Godelizing the Yablo sequence more

¨ Godelizing the Yablo sequence1 Cezary Cieslinski Institute of Philosophy University of Warsaw, Poland Rafal Urbaniak Institute of Philosophy, Sociology and Journalism Gdansk University, Poland Centre for Logic and Philosophy of Science Ghent University, Belgium rfl.urbaniak@gmail.com Abstract. We investigate what happens when ‘truth’ is replaced with ‘provability’ in Yablo’s paradox. By diagonalization, appropriate sequences of sentences can be constructed. Such sequences contain no sentence decided by the background consistent and sufficiently strong arithmetical theory. If the provability predicate satisfies the derivability conditions, each such sentence is provably equivalent to the consistency statement and to the G¨del sentence. Thus o each two such sentences are provably equivalent to each other. The same holds for the arithmetization of the existential Yablo paradox. We also look at a formulation which employs Rosser’s provability predicate. 1 1.1 Introduction Replacing truth with provability One way to look at the incompleteness of (many) mathematical theories is to compare the G¨delian sentence(s) with those used in various formulations of the o liar paradox. The liar sentence says of itself that it is not true: (λ) Sentence (λ) is not true. A G¨delian sentence (modulo syntactic encoding) says of itself that it is not o provable: (G) Sentence G is not provable (in the system). Thus, G¨delian incompleteness can be seen as a result of turning a paradox o into a theorem.2 Let’s sketch and compare the relevant arguments: Leach-Krouse cooperated with us in early stages of this work. Other commitments did not allow him to continue. 2 A fairly common phenomenon. G¨del observed the relationship between his theorem and o the liar, and seems to have in fact drawn inspiration from the paradoxes. Other semantic paradoxes have also been used in incompleteness proofs – see e.g. (Cieslinski, 2002), where Grelling’s paradox is employed in the proof of the second incompleteness theorem. In ZFC, the proof that there is no universal set can be seen as the result of formalizing Russell’s paradox, and both Hartog’s theorem, and the result that there is no set of all ordinals can been seen as the result of formalizing Burali-Forti. 1 Graham 1 ¬Tr(λ)  ¬(λ) (excluded middle) Tr(λ)  (λ) (Schema T)   Tr(λ) (content of (λ)) ¬Tr(λ) In the case of the liar, the claim that (λ) is not true turns out to be equivalent to the claim that it is true. Hence the paradox. What happens when we try to run a parallel argument about provability instead of truth, assuming the soundness of the background theory?3 ¬Prov(G) (excluded middle) Prov(G)  (soundness) ??     (G) (content of (G))  ¬Prov(G) While the right-hand case goes through, the left-hand case does not get off the ground, because we did not assume that all truths are provable (and thus cannot infer ¬φ from ¬Prov(φ)). Thus, instead of obtaining a paradox, we just conclude that (G) is not provable in the theory. The situation is similar with Prov(¬G): ¬Prov(¬G) (excluded middle) Prov(¬G)  (soundness) ??    ¬G (content of G)  ¬¬Prov(G) (classical logic)  Prov(G) (soundness)  G While from the unprovability of ¬G we cannot infer its falsehood, the assumption that it is provable leads to contradiction. Thus, neither G nor ¬G are provable in the system, but no paradox ensues. A question arises as to the extent to which this phenomenon generalizes: are 3 Of course, the soundness assumption is not needed for the original G¨del’s proof to go o through, but let’s not worry about this. 2 there any other paradoxes involving truth which yield theorems when truth is replaced with provability? We’ll take a look at a paradox related to truth which is often claimed to be importantly different from the liar: Yablo’s paradox. 1.2 Yablo’s paradox: a formulation The paradox has been formulated by Yablo (1993). Consider an infinite sequence of sentences s0 , s1 , s2 , . . . such that: s0 = ‘∀x (P1 (x) → ¬Tr(x))’, s1 = ‘∀x (P2 (x) → ¬Tr(x))’, s2 = ‘∀x (P3 (x) → ¬Tr(x))’, . . . Tr is the truth predicate, and the extension of every Pn , (for n = 1, 2, 3, . . .) is {sn , sn+1 , sn+2 , . . .}. Thus, every si says that all sj s with j > i are not true. Now suppose s0 is true. Then for any k > 0, sk is not true (s1 is among such formulas, and thus it is not true). Also, sk is not true for any k > 1. But this is exactly what s1 says – hence s1 is true after all. Contradiction. Suppose then that s0 is false. This means that there is a k > 0 such that sk is true. But we can repeat the reasoning, this time with respect to sk and reach a contradiction again. No matter whether we assume s0 to be true or false, we reach a contradiction. Hence the paradox.4 2 Yablo sequence with provability, a sketchy argument Let’s work with Peano Arithmetic (PA) and see what happens when Tr is replaced with Prov in a Yablean sequence. First, we will handwavily describe the kind of reasoning involved, without paying attention to details and without making sure the reasoning can be represented within the system. Suppose we extend the standard language of arithmetic with a new function symbol ‘f ’. We intend the function it expresses to assign to each natural number the g.n. (G¨del number) of the n-th sentence in the Yablean sequence (with o provability instead of truth): (1) f (n) = ∀x > n ¬Prov(f (x)) ¯ Now suppose PA proves the formula with g.n. f (n): (2) PA ∀x > n ¬Prov(f (x)) ¯ 4 Let us note that in contrast with the usual liar-type paradoxes based on an explicit use of direct or indirect self-reference, Yablo paradox seems to involve no self-referential loops. The question whether the paradox is really self-reference free has been much debated in recent literature. See e.g. (Beall, 2001), (Leitgeb, 2002), (Priest, 1997) and (Urbaniak, 2009). 3 It follows that PA proves a weakening of (2): (3) PA ∀x > n + 1 ¬Prov(f (x)) and it proves a particular instance of (2): (4) PA ¬Prov(f (n + 1)) But (3) simply is the formula with g.n. f (n + 1). Also, for any theorem of PA, PA proves also that it proves it (see fact 5 on page 5). So, from (3), we obtain: PA Prov(f (n + 1)) This, however, together with (4) shows that PA is inconsistent. So, if PA is consistent, (2) is false and PA doesn’t prove the formula with g.n. f (n) after all. Let’s see what would happen if PA proved its negation. Suppose (5) This yields (6) PA ∃x > n Prov(f (x)) ¯ PA ¬∀x > n ¬Prov(f (x)) ¯ Then, (since PA is ω-consistent) there should be an m which witnesses this claim: (7) PA Prov(f (m)) ¯ But then, PA also proves the formula itself (see e.g. fact 14 on page 7): (8) PA ∀x > m ¬Prov(f (x)) ¯ Now, we can apply to (8) the same reasoning we applied to (2) to reach a contradiction. So, both (2) and (5) are false and PA doesn’t decide the formula with g.n. f (n). This argument as it stands is not satisfactory. Ideally, we would like to avoid reference to a new function symbol and formulate the reasoning in the original language of arithmetic. We would also like to be clear on what assumptions are being made and on how the steps involved are justified. To this goal, in section 3 we will first survey a few facts needed for our proof (a reader familiar with the topic can safely skip this section and refer to it later if need be). Then, in section 4, we will give the full proof. 3 (Slightly) technical preliminaries We assume the reader is familiar with the general framework of G¨delian proofs, o but for the sake of accessibility, we list the main results which will be referred to later on. We do it without proofs, which are rather well-known and should not lie in the focus of our paper. For details we refer the reader to (H´jek and Pudl´k, a a 1998) or (Smith, 2007). 4 Definition 1. A theory T is p.r. axiomatized iff all the following conditions hold: (i) the numerical properties of being the g.n. (G¨del numbers) of a T-well o formed formula and a T-sentence are p.r. (primitive recursive), (ii) the numerical property of being the g.n. of an axiom is p.r., (iii) the numerical property of being the g.n. of a correct proof is p.r.. An arithmetical theory is nice iff it is consistent, p.r. axiomatized, and extends Robinson arithmetic Q. From now on, we will often fix our attention on PA (supposing it is nice), but things we will say are taken to apply to all nice theories, unless we say otherwise. Fact 2. Any nice theory T correctly decides ∆0 -sentences.5 The language of PA can express the property of being a proof in PA. In particular, there is a Σ1 formula Prf(x, y) which holds of numbers n and m just in case n is a g.n. (G¨del -number) of a sequence of formulas which is an PAo proof of the formula whose g.n. (G¨del number) is m. Moreover, Prf captures o (binumerates) PA-provability in PA: Fact 3. For any m and n: (1) If Prf(m, n), then PA ¬Prf(m, n), then PA ¬Prf(m, n). ¯ ¯ Further on, within the language of PA we can define: Prov(x) =def ∃y Prf(y, x) Thus, Prov expresses the property of PA-provability. Clearly, Prov(x) is also Σ1 . (Henceforth, we will be using Prov sometimes within the language of PA and sometimes outside of it; sometimes we will also relativize the provability predicate to a given theory. But we trust the context will make it clear what is meant.) Observe the following: Fact 4. Prov expresses provability: m is the g.n. of a PA-theorem just in case Prov(m) is true. Yet, the predicate does not capture provability in PA: it is not the case that for any m, if ¬Prov(m) then PA ¬Prov(m). Prov, standardly constructed, has certain properties which will come in handy later on: Fact 5 (Derivability condition 1). (D1) If PA φ, then PA Prov( φ ). Prf(m, n), (2) If ¯ ¯ Fact 6 (Derivability conditions 2 and 3). For any nice arithmetical theory T extending IΣ1 , there are predicates Prf and Prov such that Prf represents the relation of being a proof in T, Prov(x) is defined by ‘∃y Prf(y, x)’, and Prov satisfies (provably in T) the following derivability conditions: (D2) (D3) 5 For T Prov( φ → ψ ) → (Prov( φ ) → Prov( ψ )) T Prov( φ ) → Prov( Prov( φ ) ) a proof, see e.g. section 9.7 of (Smith, 2007). 5 Observe that fact 6 states only the existence of formulas Prf and Prov with the indicated properties. It does not state that T will prove (D2) and (D3) independently of the choice of the formula Prf representing the proof relation. Indeed, such a claim would be false. There are known examples of formulas representing the proof relation which yield provability predicates which do not satisfy (D2) and (D3) (Rosser’s provability predicate belongs to this category). However, both (D2) and (D3) hold under the standard construction of the provability predicate. We use Feferman’s dot notation. Within a quotation term, z is a term which ˙ denotes the numeral for z and depends on the particular choice of z. Also, for any φ, we abbreviate Prov(sub( φ(x) , z)) as Prov(φ(z)) or Prov(φ(z)) if we ˙ ˙ want to emphasize that z is kept free. In fact (a result due to Feferman), for the standard provability predicate, derivability conditions can be slightly generalized: Fact 7. Suppose T is nice and extends IΣ1 . If T φ(x), then T Prov( φ(x) ). ˙ Also, (D2) extends to formulas with free variables, and the third condition generalizes to T Prov( φ(x) ) → Prov( Prov( φ(x) ) ). ˙ ˙ In cases where fact 7 applies, it validates the move (within a theory) from φ → ψ to Prov( φ ) → Prov( ψ ) and to Prov( φ ) → Prov( Prov( ψ ) ) (of course, some attention has to be paid to free variables). We call these moves K1 and K2 respectively, and employ them in our proof of theorem 22. We will also need a fact which tells us, more or less, that when looking at the consistency statement, it doesn’t really make a difference which contradiction we pick. Fact 8. If T is nice and extends IΣ1 , then for any φ, T ¬Con(T).6 One more fact about Con will be useful. Fact 9. If T is nice and derivability conditions hold, then: • T • T Con ≡ G, where G is the G¨del sentence. o ¬Prov( φ ) → Con for any φ. [Prov(φ)∧Prov(¬φ)] ≡ Definition 10. An arithmetical theory T is ω-consistent iff for no open formula φ(x), T φ(m) for each standard numeral m, and yet T ¯ ¬∀x φ(x) (alternatively: iff there is no φ(x) such that T ∃x φ(x) and yet for each m, T ¬φ(m)). ¯ Definition 11. An arithmetical theory T is 1-consistent just in case for no ∆0 formula φ(x), T ∃x φ(x) and yet for each m, T ¬φ(m). ¯ The following is a straightforward observation based on definitions 10 and 11: 6 Con is defined by ¬Prov( 0 = 0 ). 6 Fact 12. ω-consistency is properly stronger than 1-consistency. Definition 13. An arithmetical theory T is Σ1 -sound just in case for any Σ1 sentence, if T φ, then φ is true in the standard model of arithmetic. Fact 14. If PA is ω-consistent, then if PA Prov( φ ), then PA φ.7 Fact 15. If an arithmetical theory T is nice, then it is 1-consistent iff is Σ1 sound. Fact 16. If φ(x) is a Σ1 -formula and T is a 1-consistent nice theory, then if T ∃x φ(x), then for some m, T φ(m). (The claim generalizes to formulas ¯ with more free variables.) Fact 16 helps us to strengthen fact 14 to the following: Fact 17. If a nice T is 1-consistent, then if T Prov( φ ), then T φ. One last thing we need will be a generalized version of the diagonal lemma: Fact 18 (Diagonal lemma). If T is nice, then for any formula φ(x, y) there is a formula ψ(x) such that T [ψ(x) ≡ φ(x, ψ(x) )]. Now we are ready to cope with the arithmetized proof. 4 Yablo’s sequence with provability, arithmetized Define φ as follows: (9) φ(x, y) =def ∀z [z > x → ¬Prov(sub(y, z))] ˙ That is, φ(x, y) says that for any number z greater than x, the formula whose g.n. is y is not provable of z. Take a nice theory T and use fact 18 and diagonalize on φ. This means there is a formula Y (x) (the provability Yablo formula) such that:8 (10) T Y (x) ≡ ∀z [z > x → ¬Prov(sub( Y (x) , z))] ˙ Theorem 19. If an arithmetical theory T is nice, then for any n, T Y (¯ ). n Moreover, if T is also 1-consistent, then for any n, T ¬Y (¯ ) (where Y (¯ ) is n n constructed following (9) and (10) using T’s provability predicate). Proof. For the first part of the theorem, suppose T show this leads to contradiction. By (10):9 (11) 7 This Y (¯ ) for some n. We’ll n T ∀z [z > n → ¬Prov(Y (z))] ¯ we have to distinguish from proving instances of reflection schema, Prov( φ ) → φ. No nice theory can prove all such instances. 8 This method of constructing a Yablo sequence (but for formulas containing the truth predicate) has been employed by Priest (1997). 9 From now on, instead of ‘Prov(sub( Y (x) , z))’ we will simply write ‘Prov(Y (z))’. ˙ 7 (11) entails two things: (12) (13) T T ¬Prov(Y (n + 1)) ∀z [z > n + 1 → ¬Prov(Y (z))] But (13) with (10) entail: (14) T Y (n + 1) (14) by the first derivability condition (D1) entails: (15) T Prov( Y (n + 1) ) This, given (12) implies that T is inconsistent. So if T is consistent, T Y (¯ ), n for any n. This ends the first half of the proof. For the second half, suppose T is 1-consistent and T ¬Y (¯ ). We employ n (10) and start as follows: T (16) ¬∀z [z > n → ¬Prov(Y (z))] ¯ T ∃z [z > n ∧ Prov(Y (z))] ¯ Since Prov is Σ1 , and > is ∆0 , the formula ∃z [z > n ∧Prov(Y (z))] is Σ1 . Thus, ¯ fact 16 applies and there is a number k such that: T (17) ¯ ¯ ¯ k > n ∧ Prov(Y (k)) ¯ T Prov(Y (k)) From this, 1-consistency and fact 17 allow us to infer: (18) T ¯ Y (k) to which we can apply the reasoning used in the first half of the proof, obtaining a contradiction. Thus T ¬Y (n). (Observe (D2) and (D3) are not used, so the result holds even for some theories without induction and is not sensitive to the construction of the proof predicate.) So, when we arithmetize Yablo’s paradox using provability instead of truth we get another incompleteness proof, which prima facie doesn’t use a formula which “says” of itself that it is not provable. (Although, if derivability conditions (D1-3) hold, they are all provably equivalent to such a sentence). 5 Equivalence of Yablo sentences First, keep Y relativized to T-provability, defined as in (10),10 and consider the following claim: 10 To remind the reader about this, we will sometimes use ‘T ’ in the subscript, but only in contexts where ambiguity may arise. 8 Fact 20. If T is nice, then for any m, n, if m > n, then T + YT (¯ ) n YT (m). ¯ Given that in the original paradox, prima facie no sentence in the sequence seems to entail a sentence earlier on the list, an obvious question comes to mind: If T is 1-consistent and nice, is it the case that for any m, n, if m > n, then T + YT (m) YT (¯ )? ¯ n This would mean T, if 1-consistent and nice, has an infinite sequence of extensions: T + YT (0), T + YT (1), T + YT (2), each weaker than the previous ones. As it turns out, the answer to the question is negative. Theorem 21. If T is nice and (D1) and (D2) hold, then T Proof. w.t.s.: Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 Q.E.D. ∀x [Y (x) → Con(T)]. T Y (x) → Con(T) (for arbitrary x) (by classical logic) T ⊥ → Y (x + 1) (from Step 1 by (D1)) T Prov( ⊥ → Y (x + 1) ) (directly from (D2)) T Prov( ⊥ → Y (x + 1) ) → [Prov( ⊥ ) → Prov( Y (x + 1) )] (from Steps 2, 3, by modus ponens) T Prov( ⊥ ) → Prov( Y (x + 1) ) (from Step 4 by contraposition) T ¬Prov( Y (x + 1) ) → ¬Prov( ⊥ ) (from Step 5 by definition of Con(T)) T ¬Prov( Y (x + 1) ) → Con(T) (by (10)) T Y (x) → ¬Prov( Y (x + 1) ) (from Steps 6 and 7) T Y (x) → Con(T) Theorem 22. For any n, if T is nice and predicate Prov used in the characterization of the Yablo sequence satisfies the derivability conditions (D1-3), then T ∀x [Con(T) → Y (x)]. Proof. We give a short sketch which uses G¨del’s second theorem, and a longer o proof which doesn’t. For the sketch, working in T, assume Con(T). For an indirect proof, assume also ¬Y (x). This yields ∃u > x Prov(Y (u)). Let’s fix such an u. Therefore by (D2), Prov( ∀y > u ¬Prov(Y (y)) ), so (D2 again) the theory proves that it doesn’t prove something. By fact 9, Prov( Con(T) ). In effect, by G¨del’s second incompleteness theorem formalized in T, ¬Con(T) (the o last move requires all three derivability conditions). 9 Before we give the longer version, the reader should check fact 7 and our comments about abbreviations and inferential moves which directly precede and follow our statement of this fact. Also, recall that K1 depends on the first and second derivability conditions while K2 depends on all three of them. w.t.s.: Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 Step 9 Step 10 Step 11 Step 12 Q.E.D. T Con(T) → Y (x) (for arbitrary x) (by (10)) T Y (x) → ∀z [z > x → ¬Prov(Y (z))] (from Step 1 by K2) T Prov( Y (x) ) → Prov( ∀z [z > x → ¬Prov(Y (z))] ) ˙ ˙ (Arithmetic) T ∀z [z > x → ¬Prov(Y (z))] → ∀z [z > x + 1 → ¬Prov(Y (z))] (from Step 3 by (10)) T ∀z [z > x → ¬Prov(Y (z))] → Y (x + 1) (from Step 4 by K2) T Prov( ∀z [z > x → ¬Prov(Y (z))] ) → Prov( Prov(Y (x + 1)) ) ˙ ˙ (Arithmetic) T ∀z [z > x → ¬Prov(Y (z))] → ¬Prov(Y (x + 1)) ˙ (from Step 6 by K1) T Prov( ∀z [z > x → ¬Prov(Y (z))] ) → Prov( ¬Prov(Y (x + 1)) ) ˙ ˙ (Logic, Steps 5 and 7) T Prov( ∀z [z > x → ¬Prov(Y (z))] ) → Prov( Prov(Y (x + 1)) )∧ ˙ ˙ Prov( ¬Prov(Y (x + 1)) ) ˙ (from Step 8 by fact 8) T Prov( ∀z [z > x → ¬Prov(Y (z))] ) → ¬Con(T) ˙ (Logic, Steps 2 and 9) T Con(T) → ¬Prov( Y (x) ) ˙ (Logic, Step 10) T Con(T) → (∀z)¬Prov( Y (z) ) (Logic, Steps 11 and (10)) T Con(T) → Y (x) From theorems 21 and 22 the following corollaries follow: Corollary 23. If T is nice and the derivability conditions are satisfied, then for any m and n, T ∀x, y [Y (x) ≡ Y (y)]. Corollary 24. The answer to our question is pretty much negative. As long as T is nice and the derivability conditions are satisfied, T + YT (m) YT (¯ ), and ¯ n it doesn’t matter whether m > n. Also, recall that according to fact 9, the G¨del sentence is provably equivalent o to the consistency statement. Thus: Corollary 25. If T is nice and the derivability conditions hold, T for any n. 10 G ≡ Y (¯ ) n 6 Existential Yablo paradox It seems we can run a similar paradox, but with existential quantification (Sorensen, 1998). Consider the infinite sequence: s0 = ‘∃x (P1 (x) ∧ ¬Tr(x))’, s1 = ‘∃x (P2 (x) ∧ ¬Tr(x))’, s2 = ‘∃x (P3 (x) ∧ ¬Tr(x))’, . . . (where the reference of Pi s is as in the original formulation). Suppose s0 is true. Then, a sentence above it in the list, say sk , has to be false. But then, it is not the case that ∃x > k ¬Tr(sx ). That is, ∀x > kTr(sx ). In particular, Tr(sk+1 ) which says that there is a false sentence above sk+1 . But it’s also above sk , and all sentences above sk are to be true. Contradiction. So s0 is not true. Suppose it is false. Then, we just run for s0 the second half of the above reasoning. What happens when we arithmetize this paradox replacing Tr with Prov? Say we introduce (19) ψ(x, y) =def ∃z > x ¬Prov(sub(y, z)) ˙ Once we diagonalize on ψ we know that there is a formula E(x) such that: (20) T E(x) ≡ ∃z > x ¬Prov(sub( E(x) , z)) ˙ ¬E(¯ ), for any standard n Theorem 26. If T is nice and 1-consistent, then T n. Proof. Suppose T (21) ¬E(¯ ). Then:11 n T T ∀z > n Prov(E(z)) ¯ Prov( E(n + 1) ) By fact 17 and 1-consistency of T: T This means: T But (21) also entails: T ∀z > n + 1 Prov(E(z)) which shows T to be contradictory. 11 ‘Prov(E(z))’ E(n + 1) ∃z > n + 1 ¬Prov(E(z)) abbreviates ‘Prov(sub( E(x) , z)’. ˙ 11 Further, say we assume for reductio that T (22) T E(¯ ). We obtain: n ∃z > n ¬Prov( E(z) ) ¯ This doesn’t allow us to infer that there is a standard witness for which this claim is provable. We can try to continue reasoning within T with an arbitrary constant (just making sure we don’t apply any principles meant to apply only to standard numerals): a > n ∧ ¬Prov( E(a) ) ¯ But here the argument is blocked. Contrary to what happened with the original proof, (22) does not entail the content of E(a). Yet, there is another, rather simple argument and the claim holds. Theorem 27. If T is nice, T E(¯ ), for any n. n Proof. If T E(¯ ), it proves that it doesn’t prove something, and thus proves n its own consistency (by fact 9). But if it is nice, it doesn’t do the latter. Now we can ask whether all E-sentences are provably equivalent the way all Y -sentences are. The answer is positive: Theorem 28. If T is nice and the derivability conditions are satisfied, all E-sentences are provably equivalent in T. That is, for any n and m: T E(¯ ) ≡ E(m). n ¯ Proof. The proof is by provable equivalence of each such sentence with Con(T). From left to right, E(¯ ) → Con(T ), this is clear because E(¯ ) states the unn n provability of a sentence, and this implies the consistency statement (fact 9). From right to left, work within T. Assume Con(T) and ¬E(¯ ). This gives n ∀z > n Prov( E(z) ). Thus Prov( E(n + 1) ). So we get Prov( Con(T) ), because we already have the implication E(k) → Con(T) within the scope of Prov for any k. But G¨del’s second theorem (formalized) gives Con(T) → o ¬Prov( Con(T) ). So ¬Prov( Con(T ) ), which gives a contradiction and ends the proof. The way the proof proceeds makes the following clear: Corollary 29. All E-sentences are provably equivalent to the consistency statement and to the G¨del sentence, if the background theory is nice and the derivo ability conditions are satisfied. 12 7 Yablo formulas with Rosser’s provability predicate A case deserving a special attention is that of Rosser’s provability predicate. We introduce two definitions:12 (23) (24) PrfR (x, y) =def Prf(x, y) ∧ ∀z < x¬Prf(z, ¬y), ProvR (y) =def ∃xPrfR (x, y). Then it is possible to construct a Yablo formula Y R (x) satisfying: (25) T Y R (x) ≡ ∀z > x¬ProvR (Y R (z))]. Since ProvR (y) satisfies (D1), theorem 19 applies: our theory T doesn’t decide Y R (n) for any natural number n. However, ProvR (y) doesn’t satisfy (D2) and (D3)13 and so corollaries 23 and 24 do not apply. The issue whether all Y R (n)-s are provably equivalent remains undecided. Is it possible to construct a model M for PAwith M |= Y R (n + 1) and M Y R (n)? We don’t know the answer, but the following is worth observing: Fact 30. Any model M of PA with M |= Y R (x + 1) and M have to satisfy ¬Con(P A). Argument. Fix x ∈ M and suppose: M |= Y R (x + 1) M |= ¬Y R (x) These (respectively) yield (in the model): (26) (27) Together, (26) and (27) entail: ProvR (Y R (x + 1)) By definition of ProvR , it follows: Prov(Y R (x + 1)) Now, work inside the scope of the provability predicate. Y R (x + 1) by definition is equivalent to: (28) 12 Of 13 This Y R (x) would ∀z > x + 1 ¬ProvR (Y R (z)) ∃z > x ProvR (Y R (z)) ∀z > x + 1 ¬ProvR (Y R (z)) course, ‘¬y’ stands for the g.n. of the negation of the formula whose g.n. is y. is the case because (D1)-(D3) are enough for the proof of the second incompleteness theorem. Otherwise we would have: T ConR (T), with ConR (T) defined by ¬ProvR (0 = 0). However, as it is well-known, PA ConR (PA). 13 (28) entails ∀z > x + 2 ¬ProvR (Y R (z)) and by the same token Y R (x+2). Thus, jumping out of the scope of the provability predicate: (29) Prov(Y R (x + 2)). On the other hand, we already have (26), which entails: ¬ProvR (Y R (x + 2)) By the construction of Rosser predicate, this means: (30) ∀x [¬Prf(x, Y R (x + 2) ) ∨ ∃z < x Prf(z, ¬Y R (x + 2) )] If the second disjunct is not satisfied, (30) holds because ∀x ¬Prf(x, Y R (x + 2) ). But then ¬Prov(Y R (x + 2)), which contradicts (29). So, the second disjunct has to be satisfied. If the second disjunct is satisfied, then Prov( ¬Y R (x + 2) ) and therefore, given (29), ¬Con(PA), which ends the proof. 8 Summary Replacing truth with provability in Yablo’s paradox yields an infinite sequence of undecided formulas, as long as the background arithmetical theory satisfies standard requirements for incompleteness. All such formulas are provably equivalent to the consistency claim and to the G¨del sentence. These properties o are preserved when the original paradox is replaced by the existential Yablo paradox. When Rosser’s provability predicate is used in the construction, independence claims hold but it remains undecided whether in such a case all sentences in a Yablean sequence (i.e. all sentences obtained from ‘Y R (x)’ by substituting numerals for ‘x’) are provably equivalent in PA. We conjecture that such sentences are not provably equivalent, even though they are all true in the standard model of arithmetic. References Beall, J. C. (2001). Is Yablo’s paradox non-circular? Analysis, 61:176–187. Cieslinski, C. (2002). Heterologicality and incompleteness. Mathematical Logic Quarterly, 48:105–110. H´jek, P. and Pudl´k, P. (1998). Metamathematics of First-Order Arithmetic. a a Springer-Verlag. Leitgeb, H. (2002). What is a self-referential sentence? critical remarks on the alleged (non-)circularity of Yablo’s paradox. Logique & Analyse, 177-178:3– 14. 14 Priest, G. (1997). Yablo’s paradox. Analysis, 57:236–242. Smith, P. (2007). An Introduction to G¨del’s Theorem’s. 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