Lecture Notes in Logic
Per Lindstrom
Aspects of Incompleteness
Springer
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Lecture Notes in Logic Editors: S. Buss, San Diego A. Lachlan, Burnaby T. Slaman, Chicago A. Urquhart, Toronto H. Woodin, Berkeley
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Per Lindstrom
Aspects of Incompleteness
Springer
Author Per Lindstrόm Department of Philosophy University of Gothenburg S-41298 Gothenburg, Sweden e-mail:
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Lindstrό'm, Per, 1936A s p e c t s of I n c o m p l e t e n e s s / Per L i n d s t r ό m . p. cm. — (Lecture notes 1n l o g i c 10) Includes b i b l i o g r a p h i c a l references and I n d e x . ISBN 3-540-63213-1 ( s o f t c o v e r : a Ik. p a p e r ) 1. I n c o m p l e t e n e s s t h e o r e m s . 2. Recursion t h e o r y . II. S e r i e s . QA9.65.L56 1997 511.3—dc21
I.
Title.
97-25726
CIP
Mathematics Subject Classification (1991): 03F30, 03F40 ISBN 3-540-63213-1 Springer-Verlag Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. © Springer-Verlag Berlin Heidelberg 1997 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: Camera-ready by the authors SPIN: 10633635 46/3142-543210 - Printed on acid-free paper
To Eva for all she's done for me
PREFACE
By GodeΓs incompleteness theorem there is no complete axiomatization of mathematics, not even of first order arithmetic. This leads naturally to the project of investigating the family of, inevitably incomplete, arithmetical theories. In this book we present some of the results obtained in pursuing this aim. A brief summary of the contents of the book is given in the introduction. GodeΓs main idea was to translate metamathematical statements, concerning formulas, formal theories, provability in such theories etc., into arithmetic - the arithmetization of metamathematics. Combining this with the technique of self-reference, he was able to construct a (formal) sentence which is undecidable, neither provable nor disprovable, in a given theory T of arithmetic, thereby demonstrating, in an entirely novel way, that T is incomplete. The ideas developed by Godel proved fruitful far beyond their original application. Later essentially new ideas were added. Most important are the basic concepts (and results) of recursion theory - these are needed for a completely general formulation (and proof) of GodeΓs theorem - notable are also the so called partial truth-definitions, the idea of formalizing the proof of the completeness of first order logic in arithmetic, and the general concept of interpretability. In this book we shall be concerned almost exclusively with properties that are common to all sufficiently strong, axiomatizable theories. Thus, for example, investigations of particular theories such as, for example, Peano Arithmetic and its fragments, requiring special, model-theoretic or proof-theoretic, methods, fall outside the scope of the book. But even so, the choice of material in some respects reflects the author's preferences; this is particularly true of Chapter 7. Credits for results, proofs, and exercises, remarks on alternative proofs, related results, etc. are given in the notes following each chapter. Results (exercises) not attributed to anyone are either easy, folklore, or due to the author. The reader is assumed to be acquainted with (first order) predicate logic including Henkin's completeness proof. We also presuppose knowledge of the elements of recursion theory. Finally, the reader may find our presentation difficult to follow unless (s)he has previously seen a more detailed development of formal arithmetic, an explicitly defined arithmetization of metamathematics, and a full proof of GδdeΓs theorem. I am grateful to Daniel Vallstrom and Rineke Verbrugge; they have very kindly and carefully read substantial portions of this book, pointed out numerous (minor) errors and suggested many improvements. The responsibility for the blemishes that remain is, of course, mine. May 1997. P.L.
CONTENTS
CHAPTER 0. Introduction
1
CHAPTER i. Preliminaries Exercises Notes
5 20 22
CHAPTER 2. Incompleteness §1. Incompleteness §2. Consistency statements §3. Independent formulas §4. The length of proofs Exercises Notes
23 23 25 31 33 35 40
CHAPTER 3. Numerations of r.e. sets §1. Numerations of r.e. sets §2. Types of independence Exercises Notes
42 42 45 50 51
CHAPTER 4. Axiomatizations §1. Finite and bounded axiomatizability; reflection principles §2. Irredundant axiomatizability Exercises Notes
52 52 57 59 61
CHAPTER 5. Partial conservativity Exercises Notes
62 70 73
CHAPTER 6. Interpretability §1. Interpretability §2. Faithful Interpretability Exercises Notes
75 75 84 88 91
CHAPTER 7. Degrees of interpretabilty §1. Algebraic properties §2. A classification of degrees
94 94 102
x
Contents
§3. Σ! and Πx degrees Exercises Notes CHAPTER 8. Generalizations §1. Incompleteness §2. Axiomatizations §3. Interpretability Notes
104 114 117 119 119 121 122 124
REFERENCES
125
INDEX
130
NOTATION
132
0. INTRODUCTION
Let T be a sufficiently strong theory formalized in the language LA of (first order) arithmetic. Following Godel, we want to show that there is a sentence φ of LA which is true (of the natural numbers) but not provable in T. GδdeΓs idea was to achieve this by constructing φ in such a way that (*) φ is true if and only if φ is not provable in T. Then, assuming (for simplicity) that all theorems of T are true, we are done. For, suppose φ is provable in T. Then, by (*), φ is not true and so, by hypothesis, φ is not provable in T. Thus, φ is not provable in T But then, by (*), φ is true. One way to achieve (*) is to find a sentence φ which, in some sense, "says" of itself that it is not provable in T. There are then three major difficulties. First of all, the sentences of LA deal with natural numbers, they do not deal with syntactical objects such as sentences (of a formal language), proofs, etc. Secondly, even if some of the sentences of LA can, somehow, be understood as dealing with syntactical objects, it is not clear that it is possible to "say" anything about provability (in T) using only the means of expression available in LA. And, finally, even if this is possible, there may be no sentence which "says" of itself that it isn't provable. Godel, however, was able to overcome these difficulties. The first problem is solved by assigning natural numbers to syntactical expressions in a certain systematic way. This is sometimes called a Godel numbering, and the number assigned to an expression, the Godel number of that expression. Thus, the numeral of the number assigned to an expression can be regarded as a name of that expression and certain number theoretic statements can be regarded as statements about syntactical objects. (In what follows "φ is a formula", "p is a proof", etc. are short for "φ is the Godel number of a formula", "p is the Godel number of a proof", etc.) To overcome the second difficulty Godel (re)invented the primitive recursive functions (sets, relations). He showed that a number of crucial properties of (Godel numbers of) expressions, such as that of being a (well-formed) formula, are primitive recursive. In particular, Godel showed that, if the set of axioms of T is primitive recursive, this is also true of the relation PRFτ(φ,p): p is a proof of the sentence φ in T φ is provable in T, PRτ(φ), if and only if ΞpPRFτ(φ,p). This property, however, is not (primitive) recursive. Godel then went on to prove that all primitive recursive functions (sets, relations) are definable in LA. Thus, in particular, there is a formula Prfτ(x,y) of LA defining PRFT(k,m). But then Prτ(x) := 5yPrfτ(x,y) defines PRT(k). (In what follows we write Th φ for PRτ(φ) ) Godel, however, proved more and this is crucial: for every sentence φ, Th φ if and only if Th Prτ(φ). (This is the first time we use the assumption that T is sufficiently strong; but, of course, if T isn't, T is incomplete for that reason.)
2
0. Introduction
This takes care of the second difficulty. So now there remains only the problem of finding a formal sentence which "says" of itself that it is not provable in T. Gδdel solved this problem in the following way. Consider the substitution function SBST(k,m) defined by: SBST(k,m) = the formula obtained from the formula m by replacing the free variable "x" by the numeral of the number k, if m is a formula, = 0 otherwise. This function is primitive recursive and so is defined in T by a formula Subst(x,y,z) in the sense that for any formula δ(x) and any number k, Th Vz(Subst(k,δ,z) z = δ(k)); in other words, for all k, δ(x), T proves that: "z satisfies Subst(k,δ,z) if and only if z is the formula δ(k)". Now consider the formula 5y(Subst(x,x,y) Λ -Prτ(y)), call it γ(x). Let θ be γ(γ). Intuitively, θ "says" that the result of replacing the variable "x" by the numeral for the number γ in the formula γ(x) is not provable in T. But this result, γ(γ), is θ itself. Thus, θ "says" that θ is not provable in T. Formalizing this argument we obtain: (**) Th θ ^ -Prτ(θ). (This is an instance of the very important fixed point lemma; Lemma 1, Chapter 1.) And now GδdeΓs proof can be completed as follows. First we show that (***) Th θ. Suppose Th θ. Then Th Prτ(θ). But then, by (**), Th - θ and so T is inconsistent (whence Th 1, where _L := ->0 = 0), contrary to assumption. Thus, (***) holds. But this is exactly what - Prτ(θ) "says". So -ιPrτ(θ) is true and consequently, by (**), θ is true. Let Conτ be the sentence ->Prτ(_L). Conτ is then a natural formalization of "T is consistent". In proving (***) we actually proved that if Thθ then Th_L and so that if T is consistent (Th_L), then Thθ. It turns out that this proof can be formalized in T provided that T is sufficiently strong. Thus, Th Conτ -» -πPrτ(θ) and so, by (**), Th Conτ -> θ. But then, since Th θ, it follows that Th Conτ; in other words, T cannot prove its own consistency. This is GόdeΓs second incompleteness theorem. This, in brief, is what Gδdel accomplished (restricted to theories in LA; generalizations to other theories containing arithmetic, for exampel set theory, are straightforward). In GδdeΓs original proofs it is assumed that the set of axioms is primitive recursive. Subsequently, when the (general) recursive functions had been defined, it turned out, however, that this assumption could, without altering the structure of the proofs, be replaced by the weaker assumption that the set of axioms is recursive. In fact, it became clear that GόdeΓs first incompleteness theorem holds for all formal systems, in the most general sense, and is actually a result belonging to recursion theory: the set of true (Π1) sentences of L A is not recursively enumerable.
0. Introduction
3
In (the above sketch of) GδdeΓs proof, and in virtually all proofs in the following chapters, the method of arithmetizing metamathematics, i.e. translating metamathematical concepts, statements, etc. into arithmetic, plays a central role. This method is based on a large number of definitions and preliminary results. In Chapter 1 we introduce the basic notation and terminology and state a number of Facts concerning these notions. These Facts will not be proved but references will be given; some of them are proved in almost every exposition of GδdeΓs theorems, others require quite extensive proofs that would be out of harmony with the rest of the book. In Chapter 1 we also prove the fixed point lemma (Lemma 1.1), the essential undecidability of Robinson's Arithmetic, Q, a very weak finite subtheory of PA, (Theorem 1.2), and the nonexistence of truth-definitions (Theorem 1.3). In Chapter 2 we present the first and most important results of the subject, GόdeΓs incompleteness theorem and his (second) theorem on the unprovability of consistency (Theorems 2.1 and 2.4). GδdeΓs results were subsequently improved in various respects and we present some of these improvements. The main result of Chapter 3 is that, assuming that T contains a minimum of arithmetic (Q), every recursively enumerable set is numerated by a (Σι) formula in T (even if not all Σ^ sentences provable in T are true) (Theorem 3.1). We also prove some refinements of this result. Given that the set of axioms of a theory T is infinite, it is natural to ask if these axioms can be replaced by a finite set of axioms. In Chapter 4 we apply the so called reflection principles to prove some negative results concerning this problem (Theorems 4.1, 4.2). For example, neither Peano Arithmetic, PA, nor any one of its consistent extensions (in LA) is finitely axiomatizable. On the other hand, every extension of PA has an axiomatization which is irredundant in the sense that none of the axioms can be derived from the other axioms (Theorem 4.6). We also prove the existence of not irredundantly axiomatizable theories (Theorem 4.7). Let Γ be a set of sentences, for example Σn or Πn. A sentence φ is Γ-conservative over T if every sentence in Γ provable in T + φ is provable already in T. Partial conservativity is studied in Chapter 5, where the basic existence theorems are proved (Theorems 5.2, 5.3,5.4); it plays an important role in Chapters 6 and 7. An interpretation of a theory S in a theory T is, roughly speaking, a function t on the set of formulas of S into the set of formulas of T such that t preserves logical form and Tht(φ) whenever Shφ. S is interpretable in T, S < T, if there is an interpretation of S in T. These concepts were introduced by Tarski. If, in addition, Sh φ whenever Tht(φ), we shall say that t is faithful and that S is faithfully interpretable inT. Interpretability was originally used as a tool in proofs of (relative) consistency and undecidability Interpretability (in arithmetical theories) is studied for its own sake in Chapter 6. The key result is that if T is an extension of PA and Cons is a sentence which (in a suitable sense) in T "says" that S is consistent, then S < T + Cons (the arithmetization of GδdeΓs completeness theorem; Theorem 6.4). From this it
4
0. Introduction
follows that S < T if and only if for every finite sub theory S' of S, Th ConS' (Lemma 6.2) and that if S, too, is an extension of PA, then S < T if and only if every Γ^ sentence provable in S is provable in T (Theorem 6.6). We also prove similar characterizations of faithful interpretability (Theorems 6.13, 6.14). Mutual interpretability is an equivalence relation; its equivalence classes will be called degrees of interpretability. Let T be a consistent extension of PA. The degrees of extensions of T, partially ordered by the relation induced by v, and for every u < v, either (y)u is a logical axiom, satisfies σ(z), or is obtained from formulas (y)w with w < u using one of the (logical) rules of derivation"; in other words "y is a proof of the sentence x from the set of sentences satisfying σ(z)". (Thus, if there are nonsentences "satisfying σ(z)", they are simply disregarded.) The fact that there is a formula Prfσ(x,y) with the desired properties (see below) follows from Facts 2 (i) and 4 (and the details of the formalization of predicate logic.) If σ(z) is Γ+, then Prfσ(x,y) is Γ+. Let Prσ(x) := ayPrfσ(x,y), Conσ := -Prσ(l), where 1 := ~Ό = 0. Thus, the intuitive meaning of Prσ(x) is: "the sentence x is provable from the set of sentences satisfying σ(z)" and Conσ intuitively says: "the set of anc sentences satisfying σ(z) is consistent". If σ(z) is Z^, then Prσ(x) is I^+l' * Conσ is Πn+1. For any formula σ(x), let (σ I y)(x) := σ(x) Λ x < y, (σ + y)(x) := σ(x) v x = y. In what follows we shall use Prfs(x,y), Prfs+z(x,y), PrfS|z(x,y), Prs(x), Cons, etc. to
14
1. Preliminaries
denote (ambiguously) any formula Prfσ(x,y)/ Prfσ+z(x,y), Prfσ |z(x,y), Prσ(x), Conσ/ etc. where σ(x) is a PR binumeration of S. If S = 0, we assume that σ(x) := -»c = x; if S is finite and nonempty, S = {φo, ,φn}/then σ ( x ) := x = Φθv
vx
= Φn
Fact 6. h Vx(σ(x) -> σ'(x)) -> Vy(Prσ(y) -> Prσ/(y)). Consequently h Vx(σ(x) -> σ'(x)) -> (Conσ, -» Conσ). Fact 7. Suppose σ(x) numerates S in T. (a) If p is a proof of φ in S, then Th Prfσ(φ,p). (b) If Sh φ, then Th Prσ(φ). (c) Suppose PAH T. Let α(xo,...,xn-l) be any formula whose free variables are XO/-/ xn-l If Sh Ot(xo/-/Xn_i), then Th Pr^V-A-i))(d) If σ(x) binumerates S in T and p is not a proof of φ in S, then Th ->Prfσ(φ,p). Fact 8. Let σ(x) be any formula, (i) PAh Prσ(x) A Prσ(x-»y) -» Prσ(y), (ii) PAh Prσ+y(z) Prσ(y->z), (iii) PAhPr σ (x)^ayPr σ|y (x). Corollary 5. Let σ(x) be any formula. (i) PAhPrσ(β(x))^Prσ(Ξxβ(x)), (ii) PAhPr σ (Vxβ(x))^Pr σ (β(x)), (iii) PAh Prσ(x) A Prσ(-x) -> -Conσ/ (iv) PAh Prσ(-ιχ) 4^ -«Conσ+x and PAh Prσ(x) ^ - Conσ+^x, (v) if PAH T, σ(x) numerates S in T, and Sh γ(x) -> δ(x), then Th Prσ(γ(x)) -^ Prσ(δ(x)), (vi) if PAH T, σ(x) numerates S in T, and Sh φ -> ψ, then Th Prσ(φ) -^ Prσ(ψ). All true Σ^ sentences are provable in Q; in fact, this is provable in PA; in other words, Q is Σ^-complete provably in PA: Fact 9. Suppose φ and δ(x0,...,xn_ι) are Σ1. (a) If φ true, then Qh φ. (b) PAh δ(x0/.../xn.1) -> PrQίδίxo,...,^)); in particular, PAh φ -» Prg(φ). By Fact 9 (a), if ψ is Γ^ and T + ψ is consistent, then ψ is true. Corollary 6. Suppose σ(x) numerates an extension of Q in PA. (a) If φ is a Σl sentence, then PAh φ -> Prσ(φ). (b) If σ(x) is Σl and τ(x) is a numeration of T (in PA), then
1. Preliminaries
15
PAhPr σ (φ)^Pr τ (Pr σ (φ)); in particular, PAh Prσ(φ) -» PrQ(Prσ(φ)). The following conditions (cf. Fact 7 (b), Fact 8 (i), and Corollary 6 (b)) are known as the Bernays-Lδb provability conditions (for PrT(x)). (BLi) if Th φ, then PAh Prτ(φ), (BLii) PAh Prτ(φ) Λ Prτ(φ-»ψ) -^ Prτ(ψ), (BLiii) PAh Prτ(φ) -> Prτ(Prτ(φ)). The construction of "self-referential" sentences and formulas will play a decisive role in what follows. Such constructions are possible in virtue of the following result, the fixed point lemma; we list a number of special cases; a completely general formulation would be needlessly complicated, φ is a fixed point of ξ(x) in T if Th φ ^ ξ(φ). Lemma 1. (a) For any Γ+ formula γ(x), we can effectively find a Γ+ sentence φ such that Qh φ γ(φ). (b) For any Γ+ formula γ(x,y), we can effectively find a Γ+ formula ξ(x) such that Qhξ(x)γ(x,ξ). (c) For any Γ+ formulas Yo(*/y) and Yι(x,y), we can effectively find Γ+ sentences φ0 and q>! such that Qh φ0 Γ'(x). Fact 10. (a) There is a Γ formula Satr(x,y) with the following properties: (i) For every Γ formula γ(x), PAh γ(x) y) -> Trr(y). (b) There is a \^+ι formula Satgn(x,y) such that for every Bn formula β(x), PAh β(x) Fact 10 (a) (i), (ii) can be used to justify self-referential constructions such as the following one. Let γ(x,y) be any Γ formula. There is then a Γ formula ξ(x) such that PAhξ(k)φ). It follows that
i- (α(χ->φ) -> (χ -»φ)) -> (χ -> φ)
Show that Th φ and Th - φ.]
5. Suppose PAH T. (a) Show that Trr(x) is not Π1^ (compare Theorem 4). (b) Show that there is a Δn+1 formula which is not B^ (compare Theorem 4). 6. (a) Show that if T is Σj^-sound, then T is Πn+1-sound. (b) Suppose PAH T and T is true. Let φ be such that PAh φ 0, is not recursive; see Exercise 2.4 (d)). The formulas Prfσ(x,y), Prσ(x), Conσ were introduced by Feferman (1960). Fact 9 is due to Feferman (1960). The Bernays-Lδb provability conditions are due to Lob (1955), simplifying the original conditions due to Bernays (cf. Hubert and Bernays (1939)). Part (a) of the fixed point lemma is implicit in Godel (1931); it was first stated explicitly by Carnap (1934) (see also Godel (1934)). The more general versions (b) - (e) were subsequently obtained by Ehrenfeucht and Feferman (1960) and Montague (1962). Lemma 2 and Theorem 2 first appeared in Tarski, Mostowski, Robinson (1953); for a stronger result, see Exercise 2.3. Theorem 3 was first published by Tarski (1933) (see also Godel (1934)). The application of (partial) truth-definitions goes back to Hubert and Bernays (1934,1939); a full proof of Fact 10 is given in Kaye (1991). For a slightly different proof of Theorem 4 and a related result, see Exercise 5. Fact 11 is essentially due to Kreisel and Wang (1955) (see also Mostowski (1952a)); for a sketch of a proof of a related result, which can easily be turned into a proof of Fact 11, see Kaye (1991), p. 140. Corollary 8 is due to Mostowski (1952a). Exercise 4 is due to Montague (1963).
2. INCOMPLETENESS
The methods of arithmetization and self-reference were originally used to prove incompleteness theorems for arithmetical theories. In this chapter we present the most important theorems of this type. A sentence φ (in the language of S) is undecidable in S if Sl^ φ and SI/ - φ. S is complete if no sentence is undecidable in S, otherwise incomplete.
§1. Incompleteness. We begin with the first and most important result of the whole subject, GόdeΓs incompleteness theorem (for theories in LA). Theorem 1. Let φ be a Π1 sentence such that (G) Qh φ φ and so Th -»φ. It follows that T is inconsistent, contrary to Convention 2. Thus, Ί\f φ. By (G), φ is true. Thus, -«φ is a false Σ^ sentence and so TI/ - φ if T is Σ1-sound. Notice the close similarity between the proofs of Theorem 1, Lemma 1.2, and Theorem 1.3 (the liar paradox). To derive the conclusion that TI/ -tφ in Theorem 1, we needed the assumption that T is Σ1-sound. We can now see that this is stronger than mere consistency: T + -ιφ is consistent but not Σ1-sound. (Note that it does not follow from Theorem 1 that T + -iφ is incomplete.) Thus, the question arises if, assuming consistency only, there is a (Π^) sentence which is undecidable in T. Our next result, known as Rosser's theorem, shows that the answer is affirmative. Theorem 2. Let θ be a Πj sentence such that (R) Qh θ o Vz(Prfτ(θ,z) -> auθ in T. Then TH θ and so, by Fact 7 (d), Qh - Prfτ(θ,q) for every q. By Fact 1 (iv), it follows that Qh z < p -> -Prfτ(θ,z), whence, by Fact 1 (v), (2) Qh Prfτ(θ,z) -» p < z. By Fact 7 (a), Qh Prfτ(- θ,p). Hence, trivially, Qh p < z -* au Ξu m < z. Now Qh p(-ιθ,m) and so Qh Vz(p(θ,z) -> 3u φ and consequently ΊV- Conτ. (b) If τ(x) is any Σ^ numeration of T, then TI/ Proof, (a) We follow closely the proof of Theorem 1 (a). By (BLiii), (1) PAhPrT(φ)^PrT(PrT(φ)). By (G) and (BLi), PAh Prτ(Prτ(φ) -> - φ) and so, by (BLii), PAhPrτ(Prτ(φ))->Prτ(-πφ). But then, by (1), PAh Prτ(φ) -» Prτ(-φ), whence, by Corollary 1.5 (iii), PAh Prτ(φ) -> -»Conτ and so, by (G), PAh Conτ -> φ. But then, assuming that Th Conτ, we get Th φ, contradicting Theorem 1 (a). It follows that ΊV Conτ. 4The proof of (b) is obtained from the above by replacing Prτ(x) by Prτ(x). In Theorem 4 (b) it is sufficient to assume that τ(x) is Σ} and numerates T in some theory S such that PAH SH T; but the assumption that τ(x) is Σ^ cannot be omitted; see Theorem 7, below. In applying Theorem 4 to an extension S of PA, we often show that there is a PR binumeration (Σ^ numeration) σ(x) of S such that Sh Conσ and conclude that S is inconsistent. A somewhat shorter proof of Theorem 4 (a) is as follows. By (G), PAh -iφ -» PrT(φ). By provable Σ1-completeness (Fact 9 (b)), PAh -iφ -> Prτ(->φ). But then, by Corollary 1.5 (iii), PAh - φ -» -«Conτ and so PAh Conτ -» φ. A similar proof yields Theorem 4 (b). Combining Theorem 4 and Corollary 1.8, we get. Corollary 1. If PAH T, then T is not finitely axiomatizable. Proof. Suppose T is finitely axiomatizable. Then there is a k such that TH T I k. Also, by Corollary 1.8, Th Conτ (k, whence T I kh Conτ ,k. But, since PAH T I k, this contradicts Theorem 4. Corollary 1 will be strengthened in Chapter 4 (Corollary 4.1) and Chapter 6 (Theorem 6.3). The proof of Theorem 4 can also be formalized in PA yielding:
§2. Consistency statements
27
Corollary 2. If PAH T, then PA + Conτh Conτ+^ConT. Proof. Let φ be as in (G). By Theorem 4 (a), (1) PAh Conτ -» φ. But then, by (BLi) and (BLii), PAh Prτ(Conτ) -> Prτ(φ) and so, by (G) (2) PAh Prτ(Conτ) -> -.φ. From (1) and (2) we get PAh Prτ(Conτ) -> -ιConτ which, by Corollary 1.5 (iv), yields the desired conclusion. The proof of our next result is another exercise in formalization, in this case of the proof of Theorem 2. Theorem 5. Let θ be a Rosser sentence for T. Then PA + Conτh -ιPrτ(θ) Λ - Prτ(-ιθ). Proof. We follow closely the above proof of Theorem 2. By Corollary 1.5 (iii), (1) PA + Conτh Prτ(θ) -> ->Prτ(- θ). It follows that PA + Conτh Prfτ(θ,z) -> - Prfτ(->θ,u) and so (2) PA + Conτh Prfτ(θ,z) -» Vu Prτ(3zγ(z)). By (R), Th Ξzγ(z) -+ -.0. But then, by (BLi) and (BLii), PAh Prτ(5zγ(z)) -» Prτ(-.θ). Combining this with (4), we get PA + Conτh Prτ(θ) -^ Prτ(-ιθ). But then, by (1), (5) PA + Conτh - Prτ(θ), as desired. Next we prove that (6) PA + Conτh-ιPrτ(-θ). From (1), we get PA + Conτh Prfτ(-.θ,u) -> - Prfτ(θ,z) and so PA + Conτh Prfτ(- θ,u) -^ Vz θ. But then (6) follows, by an argument almost the same as the proof of (5). If PAH T, this proof of Theorem 5 can be replaced by the formalization of the above short proof of Theorem 2. By (R), PAh Prτ(θ) Λ - Prτ(-θ) -> -«θ. Since ->θ is Σα, PAh -.0 -» Prτ(-ιθ). It follows that PAh Prτ(θ) -> Prτ(- θ) and so, by Corollary 1.5 (iii), PAh Conτ -> -πPrτ(θ). Next, by Corollary 1.10 (b), (R), (BLi), and (BLii), PAh Prτ(--θ) Λ --Prτ(θ) -> Prτ(θ), whence PAh Prτ(--θ) -> Prτ(θ) and so, by Corollary 1.5 (iii), PAh Conτ -> -.Prτ(- θ). Combining Theorem 5 and Corollary 1.5 (iv) we get: Corollary 3. Let θ be as in (R). Then PA + Conτh Conτ+θ Λ The sentence φ in (G) above says of itself that it is not provable in T. Let us now consider a sentence χ saying of itself that it is provable in T, i.e. such that Qh χ Prτ(χ). Is χ provable in T? In this case no simple argument in terms truth will yield an answer, not even if T is true. Nevertheless, it turns out that Th χ provided that PAH T. This follows from our next result, known as Lob's theorem. Theorem 6. Suppose PAH T and let φ be any sentence such that Th Prτ(φ) —» φ. Then Th φ. Proof. Let θ be such that (1) PAh θ (Prτ(θ) -> φ). From this, (BLi), and (BLii), we get (2) PAh Prτ(θ) -> (Prτ(Prτ(θ)HPrτ(φ)). By (BLiii), (3) PAh Prτ(θ) -> Prτ(Prτ(θ)). From (2) and (3) it follows that (4) PAh Prτ(θ) -> Prτ(φ). Since, by hypothesis, Th Prτ(φ) -> φ, this implies that (5) Th Prτ(θ) -> φ. But then, by (1), Th θ, whence, by (BLi), PAh Prτ(θ). Finally, this together with (5) yields Th φ, as desired. There is a semantic paradox related to the above proof in somewhat the same way as the liar paradox is related to the proof of Theorem 1. Let (**) If (**) is true, the earth is flat.
§2. Consistency statements
29
"Prove", by considering (**), that the earth is flat. Theorem 6 is a strengthening of Theorem 4: let φ := _L But Theorem 6 can also be derived from Theorem 4 as follows. Suppose Th Prτ(φ) —» φ. Then T + -«φh - Prτ(φ), whence, by Corollary 1.5 (iv), T + -iφh Conτ+_,φ. But then, by Theorem 4, T + -iφ is inconsistent and so Th φ. By slightly modifying the proof of Theorem 6 we can derive the stronger result that for every sentence φ, (L) PAh Prτ(Prτ(φHφ) -> Prτ(φ). In fact, from (4) we get PAh (Prτ(φ) -> φ) -> (Prτ(θ) -> φ). But then, by (1), PAh (Prτ(φ) -> φ) -» θ, whence, by (BLi) and (BLii), PAh Prτ(Prτ(φ)->φ) -> Prτ(θ). Finally, (L) follows from this and (4). Theorem 4 is sometimes informally expressed by saying that if T is as assumed, then T does not prove that T is consistent. That this must be interpreted with some care is clear from the following result. Theorem 7. Suppose PAH T. Let τ(x) be any formula binumerating T in T and let τ*(x) := τ(x) Λ Conφ. Then (i) τ*(x) binumerates T in T and (ii) PAh Con,.*. The following intuitive proof of Theorem 7 (ii) (formalizable in PA) is probably easier to understand than the formal argument below, but its formalization would be somewhat longer: "Any proof p from the set X defined by τ(x) Λ Con^ contains a greatest sentence φe X. Since φ satisfies Conτ|x, it follows that the set of members of X occurring in p is consistent. Thus, p cannot be a proof of L." Proof of Theorem 7. Note that x is free in Cort^. (i) If ke T, then Th τ(k). By Corollary 1.9 (a), Th Con^. Thus, Th τ*(k). If, on the other hand, keT, then Th -ιτ(k) and so Th - τ*(k). (ii) Trivially h τ*(x) -» τ(x). Hence, by Fact 6, (1) h Con,- -> Corv Since PA is reflexive, we have PAh Con^. (We assume that 0 is not a formula.) Also, by Fact 8 (iii), PAh VzCoriφ —> Cor^.. By the least number principle, it follows that (2) PAh -iConτ -> Ξz(-.Conτ|z+1 Λ Conψ). By Fact 6, PAh -'Conτ|z+1 -> (Conτjx -> x < z). Hence, by the definition of τ*(x), PAh ~ Conτ|z+1 -> (τ*(x) -^ τ(x) Λ x < z). Hence, again by Fact 6, PAh - Conτ|z+1 A Conτ|z But then, by (2),
30
2. Incompleteness
PAh -"Coriτ -» Conτ* and so, by (1), PAh Corv,.*, as desired. If τ(x) is PR, then τ*(x) is Π1. By Theorems 4 and 7, τ*(x) is not provably in T equivalent to a Σ^ formula. The formula τ*(x) may seem like a mere curiosity, but certain closely related formulas are actually of crucial importance in connection with interpretability (see the proof of Lemma 6.2.). By Theorems 4 and 7, there are formulas τ0(x) and τ ^x) binumerating T in T such that ConTQ and Conτ are not provably equivalent in T. We now show that this is so even if we restrict ourselves to PR formulas. Theorem 8. Suppose PAH T. Let τ(x) be any PR binumeration of T. (a) There is a PR binumeration τ'(x) of T such that (i) Th Con,. -> (ii) Ί\f- Coiv -> (b) Let π be a true Γ^ sentence such that Th π -> Conτ. There is then a PR binumeration τ'(x) of T such that Th π f(φ). Proof. Let δf(x,y) be a Σ1 formula defining f in Q (cf. Fact 3 (b)). Let φ be such that Qh φ Vy(δf(φ,y) -> Vzφ and so Th -»φ, a contradiction. Thus, φ has no proof p < f(φ) in T. But then, by Fact 1 (iv) and Fact 7 (d), Qh Vz -«Prτ*(φ) and so Th - φ.] 16. Suppose PAH T. Prove the following strengthening of Corollary 4. Suppose X is r.e. and monoconsistent with T. There is then a PR binumeration τ(x) of T such that - Conτ£ X (see Exercise 6.6 (b)). [Hint: Let τ'(x) be a PR binumeration of T, let p(x,y) be a PR binumeration of a relation R(k,m) such that X = {k: ΞmR(k,m)}, let φ be such that PAh φ o Conτ,(x)ΛVy<x_np(^φ/y), and set τ(x) := τ' 17. Suppose PAH T. Let τ0(x) and τ-^x) be PR binumerations of T. (a) Show that there is a PR binumeration τ(x) of T such that Th Con,. Con. Λ Con^. [Hint: Let τ(x) := τ0(x) v ay<xPrfτι(_L,y). See also Theorem 8 (b).] (b) Show that there is a PR binumeration τ(x) of T such that Th Con,. φ. Show that there is a PR binumeration τ(x) of T such that TI/ Coriτ —» φ (compare Theorem 8 (a)). (d) Suppose TV φ and TV - ψ. Show that there is a PR binumeration τ(x) of T such that TI/ Con,. -> φ and TY ψ -^ Co^. [Hint: Use Lemma 1 and Theorem 8 (b).] 18. Suppose PAH T Let α(x), β(x) be PR formulas and let α < β mean that there is a primitive recursive function g such that PAh Vx(Prfα(l,x) -> Prfp(JL,g(x)). (α < β implies PAh Conβ —> Conα, bur not conversely.) Let α = β mean that α < β < α. Let τ(x) be a PR binumeration of T and let α(x) be such that PAh τ(x) -> α(x). Show that there is a Γ^ (ΣI) sentence φ such that α = τ + φ. [Hint: In the Γ^ case let φ be such that PAh φ Vx(Prfα(_L,x) -> 3y<xPrfτ+φ(_L,y)). Use the fact that to every PR formula δ(x), there is a primitive recursive function h such that PAh δ(x) -> Prfτ(δ(x),h(x)).] 19. Prove the following strengthening of Theorems 3 and 9. If {Tk: ke N} is an r.e. family of theories, there is a Π^ formula which is simultaneously independent over all the theories Tk. Strengthen Theorems 10 and 11 in the same way. 20. (a) Derive Theorem 9 for extensions of PA from Theorem 11. (b) Formulate and prove a generalization of Theorem 11 which implies Theorem 10 for extensions of PA. 21. Suppose PAH T. Let σ be any Σ-^ sentence. Show that there is a ΣI sentence χ such that PAh (σ v Pr(_L)) «-» Pr(χ). Conclude that (i) for every Σ^ sentence σ such that Th Pr(_L) -> σ, there is a ΣI sentence χ such that Th σ Pr(χ) and so for any sentences φ, ψ, there is a Σj sentence χ such that Th Pr(χ) Pr(φ) v Pr(ψ), and (ii) for every Γ^ sentence π such that Th π -> Con, there is a Π1 sentence θ such that Th π Ξy(δ(y) Λ Vz s and the statement holds for s'. If Case 1.3 or Case 2 applies at s'+l, then hs/+1 = hs/ and so, by the inductive assumption, hs/+1[Q(s)] = 1. If Case 1.1 or Case 1.2 applies at s'+l, then hs,+1[Q(s')] = 1 or hs/+1[Q(s'+l)] = 1 and so, by Lemma 3 (a) (iii), hs/+1[Q(s)] = 1. Now the desired conclusion follows by induction." (Since this argument takes place in PA, Cases 1.1 and 2 cannot be ruled out.) This proves (14). Proof of Claim 1. Fix s. Argue in PA: "By (2) and (9), there is an s'> s such that for every i < s, hs,(i) = 1 iff ξ(i). By (14), hs/[Q(s)] = 1. By Lemma 3 (a) (iv), no p{ with i > s occurs in Q(s). Thus, by (1), (Q(s))(ξ)."φ Proof of Claim 2. Let m be a proof of P(ξ) in T. Let s = . By Lemma 3 (a) (i), it is sufficient to show that Q(s) —> P is a tautology. Suppose not. Let t be such that t[Q(s)] = 1 and t[P] = 0. Then ts « t. By (13), hs = ts and so hs « t. But then Case 1.1 applies at s+1 and so g(s+l,ξ) = 1, contrary to (10). Thus, Q(s) —> P is a tautology. Finally, Q(s)e F and so Pe F. + This concludes the proof of Theorem 5. For PAH T, Theorems 1, 2, 3 are, of course, special cases of Theorem 5.
50
3. Numerations of r.e. sets
Exercises for Chapter 3. 1. Suppose QH T0H Tx. Show that for every r.e. set, there is a Σx formula which numerates X in both TQ and T^. 2. We write SHpT to mean that S is a proper subtheory of T. (a) Suppose QH lΌHpIV Let XQ and Xl be r.e. sets such that XQ c X χ . Show that there is a formula ξ(x) numerating \ in Ti7 i = 0,1. [Hint: Let θ be such that TQ!^ θ and ΊI\- θ. There exist a formula ξ1(x) numerating X x in T0 and in Ί± and a formula ξo(x) numerating XQ in T0 + -»θ. Let ξ(x) := ξ1(x) Λ (θ v ξo(x)).] (b) Suppose QH T0Hp... HpTn. Let Xi/ i < n, be r.e. sets such that Xj c χ.+1 for i < n. Show that there is a formula ξ(x) numerating Xj in T^ for i < n. (c) Suppose QH T0 H Tj and suppose there is a formula σ(x) which numerates Th(S) in S for every S such that T0H S H Tα. Show that Tα H T0. [Hint: Suppose T x h θ and let φ be such that Qh φ ^-> -ισ(φvθ). Show that T0h ~ φ.] (d) Suppose QH T0/ QH T l7 and TQ and T^ are incomparable (with respect to H). Let X0 and \ι be any two r.e. sets. Show that there is a formula ξ(x) which numerates Xi in TJ, i = 0,1. 3. Suppose QH T0HpT1. Show that there is a formula ξ(x) such that for every recursive function f, the set {n: TQ|- ξ(n) & there is a proof p of ξ(n) in T1 such that ξ(n) has no proof < f(p)inT 0 ) is infinite, in fact, nonrecursive (this improves Theorem 2.13). [Hint: Let X be an r.e. nonrecursive set and let ξ(x) be a formula numerating X in TQ and N in T^.] 4. Let X0 and \ι be r.e. sets. Let ξ0(x) be a Σ^ formula numerating X0 in T. Show that there is a Σ^ formula ξ1(x) numerating X x in T such that ξ0(x) v ξχ(x) numerates X0 u Xι in T. (If n = 1 and T is Σ1-sound, this is trivial.) [Hint: Let p(x,y) be a PR formula such that Ξyp(x,y) correctly numerates X± in T, let ξ(x) be such that Qh ξ(k) ^ 3y(p(k,y) A Vzψ and so S + Xh -πψ. Thus, by (2), S + X is inconsistent. Note the obvious analogy between the proofs of Theorems 1 and 2, on the one hand, and the proof of GodeΓs theorem (Theorem 2.1), on the other. Note also that if T is Σj-sound, then X = {->Prτ(φ): TI/ φ} is a (non-r.e.) set of Πj sentences such that T + RfnτH T + X and T + X is consistent. Since PAh RFN0 (Fact 11), we have (a) of the following corollary, improving Corollary 2.1. Corollary 1. (a) There is no consistent bounded set X such that PAH X. (b) If PAH T, there is no bounded set X such that T + RFNTH T + X and T + X is consistent. If PAH S, the above proof of Theorem 2 can be replaced by the following simple argument; the proof of Theorem 1 can be simplified in a similar way. Let χ:=Vx(Γ(x)ΛPr s (x)-*Tr Γ (x)). Now let θ be any Γd sentence such that S + θh χ. Then S + θh ->Prs(^θ), whence S H- θh Cons+θ and so S + θ is inconsistent, by Theorem 2.4. This argument and (a somewhat more detailed version of) the above proof of Theorem 2 can be looked at from a different point of view which will be further
54
4. Axiomatizations
elaborated in Chapter 5: Let φ be any Γ sentence such that S + ->χh φ. Then S + ->φh χ and so Sh φ. Thus, -χh φ, then Sh φ. Next we show that if PAH T, no bounded extension of T is essentially infinite over T (and a bit more). Theorem 3. Suppose PAH T, let X be an r.e. set of Γ sentences, and let Y be any r.e. set of sentences such that T + XI/ ψ for every ψe Y. There is then a Γ sentence θ such that T + θh X and T + θ^ ψ for every ψe Y. Proof. By Craig's theorem, we may assume that X and Y are primitive recursive. Let ξ(x) and η(x) be PR binumerations of X and Y, respectively. Case 1. Γ= Πn. Let θ be such that PAh θ ^ Vy(ξ(y) A Vzu -πPrfτ+θ(z,u)) -> TrΠn(y)). Suppose ψeY and T + θh ψ. Let p be a proof of ψ in T + θ and let q = max{p,ψ}. Then (1) PAhVzu 0}. The proof of the following lemma is straightforward and left to the reader. Lemma 2. (a) If k > m > 0, then
§1. Finite and bounded axiomatizability; reflection principles.
55
PAh Con(k,S) -> Con(m,S). (b) For all k, m > 0, PAh Con(k,S + Con(m,S)) Con(k+m,S). The sets Rfn(n,S) are defined as follows: Rfn(0,S) = 0, Rfn(l,S) := RfnS, Rfn(n+l,S) := Rfn(l,S + Rfn(n,S)). Next let Rfng = U{Rfn(n,S): neN}. We write SHpS' to mean that S is a proper subtheory of S'. Theorem 4. Suppose PAH T and T is Σ1-sound. (a) T + Con^HpT + Rfnτ. (b) T + Rfn^HpT + RFNT. Lemma 3. (a) PA + Rfn τ h Rfnτ+Con . (b)PA+RFN T hRFN T+Rfnr Proof, (a) Let φ be any sentence. PA + Rfnτh Prτ(Conτ -^ φ) -» (Conτ -> φ). But, as we have already observed, PA + Rfn τ h Conτ. It follows that PA + Rfnτh Pr
τ+Conτ(φ) -» Φ^ as desired. Φ (b) We give an informal proof using the fact that Fact 10 (a) is provable in PA. We assume, as we may, that the PR binumeration p(x) of Rfnτ implicit in the notation RFNj+Rfn is such that PA proves that every sentence satisfying p(x) is of the form Prτ(θ) -> θ. Suppose Σα c Γ. Now argue in PA + RFNT: "Let ψ be any Γ sentence provable in T + Rfhτ and let PrT((pj) -> φir for i < n, be the members of Rfnτ occurring in the proof. We may assume that -iPr-j^), for i < n, since those Prτ(φ) -» φ for which Prτ(φ) are provable in T and we may add the proofs of them to the original proof. Since -iPrT((pj) —> (Prτ(φj) —»9^) is (trivially) provable in T, it follows that θ := -.Prτ(φ0) Λ...Λ - Prτ(φn) -^ ψ is provable in T. By RFNT, Trr(θ). But, by Fact 10 (a) (ii), Trrd^Pr^cpi)), for i < n. Hence, by Fact 10 (a) (iii), Trr(ψ), as desired." Proof of Theorem 4. (a) In view of Lemma 3 (a), it follows, by induction, that T + Rfn τ h Conίj!. T + Conίj! is consistent, since T is Σ1-sound, and Con^ is an r.e. set of Π1 sentences. Thus, by Corollary 2, T + Con^ \f- Rfhτ. Φ (b) By Lemma 3 (b), T + RFNτh Rfn£ Let Xk = {- Prτ+Rfn(k/τ)(φ): T + Rfn(k,T)l^ φ}. Then, by induction, T + U{Xk: k < n}h Rfn(n+l,T). Let X = U{Xk: keN}. Then X is a (non-r.e.) set of true Π^ sentences, whence T + X is consistent, and T + Xh Rfnίj?. Thus, by Theorem 2, T + Rfn^ \t- RFNT. If T is Σ1-sound then, by Theorem 4 (a), T + Conίjί is a proper subtheory of T + Rfhτ. In our next result we show that if we restrict ourselves to Π1 sentences, this is no longer true.
56
4. Axiomatizations
We write SHΠlS' to mean that S is a Uγ-subtheory of S', i.e. every Γ^ sentence 7 provable in S is provable in S . Theorem 5. If PAH T, then T + Rfn τ H Πl PA + In the proof we use the following observation. Lemma 4. If QH S, then SHΠlPA + Cons. Proof. Let π be a Πj sentence such that Sh π. Then PAh Prs(π). Since - π is Σ l 7 we also have, PAh - π -> Prs(-ιπ). It follows that PAh ->π -> -«Cons and so PA + Consh π. Proof of Theorem 5. Let φ0/ (pi, 92,-. be all sentences of LA. For every theory S, let It is sufficient to show that for every n, there is a k such that TnHΠ PA + Con(k,T). By Lemma 4, TnHΠ PA + ConTn and so we need only prove that PA + Con(k,T)h ConTn. First we note that (1) for any sentence φ, PA + Con(2,S)h Cons+Pr ^ _> φ. Argue in PA: "Suppose -iCons+Pr / φ j _> φ/ in other words, S + Prs(φ) —» φh J_. Then Sh Prs(φ) and Sh - φ. But then Sh Prs(- φ) and so Sh - Cons/ whence -τCon(2£)" This proves (1). We now show that for every n, (2) for every extension S of PA, PA + Con(2n+1,S)h ConSn. For n = 0 this holds, by (1). Suppose (2) holds for n = k. Let S be any extension of PA. Then (3) PA proves: PA + Con(2k+1,S) h ConSk, (4) PA proves: if Con(2k+1,S + ConSk), then (S + ConSk)k is consistent. Now argue in PA: "Suppose -ιConSk+1/ in other words, Sk + Prs(φk+1)->φk+1h_L Then, since SH Sk, But then, by (1), Sk + ConSkh _L and so S + ConSk + A{Prs(9i) ^Φi: i < k}h 1. It follows that S + ConSk + A{Prs+ConSk(9i) -> Φi: i < k}h 1, in otiier words, (S + ConSk)kh _L. But then, by (4), (5) -Con(2k+1,S + ConSk). By (3), we also have PA + Con(2k+1,S)h ConSk. From this and (5) we get
§2. Irredundant axiomatizability k+1
57 k+1
-πCon(2 ,S + Con(2 ,S)), k 2 and so, by Lemma 2 (b), - φ: φ is Γ}, RFNS(Γ) := Vx(Γ(x) Λ Prs(x) -> Trr(x)). (i) Improve (a) by showing that if φ is Γ, then PA + φ + Rfns(Γd)h Rfns+φ(Γd), if φ is Γ, then PA + φ + RFNs(Γd)h RFNs+^P1). (ii) Show that if QH S, then PA + Consh RίnsίΠ^, PA + RFNs(Σn)hRFNs(Πn+1). (iii) Suppose PAH T. Show that if X c Γ is r.e. and T + Xh Rfnτ(Γd), then T + X is inconsistent, if X e Γ and T + Xh REN^Γ1), then T + X is inconsistent. Define the sets Rfh®(Γ) and RFN^Γ) in the natural way. Suppose S and T are true. ^ D Conclude that
T + Rfn^ RFN^),
T + Rfn^ΣJ^ Rfnτ(Πn) for n > 2,
2. Suppose PAH T. Let RFNτ = (Vx(Γ(x) Λ Prτ(x) -» Trr(x)): Γ arbitrary}. Let φ be any sentence such that W φ. Show that there is a PR binumeration τ(x) of T such that T + RFNτ^ φ. 3. Suppose PAH T and T is Σ1-sound. (a) Show that T + RfnT(Γ) is not essentially infinite over T. (b) Let S be such that T + Rfnτ(Σ1)H SH T + Rfnτ. Show that S is infinite over T. [Hint: Use (the proof of) Theorem 5 and Theorem 2.4.] 4. (a) Suppose the formula α(x) is such that for every φ, if Th φ, then Th α(φ). Show that there is a sentence ψ such that TV α(ψ) ^ ψ. [Hint: Use Exercise 1.4.] (b) Suppose there is a formula α(x) such that for every φ, if h φ, then Th α(φ), Th α(φ) —> φ. Show that T is not finitely axiomatizable. (This also follows by the proof of Theorem 1 with S = 0.) 5. T is reducible to S if there is a recursive function g(n) such that for all sentences φ, (i) Th g(φ) and (ii) if Th φ, then Sh g(φ) -» φ. If T is a finite extension of S, T = S + θ, then T is reducible to S: let g(φ) = θ for every φ. Prove the following result, a strengthening of Theorem 1: if RfnsH T, T is not reducible to S. [Hint: Suppose T is
60
4. Axiomatizations
reducible to S and let g(n) be the relevant recursive function. Let δ(x,y) be such that for every sentence φ, Qh δ(φ,y) y = (g(φ)->φ) (cf. Fact 3). Let ψ be such that Qhψ^Ξy(δ(ψ,y)Λ-Pr s (y)). Show that Th ψ and Qh - ψ.] 6. (a) Suppose SΫ φ and S + iφ + Z is non-i.a. over S + ->φ. Show that S + {φ v ψ: ψeZ} is non-i.a. over S. (b) Suppose THpT'. Show that (i) there is a theory S such that TH SH T' and S is not i.a. over T, (ii) if T is finitely axiomatizable, there is a theory S such that TH SH T7 and S is not i.a. 7. Suppose PAH T. Let X and Y be any r.e. sets of Γ sentences such that if φe X and ψe Y, then Th φ -> ψ. Show that there is a Γ sentence θ such that if φG X and ψe Y, then Th φ -^ θ and Th θ —» ψ. [Hint: Suppose Γ = Πn. Suppose X and Y are primitive recursive. Let ξ(x) and η(x) be PR binumerations of X and Y. Let θ := Vx(η(x) A Vy<x(ξ(y) -> -TrΠn(y)) ^ TrΠn(x)).] 8. Suppose PAH T and T is Σ^-sound. (a) Show that there is a Πj formula β(x) such that for every m, T + β(m) is consistent and T + β(m)h Conτ+β(m+1). (Note that if T is true, so are the theories T + β(m).) [Hint: Let the primitive recursive function f be defined (in T and in the real world) as follows; we assume that δ(x) is a PR formula: f(δ,ξ,0) = 0, f(δ,ξ,n+l) = m if m > f(δ,ξ,n), Vz<m->δ(z), n is a proof in T of -iξ^m), if there is such a number m, = f(δ,ξ,n) otherwise. If the value of f(k,m,n) is not determined by these conditions, it is irrelevant and we may set f(k,m,n) = 0. Next let γ(z,x) be such that PAhγ(z,x)^Vy(f(z,γ,y)<x). Letg(k,s) = f(k,γ,s). Claim. If Ξlxδ(x) is true, then for every n, g(δ,n) = 0. Proof. Let k be the least number such that δ(k) is true. Then for every n, g(δ,n) < k. Thus, if the claim is false, there is a largest n such that g(δ,n) Φ g(δ,n+l). Let m = g(δ,n+l). Then n is a proof of - γ(δ,m). It follows that ^y(g(δfy) < m) is provable
Notes
6i
and so is true, a contradiction. Let δ'(x) be a PR formula such that PAh Ξxδ'(x) o Prτ(-Vy(g(δ',y) = 0)). Letβ(x):=Vy(g(δ',y)<x).] (b) Show that with each rational number a > 0, we can effectively associate a Γ^ sentence θa such that T + θa is consistent and if a < b, then T + θah Conτ+θb. [Hint: Define a function g in much the same way as in case (a) except that g may, in a sense, take rational numbers > 0 as values.]
Notes for Chapter 4. Theorems 1 and 2 are due to Kreisel and Levy (1968). The formula Prs Γ(x) and the present formulation of the proof of Theorem 2 are due to Smoryήski (1981b). Corollary 1 (a) is due to Montague (1961) and Rabin (1961). What we have called the uniform reflection principle RFNS is not quite what is usually referred to by that term, but for theories containing PA the difference is negligible. Theorem 3 is due to Lindstrom (1984a). Corollary 2 is due to Kreisel and Levy (1968). Theorem 4 (b) is a weak form of a result of Feferman (1962). For (partial) improvements of Theorems 1,2,4 and Corollaries 1,2, see Exercise 1. Theorem 5 is due to Goryachev (1986) (with a different proof); the bound 2n+1 obtained in the proof is far from optimal; using methods not explained here, it can be shown that n+2 will do (cf. also Beklemishev (1995)). More information on (transfinite) iterations of consistency statements and reflection principles, a rather technical subject which falls outside the scope of this book, can be found in Feferman (1962) and Beklemishev (1995). What we have called an irredundant axiomatization is usually called an independent axiomatization. Theorem 6 is due to Montague and Tarski (1957). Lemma 5 is due to Tarski (cf. Montague and Tarski (1957)). For a proof of the existence of an r.e. set as described in Lemma 6, a so called hypersimple set, see Soare (1987). The idea of using hypersimple sets to construct non-i.a. theories is due to Kreisel (1957). Theorem 7 is related to a result of Pour-El (1968) and Corollary 4 is Pour-EΓs result restricted to theories in LA. Theorem 8 is new; Theorem 8 with Πn replaced by Σ^ and restricted to Σn-sound theories is also true but seems to require a quite different proof. Exercise 3 (b) is due to Beklemishev (199?). Exercise 4 is due to Montague (1963). Exercise 5 is due to Kreisel and Levy (1968). Exercise 8 (a) was proved by Harvey Friedman, Smoryήski, and Solovay, independently, answering a question of Haim Gaifman; for a different proof, due to Friedman, see Smoryήski (1985), p. 179. Exercise 8 (b) is due to Alex Wilkie (with a different proof); see Simmons (1988). The present proof can be modified to yield much stronger conclusions.
5. PARTIAL CONSERVATIVITY
A sentence φ is Γ-conservatΐve over T if for every Γ sentence θ, if T + φh θ, then Th θ. In this chapter we study this phenomenon for its own sake. Results on Γ-conservativity are, however, also very useful in many contexts, in particular in connection with interpretability (see Chapters 6 and 7). Our task in this chapter is to develop general methods for constructing partially conservative sentences satisfying additional conditions such as being nonprovable in a given theory. We assume throughout that PAH T. The results of this chapter do not depend on the assumption that T is reflexive. A first example of a Π^-conservative sentence is given in the following: Theorem 1. ->Conτ is Π1-conservative over T. Proof. Suppose θ is Γ^ and (1) T + -ιConτh θ. From (1) we get PAh Prτ(--θ) -> Prτ(Conτ), whence (2) PAh Prτ(-θ) -* -Conτ^Conτ. By provable Σ1-completeness, (3) PAh -θ -» Prτ(-ιθ). By Corollary 2.2, (4) PA + ConτhConτ^Conr Combining (2), (3), (4) we get PAh -πθ -» - Conτ and so by (1), Th θ. By Corollary 2.4, Theorem 1 provides us with an example of a (Σ1) sentence φ which is Π^-conservative over T and nontrivially so, i.e. such that TM φ, even if T is not Σ1-sound. 1 If φ is Γ-conservative over T and ψ is Π , then clearly φ is Γ-conservative over T + ψ. Also note that if T is Σ1-sound and π is Πj, then π is Σ1-conservative over T iff π is true iff T + π is consistent. Let us now try to construct a sentence φ which is nontrivially Γ-conservative over T. Thus, given that (1) T + φh θ, where θ is Γ, we want to be able to conclude that Th θ. This follows if (1) implies that (2) T + - θh φ. The natural way to ensure that (1) implies (2) is to let φ be a sentence saying of itself that there is a false Γ sentence (namely θ) which φ implies in T. Thus, let φ be such that (3) PAh φ ^ 5u(Γ(u) Λ Prτ+φ(u) A -Trr(u)), where Γ(x) is a PR binumeration of the set of Γ sentences. Then (1) implies (2).
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It is, however, not generally true that T> φ. This holds if T is true, since φ is then false. But, for example, T + -ιConτh φ, and so if Th ->Conτ, then Th φ. To prevent this from happening, we redefine φ as follows: let φ be such that PAh φ ay5uv [Γ]τ(x,z), (ii) T + φ h [Γ]τ(φ,m) for all φ and m, (iii) if ψ is Γ and T + φh ψ, there is a q such that PA + [Γ]τ(φ,q)h ψ. Proof, (i) is clear, (ii) Let GQ,...^ be all Γ sentences < m provable in T + φ and whose proofs are < m. Then PAh Vuv<m(Γ(u) Λ Prfτ+φ(u,v) -^ u = Θ0 v...v u = θk). Also clearly, by Fact 10 (a) (ii), T + φh u = ΘQ v...v u = θk -> Trr(u). It follows that T + φ h [Γ]τ(φ, m). (iii) Suppose ψ is Γ and T + φh ψ. Let p be a proof of ψ in T + φ and let q = max{ψ,p}. Then PA + [Γ]τ(φ,q)h Trr(ψ) and so PA + [Γ]τ(φ,q)h ψ. S is a Γ-subtheory of T, SHΓ T, if every Γ sentence provable in S is provable in T. We write [Γ](x,y) for [Γ]τ(x,y). Lemma 2. Suppose χ(x,y) is Π1. There is then a Π* formula ξ(x) such that for all k and m, (i) T + ξ(k)hχ(k,m), (ii) T + ξ(k)H Γ T + {χ(k,q):qeN}. Proof. Case 1. Γ = Σ^. Let ξ(x) be such that (1) PAh ξ(k) Vy([Σj(ξ(k)/y) -> χ(k,y)). Then (i) follows from Lemma 1 (ii) and (1). To prove (ii), suppose ψ is Σj^ and T + ξ(k)h ψ. By Lemma 1 (iii), there is a q such that
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Hence, by Lemma 1 (i), PA + Vy χ(k,y)) and so, by (1), PA + Vy m. But then, by (1), T + ξ(k) + Vy<mχ0(k,y)h χι(k,m), as desired. (ii) Suppose ψ is Πn and (2) T + ξ(k)hψ. By Lemma 1 (iii), there is a q such that T + [Πn](ξ(k),q)h ψ and so (3) T + iψh -[Πn](ξ(k),q). By Lemma 1 (ii), for every m, (4) T + -ξ(k)h[Σn]K(k),m). By (3), (4), Lemma 1 (i), and (1), it follows that T + -ψ + -ξ(k) + VyTr Γ (q). By Lemma 2 with (Γ replaced by Π1 and) χ(x,y) := η(y) -> Trr(y), there is a Γ sentence θ such that for all φ, (2) T + θhη(φ)^Tr Γ (φ), (3) T + ΘHpd T + (η(q) -> Trr(q): qe N}. From (2) it follows that T + θh X and from (1) and (3) it follows that T + ΘHpd T + X. Φ (b) Left to the reader. So far there has been no indication that the properties of Σ^ and Πn, n > 1, in terms of partial conservativity may be different, but we shall now show that they are. Let ψ0 and ψj be Γ sentences. If (1) ThΨo v Ψl, then, trivially, (2) ψj is pd-conservative over T + -ιψι_j, i = 0, 1. If Γ = Πn, the converse of this is true. This follows from our next: Lemma 5. Let ψ0 and ψj be any Πn sentences. There are then Πn sentences Θ0 and G! such that (i)
ThθQVθ!,
(ii) (iii)
Th Ψi -> θj, i = 0, 1, Th Θ0 Λ Ql -> ψ0 A ψχ.
Proof. By Fact 5, we may assume that ψi := Vxδ^x), where δj(x) is Σ^. Let θj := Then (i), (ii), (iii) are easily verified (cf. Lemma 1.3). From (ii) and (iii) of Lemma 5 it follows that T + -ΊI^ + ψι_ih -iθj. Hence, assuming (2), T + -^ψih -«θj. It follows that Th Θ0 v θj^ -» ψ0 v ψ1 and so, by Lemma 5 (i), we get (1). We now prove that if Γ = Σ^, then (2) does not imply (1).
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Theorem 5. (a) There are 1^ sentences ψ0/ ψi such that (i) Th -.(Ψo ΛΨl), (ii) ΊV ψ0 v ψl7 l (iii) ψi is Πn-conservative over T + ~ ψι_j, i = 0, 1. (b) Suppose X is r.e. and monoconsistent with T. Then there are Σn sentences ψ0, ψ! such that (i) and (iii) hold and (iv) ψQ We derive this theorem from: Lemma 6. Let X be an r.e. set. There are then Σ^ formulas ξo(x) and ξι(x) such that for i = 0, 1, (i) Th^ξoMΛξiM), (ii) ifkeX,thenTh-.ξi(k), (iii) if kg X, then ξj(k) is Πn-conservative over T + -iξ^k). Proof. Let p(x,y) be a PR formula such that X = {k: 3mPAh p(k,m)}. For i = 0, 1, let ζi(x)/ λi(x)/ δj(x,y) be, respectively, 2^, Σ^, and Πn_ι formulas such that (1) PAh Xi(k) ay(-[Πn](ξi(k),y) Λ Vzφ, impossible. Thus, φg X and so (iv) holds, (i) and (iii) follow from Lemma 6 (i) and (iii), respectively. Theorem 5 (b) will be used in the proof of Theorem 7.7 (b), below. Note that, by Theorem 5, Lemma 5 with Πn replaced by 1^ is false. We can now partially improve Corollary 2.5 as follows:
Corollary 1. There are 1^ sentences ψ0, Ψi, such that Th ψ0 -> ~«ψι and there is no Δn sentence φ for which Th ψ0 -» φ and Th φ -> -iψj. Proof. Let ψ0, ψj be as in Theorem 5 (a). Suppose φ is Δ^ Th ψ0 —> φ, and Th φ —> -•ψ!. Then Th --ψ! -^ φ and Th -ιψ0 -» -«φ and so Th ψ0 v ψlx a contradiction. Let Cons(Γ,T) be the set of sentences Γ-conservative over T. It is clear from the definition of Cons(Γ,T) that it is a Π^ set. We now show that this classification is correct. Our next lemma follows at once from Lemma 3.2 (b) but has a simpler direct proof which we leave to the reader. Lemma 7. Let R(k,m) be any r.e. relation. There are then formulas Po(*/y) and p1(x,y) such that pg(x/y) is Σ lx Pι(x,y) is n l x po(*/y) numerates R(k,m) in T, PAh Pθ(k,m) —> p1(k,m), and if not R(k,m), then Th Theorem 6. (a) Cons(Γ,T) is a complete Π^ set. (b) If Γ Φ ΣI, then Π1 n Cons(Γ,T) is a complete n° set. Proof. Let X be any Π^ set and let R(k,m) be an r.e. relation such that X = (k: VmR(k,m)}. Let p(x,y) be a formula numerating R(k,m) in T, which is ΣI if Γ = Σn and Π^ if Γ = Πn. Let ξ(x) be as in (the proof of) Lemma 2 with χ(x,y) := p(x,y) To prove (a) it is now sufficient to show that (1) kEXiffξ(k)eCons(Γ,T). By Lemma 2, (2) T+ ξ(k)hp(k,m), (3) T + ξ(k)H Γ T + {p(k,q):qEN}. If keX, then Th p(k,q) for every q and so, by (3), ξ(k)eCons(Γ,T). If kgX, there is an m such that Th p(k,m) and so, by (2), ξ(k)£ Cons(Γ,T) (in fact, ξ(k) is not Σα- or not Π1-conservative over T, as the case may be). Thus, (1) holds.This proves (a). If Γ is Σj^ or Πn with n > 2, then ξ(x) is Γd as claimed in (b). Finally, suppose Γ = Πj. Let p0(x,y) and Pι(x,y) be as in Lemma 7. Let p(x,y) := Po(x/y) Then ξ(x) is Σα. By Lemma 7, ξ(k)£ Cons(Πl7T) if k^ X. Thus, (b) holds in this case, too. Suppose T is Σ1-sound and θ is Π^ Then θ is Σ1-conservative over T iff θ is true. Thus, H! n Cons(Σl7T) is Π°.
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We conclude this chapter with a proof of Theorem 4.8. We derive this result from the following lemma; a refinement of this lemma (for n = 1) will be proved in Chapter 7 (Lemma 7.22). Lemma 8. There is a Πn formula ξ(x) such that for every k, (i) Th ξ(k), (ii) Thξ(k+l)->ξ(k), (iii) ξ(k) is ^-conservative over T + ->ξ(k+l). Proof. In a first attempt to prove Lemma 8 it is natural to let ξ(x) be such that PAh ξ(k) ^ ξ(k+l) v Vv([ΣJK(k+l)Λξ(k),v) -> -Prfτ(ξ(k),v)). But then (i) does not follow and so we have to proceed in a more indirect way. Let δ(u) be any formula. Let κ(z,x,y) be a Πn formula such that (1) PAh -κ(z,x,0), (2) PAh κ(δ,k,y+l) ^ κ(δ,k+l,y) v Vv([ZJ(^5(k)Ak(k),v) ^ -Prfτ(ξδ(k),v)), where ξδ(x) := Vu(δ(u) -+ κ(δ,x,(u ^ x) + 1)), ηδ(x) := Vu(δ(u) -* κ(δ,x+l,u ^x)). (-i is the function such that k-im = k - m i f k > m and = 0 otherwise.) In (2) set y = u - k. Then, since neither y nor u is free in the second disjunct of (2), by predicate logic, we get (3) PAh ξδ(k) ^ ηδ(k) v Vv([ΣJ(^δ(k)Λξδ(k)Λ) -* -Prfτ(ξδ(k),v)). It follows that (4) ifThξ δ (k),thenThη δ (k). For let p be a proof of ξδ(k) in T. By Lemma 1 (ii), T + -ηδ(k)Λξδ(k)h-Prfτ(ξδ(k),p), whence T + ξδ(k)h ηδ(k) and so Th ηδ(k). Clearly (5) if Th δ(u) -+ u > k, then Th ηδ(k) ξδ(k+l). Suppose now δ(u) is PR. Then (6) if 3uδ(u) is true, then Th ξδ(0). Suppose Ξuδ(u) is true and Th ξδ(0). Let m be the least number such that δ(m) is true. Then Th δ(u) -> u > m. By (4) and (5), it follows that Th ηδ(m). But also Th δ(m) and so, by (1), Th - φ} and Xl = {k: Th θk -> - φ}. 12. Suppose T is not Σ1-sound. Show that Π^ n Cons^/Γ) is a complete Π^set. [Hint: Let R(k,m) and S(k,m,n) be an r.e. and a primitive recursive relation such that X = {k: VmR(k,m)} and R(k,m) iff 3nS(k,m,n). Let σ(x,y,z) be a PR binumeration of S(k,m,n). Let γ(x) be a PR formula such that Ξxγ(x) is false and provable in T. Let p0(x,y), Pι(x,y), and δ(x,y,z) be such that PAh p0(x,y) (p. [Hint: Let φ := Ξxγ(x) and ψ := Vxδ(x), where γ(x) and δ(x) are Π^ and Σ^, respectively. Then T + φ + ψh 3x(γ(x) A Vy<xδ(y)).] (b) Show that there is an r.e. family of consistent extensions of PA such that for no Γ does there exist a Γ sentence which is nontrivially Γd-conservative over every member of the family. [Hint: Let φ be a Π^ sentence undecidable in PA. Then {PA + - θ: PAh θ -» φ} u {PA + θ: PAh φ -> θ} is an r.e. family of extensions of PA. Suppose θ is Πn and nontrivially Σn-conservative over all members of this family. Then PA + φl/ θ. θ is Σj^-conservative over T + -ι(θ Λ φ). It follows that PA + φh θ, a contradiction. The dual case is similar.] 15. This exercise may be compared with Theorems 2.13, 2.14. (a) For each Γ, there is a primitive recursive function f such that for every Γ sentence φ, f(φ) is a proof in PA of φ f(p) for every proof q of ψ in T.
Notes for Chapter 5. The general concept Γ-conservative is due to Guaspari (1979). Theorem 1 is due to Kreisel (1962). Lemma 2 is due to Lindstrδm (1984a). Lemma 3 and Theorem 2 with X = Th(T) are due to Guaspari (1979); for somewhat stronger results, also due to Guaspari (1979), see Exercise 5 (b). The proofs of Lemma 3 and Theorem 2 are from Lindstrδm (1984a). Lemma 4 is due to Lindstrόm (1984a). (Lemmas 2 and 4 and their proofs are similar to and were inspired by results of Guaspari (1979), Solovay (cf. Guaspari (1979)), and Hajek (1971); for further applications, see e.g. Hajek and Pudlak (1993).) Theorem 3 less the references to the set X is due to Solovay (cf. Guaspari (1979); see also Jensen and Ehrenfeucht (1976); the full result is proved in Smoryriski (1981a) and Lindstrδm (1984a). The formula Prf'τ/Γ(x,y) was introduced by Smoryriski (1981a); (Sm) and the fixed point mentioned in Exercise 3.7 (a) are special cases of a very general construction due to Smoryriski (1981a); however, in the proof of his main theorem Smoryriski has to assume that the formulas %i(x,y) are PR. Theorem 4 is due to Lindstrδm (1984a). Lemma 6 and Theorem 5 are due to Bennet (1986), (1986a). Corollary 1 with Σ^ replaced by Πn is false (Exercise 3). Theorem 6 for Γ = Γ^ and for Γ = Πn+1 are essentially due to Solovay (cf. Hajek (1979)) and Hajek (1979), respectively, (in both cases with different proofs);
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Theorem 6 for Γ = Σ^ n > 1, is due to Quinsey (1980), (1981) (with a different proof); the present proof is due to Lindstrom (1984a). For more information on Cons(Γ,T) and related sets, see Exercises 12 and 13. Lemma 8 is due to Lindstrom (1993); Lemma 8 with Πn and 1^ interchanged and restricted to Σ^-sound theories is also true but the proof is quite different. An alternative concept of partial conservativity has been introduced and studied by Hajek (1984). Exercise 2 (a) is due to Smoryriski (1980); Exercise 2 (c) is due to Svejdar (cf. Hajek and Pudlak (1993)). Exercise 4 is due to Kent (1973). Exercise 5 (b) is due to Guaspari (1979). Exercise 7 is due to Bennet (1986). Exercise 10 (a) is due to Smoryriski (1981a). Exercise 12 is due to Quinsey (1981); the suggested proof is due to Bennet. Exercise 13 (c) is due to Bennet (1986). Exercise 14 is due to Misercque (1983).
6. INTERPRETABILITY
Let S and S7 be arbitrary theories. S7 is interpretable in S if, roughly speaking, the primitive concepts and the range of the variables of S7 are definable in S in such a way as to turn every theorem of S' into a theorem of S. If, in addition every nontheorem of S7 is transformed into a nontheorem of S, then S7 is faithfully interpretable in S. In this chapter, we assume that PAH T. Thus, T is essentially reflexive.
§1. Interpretability. Let S and S7 be arbitrary theories. By a translation (of the language of S' into the language of S) we understand a function t on the set of formulas (of S') into the set of formulas (of S) for which there are formulas TJO(X)/ %(*/y)/ η+(x,y,z), ηx(x,y,z) and a formula μt(x) such that t satisfies the following conditions for all formulas φ, ψ, ξ(x): (*) t(x = y) := x = y, t(x = 0) := η0(x), t(Sx = y) := ηs(x,y), t(x + y = z) := η+(x,y,z), t(x x y = z) := ηx(x,y,z), t(- φ) := -t(φ), t(φ Λ ψ) := t(φ) Λ t(ψ), t(3xξ(x)):=3x(μt(x)Λt(ξ(x))). (Here x, y, z are arbitrary variables.) We assume that V and the connectives v, ->, t(ξ(x))). Clearly t is a primitive recursive function. The translation t is an interpretation in S iff (**) Sh 3xμt(x), Sh 3x(μt(x) Λ Vy(μt(y) -> (η0(y) y = x))), Sh Vx(μt(x) -» 3y(μt(y) A Vz(μt(z) -> (ηs(x,z) z = y)))), Sh Vxy(μt(x) Λ μt(y) -> 3z(μt(z) Λ Vu(μt(u) -> (η*(x,y,u) u = z)))), * = +, x. Thus, t is an interpretation in S iff Sh t(φ) for every logically valid sentence φ. t is an interpretation o/S7 in S, t: S7< S, iff Sh t(φ) for every φ such that S 7 h φ. S7
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is interpretable in S, S'< S, if there is an interpretation of S' in S. S'< S means that S'< 7 S^S . Trivially, if S'H S, then S'< S. The reader should check that < is a transitive rela7 tion. Also note that if S'< S, then every finite subtheory of S is interpretable in a finite subtheory of S. 7 If S'< S and S is consistent, so is S'. For suppose S is not consistent. Let φ be any sentence. Then S'h φ Λ -Prσι(t(x)). To prove this, we reason (informally) in PA as follows: "Suppose φ is derivable from formulas satsifying σ0(x). Then there are ψ0/ /ψn °f formulas satisfying σQ(x), such that Λ{ψk: k < n} -> φ is provable in logic. But then, by Fact 12 (this chapter), t(Λ{ψk: k < n}) -> t(φ) is provable from the set defined by σχ(x). But t(A{ψk: k < n}) := Λ{t(ψk): k < n}). Also, by the definition of σQ(x), each t(ψk) is derivable from the set defined by σα(x). But then so is A{t(ψk): k < n}). It follows that t(φ) is derivable from the set defined by σ1(x).// This proves (1). From (1) we easily get the desired conclusion. Theorem 1 in combination with GόdeΓs second incompleteness theorem (Theorem 2.4) yields the following strengthening of GodeΓs result. For a different
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improvement of Theorem 2.4, see Theorem 8, below. Theorem 2. T + Conτ £ T. Proof. Suppose T + Conτ < T. Then, by Theorem 1, there is a Σx numeration τ'(x) of T + ConT such that T + Conτh Coiv By Theorem 2.4 it now follows that T + Conτ is inconsistent. But then, since T + Conτ < T, T is inconsistent, contrary to Convention 2. Since Conτ is Πj, Theorem 2 is also a direct consequence of Theorem 2.4 and the following: Lemma 1. If π is a Π1 sentence and Q + π < T, then Th π. Proof. There is a k such that Q + π < T I k. So, by Corollary 1, Th Conτ |k -* Cong+π. It follows that Th CoriQ+π. Since -iπ is Σl7 we have, by provable Σ1-completeness, Th -.π -> -»ConQ+π. It follows that Th π. Note that we have actually proved that Q + Con^ ^ T. In Chapter 2 (Corollary 2.1) we proved that PA is essentially infinite (in fact, PA is essentially unbounded; Corollary 4.1). This can now be improved as follows: Theorem 3. T is not interpretable in any finite subtheory of T. Proof. Let S be a finite subtheory of T and suppose T < S. By Theorem 1, there is then a Σ1 numeration τ(x) of T such that PAh Cons -» Cor^. Since, by Fact 11, T is reflexive, we have Th Cons and so Th Cor^, contradicting Theorem 2.4. Most positive results on the existence of interpretations in the sequel are applications of the following fundamental result, the arithmetization of GδdeΓs completeness theorem. Theorem 4. Let σ(x) be a formula numerating S in T. Then S < T + Conσ. Proof (informal outline). A full proof of this result would be quite long and we shall be content to give a fairly detailed sketch. The main idea is to show that (the denumerable case of) the Henkin completeness proof for first order logic can be formalized in PA. (The reader is assumed to be familiar with that proof.) We begin with an outline of Henkin's proof. Let S be a (countable) set of sentences (theory) assumed to be consistent. Let cn, ne N, be new individual constants. Let L be the language obtained from Ls by adding the constants cn. Let α^x^, ne N, be a primitive recursive enumeration of all formulas of L with one free variable. We can then form a primitive recursive set Z = {Ξx n α n (x n )^α n (c Jn ):neN} such that
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(1) for every sentence θ of S, if S + Zh θ, then Sh θ. It follows that S + Z is consistent. Now let θn/ ne N, be a primitive recursive enumeration of all sentences of L. The sentences φn are then inductively defined as follows: (2) φn = θn if S + Zh Λ{φm: m < n} -> θn/ = -«θn otherwise. (Here A{φm: m < 0} := 0 = 0.) φn is not in general a recursive function of n. Let X = {φn:nGN}. Then (3) Th(S) c X and, since S + Z is consistent, (4) X is Henkin complete in the sense that X is complete and consistent and for every formula α(x) of L with the one free variable x, if 3xα(x)e X, there is a constant ck such that α(ck)e X. We can now define a model M M M M M = (M, S , + , χ , 0 ) of X in the following way. The domain M of the model is the set {cn: neN}. (Here we ignore the minor difficulty that X may contain sentences of the form ck = cm with k Φ m and so the members of M cannot in general be the constants themselves but must instead be certain "equivalence classes" of these constants or, in the present context, members of such equivalence classes. If we disregard the trivial case where S has only finite models, this can be avoided by defining Z in a slightly different way.) M 0 = c i , CM = cn, n/Cn): Ck + Cm = Cne X},
XM = {(ck,cm,cn): ck x cm = CΠG X}, where q is the (uniquely determined) constant such that 0 = q e X. Finally, it can be shown, by induction and using the fact that X is Henkin complete, that for every sentence φ of L, (5) φ is true in M iff φe X. This is true, by the definition of M, if φ is atomic. Finally, Th(S) c X and so M is a model of Th(S). We can now transform this into a proof that S < T + Conσ in the following way. th We first define in PA a primitive recursive function c(x) (= the X new individual constant). By a c-formula we understand a formula obtained from a formula of LA by replacing each free variable v by c(v). (Thus, the c-formulas are the counterparts of the sentences of L.) Let ζ(x) be a suitably defined PR Enumeration of Z, where Z is defined as above except that we now use the function symbol c. Then (the reader will hopefully believe that) for every sentence φ of S, (6) PAhP Γσvζ (φ)^Pr σ (φ). (compare (1)). It follows that (7) PAh Conσ ->
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The inductive definition of φn can, using methods available in PA, be turned into an explicit definition. Let χ(x,y) be a suitable formalization of this explicit definition (cf. Chapter 1, p. 9). Let ξ(x) := Ξyχ(x,y). (Thus, intuitively, ξ(x) means "x is a member of X".) Then (compare (3)) (8) PAhPr σ (x)^ξ(x). Let Hcmξ be the sentence saying that the set defined by ξ(x) is Henkin complete. Thus, for all c-formulas α, β, (9) PA + Hcmξh ξ(- α) -ιξ(α). (10) PA + Hcmξh ξ(α) Λ Pr0(α-»β) -» ξ(β). Moreover, for every formula α(x) such that Ξxα(x) is a c-formula, (11) PA + Hcmξh ξ(3xα(x)) -> 3uξ(α(c(ύ))). The (inductive) proof of (4) does not use any means of proof beyond those available in PA. Thus, we get PAh Conσvς —» Hcmξ and so, by (7), (12) PAh Conσ -» Hcmξ. We can now define a translation t, corresponding to the model M, as follows. Let μt(x) := 3u(x = t(Sx = y) := 3uv(x = c(u) A y = c(v) A ξ(Sc(ύ) = c(v))), t(x + y = z) := 3uvw(x = c(u) Λ y = c(v) Λ z = c(w) Λ ξ(c(ύ) + c(v) = c(w))), t(χ x y = z) := 3uvw(x = c(u) Λ y = c(v) Λ z = c(w) Λ ξ(c(ύ) x c(v) = c(vί))). These equations uniquely determine t. The proof corresponding to the proof of (5) now yields for every formula P(XO/— /XH-I) of LA containing no free variables other than XQ/ /XR-I/ (13) PA + Hcmξh μt(xQ) Λ...Λ μt(xn_ι) -> (t(β(xo/.../Xn-i)) ** 3u0,...,un_1(x0 = c(u0) Λ...Λ xn-1 = c(un_x) Λ ξ(β(c(ύ)0,...,c(ύ)n_1)))). By the definition of t, this holds for atomic β(x0,...,xn_1). The inductive steps dealing with -i and Λ follow easily, by (9) and (10). Let us consider the step dealing with 3. For simplicity, let n = 1 and write x for x0. Let α(x,y) be such that β(x) := 3yα(x,y). Then t(β(x)) := 3y(μt(y) Λ t(α(x,y)). By the inductive hypothesis, PA + Hcmξh μt(x) A μt(y) -> (t(α(x,y)) 3uv(x = c(u) A y = c(v) A ξ(α(c(ύ),c(v)))). By (10) and (11), PA + Hcmξh 3vξ(α(c(ύ),c(v))) *-> ξ(3yα(c(ύ),y)). But then it is fairly easy to see that PA + Hcmξh μt(x) -> (3y(μt(y) A t(α(x,y)) 3u(x = c(u) A ξ(3yα(c(ύ),y)))), as desired. This proves (13). From (12) and (13), we get for every sentence φ, (14) PA + Con σ ht(φ)^ξ(φ). Finally, let φ be any sentence provable in S. Then Th Prσ(φ). Hence, by (8), Th ξ(φ) and so, by (14), T + Conσh t(φ). It follows that t: S < T + Conσ.
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This concludes our sketch of the proof of Theorem 4. If we don't insist on mimicking every detail of Henkin's proof, we can instead use the simpler interpretation t' defined in the following way: MO := x = x/
t'(x = 0) := ξ(0 = c(x)), t'(Sx = y):=ξ(Sc(x) = c(y)), f (x + y = z) := ξ(c(x) + c(y) = c(z)), t ' ( x x y = z):=ξ(c(x)χc(y) = c(z)), Thus, MX)istrivial anc*can be omitted. (This is true as long as we are dealing with theories of (elementary) arithmetic; it is not true in general.) It is via the following lemma, the (Feferman-)Orey-Hajek Lemma (and Theorem 6, below) that Theorem 4 becomes such a powerful tool in the theory of interpretability (of arithmetical theories; see also Lemma 8.4). Lemma 2. S < T iff Th Cons ( k for every k. To prove this we need the following lemma whose proof is essentially the same as that of Theorem 2.7. Lemma 3. Suppose Th Cong \k for every k. Let σ(x) be any formula binumerating S in T and let σ*(x) := σ(x) Λ Conσ)x. Then (i) σ*(x) binumerates S in T and (ii) PAh Conσ*. Proof of Lemma 2. Suppose first S < T. Let k be arbitrary. There is then an m such that SI k < T I m. By Corollary 1, PAh Conτ |m -» Cons |k. But Th Conτ ( m and so ThCon s , k . Next suppose Th Con$( k for every k. Let σ(x) be a PR binumeration of S and let σ*(x) := σ(x) Λ Conσ|x. Then, by Lemma 3, σ*(x) binumerates S in T and PAh Conσ*. Hence, by Theorem 4, S < T. There are alternative notions of interpretability more general than the one defined here. For example, we may "interpret" the equality symbol = of one theory S as a certain relation definable in another S' (and having, provably in S7, the required properties) or we may "interpret" the individuals of S as finite sequences of individuals of S' etc. It turns out, however, that if S is "interpretable" in T in some such more general, and reasonably natural, sense, then, by Lemma 2, S < T (and conversely). Thus, in the present context, there is no reason to consider these more general "interpretations". From Lemmas 2 and 3 and Theorem 4 we get the following: Corollary 2. S < T iff there is a formula σ(x) (bi)numerating S in T such that Th Conσ.
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From Lemma 2 we also obtain the following result known as Orey's compactness theorem. Theorem 5. S < T iff SI k < T for every k. In the following we use A, B, etc. to denote (consistent, primitive recursive) extensions of T. Recall that AH Γ B means that every Γ sentence provable in A is provable inB. Theorem 6. A < B iff AH Πl B. Proof. Suppose first AH Πι B. Now, Ah ConA |k for every k. It follows that Bh ConA ik for every k. But then, by Lemma 2, A < B. Suppose next A < B. Let π be any Π1 sentence such that Ah π. By Lemma 1, Bh π, as desired. By Theorem 6, A + φ < A iff φ is Π1-conservative over A. Theorem 6 has the following immediate: Corollary 3. If A < B and σ is any Σ± sentence, then A + σ < B + σ. Combining Theorem 6 and Theorem 4.5 we get: Corollary 4. T + Rfnτ < PA + Con^. In fact, this follows directly from Lemma 2 and the fact, established in the proof of Theorem 4.5, that PA + Con^h ConTn for every n. Theorem 6 can also be used to prove the following model-theoretic characterization of interpretability: Theorem 7. A < B iff for every model M of B, there is a model M' of A such that M is (isomorphic to) an initial segment of M'. Proof (sketch). "If". Let θ be any Πj sentence such that Ah θ. We show that θ holds in all models of B. Let M be any model of B. By hypothesis, there is a model M' of A such that M is isomorphic to an initial segment of M'. θ holds in M'. Since θ is Πl7 it follows that θ holds in M. Thus, θ holds in all models of B and so Bh θ. We have shown that AH Π B and so A < B, by Theorem 6. "Only if". Let t: A < B. Let M be any model of B and let M' be the structure defined by t in M. M' is a model of A. Since induction holds in M, we can in M define a function f on M satisfying the following conditions: f(0M) = 0M', f(SM(a)) = SM (f(a)). f maps M isomorphically onto an initial segment of M'. Given Theorem 6, we can now derive Theorems 8-12 below as corollaries to
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results from Chapter 5. Like Theorem 2 the following result is a sharpening of GodeΓs second incompleteness theorem. Theorem 8. T + ->Conτ < T. Proof. This follows from Theorem 5.1 and Theorem 6. A more direct proof of Theorem 8 is as follows. We need the following: Lemma 4. If S < S'+ φ0 and S < S'+ φα/ then S < S'+ φ0 v q>r Thus, if S + φ < S + -ιφ, then S + φ < S. Proof. Suppose tji S < S'+ φ^ i = 0,1. Let t be the translation which coincides with t0 if φ0 and with tj if -i(pQ Λ φ1. Thus, for example,
μt(χ) := (ΦQ Λ MtoM)v (""ΦoA ΦiA m^*))-
It follows that for all φ, (1) S' + φ 0 ht(φ)^t 0 (φ), (2) S/ + -φ 0 Λφ 1 ht(φ)t 1 (φ). Now, suppose Sh φ. Then S' + φQh t0(φ) and so, by (1), S' + φQh t(φ). Also S' + φ1h t^φ) and so, by (2), S' + - φ0 Λ φ1h t(φ). It follows that S'+ φ0 v q^h t(φ). Thus, t: S < S'+ φ0 v φ1, as desired. By Corollary 2.2, T + Conτh Conτ+_,ConT. But then, by Theorem 4, T + - Conτ < T + Conτ and so, by Lemma 4, T + -«Conτ < T, as desired. (In this proof of Theorem 8 it is not necessary to assume that T is (essentially) reflexive.) Theorem 9. Suppose X is r.e. and monoconsistent with T. There is then a Σ1 sentence φ such that T + φ < T and φ£ X. Proof. This follows from Theorem 5.2 and Theorem 6. Corollary 5. Let τ(x) be a formula numerating T in T such that Tl^ -"Co^. There is then a (Σ{) sentence φ such that T + φ < T and Tl^ Conτ —»Conτ+φ. Proof. Let X = {ψ: Th Coi^ -» Con^} and use Theorem 9. Theorem 10. Suppose X is r.e. and monoconsistent with T. There is then a sentence φ such that T + φ < T, T + -.φ < T, φ£X, -κp£X. Proof. By Theorem 5.3 we can take φ to be, say, a Σ2 sentence such that φ is Π2-conservative and ->φ is Σ2-conservative over T. Now use Theorem 6. A sentence φ such that T + φ < T, T + -«φ < T is known as an Orey sentence for T. Clearly, any Orey sentence for T is undecidable in T.
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The intended applications of Theorems 9 and 10 are as follows. There are consistent finitely axiomatized extensions U of T in languages extending LA. In fact, U may chosen to be a conservative extension of T in the sense that for every sentence φ of LA, Uh φ iff Th φ. Thus, U and T are equivalent in terms of provability of sentences of LA. So it is natural to ask if U and T are (ever) equivalent in terms of interpretability of sentences of LA in the sense that for every sentence φ of L A/ T + φ < T iff U + φ < U. (We assume the reader can extend the defintion of "interpretation" and "interpretable in" to the case where the theories need not be formalized in LA.) The answer is a resounding "no" (see also Corollary 8.8). To prove this we need the following essentially trivial lemma whose proof is left to the reader. Lemma 5. Let V be any r.e. theory, not necessarily in LA. Then the set {φ: U + φ < V} is r.e. Corollary 6. There is a ΣI sentence φ such that T + φ < T and U + φ ^ U. Proof. The set {φ: U + φ < U} is clearly monoconsistent with T and, by Lemma 5, it is r.e. Now apply Theorem 9. By a similar proof, but using Theorem 10 in place of Theorem 9, we get: Corollary 7. There is a sentence φ such that T + φ < T, T + -«φ < T, U + φ ^ U , and U + -φ i U. As we saw in Chapter 4, speaking in terms of provability, we have to distinguish between finite, infinite, and unbounded extensions of a given theory T. In terms of interpretability the situation is quite different. We write S = S' to mean that S < S'< S. Theorem 11. (a) If AH B, then there is a sentence φ such that A + φ = B. (b) Let X be an r.e. set of Σj sentences. Then there is a Σ^ sentence σ such that
Proof, (a) Let X = Th(B) n nx. Then, by Theorem 6, A + X = B. By Theorem 5.4 (a), there is a sentence φ such that A + φ is a Γ^- conservative extension of A + X. By Theorem 6, A + φ Ξ A + X and so A + φ = B. 4 (b) This follows from Theorems 5.4 (a) and 6. Finally, we have a result which proves the claim made earlier that the fact that, for example, A + φ < B does not imply that this is provable in PA, or in any other preassigned consistent axiomatizable theory. From the definition of < it is clear that the set {φ: A + φ < B} is Σ^. From Theorem 6, it follows, however, that {φ: A + φ < B} is Π^. That this cannot be improved follows from:
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Theorem 12. Suppose A < B. Then the set (ψE Σα: A + φ < B} is a complete Π^ set. Proof. For A = B, this follows from Theorem 5.6 and Theorem 6; we leave the proof of the general case to the reader. A translation t is given by a finite amount of information which can certainly be coded by a natural number; thus we may "identify" t with that number. Let IntAB be the set of interpretations of A in B. Corollary 8. If A < B, then IntA/B is n° but not Σ°. Proof. Clearly IntA B is Π2. Suppose it is Σ 2 . Evidently A + φ < B iff Ξte IntA B(Bh t(φ)). It follows that {φ: A + φ < B} is Σ!?, contradicting Theorem 12. π In the next § we are going to prove that Int AB is, in fact, a complete Π2 set (Corollary 12).
§2. Faithful interpretability. Let t: S'< S. t is a faithful interpretation of S' in S, t: S'^ S, if for every sentence φ, if Sh t(φ), then S'h φ. S' is faithfully interpretable in S, S'^ S, if there is a t such that t: S'^3 S. Most of the differences between < and ^ are explained by the following lemma; for example, it is not true in general that if SH T, then S ^ T. Lemma 6. If QH S ^ T, then THΣιS. Proof. Suppose t: S ^ T. Let σ be any ΣI sentence such that Th σ. Clearly t: Q + -*σ < T + - t(σ). But then, by Lemma 1, T + - t(σ)h ->σ, and so Th t(σ). Since t is faithful, it follows that Sh σ. Our main aim in this § is to prove the following characterizations of ^. Theorem 13. S ^ T iff S < T and for every φ, if Th Pr0(φ), then Sh φ. Theorem 14. A ^ B iff AH Πl BH Σι A. Corollary 9. (a) S ^ T iff for every k, Th Cons |k and for every φ, if Th Pr0(φ), then Shφ. (b) If T is Σ^sound, then S ^ T iff S < T. (c) If S < TH S, then S ^ T. Proof, (a) and (b) follow at once from Theorem 13 and Lemma 2. 4 (c) Suppose Th Pr0(φ). Then, since T is essentially reflexive (Fact 11), Th φ and so, by assumption, Sh φ. Now use Theorem 13.
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By Corollary 9 (c), Theorems 8, 9,10 remain true when < is replaced by ^. Theorem 13 will be derived from the following two lemmas: Lemma 7. Let σ'(x) be a (Σj) formula binumerating S in T. There is then a (Σ1) formula σ(x) binumerating S in T and such that (i) h σ(x) —> σ'(x), whence h Conσ/ -> Conσ/ (ii) for every sentence φ, if Th Prσ(φ), then there is a q such that Th Prs \ q(φ). Lemma 8. Suppose σ(x) numerates S in T and Th Conσ. There is then an interpretation t: S < T such that for every φ, if Th t(φ), then Th Prσ(φ). Proof of Lemma 7. For simplicity we assume, as we clearly may, that if p is a proof of φ in T, then φ < p. Let σ(x) be such that PAh σ(x) o σ'(x) Λ Vyz<x(Prfτ(Prσ(y),z) -> PrσΊz(y)). Then (i) is trivial. We now show that (1) if p is a proof of Prσ(φ) in T, then Th Prσ,,p(φ). Let p and φ be as assumed. Then, since φ < p, Th -Prσ, i p(φ) -> (σ(x) -> σ'(x) A x < p). It follows that Th -πPrσ, i p(φ) -» (Prσ(φ) -> Prσ,,p(φ)). But then, since Th Prσ(φ), we get Th Prσ, |p(φ), as desired. Since σ'(x) binumerates S in T, it follows from (1) that (ii) holds. To show that σ(x) binumerates S in T it suffices to show that for all φ and p, Th Prfτ(Prσ(φ),p) -> Prσ, ,p(φ). But this, too, follows at once from (1). Proof of Lemma 8. The following proof is a modification of the proof of Theorem 4. The interpretation t constructed in that proof does not necessarily have the additional property that (1) Th t(φ) implies Th Prσ(φ). To achieve this we proceed as follows. The function c, the set Z, and the formula ζ(x) are the same as before, but the definition of φn is different. Here we put (2) φn := θn if S + ZhΛ{φm: m < n}->θn or (S + ZhΛ{φm: m < n}-> - θn & ne Y), := -»θn otherwise, where Y is any set of natural numbers. As before let X = {φn: ne N}. Either θn or -«θn is put in X. We put θn in X if putting - θn in X would make X inconsistent, and similarly for -«θn. Otherwise we put θn in X iff nE Y. The idea is to achieve (1) by letting Y be formally represented by a sufficiently independent formula η(x). Let γ(x) := σ(x) v ζ(x). Let η(x) be as in Theorem 2.10 with δ(x) := Prγ(x). Next, as in the proof of Theorem 4, let χ(x,y) be the formalization of the result of turning the
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inductive definition of φn into an explicit definition using η(x) to represent Y. Let Letξ(x):=Ξyχ(x,y). As in the proof of Theorem 4 we can now define an interpretation t of S in T such that (3) Tht(φ)^ξ(φ). It remains to be shown that (1) holds. Suppose ΊV Prσ(φ). We must then show that ΊV t(φ). We have ΊV PΓγ(φ) (see (6) in the proof of Theorem 4). For any fe 2N, let Y f = (PΓγ(n)f(n): ne N}. Now let f(n) be such that f(φ) = 1 and (4) T + Yf is consistent. Next we define ψn as follows (compare (2)). ψn := θn if Prγ(Λ{ψm:m φ. Let ge 2N be such that and set Y f/g = Yf+{η(n)S( n ):neN}. Then, by (4) and the choice of η(x), (5) T + Yf^g is consistent. Recalling the definition of χ(x,y), we can now show, by induction, that for every n, T + Y f / g h χ(ψn,n) and so (6) T + Y f/g hξ(ψ n ). Next we show, by induction, that for every n, (7) PΓγ( Λ{ψm:m a. If a cups to b, there is a Πj degree c < a which cups to b. But then c cups to a, contrary to assumption. Finally, we prove Theorem 5 (and a bit more). We have already observed that d(-ιπ) is the p.c. of d(π). Thus, every Γ^ degree has a p.c. It follows that, in terms of our classification of degrees, the following result is the best we can do.
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Theorem 17. There is a Σ^ degree which has no p.c. This is a consequence of the following strengthening of Lemma 20. Lemma 23. There is a sentence σ such that {b > d(-ισ): b is Σ-J has no g.l.b. To prove this, we need another: Lemma 24. Suppose {πk: keN} is r.e. and let G = {d(πk): keN}. Suppose there is no finite subset H of G such that OH is a lower bound of G. Then G has no g.l.b. Proof. Let X = {π: T + π k h π for every k}. X is not r.e. This can be seen as follows. Let R(k,m) be a primitive recursive relation such that Y = {k: VmR(k,m)} is not r.e. and let p(x,y) be a PR binumeration of R(k,m). We may assume that Z = {πk: ke N} is primitive recursive; let ζ(x) be a PR binumeration of Z. Finally, let η(x) := Vz(-p(x,z) -> 3uπk. Then T < T + πk v ψ < T + πk. Let σk := g(πkvψ) and πk+1 := f(πkvψ). Then ΊV πk+1. For every k, (1) π k + 1 d(πk). For suppose b = d(χ) > d(-ισ), where χ is Σx. Then T + χh -,σ, whence T + σh -.χ. But then, by (2), there is a k such that T+-,πkh-πχ, whence T+χh πk and so b > d(πk).
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Let G = (d(πk): keN}. If {b > d(-ισ): b is Σα} has a g.l.b. c, then, by (1), (3), (4), c is the g.l.b. of G. But from (1) it follows that no d(πk) is a lower bound of G. Hence, by Lemma 24, G has no g.l.b. and so {b > d(-.σ): b is Σ^ has no g.l.b. Proof of Theorem 17. Let σ be as in Lemma 23. By Lemma 6, for all B, (T + σ)iB < T iff B < T + χ for all Σl sentences χ such that T + χh -πσ. But then the p.c. of d(σ), if it had one, would also be the g.l.b. of {b > d(-.σ): b is Σj}. Thus, by Lemma 23, d(σ) has no p.c. Every ΣI degree is the p.c. of some degree. It is an open problem if the converse of this is true. If it is, the Σ1 degrees can be characterized in a purely algebraic way as those degrees that are p.c.s.
Exercises for Chapter 7. In the following exercises we assume that PAH T and that A, B, etc. are extensions ofT 1. Suppose G £ Dτ. G is independent if for any disjoint finite subsets G0 and G^ of G, OG0 ^ LJG^ (O0 = lτ, LJ0 = Oτ.) (Thus, for example, 0 is independent and {a} is independent iff Oj < a < lτ.) Show that for every finite independent set G, there are degrees b0, bj such that G u {bj is independent, i - Ό, 1, and b0 n b1 = Oτ. Conclude that every finite independent set is included in 2 κ o many maximal independent sets. 2. Suppose a < b. (a) c cups to b above a if there is a d such that a < d < b and c u d = b. Show that there is a ce (a,b] which doesn't cup to b above a. (b) c caps to a below b if there is a d such that a < d < b and c n d = a. Show that there is a ce [a,b) which doesn't cap to a below b. 3. Suppose a < b and b < l τ if T is Σ1-sound. For ce [a,b], let c* be the complement of c in [a,b] if it exists, i.e. c n c* = a and c u c* = b. (Complements are unique.) Let Cpla/b be the set of degrees in [a,b] having complements in [a,b]. (a) Show that Cpla/b is closed under n, u, and *. Let Cplab = (Cpla/b, π, u, *, a, b). Then Cplab is a Boolean algebra. (b) Show that if c, de Cpla^b and c < d, there is an ee Cplab such that c < e < d. (It follows that the Boolean algebras Cpla/b are (denumerable and) atomless and therefore isomorphic.) (c) Show that if a < c < d < b, there is an ee [c,d) such that Cpl ab n [e,d) = 0. [Hint:Cpl a/b n[c,d]CCpl c/d .] (d) Show that i f a < c < e < d < b and e£ Cpla/b, there are c', d' such that c < cr < e < d r < d and Cpla/b n [c'^7] = 0.
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4. Suppose a is Σ1. (a) Show that if a < b < l τ/ then a caps to Oτ below b. (b) Show that if a < b and b is high, then a «b. Conclude that if bt > a, i = 0,1, and b0 n bx = a, then b0 and b1 are low. ((a) and (b) are true of every a which is the p.c. of some degree,) 5. Show that for every low degree a, there is a low Γ^ degree > a. [Hint: Let B = T + σ and σ := Ξxδ(x), where δ(x) is PR, be such that a < d(B) < lτ. We may assume that BV - πk/ where κ(x) is as in (a), and KQ Λ...Λ πk_ι Λ σk Ξ= πg Λ...Λ πk. Let a = d({σk:ke N}).] 8. Suppose AH B. Show that there is a Δ2 sentence φ such that A + φ - B (compare Corollary 6.10 and Theorem 8). 9. Show that there is a Σ^ sentence σ such that Oj < d(σ) < lj and for every Σ^ sentence χ, if σ < χ, then T + χh σ. [Hint: Let σ be such that d(-^σ) is Σj.] 10. Let k c. 15. (a) Verify that in any distributive lattice, for any a, b, the intervals [a n b,a] and [b,a u b] are isomorphic (b) Show that there are degrees a, b, c, d such that a «b, c < d, not c «d, and [a,b] and [c,d] are isomorphic. [Hint: Use Exercises 4 (b) and 11 (a).] 16. (a) Verify that in any distributive lattice, if a < b < c and [a,c] satisfies the reduction principle, so does [b,c]. (b) Show that for each degree a < lγ, there is a b such that a < b < 1 j and [a,b] does not satisfy the reduction principle. (c) The non-r.p. degree a defined in the proof of Lemma 12 is high (cf. Exercise 11 (a)). Show that there is a Σ^ degree which is not r.p. Conclude from Exercise 7 (b) that there are non-Γ^ Σ± degrees such that [Oτ,a] and [Oτ,b] are not isomorphic. [Hint: Use Theorem 14' (a).] 17. (a) Suppose φ and X are as in Lemma 16. Show that if φ < X, then φ «X. (b) Suppose a < b. Show that there are c, d such that a < c < d < b and [c,d] contains no B! degree. 18. (a) Show, by combining the proofs of Theorem 4 and Lemma 15, that there are cupping degrees a0 and a^ which are Σ1 and Γ^ and such that a0 n al = Oτ. Conclude that there are low cupping degrees. (This also follows from Theorem 14' (b) ) (b) Show that there is a high (Γ^) degree a which is not cupping. [Hint: Suppose d(Conτ) < lτ. Let a = d(π) where π is Σ1-conservative over T + -»Conτ and -«π is Γ^-conservative over T + -iConp.] (c) Show that there is a low (Π1) degree a such that for every degree b, if a n b = Oτ/ then a u b is not cupping (compare Exercise 13). [Hint: Let d(π) be as in (b). Define a sentence σ such that d(σ) > Oτ and d(σ) u d(- σ) < d(π); use Theorem 11. Let a = d(- σ).] 19. Show that there are degrees a, b such that a is Σ l7 b is both Σx and n l7 and a u b is not B L. 20. Prove Lemma 15 by letting Θ0 be a Γ^ Rosser sentence for T and θ x := Vu(Prfτ(-θ0,u) -> Ξz Oτ. Show that there is a degree b < a such that [b,a] c
Exercises
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Eτ. [Hint: We may assume that a ^ d(Conτ). Let b = a n d(Conτ) and use Theorem 14' (b). (By Lemma 17, no member of [b,a) is Σ^.] (b) Suppose there is a Σ1 degree which cups to a. Show that there is a b < a such that for every CE [b,a], there is a Σj degree which cups to c. 22. (a) Let E^be the set of degrees obtained from Oτ by taking l.u.b.s, g.l.b.s, and Σ1-extensions. Show that if ae E^ there is a least Σj degree > a. Conclude that there is a Π1 degree not in E^ (This improves Theorem 12.) (b) Let Ej,be the set of degrees obtained from E^ and the Γ^ degrees by taking l.u.b.s and Σ1-extensions. Show that the degree defined in the proof of Theorem 13 is not in F^ Conclude that there is a degree which is not the l.u.b. of a finite set of degrees of the form d(πΛσ). (This improves Theorem 13.) 23. Show that for any a, if there is a member of Gj which cups to a, then there is a ΣI degree which cups to a. (This improves Theorem 15.) 24. (a) Show that not all non-Γ^ ΣI degrees are as stated in Theorem 16. (b) Improve Theorem 16 by showing that for every degree b > 0^, there is a Σ^ degree a such that Oτ < a < b and no Γ^ degree cups to a. [Hint: By Theorem 11, there are sentences π and σ such that Oj < d(π) = d(σ) < b. Let C = T + ->π. By the proof of Lemma 22, with T replaced by C, there are Γ^ formulas ξ(x), η(x) and ΣI sentences χ^ such that (i) - (iv) hold with T replaced by C and C + (ξ(k): ke N} = C + {χk: keN}. Let a = d({ξ(k)vπ:keN}).] 25. Show that in contrast to Lemma 24 we have the following: There is a set G = {d(σ^): kE N} of Σ! degrees, where {σk: ke N} is (primitive) recursive, such that OH > Oj for every finite subset H of G and OG = Oτ. [Hint: Let a be high and such that there is no high Π^ degree < a (cf. Exercise 11 (b)). Let AE a and let σk := -«ConA j ^.] 26. (a) Show that there is a PR formula δ(u) such that if θ is defined as in the proof of Theorem 11, then d(- θ) isn't Πx. (b) Let θ be as in (a). Show that d(->θ) has a p.c. Conclude that there is a non-Γ^ ΣI degree which has a p.c. (c) Let θ be as in (a). Show that there is a Γ^ degree < d(- θ) which does not cap to Of below d(-ιθ) (compare Exercise 10).
Notes for Chapter 7. The lattice Dτ was introduced by Lindstrόm (1979), (1984b); a related lattice V τ (degrees of finite extensions of T) has been defined by Svejdar (1978) (see also Jeroslow (1971a)). (By Theorem 6.11 (a), V τ and Dτ are isomorphic.) Theorem 1 is due to Lindstrom (1979), (1984b) and (for Vτ) to Svejdar (1978). Corollary 1 is,
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modulo Theorem 6.6, a restatement of the equivalence of Exercise 2.22 (i) and (ii). The proof of Theorem 4 was suggested by the proof of a related result in Hajkova II (1971). Theorem 7 is new; the term "reduction principle" is borrowed from descriptive set theory and recursion theory (cf. Soare (1987)). (The only way of showing that intervals are isomorphic known so far is given in Exercise 15 (a) and works in all distributive lattices.) The remaining results of § 1 are due to Lindstrom (1979), (1984b). In connection with the proof of Theorem 4, see Exercise 6. Lemmas 11 and 12 lead to the question if there is a non-Πj r.p. degree; this question is answered in Exercise 7. Theorem 8 (with a slightly different proof; see Exercise 6.12 (a)) is due to Montagna (cf. Lindstrom (1993)). Theorem 9 is due to Lindstrom (1979), (1984b), (1993); (a) and (c) were also proved by Svejdar (1978); for a different proof of Theorem 9 (d), see Exercise 12. Theorem 10 is due to Lindstrom (1979), (1984b); (a) and the first half of (b) were also proved by Svejdar (1978). Theorems 14 and 16 are new, they were announced in Lindstrom (1993), where a weaker form of Theorem 16 is proved; Theorem 16 leads to the question if there is a Σ1 degree a such that no Π^ degree caps to a; this is answered negatively in Exercise 5; in connection with Theorem 16, see also Exercise 24. The remaining results of § 3 are due to Lindstrom (1984b), (1993). The definition of the sentences φn and ψn in the proof of Theorem 14 (a) and the observations concerning these sentences, except (8), were first used by Misercque (1982) in a different context. For improvements of Theorems 12, 13, 15, and 16, see Exercises 22 (a), 22 (b), 23, 24 (b). Theorem 17 leads to the question if no non-Γ^ Σl degree has a p.c.; this question is answered in Exercise 26 (b). For a proof of Exercise 26 (a), see Lindstrom (1993).
8. GENERALIZATIONS
So far our results have been explicitly stated (and proved) only for theories of first order arithmetic. But, as mentioned in the introduction, they hold, after suitable reformulation, in a much more general setting. Needless to say, we are not going to show this in every detail. In fact, we shall skip Chapters 3,5, 7 altogether and concentrate on some of the main results of Chapters 2,4, and 6. These examples should enable the reader to generalize (most of) the results of the preceding chapters. In this chapter he theories S, T, etc. are no longer arithmetical theories, but they are still consistent and primitive recursive and we assume that the languages of these theories are always finite. Lτ is the language of T. T is a pure extension of S if SH T and Lτ = Ls. Lower case Greek letters are now used for formulas of Lτ as well as for formulas of LA. We assume that the reader can extend the definition of t: S < T to the present more general setting. Let t~a(T) = {φ: Th t(φ)}. Then t-1(T)h ψ iff Th t(ψ). Since Ls is finite, t is primitive recursive. The following lemma is immediate. Lemma 1. (a) t: S < T iff SH Γ^T). (b) t: f XT) < T and so t-1(T) < T; in fact, t: t-1(T) ^ T; it follows that t-1(T) is consistent. (c) t-\Ί+ t(φ))Hh Γ'CΓ) + φ.
§1. Incompleteness. Our first result, GodeΓs incompleteness theorem, is a straightforward generalization of Theorem 2.1; 6t(x,y) is a formula defining t as in Fact 2. Theorem 1. Suppose t: Q < T. Let φ be such that (Gt) Qh φ ^ -3y(δt(φ,y) Λ Prτ(y)). Then φ is a true Πj sentence such that TI/ t(φ). Hence if t~\Ί) is Σ1-sound, then also Th - t(φ). By Theorem 1, for each t: Q < T, there is a true Γ^ sentence φt such that Th t(φt). By a similar generalization of Rosser's theorem, we obtain a Γ^ sentence θt such that W t(θt) and TI/ - t(θt). This result can be improved by showing that there is a single Πα sentence ψ such that W t(ψ) and W - t(ψ) for every t: Q < T: Theorem 2. There is a (true) Γ^ sentence ψ, such that Q + ψ i T and Q + - ψ i T.
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Proof, {φ: Q + φ < T} is r.e. (Lemma 6.5) and monoconsistent with Q. Now use Lemma 2.1. Our next result, GόdeΓs second incompleteness theorem, is a generalization of Theorem 2.4 (a). Since each t is primitive recursive, in PA we may use t as a function symbol. Theorem 3. Suppose t: PA < T. (a) Th t(Conτ). (b) If τ(x) is any Σ^ numeration of T, then Th ^Cor^). Proof. We prove (a); the proof of (b) is almost the same. Let φ be as in the proof of Theorem 1. Then PAh ->φ -> PrT(t(φ)). Moreover, -ι
