Interpreting the compositional truth predicate in models of arithmetic∗ Cezary Cieśliński September 14, 2019 Abstract We present a construction of a truth class (an interpretation of a compositional truth predicate) in an arbitrary countable recursively saturated model of first-order arithmetic. The construction is fully classical in that it employs nothing more than the classical techniques of formal proof theory. 1 Introduction The goal of this paper is to sketch a fully classical construction of truth classes in models of first-order arithmetic. The initial introductory remarks describe the background and the motivation for this endeavor. It is well-known that every non-standard model of arithmetic will contain nonstandard arithmetical formulas. In other words, for an arbitrary nonstandard model M there will be an element s of M such that M |= 's is an arithmetical formula' (all such objects will be called here 'formulas in the sense of the model' or 'M -formulas') even though in the real world s is not a formula at all.1 In this situation it is natural to ask whether semantics for formulas in the sense of the model can be developed. First attempts in this direction were made by ∗An extended abstract (Cieśliński, 2019) summarizes the basic ideas presented in this paper. Here we provide both the technical material (proofs) absent in the abstract and a more comprehensive discussion of the motivations and the background of our approach to the semantics of truth theories. 1Thus, for example, M will think that expressions of the form '0 = 0 ∧ . . . 0 = 0' are arithmetical formulas no matter how many times the conjunct '0 = 0' is repeated. In effect, given an arbitrary non-standard element a of M , M will think that the expression '0 = 0 ∧ . . . 0 = 0' with the conjunct repeated a times is an arithmetical formula. However, since a is non-standard, the expression in question is not a formula in the real world. 1 Robinson (1963) and Krajewski (1976), with the notion of a satisfaction class playing the key role. Namely, a satisfaction class inM is characterized as an arbitrary subset of the model which can be treated as a reasonable interpretation of the satisfaction predicate: roughly, it is a set of pairs (φ, v) which satisfies the usual Tarski-style compositional clauses for satisfaction (the assumption here is that φ is an M -formula and v is a variable assignment for φ.) One of the most remarkable results in the theory of satisfaction classes is that a (non-inductive) satisfaction class can be constructed in an arbitrary countable recursively saturated model of arithmetic.2 Since every model of arithmetic has an elementarily equivalent recursively saturated model, it immediately follows that the compositional axioms of satisfaction are proof-theoretically conservative over first-order arithmetic.3 Later it transpired that the conservativity of compositional satisfaction (or truth) is an interesting property not just to the mathematical logicians but also to philosophers. In particular, in recent philosophical debates on the so-called 'deflationism about truth' conservativity has been explicitly postulated as a desirable trait of axiomatic truth theories.4 Independently of the outcome of these philosophical discussions, the upshot is that proofs of the conservativity results became important and interesting for a quite wide and diverse community of researchers investigating the properties of semantic notions. However, the original proof of the theorem uses the (so-called) 'technique of approximations' which many readers found exotic. From the author's experience, the machinery of 'approximations', developed by KKL in their paper, remains one of the main stumbling blocks in the wider dissemination of this important result. Accordingly, the question has been asked whether the result can be proved by purely classical methods. One successful attempt in this direction has been recently made by Enayat and Visser (2015). In their paper, they showed how to construct a satisfaction class using classical techniques of formal semantics, that is, compactness and the union of elementary chain theorem.5 In the present paper I propose to prove the theorem by the classical techniques of formal proof theory, namely, by cut elimination. Coupled with Enayat 2This theorem is due to Kotlarski, Krajewski and Lachlan (KKL in short), see (Kotlarski et al., 1981). For an overview, see also (Kotlarski, 1991) and (Kotlarski and Ratajczyk, 1990). 3For the statement and the proof of this important fact about the recursively saturated models, see (Kaye, 1991), p. 148-9. 4See (Horsten, 1995), where the conservativity postulate was introduced for the first time in the context of a philosophical discussion of deflationary views. For further philosophical developments in this direction, see, for example, (Ketland, 1999), (Tennant, 2002) and (Cieśliński, 2015). For a synthetic presentation of the conservativeness debate, see Chapters 9 and 11 of (Cieśliński, 2017). 5In their proof it is assumed that arithmetic is formulated in the relational language. See (Cieśliński, 2017) for extending the result to the language with function symbols. 2 and Visser construction, this makes the fascinating field of satisfaction classes accessible to the students and the logicians, whose primary interest is either model theory or proof theory. 2 Preliminaries 2.1 Basic notions The original construction of KKL takes as a starting point first-order arithmetic as formulated in purely relational language (all the function symbols are replaced by predicates). In contrast, here the language of first-order arithmetic (denoted as LAr) will be assumed to contain function symbols, namely '+', '×', '0' and 'S' for addition, multiplication, zero and the successor operation. The expressions V ar, Tm, Tmc, FmLAr and SentLAr will be used here in a double role. Firstly, they will be treated as referring (respectively) to the sets of variables, terms, constant terms, formulas and sentences of LAr. In addition, they will also be used as shorthands for arithmetical predicates representing the relevant sets in Robinson's arithmetic.6 Given a model M , we write SentLAr (M) for the set of all objects a such that a ∈M and M |= SentLAr (a). The perspective adopted in this paper is that of truth, not satisfaction. Accordingly, we will consider the language LT obtained from LAr by adding the unary truth predicate 'T (x)' (instead of a binary satisfaction predicate). SentLT is the set of sentences of LT . We introduce now the basic theory of truth, denoted as CT−. Definition 1 Let Ax be a set of axioms of an arithmetical theory Th. Then CT−(Ax) is defined as the theory in the language LT axiomatized by Ax together with the following truth axioms: • ∀s, t ∈ Tmc ( T (s = t) ≡ val(s) = val(t) ) • ∀φ ( SentLAr (φ)→ (T¬φ ≡ ¬Tφ) ) • ∀φ∀ψ ( SentLAr (φ ∨ ψ)→ (T (φ ∨ ψ) ≡ (Tφ ∨ Tψ)) ) • ∀v∀φ(x) ( SentLAr (∃vφ(v))→ (T (∃vφ(v)) ≡ ∃xT (φ(ẋ))) ) The acronym 'CT ' stands for 'compositional truth'. The natural truth axioms listed above follow the familiar pattern of Tarski's inductive truth definition. A theory Th axiomatized by Ax is called the base theory of CT−(Ax). It is usually assumed that Th will play the role of a theory of syntax, so it should be strong enough to formalize syntactic operations. The superscript in 6Thus, for example, 'M |= a ∈ V ar' means that the object a satisfies in M an arithmetical formula 'x ∈ V ar' representing the set of arithmetical variables. 3 CT− indicates that if Th is schematically axiomatized, we are not allowed to substitute formulas of LT in the schemas. Thus, for example, if Th is Peano arithmetic axiomatized by means of the schema of induction, then in CT−(Ax) there will be no induction available for formulas containing the truth predicate. With an axiomatization of Th being fixed, we will write CT−(Th) instead of CT−(Ax).7 A 'truth class' in a model M of Th is a subset T of M such that (M,T ) |= CT−(Th). Let us emphasize that the quantifier axiom of CT− employs numerals. A numeral is an arithmetical constant term of the form 'S . . . S(0)'; in other words, numerals are expressions obtained by preceding the symbol '0' with arbitrarily many successor symbols. Accordingly, the intended meaning of the quantifier axiom is that '∃vφ(v)' is true iff the result of substituting some numeral for v in φ(v) is true.8 One of the key results in the area of axiomatic truth theories is the conservativity theorem, stating that the truth axioms of CT− are conservative over mathematical theories sufficient for reconstructing basic theory of syntax. In particular, adding the compositional truth axioms to an arithmetical theory Th containing IΣ1 (the last one being typically axiomatized by the usual axioms of Robinson's arithmetic together with the induction schema restricted to Σ1 arithmetical formulas) produces a conservative extension of Th. Conservativity is a direct corollary of the KKL theorem, which can be formulated as an expandability result concerning countable recursively saturated models of theories containing IΣ1. Two definitions below introduce the notion of a recursive type and the concept of a recursively saturated model. Definition 2 Let Z be a set of formulas with one free variable x and with parameters a1...an from a model M . We say that: (a) Z is realized in M iff there is an s ∈ M such that every formula in Z is satisfied in M under a valuation assigning s to x. (b) Z is a type of M iff every finite subset of Z is realized in M . 7The choice of axiomatization is not always innocent. Thus, for example, let Ax be a set of axioms of PA. Is CT−(Ax)+ 'All elements of Ax are true' a conservative extension of PA? It transpires that the answer to this question depends on the choice of our axiomatization of Peano arithmetic. 8An alternative axiomatization would employ constant terms instead of numerals. In this version, the quantifier axiom would read: ∀v∀φ(x) ( SentLAr (∃vφ(v)) → (T (∃vφ(v)) ≡ ∃t ∈ TmcT (φ(t))) ) . Both versions turn out to be equivalent if induction for formulas of LT is added. This is not so without extended induction, hence in a non-inductive setting one should be careful about choosing the shape of the quantifier axiom. 4 (c) Z is a recursive type of M iff apart from being a type of M , Z is also recursive. Definition 3 M is recursively saturated iff every recursive type of M is realized in M . The KKL theorem can now be formulated as follows. Theorem 4 Let Th be an arithmetical theory containing IΣ1. For every M |= Th, if M is countable and recursively saturated, then there is a set T ⊂M such that (M,T ) |= CT−(Th). It immediately follows that CT−(Th) is syntactically conservative over Th. 2.2 Truth and satisfaction In our presentation we take the notion of truth, and not of satisfaction, as basic.9 In this context, let us emphasize that in a non-inductive setting the choice of the basic semantic notion is not entirely innocent. In general, proofs of results about non-inductive satisfaction classes do not automatically deliver corresponding results about truth classes, nor the other way round. In order to appreciate the differences between truth and satisfaction, let us introduce properly the basic theory of satisfaction, denoted as CS−. This time we extend the arithmetical language with a new binary predicate S(x, y) (the satisfaction predicate). In what follows 'v ∈ Asn(x)' is an arithmetical formula which reads 'v is an assignment for a formula x' (roughly, 'v ∈ Asn(x)' states that v is a finite function which assigns numbers to variables which are free in x). The expression 'x = val(t, v)' reads 'x is a value of the term t under the assignment v. Definition 5 Let Ax be a set of axioms of an arithmetical theory Th. Then CS−(Ax) is defined as the theory in the language LS axiomatized by Ax together with the following satisfaction axioms: • ∀s, t ∈ Tm ∀v ∈ Asn(pt = sq) ( S(pt = sq, v) ≡ val(s, v) = val(t, v) ) • ∀φ ∈ Fm ∀v ∈ Asn(φ) ( S(¬φ, v) ≡ ¬S(φ, v) ) • ∀φ,ψ ∈ Fm ∀v ∈ Asn(pφ ∨ ψq) ( S(φ ∨ ψ, v) ≡ (S(φ, v) ∨ S(ψ, v))) ) • ∀a ∈ V ar ∀φ(x) ∈ Fm ∀v ∈ Asn(p∃aφ(a)q) ( S(p∃aφ(a)q, v) ≡ ∃x S(φ, v[x/a])) ) 9In most discussions of the KKL theorem and related results, satisfaction is employed as the basic semantic concept; see, for example, (Krajewski, 1976), (Kotlarski et al., 1981), (Kaye, 1991) and (Enayat and Visser, 2015). The exceptions are (Engström, 2002) and (Leigh, 2015), where the discussion is carried out in terms of truth, not satisfaction. 5 As in the case of CT−, the axioms of CS− contain only arithmetical substitutions of the axiom schemata of Ax (i.e., no such substitution by a formula containing 'S' is an axiom of CS−). A satisfaction class in a model M is a subset S of the model such that (M,S) |= CS−. The KKL theorem for CS− can be formulated as follows. Theorem 6 Let Th contain IΣ1. For every countable, recursively saturated model M of Th there is a set S ⊆M such that (M,S) |= CS−(Th).10 What we want to emphasize is that it is not obvious at all how to derive Theorem 4 from Theorem 6. The derivation would be trivial if CS− permitted us to define the truth predicate of CT−, that is, if for some τ(x) ∈ LS , CS− proved every formula obtained by replacing 'T ' with τ in some axiom of CT−. However, it seems implausible that CS− can do that.11 Anyway, all the usual methods of defining truth from satisfaction (truth as satisfaction under all assignments, under some assignment or under the empty assignment) fail to deliver the truth predicate of CT− in the context of our non-inductive satisfaction theory.12 In view of this, the transition from satisfaction to truth is sometimes made by postulating stronger properties of a satisfaction class. A prominent example of this strategy can be found in (Enayat and Visser, 2015), where results are obtained about satisfaction classes satisfying not just the usual compositional axioms of CS−, but also the so-called 'extensionality condition', guaranteeing the possibility of defining the truth predicate of CT− in the theory of satisfaction.13 The approach adopted in this paper is different in that we deal directly with truth, without a detour via satisfaction. Finally, let us mention in passing that the transition in the opposite direction, that is, from results about truth to results about satisfaction, can also be 10KKL considered satisfaction, not truth, hence this formulation is closer than Theorem 4 to what they actually proved. Still, the difference remains that the language they considered was relational, that is, without function symbols. For the KKL-style proof of Theorem 6 as stated here (for the arithmetical language with function symbols), see (Kaye, 1991). 11To my knowledge, the question whether CS− defines the truth predicate of CT− has no answer in the literature. 12For example, define τ(x) as the formula φ ∈ SentLAr ∧ S(φ, ε), with ε being the empty assignment. Is τ a CT− truth predicate provably in CS−? If so, then CS− should prove (among other things) the τ -version of the truth axiom for existential sentences, that is, it should prove that τ ( ∃xφ(x) ) ≡ ∃xτ ( φ(ẋ) ) , which boils down to proving that S(∃xφ(x), ε) ≡ ∃xS(φ(ẋ), ε). However, given that S(∃xφ(x), ε), the appropriate satisfaction axiom delivers only ∃aS(φ(x), a) (the formula φ(x) is satisfied by a). How to prove that then the sentence φ(a), with the numeral for a substituted for the free variable, will be satisfied by the empty assignment? It seems that extended induction (for formulas with the satisfaction predicate) is needed for this. 13For the formulation of the extensionality condition, see (Enayat and Visser, 2015, p. 329-330). 6 problematic and it might depend on the choice of the quantifier truth axioms. In the version of CT− adopted here, with the quantifier axiom employing numerals, the satisfaction predicate of CS− can indeed be defined.14 However, it is not obvious at all how to define the satisfaction predicate if we switch to the version with the truth axioms for quantified sentences which employ constant terms instead of numerals. 3 From consistent M-logic to a truth class From now on, we will work with a fixed countable and recursively saturated model M of IΣ1. Following the original strategy of KKL, our first step is the development of a proof system called 'M -logic' (ML in short). Intuitively, ML is a system which permits us to process arbitrary sentences in the sense of M , including the nonstandard ones. The system is described externally (not in the model) in the form of a sequent calculus.15 We will use '⇒' for the sequent arrow, with expressions of the form 'Γ ⇒ ∆' referring to sequents. We shall always assume that both Γ and ∆ are externally finite sequences of M -sentences. Note that, unlike in Gentzen's original system, we do not admit formulas with free variables in the sequents. This deficiency will be compensated by the presence of infinitary rules of inference in ML. The definition of M -logic is framed after Gentzen's original system LK (see (Gentzen, 1964)). All the initial sequents have the form φ⇒ φ, for an arbitrary φ ∈ SentLAr (M). The following rules ofML are copied directly from Gentzen's system: • Weakening, left and right (W-left and W-right): Γ⇒ ∆ Γ⇒ ∆, φ Γ⇒ ∆ φ,Γ⇒ ∆ • Exchange, left and right (E-left and E-right): Γ, ψ, φ,Γ′ ⇒ ∆ Γ, φ, ψ,Γ′ ⇒ ∆ Γ⇒ ∆, ψ, φ,∆′ Γ⇒ ∆, φ, ψ,∆′ • Contraction, left and right (C-left and C-right): φ,φ,Γ⇒ ∆ φ,Γ⇒ ∆ Γ⇒ ∆, φ, φ Γ⇒ ∆, φ • Cut: Γ⇒ ∆, φ φ,Σ⇒ Λ Γ,Σ⇒ ∆,Λ 14Namely, let S(φ, v) be the formula of LT stating that the result of substituting numerals for free variables in φ in accordance with the assignment v is true. 15This is the first difference between our proof and the original KKL's construction. 7 • ¬-left and ¬-right: Γ⇒ ∆, φ ¬φ,Γ⇒ ∆ φ,Γ⇒ ∆ Γ⇒ ∆,¬φ • ∧-left and ∧-right (for arbitrary sentences A and B such that one of them is φ): φ,Γ⇒ ∆ A ∧B,Γ⇒ ∆ Γ⇒ ∆, φ Γ⇒ ∆, ψ Γ⇒ ∆, φ ∧ ψ • ∨-left and ∨-right (for arbitrary sentences A and B such that one of them is φ): φ,Γ⇒ ∆ ψ,Γ⇒ ∆, φ ∨ ψ,Γ⇒ ∆ Γ⇒ ∆, φ Γ⇒ ∆, A ∨B • →-left and →-right: Γ⇒ ∆, φ ψ,Σ⇒ Λ φ→ ψ,Γ,Σ⇒ ∆,Λ φ,Γ⇒ ∆, ψ Γ⇒ ∆, φ→ ψ In addition, M -logic has the following rules of inference: • The truth rule for literals (Tr-lit). Let φ be of the form t = s with M |= val(t) = val(s) or of the form t 6= s with M |= val(t) 6= val(s): φ,Γ⇒ ∆ Γ⇒ ∆ • The M -rule, left and right (M -left, M -right): {φ(a),Γ⇒ ∆ : a ∈M} ∃xφ(x),Γ⇒ ∆ {Γ⇒ ∆, φ(a) : a ∈M} Γ⇒ ∆,∀xφ(x) • ∃-right and ∀-left: Γ⇒ ∆, φ(a) Γ⇒ ∆,∃xφ(x) φ(a),Γ⇒ ∆ ∀xφ(x),Γ⇒ ∆ Proofs in ML are (possibly infinite) trees of finite height, where the height of a proof is defined (as usual) as the length of the maximal path. By definition, trees with no maximal finite path do not qualify as proofs in ML.16 Observe that in ML, the infinitary rules M -left and M -right replace the original rules ∃-left and ∀-right of Gentzen.17 It should be also emphasized that in all the 16This is a minor difference between our construction and the one of KKL, who work without such a finiteness restriction. As a consequence, they employ the assumption that M is recursively saturated in the consistency proof of their version of M -logic. Here the recursive saturation assumption is not needed in the consistency proof; it is employed instead to guarantee the transition from the consistent M -logic to the truth class (see Lemma 7 and its proof). 17Proof systems with similar infinitary rules have already been studied in the literature in the context of cut elimination. See, for example, (Yasugi, 1970). 8 quantifier rules ofML we employ numerals. Thus, for example, in order to apply ∃-right, we need a sentence φ(a) with a numeral for a. In contrast, in Gentzen's original system the rule ∃-right would permit us to derive Γ⇒ ∆,∃xφ(x) from Γ⇒ ∆, φ(t) for an arbitrary term t, not necessarily a numeral. The effect of this modification of Gentzen's system is that the truth class which we construct can contain term pathologies. Thus, in a model (M,T ) of CT− which we eventually obtain there can exist a nonstandard formula φ(x) such that for some term t, φ(t) belongs to T (so that, loosely speaking, the model thinks that φ(t) is true), while the sentence ¬∃xφ(x) also belongs to T . In this way we obtain a disconcerting effect: the model thinks that ¬∃xφ(x) is true even though it considers as true some term instantiation of φ(x).18 (This will happen if all the numerical instantiations of φ(x) are seen as false by the model, that is, if for all numerals a, the sentence ¬φ(a) belongs to T .) However, (M,T ) can still be a model of CT−, since the quantifier axiom of CT− employs numerals. The expression 'M -logic' is a bit of a misnomer, since it is clearly not a system of pure logic for sentences in the sense of M . The extralogical intrusions are not just the infinitary rules (the M -rules given above). In addition, thanks to the (Tr-lit) rule, the system contains the means permitting it to recognize the truth of literals, thus going beyond pure logic also in this respect. We write 'ML ` φ' as an abbreviation of 'ML `⇒ φ' (in other words, 'ML ` φ' means that M -logic proves the sequent whose antecedent is empty and the succedent contains just φ). Now, if φ is a true literal (that is, if φ = pt = sq and M |= val(t) = val(s) or φ = pt 6= sq with M |= val(t) 6= val(s)), then ML ` φ, since the sequent⇒ φ can be derived from the initial sequent φ⇒ φ by (Tr-lit). The lemma below establishes a connection betweenM -logic and truth classes. Lemma 7 Let M |= IΣ1 be countable and recursively saturated. If M -logic is consistent, then M can be expanded to a model of CT−. For the proof of the lemma, we introduce first the family of unary arithmetical predicates 'Prn(S)' with the intuitive reading 'sequent S has a proof inM -logic of height at most n' (in short, S is n-provable). Observe that for each rule R of M -logic, the relation 'S can be obtained by R from n-provable sequents' can always be expressed by an arithmetical formula, provided that n-provability is arithmetically expressible. Thus, for example, 'S can be obtained by M -left from n-provable sequents' can be written down as: ∃Γ,∆, φ(x) ( S = p∃xφ(x),Γ⇒ ∆q ∧ ∀aPrn(φ(a),Γ⇒ ∆) ) . In view of this, we introduce the following definition. 18For more information about the related pathological phenomena in satisfaction classes, see (Cieśliński, 2010). 9 Definition 8 • Pr0(S) := S is an initial sequent, • Prn+1(S) := Prn(S) ∨ ∨ R∈ML (S can be obtained by R from n-provable sequents). By external induction on natural numbers it can be demonstrated that: Observation 9 ∀k ∈ ω ∀S [ ML `k S ≡M |= Prk(S) ] We can now turn to the proof of Lemma 7. Proof. Let φ0, φ1, . . . be an enumeration of the set of M−sentences (this is the only place where the countability assumption is used). We define: T0 = ∅ Tn+1 =  Tn ∪ {φn} if ML 0 (Tn → ¬φn) and φn is not existential, Tn ∪ {∃xψ(x)} ∪ {ψ(a)} if φn = ∃xψ(x) and ML 0 (Tn → ¬φn), for an a ∈M such that ML 0 (Tn → ¬ψ(a)), Tn ∪ {¬φn} otherwise. The above definition strongly resembles the one typically employed in the proof of Lindenbaum's lemma. One difference is that in the familiar Lindenbaum's construction, the extension Tn+1 of Tn depends on whether Tn proves the negation of the n-th sentence in the enumeration. However, this would not make sense in our setting,19 hence we replace it with the condition stating that an appropriate implication is not provable in ML. The intended meaning is that the expression 'Tn' on the right side of the definition (as in 'ML 0 (Tn → ¬φn)') stands for the conjunction of all the sentences φi or their negations, whichever of them were added on previous levels. The second difference is that, unlike in Lindenbaum's construction, existential statements are always added to Tn together with the witnessing formulas. In view of this, we need to verify that whenever ML 0 (Tn → ¬∃xψ(x)), there will exist an a ∈M such that ML 0 (Tn → ¬ψ(a)). There is no analogous step in the proof of Lindenbaum's lemma; this is also the only place in the whole proof where recursive saturation of the model is important. Thus, assume that ML 0 (Tn → ¬∃xψ(x)). Define: p(x) = {¬Prk(Tn → ¬ψ(x)) : k ∈ ω}. 19Unlike the notion of provability in ML, the notion of provability from Tn is not welldefined in the context of our sequent proof system. 10 We observe that p(x) is a type. Otherwise there is a natural number k such that M |= ∀aPrk(Tn → ¬ψ(a)). Hence for all a, ML `k Tn → ¬ψ(a). But then by the M -rule and cut, ML ` Tn → ¬∃xψ(x), which is a contradiction. Since p(x) is a type, by recursive saturation there is an a ∈M which realizes it and we have: ∀kM |= ¬Prk(Tn → ¬ψ(a)), hence the sentence Tn → ¬ψ(a) is not provable in M -logic, as required. Now, define T as ⋃ n∈ω Tn. The proof of Lemma 7 is completed by demonstrating that (M,T ) |= CT− provided that M -logic is consistent. The set T is clearly complete (for every M -sentence ψ, either ψ or negation of ψ belongs to T ) and it contains a numerical example for every existential statement which belongs to T . In addition, since by assumption M -logic is consistent, there is no ψ such that both ψ and ¬ψ belongs to T . In this setting, checking that all the axioms of CT− are true in (M,T ) is fairly easy and we consider just one example, namely, the axiom for the quantifier. In other words, we verify that (M,T ) |= ∀v∀φ(x) ( SentLAr (∃vφ(v))→ (T (∃vφ(v)) ≡ ∃xT (φ(ẋ))) ) . Observe that in the proof of both implications the assumption of the consistency of M -logic will be used. Assume that (M,T ) |= T (∃vφ(v)). Let p∃vφ(v)q be φn (that is, let it be the n-th sentence in our enumeration of M -sentences). Then on the level n + 1 of the construction, φn must have been added to Tn together with the witnessing statement φ(a) for some numeral a (otherwise the negation of φn was added, but this is impossible since it would make T inconsistent). Therefore (M,T ) |= ∃xT (φ(ẋ)). For the opposite implication, assume that (M,T ) |= ∃xT (φ(ẋ)), then for some a ∈ M , φ(a) ∈ T . Assuming for the indirect proof that ∃vφ(v) /∈ T , pick a natural number n such that both φ(a) and ¬∃vφ(v) belong to Tn. But then all Tk-s for k ≤ n are inconsistent in M -logic, meaning that for every k ≤ n, ML ` Tk → ψ for every M -sentence ψ. In effect T would have to contain a pair of contradictory statements, which is impossible. 2 4 Consistency of M-logic At this stage all that is missing for the proof of Theorem 4 is the argument for the consistency ofM -logic. In (Kotlarski et al., 1981) the consistency ofM -logic is proved by the technique of approximations. Here we propose cut elimination as the proof method. Let us start by the following simple observation. 11 Observation 10 If every sequent provable in M -logic has a cut-free proof, then M -logic is consistent. Proof. If M -logic is inconsistent, then it proves that 0 = 1. By cut elimination, take a cut-free proof P of 0 = 1. It is easy to observe that every sentence in P has to be either atomic or negated atomic (the reason is that without cut, (Tr-lit) is the only rule that permits us to eliminate sentences in the proof and (Tr-lit) can eliminate literals only.) For a sequent S belonging to P , let the level of S in P be defined as the length of maximal path generated by S in P .20 Let Tr0(x) be the arithmetical truth predicate for atomic sentences and their negations. By external induction on the level of sequents in P , it can be demonstrated that for every sequent S in P , if all sentences in the antecedent of S are Tr0, then some sentence in the succedent of S is Tr0.21 This trivially holds for sequents of level 0 (that is, for the initial sequents). In the inductive part, observe that any sequent S of level n + 1 must have been obtained in P from sequents of lower level by weakening, contraction, exchange, (Tr-lit) or by the rules for negation applied to atomic sentences; this is so by assumption that P is cut-free and the application of any other rule would introduce a superfluous logical symbol. In effect, very weak resources are enough to verify that if all sentences in the antedecent of S are Tr0 then some element of the succedent of S is Tr0 (in particular, the argument does not require that the model M satisfies any stronger arithmetical theory than IΣ1). It immediately follows thatM |= Tr0(0 = 1), which is impossible for a model of IΣ1. 2 The next lemma states that cut can be eliminated in all proofs in ML. Lemma 11 For every sequent S, if S is provable in ML, then S has a cut-free proof in ML. The aim of the remaining part of the paper is to lead the proof of Lemma 11 to the point at which it can be completed simply by repeating Gentzen's original argument for cut elimination. It should be emphasized that we are not there yet. Our setting is that of possibly nonstandard sentences (sentences in the sense of M) and this generates an obstacle which first has to be removed. 20Thus, sequents which are initial in P have level 0 and the maximal level of a sequent in P is not larger than the height of P . 21Note that we are using here the resources of the model M . Strictly speaking, we demonstrate that: for every sequent S in P , if for every ψ in the antecedent of S M |= Tr0(ψ), then for some sentence ψ in the succedent of S M |= Tr0(ψ). 12 In order to see the obstacle, let us recap the classical argument. The aim is to show that the system with the following mix rule (which is a generalized version of cut) admits mix elimination: Γ⇒ ∆ Σ⇒ Λ Γ,Σ∗ ⇒ ∆∗,Λ (φ) where Σ and ∆ contain φ (the mix formula); Σ∗ and ∆∗ differ from Σ and ∆ only in that they do not contain any occurrence of φ. Since mix and cut produce equivalent proof systems, mix elimination gives us the desired result. In the next stage it is demonstrated that mix can be eliminated from any proof which contains only a single application of the mix rule in the last step. This is done by double induction on the degree of proofs (main induction) and on the rank of proofs (subinduction). For proofs with mix used only in the last step, we define: • The left rank of the proof is the largest number of consecutive sequents in a path starting with the left-hand upper sequent of the mix and such that every sequent in the path contains the mix formula in the succedent. • The right rank of the proof is the largest number of consecutive sequents in a path starting with the right-hand upper sequent of the mix and such that every sequent in the path contains the mix formula in the antecedent. • The rank of the proof is the left rank of the proof + the right rank of the proof. • The degree of the proof is the syntactic complexity of the mix formula. After this is done, it follows that mix can be eliminated from an arbitrary proof (not just from proofs which contain only a single application of the mix rule in the last step). Namely, given an arbitrary proof P , we can eliminate all the applications of mix stage by stage, by considering subproofs of P which contain mix only in the last step. When applying this strategy to the case ofM -logic, one immediate difference is that in the final part of the reasoning (the one in which mix is eliminated from an arbitrary proof) we have to make sure that the height of the mix-free proof remains finite.22 As for the earlier parts, there is no problem in our setting with induction on the rank of proofs, since both the left and the right rank of the proof in ML will always be a (standard) natural number, restricted by the height of the proof. However, the induction on the degree of proofs in ML is quite problematic. Since the mix formula might be a non-standard element 22Unlike Gentzen's original system, ML contains infinitary rules. In effect, we must make sure that there are no infinitely many sequents S0, S1 . . . in a proof P inML such that each of them is obtained in P by mix but the elimination of all these mixes requires an infinite series of higher and higher mix-free proofs. If this happened, then P could not be transformed into a single mix-free proof of finite height. 13 of the model M , its syntactic complexity might be a non-standard number. Arguing externally by induction on non-standard numbers is clearly an invalid move and this is the main obstacle complicating the situation. Our remedy is to replace the general notion of a degree with a notion relativized to a proof. Assume that we are given a proof P with mix applied only in the last step, that eliminates the (possibly non-standard) mix formula φ. The guiding intuition to be formalized below is that in the mix-elimination proof the syntactic shape of φ matters only comparatively. For example, φ might have the form ¬ψ. The intuition is that this will matter only provided that ψ itself (without negation) appears in P ; otherwise in the context of a mix-elimination proof φ might just as well be treated as a formula of syntactic complexity 0, even if it is non-standard. The underlying reason is that in a mix-elimination proof the notion of a degree of a mix formula φ is used only in analysing the case of φ being obtained in the proof by a logical rule (thus, in our example, φ would be obtained by one of the rules for negation, which means that ψ itself must appear in the proof). Our objective is to make these ideas precise. In what follows the word 'sequence' should always be interpreted externally; in other words, sequences are finite or infinite objects in the real world, not necessarily elements of M . The length of a finite sequence a = (a0 . . . ak) is the number of its elements, that is, lh(a) = k + 1. For an infinite sequence a we define lh(a) as ω. Definition 12 • x C y ('x is a direct subsentence of y') is an abbreviation of the following arithmetical formula: SentLAr (x) ∧ SentLAr (y) ∧( ∃ψ ∈ SentLAr (y = p¬ψq ∧ x = ψ) ∨ ∃φ,ψ ∈ SentLAr (y = pφ ◦ ψq ∧ x = φ ∨ x = φ) ∨ ∃θ(x) ∈ FmLAr∃a∃v ∈ V ar(y = pQvθ(v)q ∧ x = pθ(a)q) ) . • Let φ ∈ SentLAr (M). We say that s is a C-sequence for φ iff s0 = φ and for every k < lh(s)− 1 sk+1 C sk. The notion of a degree can now be defined in the following way. Definition 13 Let P be an arbitrary proof in ML with mix used only in the last step. Let φ be the mix formula in P . We define: • d(φ, P ) (the degree of φ in P ) = sup{lh(s) : s is a C-sequence for φ such that for every k < lh(s) sk ∈ P}. • d(P ) (the degree of P ) is defined as d(φ, P ). 14 In effect, given a proof P with a mix formula φ, its degree is identified with the maximal length of a C-sequence generated by φ and containing just the sentences used anywhere in P (not necessarily on a single path in P ). The actual length of this sequence is left open by the definition; in particular, it is not decided whether the relevant sequence is finite or not. Nevertheless, the next key lemma states that proofs with mix used only in the last step always have finite degrees. Lemma 14 Let P be an arbitrary proof in ML with mix used only in the last step. Then d(P ) is a natural number (in other words, it is never ω). In order to prove the lemma, we introduce first the function str(x) ('the structure of a formula x'). Let the letter p be a new symbol, which will be treated as a propositional variable. The function is defined as follows (Q is either ∃ or ∀ and ◦ is an arbitrary binary connective). Definition 15 • str(pt = sq) = ppq • str(p¬ψq) = ¬str(pψq) • str(pφ ◦ ψq) = str(pφq) ◦ str(pψq) • str(pQxφq) = Qx str(pφq) Intuitively, given a formula φ, the function produces a formula which is exactly like φ, except that the letter 'p' is substituted for all occurences of atomic formulas in φ.23 Abbreviate str(φ) = str(ψ) as φ ∼ ψ. The key property of the equivalence relation ∼ is encapsulated in the following observation. Observation 16 Let Z ⊆ SentLAr (M). For every s, if s is a C-sequence with elements from Z, then lh(s) ≤ card ( {[φ]∼ : φ ∈ Z} ) , where [φ]∼ is a class of sentences ψ from Z such that φ ∼ ψ. The intuitive reason behind Observation 16 is that formulas in a C-sequence s have always larger syntactic complexity than formulas which appear later in s and therefore they cannot belong to the same equivalence class of the ∼ relation. We formalize this argument below. Let compl(φ) be the number of connectives and quantifiers in φ. Observation 16 follows from the following fact (we use C∗ for the transitive closure of C). Fact 17 23Thus, for example, str(p∃x(x+ x = 0 ∨ ∀y¬x× y = 0)q) is p∃x(p ∨ ∀y¬p)q. 15 (a) ∀φ,ψ ( φ C∗ ψ → compl(φ) < compl(ψ) ) . (b) ∀φ,ψ ( φ ∼ ψ → compl(φ) = compl(ψ) ) . The proof of Fact 17 is done by easy induction and does not contain any surprises. For part (a), proceed with induction on the length of the C-sequence s leading from ψ to φ. Part (b) can be done by induction on the complexity of ψ. Proof of Observation 16 Let s be a C-sequence with elements from Z. It is enough to observe that for no φ, ψ ∈ s we will have: φ ∼ ψ (in effect, no two different elements of s belong to the same equivalence class of the ∼ relation, hence the length of s is not greater than the number of equivalence classes). Fix φ and ψ ∈ s and assume (without loss of generality) that φ C∗ ψ. Then by Fact 17(a), compl(φ) < compl(ψ) and therefore by Fact 17(b) not φ ∼ ψ. 2 We can now move to the proof of Lemma 14 Proof of Lemma 14 Fix a proof P in ML which contains mix only in the last step. Let Z be the set of all sentences which appear in P . We demonstrate that {[φ]∼ : φ ∈ Z} is finite, which by Observation 16 guarantees the conclusion of Lemma 14. For an arbitrary sequent S in P , let l(S) (the level of S) be the length of the path leading from S to the end sequent of P . We denote by Si the set of all sequents in P whose level is not greater than i. Let Senti be defined as the set of all sentences which appear in some element of Si. Let k be the height of P . The task is to show that: ∀i ≤ k{[φ]∼ : φ ∈ Senti} is finite. This will end the proof, since Sentk = Z. We proceed by induction. For i = 0 the conclusion is trivial, as Sent0 itself is finite (Sent0 is the set of sentences which appear in the end sequent of P ). Now, assuming that {[φ]∼ : φ ∈ Senti} is finite, we claim that {[φ]∼ : φ ∈ Senti+1} is also finite. Observe that Senti+1 can contain more sentences than Senti only provided that some sequent of the level i is obtained in P either by mix or by (Tr-lit) or by a logical rule (structural rules other than mix do not generate any new sentences on level i+ 1). If some sequent of the level i is obtained in P by mix, then i = 0, since by assumption P contains mix only in the last step. Then the conclusion easily follows, since in such a case Senti+1 is finite (it is simply Sent0 together with the mix sentence). In effect, we can now assume that i 6= 0, with no sequent of the level i being obtained in P by mix. Assume that the set Senti+1 \ Senti of new sentences at the level i + 1 is not empty. Then each element of this set is either eliminated by (Tr-lit) in 16 the next stage of the proof or it is a side formula of the logical inference with the main formula belonging to Senti. Let us note first that all the elements of Senti+1 \ Senti which are eliminated by (Tr-lit) fall into just two possible ∼-classes (one for atomic sentences and one for their negations). This leaves us with the second case that of sentences new on the level i+1 which are obtained by logical rules. By inductive assumption, all the main formulas of logical rules belong to finitely many ∼-classes. We will demonstrate that in such a case all side formulas also belong to finitely many ∼-classes, which will finish the proof. Let [φ]∼ be a ∼-class on Senti. We claim that there are at most two ∼classes x and y such that for every ψ ∼ φ, if ψ is the main formula of a logical inference rule which produces in P the sequent S belonging to Si, then every side formula of this inference rules belongs either to x or to y. Let us analyse cases. • Case 1: φ = p¬θq. Define x as [θ]∼. Take an arbitrary ψ ∼ φ such that ψ is the main formula of a logical inference rule which produces the sequent S. Then ψ has the form ¬γ and the side formula must be γ. Since ¬γ ∼ ¬θ, we have γ ∼ θ, therefore γ ∈ x. • Case 2: φ = pχ ◦ θq. Define x as [χ]∼ and y as [θ]∼. Take an arbitrary ψ ∼ φ such that ψ is the main formula of a logical inference rule which produces the sequent S. Then ψ has the form χ′ ◦ θ′ and the side formula must be χ′ or θ′. Since χ ∼ χ′ and θ ∼ θ′, we have: all side formulas belong either to x or to y. • Case 3: φ = pQxθq. Define x as [θ]∼. Take an arbitrary ψ ∼ φ such that ψ is the main formula of a logical inference rule which produces the sequent S. Then ψ has the form Qxχ and the side formulas must be of the form χ(a). Since χ ∼ θ and for every a, χ(a) ∼ χ,24 we have: all side formulas belong to x. 2 In the proof of the cut elimination lemma one more property of degrees of proofs will be important. In the following observation we use the notation Sent(P ) for the set of all sentences which appear in a proof P . Observation 18 Let φ and ψ be M -sentences such that ψ C φ. Let P and P ′ be proofs in M -logic such that: • Both P and P ′ contain mix only in the last step, • The mix formula in P is φ, • The mix formula in P ′ is ψ, • Sent(P ′) ⊆ Sent(P ). 24We remind the reader that ∼ ignores the term differences between formulas. 17 Then d(P ′) < d(P ). Proof. Since the degree of a proof has been defined as the maximal length of a C-sequence of sentences belonging to the proof generated by the mix formula, let (ψ, θ0 . . . θk) be a maximal such sequence for P ′. Since Sent(P ′) ⊆ Sent(P ), it is easy to observe that the sequence (φ,ψ, θ0 . . . θk) is a a C-sequence of sentences belonging to P generated by φ. Hence, d(P ′) < d(P ). 2 In effect, Definition 13, Lemma 14 and Observation 18 give us a notion of a degree such that degrees of proofs (with mix used only in the last step) are always standard natural numbers. Hence, induction on the degree of such proofs can be applied and the way to proving cut elimination theorem for ML is now open. I will not present the whole cut elimination proof, since it is mostly a repetition of Gentzen's reasoning. Instead, I will mostly restrict myself to discussing the cases of the new rules (the ones not present in the original Gentzen's system). Proof of Lemma 11 (outline and chosen cases). It is demonstrated that: (1) mix can be eliminated from any proof which contains only a single application of the mix rule in the last step, (2) given a proof P with mix only in the last step, the new mix-free proof will employ only sentences used in P , (3) the height of the new mix-free proof P ′ is determined by the height of the initial proof P . Let us assume (main induction) that mix can be eliminated in this way in every proof of a degree < n. Let us also assume (subinduction) that mix can be eliminated in this way in every proof of a degree n but with rank < k. Our task is to show that mix can be eliminated in this way in proofs of degree n and rank k. The proof starts with the case of k = 2 (the lowest possible rank) and proceeds by analysing subcases. Here we analyse only two subcases, with the first one corresponding to a rule of ML absent in LK.25 Namely, let us assume for a start that the mix formula ∀xφ(x)26 is obtained in P by a logical rule in both the succedent of the left-hand upper sequent of the mix and in the antecedent of the right-hand upper sequent of the mix. Then the last stage of the proof runs as follows: {Γ⇒ ∆, φ(a) : a ∈M} M -right Γ⇒ ∆,∀xφ(x) φ(c),Σ⇒ Λ ∀-left∀xφ(x),Σ⇒ Λ mix Γ,Σ⇒ ∆,Λ 25Note that for a proof of rank 2, neither the left-hand upper sequent nor the right-hand upper sequent of the mix can be the lower sequent of (Tr-lit). 26The case of the existential mix formula is closely analogous and I do not discuss it separately. 18 We can then eliminate mix constructing P ′ in the following way: Γ⇒ ∆, φ(c) φ(c),Σ⇒ Λ mix Γ,Σ∗ ⇒ ∆∗,Λ possibly, some weakenings and exchanges Γ,Σ⇒ ∆,Λ Here the observation is that the same end sequent can be obtained by applying mix to the formula φ(c); in effect, we build a proof P ′ of Γ,Σ∗ ⇒ ∆∗,Λ which contains mix only in the last step and whose degree is smaller than that of P (the last property follows by Observation 18, since Sent(P ′) ⊆ Sent(P )). Hence, applying the inductive assumption we conclude that mix in P ′ can be eliminated without introducing any new sentences absent from P ′. The final modifications (weakening and exchanges) leading to the end sequent Γ,Σ⇒ ∆,Λ also do not involve adding to the proof any new formulas (in general: in the present setting new proofs without mix are produced from sentences belonging to the initial proof P ). Moreover, the number of weakenings/exchanges to be performed can be showed to depend on the lengths (i.e., number of elements) of sequences Γ,Σ and ∆,Λ, which in turn depend on the height of the initial proof.27 In effect, the height of the mix-free proof will be bounded by a function taking the height of the initial proof as an argument. The second case we consider is that of the mix formula φ→ ψ.28 Then the last stage of the proof runs as follows: φ,Γ⇒ ∆, ψ→-right Γ⇒ ∆, φ→ ψ Σ⇒ Λ, φ ψ,Σ1 ⇒ Λ1 →-left φ→ ψ,Σ,Σ1 ⇒ Λ,Λ1 mix Γ,Σ,Σ1 ⇒ ∆,Λ,Λ1 We can then eliminate mix in the following way: Σ⇒ Λ, φ φ,Γ⇒ ∆, ψ ψ,Σ1 ⇒ ∆1 mix φ,Γ,Σ∗1 ⇒ ∆∗,∆1 mix Σ,Γ∗,Σ∗∗1 ⇒ Λ∗,∆∗,Λ1 possible weakenings and exchanges Γ,Σ,Σ1 ⇒ ∆,Λ,Λ1 Note that for this to work, we need the information that at each stage, eliminating mix from a proof P produces a proof P ′ such that Sent(P ′) ⊆ Sent(P ) (mix elimination introduces no new sentences, cf. Observation 18). When k > 2, we have in addition the case of (Tr-lit) to analyse. Thus, in the last stage (Tr-lit) could be used to obtain the right-hand upper sequent of the mix. In effect, the last stage of the proof might run as follows: 27Only a limited number of sentences can belong to a sequent provable in ML by a proof of a given (fixed) height n. 28The reasoning in this case is essentially the classical argument due to Gentzen. The case of implication is considered here just in order to emphasise once again the crucial role of Observation 18. 19 Γ⇒ ∆ φ,Σ⇒ Λ Tr-lit Σ⇒ Λmix Γ,Σ∗ ⇒ ∆∗,Λ If φ is the mix formula, we can omit (Tr-lit) and use the mix rule instead: Γ⇒ ∆ φ,Σ⇒ Λ mix Γ,Σ∗ ⇒ ∆∗,Λ If the mix formula is not φ, we can build the following proof: Γ⇒ ∆ φ,Σ⇒ Λ mix Γ, φ,Σ∗ ⇒ ∆∗,Λ some exchanges φ,Γ,Σ∗ ⇒ ∆∗,Λ Tr-lit Γ,Σ∗ ⇒ ∆∗,Λ Since in both cases the new proof has lower rank than k (we moved the mix up the derivation), the inductive hypothesis applies and the mix rule is eliminable. The case of (Tr-lit) being used to obtain the left-hand upper sequent of the mix is very similar. 2 Acknowledgements. 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