A Kripke-style semantics for paradox-tolerant, nontransitive intuitionistic logic. Nick Thomas February 20, 2014 Ripley [201+a,b] explores an approach to adding a transparent truth predicate to classical logic. A proof-theoretic way of describing the approach is to say that we take a standard, cut-eliminable sequent system for classical logic, drop the rule of cut, and add the truth predicate rules Γ ` φ, ∆ . Γ ` T hφi, ∆

Γ, φ ` ∆ Γ, T hφi ` ∆

The resulting logic is called STT. Any classically valid inference (even one mentioning the truth predicate, though not using the truth rules) is valid in STT, for the following reason. If Γ ` ∆ is a classically valid inference, then there is a derivation of this sequent in the sequent system for classical logic. Since this sequent system enjoys cut elimination, there is a cut-free derivation of the sequent, and this will be a valid STT derivation of the sequent. In fact, STT is a conservative extension of classical logic. This means that given a theory T in classical vocabulary (i.e., not mentioning the truth predicate), the classical-vocabulary consequences of T in STT are only the classical consequences — so that in particular T is nontrivial in STT if it is consistent in classical logic. Moreover, T is genuinely transparent. φ and T hφi are intersubstitutable in every context, in the sense that if Γ ` ∆ is a derivable sequent and Γ∗ , ∆∗ are the result of substituting formulas φ for the corresponding formulas T hφi any number of times in Γ, ∆, then Γ∗ ` ∆∗ is a derivable sequent. This fact, and the fact of the previous paragraph, are shown proof theoretically in Ripley [201+b]. STT also has an elegant model theoretic presentation. STT models are just standard three-valued K3 models; but satisfaction is defined in a non-standard way on them. Let {0, .5, 1} be the set of K3 truth values, and given a K3 model M , let υM (φ) denote the truth value of φ in M . An “admissible” K3 model is one in which υM (T hφi) = υM (φ) for all φ. Then we say that Γ |= ∆ iff for all admissible models M , if υM (φ) = 1 for all φ ∈ Γ, then υM (ψ) 6= 0 for some ψ ∈ ∆. The proof theoretic definition of STT is sound and complete with respect to this notion of semantic consequence. STT gives us a way of retaining classical logic while adding a transparent truth predicate. Here we consider how the same approach might be used to add 1

a transparent truth predicate to intuitionistic logic. One reason one might be interested in this question is if one is an intuitionist, and rejects classical logic, but still wishes to use an STT-like approach to add a truth predicate to one’s language. It is actually quite easy to do this. We just have to start with a standard, cut-eliminable sequent system for intuitionistic logic, and add the truth rules while removing the cut rule; the arguments given by Ripley [201+b] for the classical case will go through unchanged to show that the result conservatively extends intuitionistic logic, and that the truth predicate is transparent. A natural follow-up question is the following: what kind of model theory is appropriate to this variant of intuitionistic logic? This question is interesting from a technical standpoint, and in particular it seems likely to shed light on the broader question of what kinds of model theories are appropriate to nontransitive logics in general. Here we shall develop a Kripke-style semantics for cut-free intuitionistic propositional logic with truth-like rules, and prove it sound and complete.

1

Syntax and proof theory

We will work with intuitionistic propositional calculus. Formulas are built up from a countably infinite set V = {p, q, r, ...} of propositional variables, and the connectives ∧, ∨, →, ⊥. Instead of a truth predicate, we consider the more general idea of a “system of definitions.” The idea of a system of definitions is that we can define atomic formulas to be equal to (possibly non-atomic) formulas. Formally, a system of definitions is a set D of expressions (called “definitions”) of the form p = φ, where p is a propositional variable (the definiendum), and φ a formula (the definiens), such that no variable occurs as the definiendum of more than one definition. Constructions such as p = p → ⊥ are legal, so that systems of definitions can produce self-referential paradoxes. A “sequent” is an expression of the form Γ `D φ, where φ is a formula, Γ is a finite set of formulas, D is a finite system of definitions. We omit the subscript where clear. Our sequent system is defined as follows. When a subscript is absent, we mean that the rule holds regardless of what the system of definitions is. Basic φ`φ Structural Γ`φ Γ, ψ ` φ Conjunction Γ, φ ` ζ Γ, φ ∧ ψ ` ζ

Γ, ψ ` ζ Γ, φ ∧ ψ ` ζ 2

Γ ` φ Γ0 ` ψ Γ, Γ0 ` φ ∧ ψ Disjunction Γ`φ Γ`φ∨ψ

Γ`ψ Γ`φ∨ψ

Γ, φ ` ζ Γ0 , ψ ` ζ Γ, Γ0 , φ ∨ ψ ` ζ Implication Γ, φ ` ψ Γ`φ→ψ

Γ ` φ Γ, ψ ` ζ Γ, φ → ψ ` ζ Bottom Γ`⊥ Γ`φ

Definitions Γ `D φ Γ `D p

Γ, φ `D ψ Γ, p `D ψ

whenever (p = φ) ∈ D. Given a system of definitions D, a theory T , and a formula φ, we say that T `D φ iff there are finite Γ ⊆ T and D0 ⊆ D such that Γ `D0 φ is a derivable sequent. Call this system IST, for “intuitionistic ST.” We have the following facts, like the ones given by Ripley [201+b] for STT: Theorem 1. If T ` φ in intuitionistic logic, then T `D φ. Proof. Any intuitionistically valid derivation of T ` φ can be transformed into a cut-free derivation, which will be a valid IST derivation. It is also easy to see that T `∅ φ iff T ` φ in intuitionistic logic, and that if D0 ⊆ D and T `D0 φ, then T `D φ. Theorem 2 (Conservativeness). If T `D φ and T, φ do not contain (as subformulas) any atomic formulas which are the LHS of a definition in D, then T `∅ φ. Proof. Any application of a definition rule introduces a definition-LHS as a toplevel formula in the premise set or as the conclusion, and all of the proof rules preserve the property of “having a definition-LHS as a subformula of a premise or the conclusion.” So any derivation using the definition rules will have as its conclusion a sequent containing a definition-LHS. So any derivation of T `D φ will not have used the definition rules, and will thus be a valid derivation over the empty set of definitions; so T `∅ φ. 3

Theorem 3 (Weak transparency.). If T `D ζ and (p = φ) ∈ D, then we can substitute any instance, as a premise or the conclusion, of p or φ for the other formula and produce a valid sequent. Proof. It suffices to show that we can replace one instance of p or φ with an instance of the other, and the replacement of φ with p follows by the definition rules. It can be shown by induction on proof length that any derivation of T `D ζ can be transformed into a derivation of a variant on T `D ζ which replaces one instance of p with φ. The stronger version of transparency, that p and φ are intersubtitutable when they occur as subformulas of formulas in derivable sequents, will follow from the completeness theorem.

2

Semantic definitions

The task of this paper is to formula a sound and complete semantics for IST. Our semantics is a variant of Kripke semantics. At each point, we distinguish between the formulas true at the left of the turnstile, and the formulas true at the right of the turnstile. Conditionals are defined in a Routley-Meyer type way which bounces back between these two aspects of the point. Henceforth we assume a fixed system of definitions D. When we write T ` φ, we mean T `A D φ, unless otherwise specified. Definition 1. Given a theory (i.e., set of formulas) A, we write will sometimes write A |= φ as syntactic sugar for φ ∈ A. A “theory pair” is just a pair (A, A0 ) of theories. We write (A, A0 ) ⊆ (B, B 0 ) to mean A ⊆ A0 and B ⊆ B 0 . A “D-valuation” is a theory A such that A 2 ⊥ and for all formulas φ, ψ: 1. A |= φ ∧ ψ iff A |= φ and A |= ψ. 2. A |= φ ∨ ψ iff A |= φ or A |= ψ. 3. If (p = φ) ∈ D, then A |= p iff A |= φ. A “D-point” is a pair (A, A0 ) of D-valuations such that A ⊆ A0 . A “D-model” is a pair (K, ≤) where K is a set of D-points and ≤ is a partial order on K such that: 1. If (A, A0 ) ≤ (B, B 0 ) then (A, A0 ) ⊆ (B, B 0 ). 2. For all (A, A0 ) ∈ K and all φ, ψ, if A |= φ → ψ and A0 |= φ, then A |= ψ. 3. For all (A, A0 ) ∈ K and all φ, ψ, A0 |= φ → ψ iff for all (B, B 0 ) ∈ K, if (A, A0 ) ≤ (B, B 0 ) and B |= φ, then B 0 |= ψ. Given a theory T and a formula φ, we say T |=D φ iff for all D-models M = (K, ≤) and all (A, A0 ) ∈ K, if A |= ψ for all ψ ∈ T then A0 |= φ.

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3

Philosophical interpretation

Before proving IST sound and complete for this model theory, I want to take a moment to attempt a philosophical interpretation of the model theory. My interpretation of the model theory follows Ripley [201+b]’s understanding of the entailment relation for STT. Ripley, following Restall [2005], understands a (multiple-conclusion) STT entailment Γ ` ∆ as meaning “it is out of bounds to assert all members of Γ and simultaneously deny all members of ∆.” Following Kripke [1965], we may understand points as states of information. Let (A, A0 ) be a point. We shall understand A |= φ as meaning “given the current state of information, φ is assertible.” We shall understand A0 |= φ as meaning “given the current state of information, φ is undeniable.” This is related to Ripley’s interpretation of the turnstile because the left hand side A of a point is where statements on the left hand side of a turnstile are evaluated, and the right hand side A0 of a point is where statements on the right hand side of a turnstile are evaluated. The truth conditions then say that φ ∧ ψ is assertible iff φ is assertible and ψ is assertible; φ ∧ ψ is undeniable iff φ is undeniable and ψ is undeniable; and so forth. Of particular interest is to interpret the conditional. Just reading off the truth conditions, we have: 1. If φ → ψ is assertible and ψ is undeniable, then φ is assertible. 2. φ → ψ is undeniable iff it is impossible that φ is assertible while ψ is deniable. The condition for undeniability of a conditional aligns with the condition for validity of an entailment. The condition for assertibility of a conditional is something new. What to make of it is an interesting question.

4

IST0 : Adding axiom rules

In order to prove IST sound and complete for the semantics we have just given, we will look at a theory IST0 which extends IST. We are interested in IST0 for technical reasons: it gives us some tools which let us prove the completeness theorem. We will prove IST0 complete with respect to its variant of IST’s model theory, and establish a proof theoretic connection between IST0 and IST which gives the completeness theorem for IST. What IST0 gives is the ability to add rules of the form ` φ to the theory. We call these “axiom rules” (not to be confused with rules of the form φ ` φ, which are also sometimes called by this name). In addition, IST0 adds several inference rules which are redundant in the presence of cut, and still eliminable in IST, but apparently not eliminable in the presence of axiom rules, and needed to make IST0 complete with respect to the target model theory. An IST0 sequent is an expression of the form Γ R D φ, where Γ is a finite set of formulas, R is a finite set of formulas, D is a finite system of definitions, and φ

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is a formula. R is to be understood as the set of formulas for which we stipulate axiom rules. Our sequent rules are the rules of IST (where the superscript R is allowed to be anything), plus the following rules.

R φ, whenever φ ∈ R. Γ, φ ∧ ψ ζ . Γ, φ, ψ ζ Γ φ∧ψ Γ ψ

Γ φ∧ψ Γ φ Γ, φ ∨ ψ ζ Γ, φ ζ

Γ, φ ∨ ψ ζ Γ, ψ ζ

Γ φ→ψ Γ, φ ψ Γ D p Γ D φ

Γ, p D ψ , Γ, φ D ψ

whenever (p = φ) ∈ D. 0

0 0 R We say T R D φ iff there are finite Γ ⊆ T, R ⊆ R, D ⊆ D such that Γ D 0 φ is a derivable sequent. Given a system of definitions D and a set R of formulas, a “D, R-point” is a D-point (A, A0 ) where R ⊆ A0 . A “D, R-model” is a D-model (K, ≤) wherein all points are D, R-points. We say T |=R D φ iff for all D, R-models M = (K, ≤) and all (A, A0 ) ∈ K, if A |= ψ for all ψ ∈ T then A0 |= φ.

5

Eliminability of IST0 rules

The extra rules (other than the axiom rules) of IST0 are eliminable in IST. This lets us establish the basic relationship between IST0 and IST: that T `D φ iff T ∅D φ. We first note that all of the rules of IST0 are eliminable in IST: Lemma 1.

1. If Γ, φ ∧ ψ ` ζ then Γ, φ, ψ ` ζ.

2. If Γ ` φ ∧ ψ, then Γ ` φ and Γ ` ψ. 3. If Γ ` φ → ψ then Γ, φ ` ψ. 4. If (p ∈ φ) ∈ D and Γ `D p then Γ ` φ, and if Γ, p `D ψ then Γ, φ `D ψ. Proof. Each of these facts may be proven by induction on proof length. Now we have the basic fact connecting IST and IST0 . 6

Theorem 4. Γ `D φ iff Γ ∅D φ. Proof. If Γ ∅D φ then Γ `D φ, since every valid IST0 derivation is a valid IST derivation. If Γ ∅D φ then Γ `D φ because all of the rules of IST0 except for the axiom rules are eliminable in IST, so that any valid IST0 derivation without axiom rules can be transformed into a valid IST derivation. It is worth noting (and easy to check) that the same connection holds at the semantic level; that is, Theorem 5. Γ |=D φ iff Γ |=∅D φ.

6

Soundness and completeness

R Theorem 6 (Soundness). If T R D φ then T |=D φ.

Proof. Check in the usual way. Definition 2. Let (A, A0 ) be a theory pair, and φ a formula. We say “φ is 0 eliminable over (A, A0 )” iff for all (B, B 0 ) ⊇ (A, A0 ) and all ζ, if B, φ B ζ then B0 B ζ. Definition 3. A theory pair (A, A0 ) is “saturated” iff A ⊆ A0 , A0 2 ⊥, and: 0

1. For all φ, A0 |= φ iff A A D φ. 2. For all φ, A |= φ iff φ is eliminable over (A, A0 ). 3. If A |= φ ∨ ψ then A |= φ or A |= ψ. 4. If A0 |= φ ∨ ψ then A0 |= φ or A0 |= ψ. We note a few proof theoretic facts that will be useful. Lemma 2.

1. If Γ R,φ ζ, then Γ, φ R ζ.

2. If Γ R,φ ζ and Γ R φ, then Γ R ζ. 3. If T R∪S ζ and T R φ for all φ ∈ S, then T R ζ. 4. If φ → ψ is eliminable over (A, A0 ) then ψ is eliminable over (A, A0 + φ). Proof. (1) and (2) may be proven by induction on proof length. (3) follows by repeated application of (2). Let us prove (4). Suppose φ → ψ is eliminable over (A, A0 ). Let (B, B 0 ) ⊇ (A, A0 +φ). Suppose 0 0 0 B, ψ B ζ. B B φ, since φ ∈ B 0 . So B, φ → ψ B ζ. By eliminability of 0 φ → ψ, B B ζ. So ψ is eliminable over (A, A0 + φ). Lemma 3. Let (A, A0 ) be a saturated theory pair with R ⊆ A0 . Then (A, A0 ) is a D, R-point.

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Proof. It suffices to check that A, A0 are D-valuations, and we already have that A 2 ⊥ and A0 2 ⊥. 0

0

0

1. A0 |= φ ∧ ψ iff A A φ ∧ ψ iff A A φ and A A ψ iff A0 |= φ and A0 |= ψ. 0

2. If A0 |= φ ∨ ψ, then A0 |= φ or A0 |= ψ. If A0 |= φ or A0 |= ψ, then A A φ 0 0 or A A ψ, then A A φ ∨ ψ, then A0 |= φ ∨ ψ. 0

0

3. Let (p = φ) ∈ D. A0 |= φ iff A A φ iff (by weak transparency) A A p iff A0 |= p. 4. Suppose A |= φ ∧ ψ. Then φ ∧ ψ is eliminable over (A, A0 ). Let (B, B 0 ) ⊇ 0 0 0 (A, A0 ). Suppose B, φ B ζ. Then B, φ ∧ ψ B ζ. Then B B ζ. So φ is eliminable over (A, A0 ). So A |= φ. By similar reasoning, A |= ψ. Suppose A |= φ and A |= ψ. φ and ψ are both eliminable over (A, A0 ). 0 0 Let (B, B 0 ) ⊇ (A, A0 ), and suppose B, φ ∧ ψ B ζ. Then B, φ, ψ B ζ. 0 0 Then B, φ B ζ. Then B B ζ. So φ ∧ ψ is eliminable over (A, A0 ). So A |= φ ∧ ψ. 5. If A |= φ ∨ ψ, then A |= φ or A |= ψ. Suppose A |= φ; we want to prove that A |= φ ∨ ψ. The case when A |= ψ will proceed analogously. φ is 0 eliminable over A. Let (B, B 0 ) ⊇ (A, A0 ), and suppose B, φ ∨ ψ B ζ. 0 0 Then B, φ B ζ. Then B B ζ. So φ ∨ ψ is eliminable over A. So A |= φ ∨ ψ. 6. Let (p = φ) ∈ D. Suppose A |= p. Then p is eliminable over (A, A0 ). 0 Let (B, B 0 ) ⊇ (A, A0 ), and suppose B, φ B ζ. By weak transparency, 0 0 B, p B ζ. By eliminability of p, B B ζ. So φ is eliminable over (A, A0 ). So A |= φ. By similar reasoning, if A |= φ then A |= p.

0

Lemma 4 (Point existence). Let (T, T 0 ) be a theory pair such that T 1T χ. There is a saturated theory pair (A, A0 ) ⊇ (T, T 0 ) such that A0 2 χ. Proof. We construct a transfinite sequence of theory pairs (A0 , A00 ), (A1 , A01 ), ... running through the ordinals. We let A0 S = T and A00 = T 0 . For α a limit S 0 ordinal, we say Aα = γ<α Aγ and Aα = γ<α A0γ . Suppose α is a successor ordinal. For some β, n, i, where β is zero or a limit ordinal, n ∈ N, and 1 ≤ i ≤ 4, def α = β + 4n + i. We let γ = α + 4n, and split into cases according to the value of i. 0

1. Suppose i = 1. We let Aγ+1 = Aγ , and let A0γ+1 = {φ : Aγ Aγ φ}. 2. Suppose i = 2. We let A0γ+2 = A0γ+1 . If there is no formula φ such that φ is eliminable over (Aγ+1 , A0γ+1 ) and Aγ+1 2 φ, let Aγ+2 = Aγ+1 . Otherwise choose such a φ, and let Aγ+2 = Aγ+1 + φ.

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3. Suppose i = 3. Let A0γ+3 = A0γ+2 . If there is no formula φ ∨ ψ such that A |= φ ∨ ψ but A 2 φ and A 2 ψ, let Aγ+3 = Aγ+2 . Otherwise choose 0 such a φ ∨ ψ. If A, φ A χ, let Aγ+3 = Aγ+2 + ψ, and otherwise let Aγ+3 = Aγ+2 + φ. 4. Suppose i = 4. Let Aγ+4 = Aγ+3 . If there is no formula φ ∨ ψ such that A0 |= φ ∨ ψ but A0 2 φ and A0 2 ψ, let A0γ+4 = A0γ+3 . Otherwise choose 0 such a φ ∨ ψ. If A A +φ χ, let A0γ+4 = A0γ+3 + ψ, and otherwise let A0γ+4 = A0γ+3 + φ. 0

We prove, by induction, that for all α, Aα 1Aα χ. This is true by hypothesis when n = 0. This is true by compactness of deducibility when α is a limit ordinal. Now suppose α is a successor ordinal. For some β, n, i, where β is zero or a limit ordinal, n ∈ N, and 1 ≤ i ≤ 4, α = β + 4n + i. We let γ = β + 4n, so α = γ + i. 0

0

1. Suppose i = 1. Suppose (Aγ =)Aγ+1 Aγ+1 χ. For all φ ∈ A0γ+1 , Aγ Aγ 0 φ; so by Lemma 2, Aγ Aγ χ. This is a contradiction. 0

0

2. Suppose i = 2. Suppose Aγ+2 Aγ+2 (=Aγ+1 ) χ. We can assume Aγ+2 6= Aγ+1 . For some φ, Aγ+2 = Aγ+1 +φ, where φ is eliminable over (Aγ+1 , A0γ+1 ). 0 0 Aγ+1 , φ Aγ+1 χ. By eliminability, Aγ+1 Aγ+1 χ, a contradiction. 0

3. Suppose i = 3. Suppose Aγ+3 Aγ+3 χ. We can assume Aγ+3 6= A0γ+3 . So there is φ ∨ ψ such that Aγ+2 |= φ ∨ ψ, and Aγ+3 is Aγ+2 plus one of 0 φ or ψ. Say it is φ. Using that A0γ+3 = A0γ+2 , we have Aγ+2 , φ Aγ+2 χ. But our choosing φ meant that this failed; so in fact we chose ψ. This 0 means that Aγ+2 , φ Aγ+2 χ. Moreover, since Aγ+3 = Aγ+2 + ψ, we have 0 0 Aγ+2 , ψ Aγ+2 χ. Therefore Aγ+2 Aγ+2 χ, and we have a contradiction. 0

4. Suppose i = 4. Suppose Aγ+4 Aγ+4 χ. Again we can assume there is φ ∨ ψ such that A0γ+3 |= φ ∨ ψ, and A0γ+4 is A0γ+3 plus one of φ or ψ. 0 If it is φ, we have Aγ+3 Aγ+3 ,φ χ, contradicting that we chose φ. So 0 0 we chose ψ, and Aγ+3 Aγ+3 ,φ χ. Moreover we have Aγ+3 Aγ+3 ,ψ χ, 0 0 and by Lemma 2 we have Aγ+3 , φ Aγ+3 χ and Aγ+3 , ψ Aγ+3 χ, so 0 Aγ+3 Aγ+3 χ, a contradiction. Since the sequence (A0 , A00 ), (A1 , A01 ), ... is monotone and running through a countable space, it eventually reaches a fixed point (Aλ , A0λ ) = (A, A0 ). The closure properties of this fixed point tell us that: 0

1. If A A φ then A0 |= φ. 2. If φ is eliminable over (A, A0 ) then A |= φ. 3. If A |= φ ∨ ψ then A |= φ or A |= ψ. 4. If A0 |= φ ∨ ψ then A0 |= φ or A0 |= ψ. 9

Now we check that (A, A0 ) is saturated. 0

1. Suppose A |= φ. Then A A φ. Then A0 |= φ. So A ⊆ A0 . 0

0

0

2. Suppose A0 |= φ. Then A φ. Then A A φ. If A A φ, then A0 |= φ by the first closure property. 0

3. Suppose A |= φ. Let (B, B 0 ) ⊇ (A, A0 ). Suppose B, φ B ζ. Since φ ∈ B, 0 B B ζ. So φ is eliminable over (A, A0 ). If φ is eliminable over (A, A0 ), then φ ∈ A by the second closure property. 4. If A |= φ ∨ ψ then A |= φ or A |= ψ by the third closure property, and if A0 |= φ ∨ ψ then A0 |= φ or A0 |= ψ by the fourth closure property. 0

0

5. Suppose A0 |= ⊥. Then A A ⊥. Then A A χ, a contradiction. So A0 2 ⊥. Now we have all the properties needed to conclude that (A, A0 ) is saturated. Let R be a set of formulas (axiom rules). We define the canonical D, Rmodel M = (K, ⊆), where K is the set of all saturated theory pairs (A, A0 ) with R ⊆ A0 , and ⊆ is the usual ⊆ relation on theory pairs. Lemma 5. M is a D, R-model. Proof. We just need to check conditions (2) and (3) on being a D-model. 2. Let (A, A0 ) ∈ K. Suppose A |= φ → ψ, and A0 |= φ. φ → ψ is eliminable over (A, A0 ). By Lemma 2, ψ is eliminable over (A, A0 + φ). Since φ ∈ A0 , ψ is eliminable over (A, A0 ). So A |= ψ. 0

3. Let (A, A0 ) ∈ K. Suppose A0 |= φ → ψ. Then A A φ → ψ. Let 0 (B, B 0 ) ∈ K, and suppose (A, A0 ) ⊆ (B, B 0 ) and B |= φ. B, φ B ψ. 0 Since φ is eliminable over (B, B 0 ), B B ψ. So B 0 |= ψ. We prove the other direction by contrapositive. Suppose A0 2 φ → ψ. 0 0 Then A 1A φ → ψ. Then A, φ 1A ψ. There is a saturated theory pair 0 (B, B 0 ) ⊇ (A + φ, A0 ) such that B 1B ψ. (B, B 0 ) ∈ K, B |= φ, and B 0 2 ψ, so this is what we wanted.

R Theorem 7 (Completeness). If T |=R D φ then T `D φ. 0 0 Proof. Suppose T 1R D φ. There is saturated (A, A ) ⊇ (T, R) such that A 2 φ. 0 0 R (A, A ) ∈ K, so M and (A, A ) form a counterexample to T |=D φ.

Corollary 1. T `D φ iff T |=D φ. Proof. Theorems 4, 5, 6, and 7.

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7

Underdetermined truth conditions

Our model theory does not completely determine the truth values of conditionals in the left hand parts of points. This is because the truth conditions are a oneway conditional, rather than a biconditional. The natural two-way truth condition would be that in a point (A, A0 ) in a model M = (K, ≤), A |= φ → ψ iff for all (B, B 0 ) ∈ K such that (A, A0 ) ≤ (B, B 0 ), if B 0 |= φ then B |= ψ. The reason I did not use this truth condition is that I do not know how to prove completeness when it is present. I suspect that completeness is still true in the presence of this condition, but again I do not know this. It is interesting that IST allows for some connectives to have underdetermined truth values, and there is in fact an additional example of this. Consider the model theory where we drop the condition that in a point (A, A0 ), if A0 |= φ∨ψ then A0 |= φ or A0 |= ψ. Then disjunctions on the right hand side are defined by the one-way truth condition: if A0 |= φ or A0 |= ψ then A0 |= φ ∨ ψ. IST is still sound with respect to this model theory, as you can check. It is also still complete, because adding more models cannot hurt completeness. That IST is sound and complete with respect to this semantics, where two connectives have underdetermined truth values — namely conditionals on the left and disjunctions on the right — shows that the complete truth conditions for disjunction on the right, and the complete truth conditions for conditionals on the left (if they leave IST complete), are in some sense purely cosmetic features of the model theory which do not reflect any inferential features of IST’s proof theory. This is an interesting fact in need of explanation. Why doesn’t IST enforce these kinds of truth conditions on its models? I have no explanation to offer, but I think it is an interesting phenomenon whose understanding might shed light on this system, and potentially on non-classical cut-free logics more broadly.

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Acknowledgements

Thanks to Nathan Kellen for inspiring my interest in this topic. Thanks to Dave Ripley for helpful discussions of the paper: in particular for help defining the truth conditions for conditionals on the left hand side of points, and help understanding the conditional rules of intuitionistic sequent calculus.

References Saul Kripke. Semantical analysis of intuitionistic logic I. Formal systems and recursive functions, pages 92–130, 1965. Greg Restall. Multiple conclusions. In Petr Hajek, Luis Valdes-Villaneuva, and Dag Westerstahl, editors, Logic, Methodology and Philosophy of Science: Pro-

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ceedings of the Twelfth International Congress, pages 189–205. Kings’ College Publications, 2005. David Ripley. Conservatively extending classical logic with transparent truth, 201+a. Review of Symbolic Logic. To appear. David Ripley. Paradoxes and failures of cut, 201+b. Australasian Journal of Philosophy. To appear.

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