SSn and 11s(Sn) = 11s(n) ·SSn

which proves that (Sn)!=11s(n). 2.91

From

(l~c)

{I .z; (l-'-c)}=O and (l-'-b)=Ofollow the two equations {l-'-(l~a) (l-'-b)) (I-"--c) {(I~(I-'-c»-'-b}=O {I-'-(I~a) (l-'-b)) (I-'-a) {(l-'-b)-'-(l-'-c)}=O whence by example 2.47 {l-'-(l~a) (I~b)) (l-'-a)

{I ~ (1 .z; a) (I -"-- b)} (1-'- ac) {I -"-- (b + (I

.z:

c»} = 0

Solutions to Examples III 3.01

Use examples 2.25 and 2.26.

3.02

Examples 2.25, 2.47 and 2.26

3.03

Examples 2.351, 2.1

3.031

Examples 2.25, 2.1

3.032

Associative property of addition.

3.04

Example 2.1

3.041

Examples 2.25, 2.1

3.05

Commutative property of multiplication

3.051

Example 2.25

3.052

Example 2.25

3.06

Commutative property of multiplication

3.0601 Example 2.31 3.061

Example 2.36

3.0611 Example 2.25 3.062

Example 2.26 (last part)

3.063

Example 2.471

3.064

Example 2.351

3.07

Example 2.36

3.071

As for 3.07

3.08

As for 3.07

3.081

As for 3.07 170

SOLUTIONS TO EXAMPLES

3.082

Example 2.351

3.083

Example 2.48

3.084

Example 2.351

3.085

By examples 3.081, 3.082, 3.083 and 3.084

3.09

Example 2.25

3.091

As in 3.09

3.0911 Examples 2.25, 2.36 and 2.232 3.092

Example 2.232

3.093

Examples 2.25 and 2.232

3.1

Example 2.242

3.11

Examples 2.331 and 2.244

3.12

Example 2.244

3.121

The formula is equivalent to

(Sb~a) (Sc-'-b) (a-=--c)

= (Sb -'-a) (Be-'-b)[{(b -'-c) + (a-'- (c-'- b))}-'- b] = (Sb -'- a) (Be-'- b)· {(a ~ b) -'- (c -'- b)}= O.

3.13

Equivalent to 3.12

3.131

Theorem 3.43 and example 3.052

3.132

Theorem 2.63 and example 2.35

3.14

Example 2.1

3.15

Example 2.4701

3.16

If l(x)=2x-=--(1+x 2 ) then 1(0)=0, 1(Sx) = (2x+ 2) -'-(2 + 2x+x 2 ) = O.

3.161

If [ta, b)=2ab-'-(a2+b2 ) then [ia, 0)=/(0, b)=O and I(Sa, Sb) = I(a, b) so that [ta, b) = O.

3.17

Example 2.43

171

172

3.171

SOLUTIONS TO EXAMPLES

By theorem 2.63 and example 2.25 {1-'--la, AI}'!f(a, b), f(A, b)I=O

and {1-'--lb, BI}·lf(A,b),f(A, B)I=O; multiplying the first equation by 1-'--lb, BI and the second by l-'--Ia, AI and using example 2.453 {l-'--la, AI} {l-'--Ib, Bl}lf(a, b), f(A, B)I = 0 whence the result follows by example 2.25. 3.18

Since k-cb -+ b=k+(b-'--k) and l
3.19

b z:c

-+ b=c+(b-'--c)

and

b=c+(b-'-c) -+ (a+b)-'--c={a+(b-'--c)+c}-'--c 3.191

l-e-n. -+ n=l+ (n-'--l) , m;;;;;;,n -+ n=m+ (n-'-m) and so (by example 3.131) l-e-n. 8< m
but n--'-l
+ (n --'- m)}-'-S(n-'--l) =Sm+ {(n-'--m)-'--8(n--'-l)}

-+ {8m

and therefore l-c n. 8< m
3.192

p>q 88-+ {p=q+(p-'--q)} 8< {r=s+(r-'--s)} -+ {(P+8) .s: (q+ r)}= {(q +8+ (p-'--q» -'-- (q+ s+ (r-'--s»} = (p-'-q) -'-- (r -'-- s)

3.193

Denote the implication by P(n). P(O) is simply (l-'--x){f(x)-'-- f(O)}=O. From (x=Sn) -+ {f(x)=f(Sn)} and a=b -+ a
SOLUTIONS TO EXAMPLES

173

a-cb --+ a
(p -=- r)

.s: (q -=- r);=

p -=- (r+ (q .s: r)) = p-=- (q+ (r -=-q» = (p-=-q)-=- (r-=-q) and so {1-'--(r-'--q)} {(p-'--r)-=-(q-=-r)}={I-'--(r-=-q)}(p-'--q) whence {1-=-(r-=-q)}l(p-=-r)-=-(q-=-r), (p-=-q)1 ~O

3.2

Let p(x, a) denote the equation /(x, a)=O; from a proved equation /(x, a)=O we derive /(0, a)=O and /(Sn, a)=O and therefore {l-=- E/(n)}I/(Sn) = {I-=- Et(n)} {If(n) + /(Sn, a)} = O. For the second schema we observe that {1-'--j(x)}flAx)=O by example 2.741

3.31

Substitute 0 and Sn in turn for n.

3.32

Since

v-> ,.....,

p

** p

we have to prove

cp(n) ~ It(n). IIl~t(n)= 0 (where (I-=- f) (n) is taken to denote the function 1 -=- / (n)).

The result follows from the equations cp(O) = /(0)· (I-=- /(0»= 0, {I-=- cp(n)}cp(Sn) = {1-'- cp(n) }{It(n) + /(Sn)}IIl~t(n)· (I-=- /(Sn») = {1-=-tp(n)}cp(n) (l-=-/(Sn»=O.

3.321

If tp(n)= {I-=- IIf(n)} {1-=-I1

-O-

1(n)}

then

'1'(0) = {l-'- /(O)} {1-'- (I-=- /(O»)= 0

and {1-=-tp(n)}tp(Sn) =

{l-'--(l-=-IIt(n)) (I-=-I1_ct(n)} {1-=-II/(n)-!(Sn)} . . [I-=- {E1-,-/(n) + (I-=- /(Sn»))]

=0, by example 2.91. 3.322

This follows from the previous two examples.

3.33

{l-=-I/(O)}/ (0) =0 and {I ~I/(Sn)}/(Sn) ~ {I-=- /(Sn)}/(Sn)-=-It(n)/(Sn) =0

174

SOLUTIONS TO EXAMPLES

3.34

Induction over n, using example 3.041

3.35

Let Z=L;p(z) then Ezp(z) -+p ( l ) l < n lgn-+l=nz(n-l) Z=nz(n-11) + (p(1)+ p ( n ~ ( n ~ 1 ) ) ) and so (using example 3.092) E,"p(z)-+ p(nA (n2 1 ) ) --f

E;p(n r ) .

3.36

This follows from 3.33 and 3.34

3.37

Write ~ ( az)=f(z, , a); by example 3.36 we have both AP'"f(t, cc+c)=O) + / ( a , a + c ) = O and A:+"(& a +c ) = O } -+ y(a, a + c) =0 i.e. AP+'{/(a+c,t ) = O } -+f(a+c, a)=o Thus if P(z, a) denotes A;{/ (t, a )= O } & A;(/ (2,u)= 0) -+ f (z,a )=0 we have proved both P(a+c, a) and P(a, a+c) whence P(z, a) follows by example 2.272

3.38

Denote L:(f(z)=a) by b, so that E;(f(z)=a) -+ f ( b ) = a . Sime f ( b )=a -+ ( p ( 4 + P ( f ( b ) ) } i.e. (f( b )=a} p ( 4 -+p ( / ( b ) ) , and b <m, therefore e { f ( z ) = a ) Or p(a) -+c p ( / ( z ) ) ,by example 3.42.

3.39

f(a)=O follows from {1-1lu, al}f(a)=o

3.391

l-f(a)=O follows from ( l ~ f ( a ) {lAlu, } al}=O

3.392 From p ( a + ~-+ ) q(a) we derive ( b2- a))-j a@); by theorem 3.43

+

+

+

(b =a (b a)}-+ (p(b)-+ p(a (b2L a))) and so @=a + (b 4 ) {p(b) a ( 4 ) -+

-+

SOLUTIONS TO EXAMPLES

175

whence taking b l r for a, since ( b A T ) + {b- (b A ~ ) } = + b { ( bA T ) 2 b ) = b , @ =b ) {P(b) + Q(b .>I and so P(b) + P(b r ) . +

3.41

{ l ~ l I , ( n ) ) l T , ( S n ) = { l - R ; ( n ) } l T ,( n ) .f ( S n ) = O

3.42

Let P(b) denote the formula a ~ bp ( a ) -@ip(x) where p ( s ) is the propositional function f (s)= 0. P(0) holds by theorem 3.43 Since E b ( z )-+ E:bp(z) (by examples 3.322, 3.34) therefore P(b)-+ {a
3.5

Let Q(r) denote A:’s’q(z) then Q (0) follows from AEq(z) and q(p+l); since &(ST)is equivalent to Q(r) &q(p+SSr) then q(p+XXr) --f ( & ( T ) --f &(fir))and so Q(r)-+Q(Sr)follows from q(p+XXr) which proves Q(r), that is A;+s‘q(z).By examples 3.34 and 3.392 we have A!”&) follows from A:q(z), and so Azq(x) follows by example 2,7301.

3.6

We have b < l --f p(n-=b) by a characteristic property of the L-operator. Since g c a & a< n + n i a t I , (takingn i a for M in example 3.191), therefore g < a & a
3.7

Since a=O+ab=O therefore a b > O + a > l & b > l . Moreover a 2 1 + ( y + l ) a > ( y + I ) > y . From (y + 1)a >y follows Ey,+ l- {nb<5 & na Q y] and there-

-

176

SOLUTIONS TO EXAMPLES

fore if N =£:+1 "" {nb<x Br naO and 80 N -'--I
From a proved equation (l~p)q(x)=O

we derive both (l~p)q(O)=O and (l-'--p)q{x+I)=O and therefore (1-'--(I~p)IQ(x» (1~p)Iq(x+I)= (l .s: (l ~ p)Iq{x» (1-'-- p) Iq(x) = 0 so that, by induction, (1-'-- p)IQ(x) = 0 follows. 3.82

From a proved equation (l~q(x»p=O

follows (I ~q(O»p=O and (l ~q(x+ 1»p=O and hence, using example 2.272, {1-'-- (I ~ l1Q(x»p} {1-'--l1q(x+ I»p = {I-'-- (l-'--l1Q(x»p} (I-'--q(x+ I»p= 0 which completes a proof by induction. 3.831

From the true formula (n-'--r»n -+ "" (0=0) and the previous example, follows E:«n-'--r) >n) -+ "" (0= 0) whence (0=0) -+ "" E:{{n-'--r»n) -+ ""E:(x>n) , by example 3.35, -+ A:(x
3.832

In the form "" (y<.n -+ p{y» -+ "" A:p(x) this is equivalent to example 3.42.

3.833

Taking r{x) to be the function p·q(x) we have to prove (l-'-- L'r{n»pEQ(n) = O. This follows from the equation Er{n) = pL'q{n)

SOLUTIONS TO EXAMPLES

177

which is proved by observing it is true for n=O and that E,(n+ 1) == E,(n)+p·q(n+ 1)

and p·EIl(n+ 1)=p·EIl(n)+p·q(n+ 1). 3.834

Take,...., p for p in 3.833.

3.84

By 3.81, P -+ A:(q -i>- r(x» follows from p -+ (q -+ r(x», and by 3.834, A~(q -+ r(x» -+ (q -+ A:r(x», whence the truth of the schema follows.

3.91

As in 3.833, if r(x)=p.q(x), E,(n) =p' EIl(n), and so (l-'--p)E,(n)=(I-'--p)pEIl(n)=O

3.92

Take r(x)=p+q(x), then E,(n)=(n+ l)p+EIl(n), the functions on either side of the equation satisfying the same introductory equations, and therefore {1.:- E,(n)} {p+ EIl(n)} = {(1-'-- (p+ l:1l(n») -'--np}(p+ EIl(n» = 0 and {l-'--(p+ EIl(n»}E,(n) = {l-'-- (p+ EIl(n»}np = {(1-'--p)-'--l:Il(n)}pn=O.

3.93

The functions l:2)+Il(n) and E2)(n)+EIl(n) are readily shown to satisfy the same introductory equations and from the equality E1l+Il(n) = l:2)(n) + EIl(n)

the example follows. 3.931

Negate both sides of 3.93, with

3.94

Let r(x) = {l..:.. p(x)}q(x), so that we must prove tp(n) = (1":" E,(n» (I-=- E2)(n»EIl(n) = O. Clearly tp(O)=O, and tp(n + 1) = (1...:.. E,(n + 1» (I-=- Ep(n + 1»EIl(n + 1) ={(I..:..r(n+ 1»-=-E,(n)} {(1...:..p(n+ 1»...:.. E2)(n)} {q(n+ 1)+ +EIl(n)} = {( 1..:.. E,(n»...:.. r(n + I)} {(1':'-l:2)(n» -=- p(n + I)} EIl(n), since (l-=-r(n+ 1» (l..:..p(n+ l»q(n+ 1) = 0, =tp(n)-=-{(I"':" E,(n»p(n+ 1) + (1':'-l:p(n+ l»r(n+ I)}EIl(n) and therefore (l':'-tp(n»tp(n+ 1)=0, proving tp(n)=O.

3.95

A:(x
i--;

p, ,...., q in place of p, q.

178

SOLUTIONS TO EXAMPLES

and so A:(x..;;n) ~ (A:(x..;;n ~ p(x» ~ A:P(x» whence one part of the example may be derived from example 3.831. For the second part, by 3.832, A:p(x) ~ (y..;;n ~ p(y» and so A=p(x) ~ A:(y..;;n ~ p(y» ~ A:(x..;;n ~ p(x».

3.96

Negate both sides of 3.95, with f"'oJ P in place of p.

3.97

From the equation (l-'-(l-'-p)q) (1-'-(1-'-q» (1-'-p)=O (proved by taking 0 and Sp in turn for p) we derive (p(x) ~ q(x» ~ (f"'oJ q(x) ~ f"'oJ p(x»

and from this proved equation we obtain A:(p(x} ~ q(x» ~ A;(f"'oJ q(x) ~ -- p(x» ~ (A:(-- q(x» -+ A;(-- p(x)))

~ (-- A;(-- p(x» ~ -- A:(-- q(x))) ~ (E:P(x) ~ E;q(x».

3.98

Combine 3.97 with the formula obtained from it by interchanging p(x), q(x) and then apply example 3.93.

3.981

A:(p(x)

~

q)

B-

A;(-- p(x)

v

q)

B-

~

3.982

A:A:p(x, y)

B-

~ ~

(f"'oJ E;p(x) y q) (E:p(x) ~ q).

A:A:p(u, v) A:(y <; n ~ p(u, y» (x..;;n) ~ (y..;;n ~ p(x,y»

and so (A:A:p(u, v) 8< y..;;n) ~ «x..;;n) ~ p(x, y»

therefore and so

(A:A~p(u, v) 8<

y<.n) ~ A:p(x, y),

A:A:p(u, v) ~(y<;n ~ A:p(x, y» A:A:p(u, v) ~ A:A:p(x, y).

3.983

Negate both sides of 3.982 with

>«

p in place of p.

SOLUTIONS TO EXAMPLES

3.984

From (x
&

179

p(x, y) -» E:p(z, y)

we derive in turn A:((x<;n) & p(x, y» -» A:E:p(z, y) E:(x
E~:(p(x) &

p(y} & x>y) E:E:(p(u} & p(v) & u>v) ~ E"vE:(p(y) & p(x) & y>x) ~E~=(p(x) &p(y) &y>x), by 3.983; ~

from the equation (1-'-(a+bc)) (a+b) (a+c)=(a+c) ((l-'-a)-'-bc)b= = «1-'-bc) .s: a)bc = 0 we derive p(x) & p(y) & «x>y) v (x
(p(x) & p(y) & (x>y)) v (p(x) & p(y) & (x
and so, by two applications of example 3.931, and the equation (1-'- (l-'-b)a) (1-'-ab)a=O, E:E:(p(x) & p(y) & x#y) ~ E".,E:(p(x) & p(y) & x>y).

Solutions to Examples IV 4.

R(O, O} = 0 and, by theorem 4.03, n>O & n=n·1+0 -+ R(n, n)=O.

4.01

If
4.0B

x>o & y=O -+ R(x, y)=x>O.

4.012

R(a, b)=O -+ a=b·Q(a, b) R(b, e)=O -+ b=e·Q(b, e) and so R(a, b)=O & R(b, e}=O -+ a=c{Q(a, b)·Q(b, e)} whence e>O -+ {R(a, b)=O & R(b, e)=O -+ R(a, e)=O}. But e=O 8< R(b,c)=O -+ b=O b=O & R(a, b)=O -+ a=O and so e=O -+ {R(a, b)=O 8
which completes the proof. 4.013

x=O&R(a,x}=O-+a=O

so that x=o -+ {R(a, x)=O -+ R(ab, x)=R(O, O)=O} x>o

8<

R(a, x}=O -+ ab=x{b.Q(a, x)} -+ R(ab, x)=O.

4.014 It suffices to prove the equivalent formula R(ax, bx)=O 8< R(a, b»O -+ x=O. Since a = bQ(a, b) + R(a, b) therefore ax = bxQ(a, b)+xR(a, b); if b>O, R(a, b)O -+ {x>O -+ xR(a, b)O -+ {x>O -+ xR(a, b)=R(ax, bx)} whence b>O -+ {x>O -+ R(a, b)=O v R(ax, bx»O} so that b>O -+ {R(ax, bx)=O & Ria, b»O -+ x=O}. Furthermore b=o 8< R(ax, bx)=O -+ ax=O; b=O & R(a, b»O -+ a>O so that b=O -+ {R(ax, bx}=O 8< R(a, b»O -+ x=O}. 180

SOLUTIONS TO EXAMPLES

4.02

lSI

{R(p, a)=O Be R(q, b)=O -+ pq=ab{Q(p, a)·Q(q, b)}} that ab>O -+ {R(p, a)=O Be R(q, b)=O __ R(pq, ab)=O}. ab=O -+ a=O v b=O; a=O Be R(p, a)=O -+ p=O and b=O Be R(q, b)=O __ q=O so that ab=O -+ {R(p, a)=O Be R(q, b)=O -+ R(pq, ab)=R(O,O)=O}.

80

4.03

R(O, 1)=0, x=x·l +0 Be x>O -+ R(x, 1)=0.

4.04

R(a, be)=O -+ a=b{e·Q(a, be)} and so b>O -+ {R(a, be)=O -+ R(a, b)=O}; but b=O Be R(a, be)=O -+ a=O so that b=O -+ {R(a, be)=O -+ R(a, b)=R(O, O)=O}

4.1

R(a+b, e)=O __ a+b=e·Q(a+b, e) R(a, e)=O -+ a =e·Q(a, e) and 80 R(a+b, e)=O Be R(a, e)=O -+ b=e{Q(a+b, e)-=-Q(a, e)} which completes the proof for the case e> O. If e = 0 we deduce as above that b=O and so R(b, e)=O.

4.2

ah--b-a-it: 80 that b>O -+ R(ab, b)=O; if b=O, R(ab, b)=R(O, 0)=0.

4.21

a+l=b·Q(a, b)+R(a, b)+I=b·Q(a, b)+1 if R(a, b)=O, so that b » 1 -+ R(a+ 1, b) = 1.

4.3

This follows from 4.2 and 4.21.

4.31

By example 2.473, b>O -)- a-sb s-cd and so b>O -+ a=d(e+Q(b, d))+R(b, d) Be R(b, d)O -+ R(a, d)=R(b, d).

4.32

a={Q(b, d)+x}d+R(b, d) whence R(a, d)=R(b, d).

4.321

R(e, d) = R(b, d) = R(a, d).

4.322

Since ar=Q(a, d)dr+R(a, d)r br=Q(b, d)dr+R(b, d)r therefore, by 4.32, ar = R(a, d)r (mod d) br = R(b, d)r (mod d) whence, since R(a, d) = R(b, d) the result follows by 4.321.

182

4.323

SOLUTIONS TO EXAMPLES

If r 1 denotes the common value of R(a1 , d), R(b 1 , d) and r s that of R(as, d), R(b s, d) then a1 +a s= {Q(~, d) +Q(a2 , d)}d+r1 +r2 ht + b2= {Q(b 1 , d)+Q(b2 , d)}d+r1 +r2 whence ~ +~= (r 1 +r2 ) (mod d) =b1 +b2

•

4.4

If q;(n)=R(IIj(n), f(n», q;(O) =0 and q;(Sn)=O by 4.2.

4.41

If q;(n)=R(IIf(r+n), IIj(r» then q;(O)=O and, by 4.013, q;(n)=O _ q;(Sn)=O.

4.42

IIj(a+ (n-=--a» is divisible by IIj(a), and a.;;;;;n _ a+(n-=--a)=n.

4.43

If P(n) denotes the given formula, then P(O) holds and since (Sn)! =IIs(n) x.;;;;;n_R((Sn)!,Sx)=O (by 4.42) i.e. Sx.;;;;;Sn_R((Sn)!,Sx)=O so that O
4.44

Examples 4.21 and 4.43.

4.45

q(2)';;;;; 2 , q(2) > 1 so that q(2) = 2 and so p(2)

4.46

q(3)';;;;;3 ,q(2»I; since 3=2.1+1 therefore R(3, 2)=1 and since R(3, q(3» = 0 it follows by theorem 3.43 that "'" (q(3) = 2) holds, and therefore q(3) = 3, which proves that

=

O.

3 is prime. 4.5

lJ(O) = 2 ,lJ(I»lJ(O) so that lJ(I»2; from lJ(n+2»lJ(n+l) and lJ(n+2»lJ(n-t- 1) &lJ(n-t-1»2 -lJ(n-t-2»2 follows lJ(n -t- 1) > 2 by induction.

4.51

lJ(0)=2, 2>0; and lJ(n-t- 1);;;"SlJ(n) & p(n»n _ p(n-t- 1»n-t- 1.

4.6

Let P(a) denote the formula (a.;;;;; 1) v (a=n) v R(n, a»O

183

SOLUTIONS TO EXAMPLES

then n>2 Br R(n, 2)=0 -+ '"-' P(2); but n> 2 Br '"-' P(2) -+ E:{'"-' pea)} -+ '"-' A:{P(a)} and so n> 2 Br R(n, 2)= 0 -+ pen) > o. 4.61

By 4.6, since p(n)=O

4.7

If lp(m)=(I+mx)--"-(I+x)m then 1p(0)=0 and (1 +x)lp(m) = {I +(m+ l)x+X 2 } --"- (l+x)m+l =Ip(m+ 1) + [x 2 --"- {(I +x)m+l--"- (1 + (m + l)x)}]

-+ n>

1, and R(n, 2).;;; 1.

so that, multiplying by 1-'-Ip(m) we find {1--"-Ip(m)}Ip(m+ 1)=0 which proves lp(m)=O 4.701

By 4.7, a = 1 + x Br x;;,;. 1 -)- o":» 1 + mx Br 1 + mx;;,;. 1 + m and so a> 1 -+ am>m.

4.71

a=O·b+a Br a-cb -+ R(a, b)=a>O.

4.711

By 4.702, a-cb Br x> 1 -+ xb>x u -+ R(x xb) > O. G

,

4.8

If P(k) denotes the given formula, then P(O) holds, and R(xS k+ l, p)=O -+ R(xSk, p)=O v R(x, p)=O . (l--"-a)bc=O

and so, usmg the schema

{1--"-(l--"-b)c}(l-'-a)c=O

which

is proved by observing that if (l--"-a)bc=O then (1 -'- (1 -'- b)c) (1 --"- a )c = (1 .z: a )c .z: {( 1 .s: a )c2 -'- (1 .z; a )bc2 } =(I--"-a)c(l--"-c)=O, it follows that P(k) -+ P(Sk), which completes the proof.

4.801 If P(k) denotes R(a, p»O Br R(ab, P')=O -+ R(b, P')=O then P(O) holds (by 4.03) and since (by 4.04) R(ab, pk+l)=O -+ R(ab, pk)=O

therefore P(k) -+ {R(a, p»O" R(ab, pk+l)=O -)- R(b, pk)=O};

184

SOLUTIONS TO EXAMPLES

but, if c=Q(b, pk), R(b, Pk)=O-+b=cp"', and so (by 4.014) P(k) & R(a, p»O & R(ab, pk+I)=O -+ R(ac, p)=O. However R(a, p»O & R(ac, p)=O -+ R(c, p)=O

and R(c, p)=O -+ c=pQ(c, p).

Accordingly P(k) & R(a, p»O & R(ab, pk+I)

= 0-+ b=pk+IQ(C, p) -+ R(b, pHI) = 0

that is, P(k) -+ P(k+ 1), which proves P(k). 4.802

R(m, a)=O -+ m=a.Q(m, a), and, by 4.801, R(Q(m, a), pk)=O.

4.81

Since

~k

is prime R(~k' x)=O -+ x= 1 V X=~k and so, since ~l> 1, R(~k' ~1)=O -+ (~l=~k) -+ (k=l).

4.82

For R(~~k, ~l) = 0-+ R(~k' ~l) = 0 -+ (k= 1).

Since

q(m)<;m,p(q(m»=O-+E::(q(m)=~",);

4.83

but R(m, q(m»=O and m:» 1 -+ p(q(m» = 0 and so m:» 1 -+ Er;:R(m, ~",)=O, by example 3.38.

4.84

By theorem 3.91 we derive Ar;:R(m, V",» 0, i.e. E";R(m, ~"') = 0 ,from the condition x<;m -+ R(m, ~"'»O. From 4.83 Er;:R(m, ~"') = 0 -+ m «; 1. Since m=O -+ R(m, ~o)=O therefore m=O -+ E';R(m, ~",)=O, whence the result follows. ro..J

ro..J

4.85

Denote the given formula by F(k); clearly F(O) holds. Furthermore R( II !(x), p»O & R( II !(x), p)=O -+ R(f(Sk), p)=O ~~k

x~Sk

SOLUTIONS TO EXAMPLES

185

and .A~kR(f(x), p»O -+ .A~(t(x), p»O & R(t(Sk), p»O whence F(k) & A~kR(f(x), p»O-+R( II t(x),p»O &R(f(Sk),p»O "'''k

-+R( II t(x), p»O, by theorem "'''k

4.51,

i.e. F(k) -+ F(Sk), which proves F(k). 4.86

We have R(m, ~;(m.",» = 0; let P(k) denote the given formula (4.86). Then P(O) holds. By 4.82 and 4.85 R( II ~;(m.",) , ~k+ 1)> 0 z"k and so by 4.802, since R(m, ~~(~ik+l» = 0, P(k) -+ R(m, { II ~;(m.z)}-Pi<.':ik+l» = 0 z"k i.e. P(k) -+ P(k+ 1), so that P(k) is proved.

4.87

Denoting L':{R(b, -Pz)=O} by ~ we have N;{R(b, ~z)=O} -+ R(b, ~e)=O & ~<,m. From x<,m -+ Ria, ~:Z)=O and N;{R(b, ~z)=O} we deduce therefore R(a, ~~al;)=O and so, by 4.02, R(ab, ~~ae+l) = 0 which in conjunction with

~ <, m

yields

E';{R(ab, ~",a",+l)=O}.

In particular, it c =

{

II ~za"'}b and if E';{R(b, ~z) = O} a:~m

a", then N;{R(c, -Pz + 1) = O} .

4.9

We have v(N, l»O and l-c k -+ v(N, k)=O and, since N> 1, v(N, g(N» > 0 and g(N) < k -+ v(N, k) = O. Hence g(N) 0 -+ g(N):> l and similarly, since lO -+ g(N)<,l, whence it follows that l=g(N).

186

SOLUTIONS TO EXAMPLES

4.91

Denoting R(a, be) by r we have a=bcQ(a, bc)+r, r=bQ(r, b)+R(r, b), where R(r, bs-cb since b>O, and so a=b(Q(r, b)+cQ(a, bc»+R(r, b) whence, since R(r, bi-cb, R(r, b)=R(a, b) which completes the proof.

4.92

H divides b, c and so H divides a, band c. If f divides a, b, c then it divides h, c and so f divides H, so that f < H. Thus H is the greatest common factor of a, b, c.

4.93

We define L(x, y) = L:lI{a > 0" R(a, x) = 0" R(a, y) = O}. Since xy>O --+ {R(xy, x) = R(xy, y)=O} therefore xy>O --+ {R(L(x, y), x)=O=R(L(x, y), y)} and a-cLtx, y) --+ a=O v R(a, x»O v R(a, y»O so that L(x, y) is the least common multiple of x and y, for xy>O.

Bibliographical Notes Recursive arithmetic was discovered by the Norwegian mathematician Thoralf Skolem and his first account (in German) was published in 1923 in Skolem [1] *). Skolem's work is described in Hilbert-Bernays [1], pp. 286-346, which contains also further developments of recursive arithmetic. The first account of the development of logic from arithmetic was given by R. L. Goodstein in Goodstein (1]. Though printed in 1945 this paper was presented to the London Mathematical Society in June 1941. A system somewhat similar to the equation calculus was developed independently by H. B. Curry, in Curry [1] The most complete account of the properties of recursive functions, the reduction to normal forms and the elimination of parameters is given by their discoverer Rozsa Peter in Peter (1]. This book contains a comprehensive bibliography of writings on recursive functions up to 1950, and is the only book on the subject. The proof of the schema of Generalised Induction was given by Th. Skolem in Skolem [2], and completed by R. Peter in her review of the paper in Peter (2]. For the elimination of parameters we have followed R. M. Robinson [1]. The discovery that arithmetic has a non-standard model (i.e. an interpretation in which a richer class than the natural numbers plays the number role) was presented by Th. Skolem in Skolem [3]. The discovery by Kurt Godel of undecidable propositions was first published in Godel [1] and was applied by him to prove that no system of number theory can be proved consistent using only methods formalisable within the system. Accounts of Godel's work are given in Hilbert-Bernays [1]; in Kleene [1]; in Mostowski [1] and (in outline only) in Goodstein [2]. The theory of general recursive functions which finds no part in the present book is dealt with in Peter's and Kleene's books listed above. Peter has also made a study of transfinite recursions, which were introduced by Ackermann, in Ackermann [1]. A brief survey of recursive arithmetic is given in Skolem [4]. *) The numbers in square brackets refer to the bibliography on the next page. 187

Bibliography ACKERMANN, W. [1] Zur Widerspruchsfreiheit der Zahlentheorie, Math. Annalen, 117 (1940) pp. 162-194. BERNAYS, P. [1] Uber dae IndukstwnB-schema in del' rekursiven Zahlentheorie, in Kontrolliertes Denken, Miinchen 1951. CURRY, H. B. [1] A formalisation of recursive arithmetic, American Journal of Mathematics, 63 (1941) pp. 263-282. GOODSTEIN, R. L. [1] Function theory in an axiom-free equation calculus. Proceedings of the London Math. Society (2) 48, pp. 401-434. [2] Constructive Formalism, University College Leicester 1951. GOODSTEIN, R. L. [3] Permutation in recursive arithmetic, Math. Scandinavica, Vol. 1 (1953), pp. 222-226. GOODSTEIN, R. L. [4] Logic-free jormaiisatione of recureioe arithmetic, Math. Seandinaviea, Vol. 2 (1954), pp. 247-261. GODEL, K. [1] tn« formal unentscheidbare Siitze del' Principia Mathematica und oeruxmdter Systeme I. Monatshefte fur Math. und Phys.• Vol. 38, pp. 173-198. HILBERT, D. and BERNAYS, P. [1] Grundlagen der Mathematik. Vol. I, Berlin 1934, Vol. II, Berlin 1939. KLEENE, S. C. [1] Introduction to Metamathematics. Amsterdam. 1952. MOSTOWSKI, A. [1] Sentences Undecidable in Formalised Arithmetic. Amsterdam. 1952. PETER, R. [1] Rekursive Funktionen. BUdapest 1951. [2] Review of Skolem. [2], Journal of Symbolic Logic, Vol. 5 (1940), pp. 34-35. ROBINSON, R. M. [1] Primitive recursive functions. Bulletin of the American Math. Soc. Vol. 53 (1947), pp. 925-942. SKOLEM, TH. [1] Begrnndung der eiementaren Arithmetik durch. die rekurrierende Denknoeiee ohne Anwendung scheinbarer Veranderlichen mit unendlichem Ausdehnungsbereich. Skrifter Norske Videnskaps-akademi, I, No. 6b, Oslo. SKOLEM, TH. [2] Eine Bemerlcunq uber die Lnduktionsschemata in der rekursioen. Zahlentheorie. Monatshefte fur Math. und Phys., 48 (1939) pp. 268-276. [3] Uber die Nicht-charakterieierbarkeit der Zahlenreihe mittels endlich oderabzi.ihlbarunendlichviclerAussagen mit ausschliesslichZahlenvariablen. Fundamenta Mathematicae, Vol. 23, (1934) pp. 150-161. [4] The development of recursive arithmetic. Xth Congress of Scandinavian Mathematicians. Copenhagen 1946. 188

INDEX Page

Page

1" 29

Addition -, associativity of . -, commutativity of . axiom A. - P

1,1·

28

1X(X)

Alt(x)

115

E(x) . g(n) h(a,b) H(p,n, a,) Hf(x)

65

Commutativity of addition of multiplication constants: 0

28 30

1

e

&

57 57 57 57 57 7

V -)0

-

counting operator course-of-values function - recursion.

.9,

l(x)

o

J(u,v) If flf

»;

v(n,k) p(n) or

n

Deduction theorem divisor, least. double recursion

112 89 22

Enumeration of primitive recursive functions. equation calculus introductory -, provable. -, verifiable exponentiation .

140 27 27 27 47 16

Q(a,b) R(a,b)

Rt(x)

e(x) Sx. Ef .

6If(x) U(x) .

V(x) Z(x) A.

-, course-of-values. -, initial. -, propositional.

99 93 2 10 17 .14, 18, 28 .15, 18, 30

Factor, greatest common -, prime Frege, G. formal system functions

-,+ .

Pn

IIf · IIA

119 119

189

31 33 37 20 134 96 97 22 20 21 .134, 136 146 38 150 93 90 35 147 86 86 20 38 7 35 38 134, 136 134, 136 21 146 119 21 62 17,

115 11

Boole, G. bound variable.

17,

--'-

.)

Godel, K. - number greatest common factor

143 144 97

Inequalities induction schema I g

46 66 123

190

INDEX

Page

125 127 21 27

induction schema I g ' ~"

__ ~_

Tgif . . . . .

initial functions introductory equations iteration. . .

18

15

Multiplicat ion ", cornmut.at ivity of .

an

Number . . variable , Glidel , prime numeral . Operator, counting operators: A~

I

6 ]44 89 3

9,

1"o

L~.

64 64 64

iV;.

7a

}j,~.

Page

proof schema proposition provable equation

56 27

Quotient,

86

:~4

Recursion, course-of-values JI9 , double. 22 , primitive 19, 119 . , single 19 with parameter substitut.ion . 120 remainder. 86 Russell, B.A.W. 3 --

-

Schema, proof schemata: A . R ~~1' I~2 1~~3

.

II

Parameter substitution, -, recursion with permutation pentation . . . Pl~ter,

n. . ..

predicates: E(n) E(m,n,p) Pf(n) . . . . Pr(m,n) SI(m,n), S2(m,n) Stj(v/n) Subj(v/n) '1'(n) . .

T(m,n,l l ) prime numbers . -, infinitude of. - -, sequence of "- factors . . . . primitive recursion proof .

12,13 ,

120 129 16, 17 24, 121 148

rso 150 150 149 150 147 148 149 89 90 90 93 22 27

K

s

-f~~~ \ ~

Hbl' Sb 2 T U UI · single recursion. Skolem, Th. system, formal. s : fJt . fJt*

&1+ Tetration transposition. Variable. - , bound. --, number verifiable equation

34 115 115 105 108 109 110 105 118 104 104 104 105 19 143 10 112 115 152 16 129 13 65 6

47