The Laws of Thought and the Laws of Truth as Two Sides of One Coin

Abstract

Some think that logic concerns the “laws of truth”; others that logic concerns the “laws of thought.” This paper presents a way to reconcile both views by building a bridge between truth-maker theory, à la Fine, and normative bilateralism, à la Restall and Ripley. The paper suggests a novel way of understanding consequence in truth-maker theory and shows that this allows us to identify a common structure shared by truth-maker theory and normative bilateralism. We can thus transfer ideas from normative bilateralism to truth-maker theory, such as non-transitive solutions to paradox, and vice versa, such as notions of factual equivalence and containment.

Appendix A: Proofs and Technical Details

1.1 A.1 Classical Logic

Definition 13 (Proof-search)

A root-first proof-search produces a proof-tree from a sequent , which is the root of the tree, by recursively applying the following procedure until the process terminates: (i) If is the leaf of a branch of the tree at the current stage and all the sentences in Γ and Δ are atomic, then the branch remains unchanged. (ii) Otherwise, we look for the first complex sentence in (starting on the left, ordering the sentences in Γ and Δ alphabetically) and build the branch up from that leaf by using the appropriate rule of CL. For example, we apply the top-to-bottom version of [L∨] (moving upwards in the tree) if the left-most complex sentence in our sequent is a disjunction, etc. (Although we work with sets (and so contraction is built in), we represent the sets in our sequents with the number of copies of sentences that we get by applying this procedure to the given representation of the root, thus treating our sets (in how we represent them) like multi-sets.)

Someone might worry that this definition will yield different results for representations of the root sequent that differ in the numbers of copies of sentences. In fact, however, this doesn’t happen.

Proposition 14

Proof-searches on and yield the same results, and the same holds for proof-searches on and

Proof

By induction on the complexity of A. □

Proposition 15

Proof-searches terminate, and their results are the same if we change the order of the sentences in Γ and Δ.

Proof

Proof-searches terminate because the root contains finitely many logical connectives, and the children of a node always contain one fewer connective than the parent node. To show that the order doesn’t matter, it suffices to show that for each pair of rules, the order in which they are applied doesn’t matter. If we have, e.g., , applying our procedure to the first two sentences yields: and and . This result is the same whether we use [L¬] first and then [L∨] or the other way around. The same holds for all pairs of rules. □

Proposition 16

In CL, [weakening] is redundant.

Proof

We can add the desired additional context to every application of [ID]. □

Proposition 17

In CL, the bottom-to-top operational rules can be eliminated, i.e., omitting these rules does not change which sequents are derivable.

Proof

We argue by induction on proof-height, and look at each bottom-to-top rule in turn. Since [weakening] can be eliminated, it suffices to look at proof-trees without [weakening]. I will give the proof for [L∧]; the other cases are analogous. Suppose we have a derivation of . We must show that is derivable. If was derived via [L∧], we’re done. For all the other rules by which may come, A ∧ B must have been in the left context of the rule-application. We can apply our induction hypothesis and replace the conjunction with the two conjuncts. We then get by applying the rule by which was derived in our initial proof-tree. □

Proposition 18

In CL, [cut] can be eliminated.

Proof

From the proof of Proposition 26 below, it is easy to see that CL without [cut] is equivalent to the sequent calculus of ST without the truth-rules, and it is well-known that [cut] is admissible in that sequent calculus. Hence, [cut] is admissible in CL without [cut]. □

Proposition 19

CL is sound and complete with respect to classical propositional logic, i.e., iff .

Proof

For soundness, it suffices to note that every classical truth-assignment satisfies (i.e., is not a counter-model to) any instance of [ID] and that all rules of CL preserve that property. For completeness, suppose that cannot be derived. Hence, a proof-search for yields at least one atomic sequent, , such that Γ₀ ∩Δ₀ = ∅. So, there is a counterexample to . Any counterexample to is also a counterexample to . By the contrapositive of [cut], for any atomic sentence, p, either or is not derivable. If is not derivable, we make p false; otherwise is not derivable, and we make p true. In this way, we can extend our counterexample by assigning truth-values to all atomic sentences. Hence, . □

Proposition 20

The rules [ID], [weakening], and [cut] are valid for iff possible states obey Exclusivity, Downward-Closure, and Exhaustivity respectively.

Proof

Downward-Closure and [weakening]: Downward-Closure says that if s ∈ S^♢ and $t\sqsubseteq s$, then t ∈ S^♢. Now, if , then any fusion of verifiers of everything in Γ and falsifiers of everything in Δ is impossible. By Downward-Closure, all states that include any such fusion as a part are also impossible. Hence, . For the other direction, suppose that [weakening] is valid and that s ∈ S^♢ and $t\sqsubseteq s$. In accordance with Assumption 1, let Γ, Θ, Δ, and Σ be such that s is the unique state that is a fusion of verifiers for everything in Γ ∪Θ and falsifiers for everything in Δ ∪Σ. Since s ∈ S^♢, we know that . If [weakening] is valid for , it follows that . Since $t\sqsubseteq s$, without loss of generality, let t be the state that results from s by omitting the verifiers for Θ and the falsifiers for Σ. Then t is the unique state that is a fusion of verifiers for everything in Γ and falsifiers for everything in Δ. Hence, t ∈ S^♢.

Exclusivity and [ID]: Exclusivity says that if s ∈|p|⁺ and t ∈|p|⁻, then ∀u(s ⊔ t ⊔ u∉S^♢). So, . For the other direction, suppose [ID] is valid and let s ∈|p|⁺ and t ∈|p|⁻. By [ID], for any Γ and Δ, we have . So every state that includes a truth-maker and a falsity-maker of p is impossible, i.e., ∀u(s ⊔ t ⊔ u∉S^♢).

Exhaustivity and [cut]: Suppose that and let u be a state witnessing this fact, i.e., a state that is a fusion of verifiers of everything in Γ and falsifiers of everything in Δ such that u ∈ S^♢. By Exhaustivity, ∃s ∈|p|⁺(u ⊔ s ∈ S^♢) or ∃t ∈|p|⁻(u ⊔ t ∈ S^♢). Therefore, either or . But that is just what is required for the contrapositive of [cut]. For the other direction, suppose that [cut] is valid for . Let u be possible; and, in accordance with Assumption 1, let Γ be a set such that u is the unique state that is a fusion of verifiers for everything in Γ. Hence, . By the validity of [cut], either or . Hence, either ∃s ∈|p|⁺(u ⊔ s ∈ S^♢) or ∃t ∈|p|⁻(u ⊔ t ∈ S^♢). □

Lemma 1

For every top-to-bottom application of an operational rule of CL, the set of states deemed impossible by the bottom-sequent is the union of the states deemed impossible by the top-sequents.

Proof

I do the case for conjunction; the proofs for negation and disjunction are analogous. For [L∧]: Note that by our semantic clauses |A ∧ B|⁺ = {s : ∃a ∈|A|⁺∃b ∈|B|⁺(s = a ⊔ b)}. Hence, for any Γ and Δ, we have $\{g\sqcup d\sqcup a\sqcup b:g\in \vert \bigwedge {\Gamma }\vert ^{+}$ and $d\in \vert \bigvee {\Delta }\vert ^{-}$ and a ∈|A|⁺ and b ∈|B|⁺} = $\{g\sqcup d\sqcup s:g\in \vert \bigwedge {\Gamma }\vert ^{+}$ and $d\in \vert \bigvee {\Delta }\vert ^{-}$ and s ∈|A ∧ B|⁺}.

Similarly for [R∧], note that |A ∧ B|⁻ = |A|⁻∪|B|⁻∪{s : ∃a ∈|A|⁻∃b ∈|B|⁻(s = a ⊔ b)}. Therefore, $\{g\sqcup d\sqcup s:g\in \vert \bigwedge {\Gamma }\vert ^{+}$ and $d\in \vert \bigvee {\Delta }\vert ^{-}$ and s ∈|A ∧ B|⁻} $=\{g\sqcup d\sqcup a:g\in \vert \bigwedge {\Gamma }\vert ^{+}$ and $d\in \vert \bigvee {\Delta }\vert ^{-}$ and a ∈|A|⁻} $\cup \{g\sqcup d\sqcup b:g\in \vert \bigwedge {\Gamma }\vert ^{+}$ and $d\in \vert \bigvee {\Delta }\vert ^{-}$ and b ∈|B|⁻} $\cup \{g\sqcup d\sqcup c:g\in \vert \bigwedge {\Gamma }\vert ^{+}$ and $d\in \vert \bigvee {\Delta }\vert ^{-}$ and c ∈{x : ∃a ∈|A|⁻∃b ∈|B|⁻(x = a ⊔ b)}}. □

Proposition 21

All operational rules of CL are valid for .

Proof

Immediate from Lemma 2. □

Proposition 22

If a model is a counterexample to a top-sequent of a top-to-bottom application of an operational rule of CL, then the model is also a counterexample to the bottom-sequent.

Proof

By Lemma 2, if a state deemed impossible by a top-sequent is possible in ${\mathscr{M}}$, then a state deemed impossible by the bottom sequent is possible in ${\mathscr{M}}$. □

Proposition 23

If we impose Exclusivity, Exhaustivity, and Downward-Closure, then iff iff .

Proof

We show that is sound and complete with respect to and to . By Proposition 19, CL is sound and complete with respect to . For , we know soundness from Propositions 20 and 21. For completeness, suppose that there is no proof of . Hence, a proof-search for yields an atomic sequent, , where Γ₀ ∩Δ₀ = ∅. Let ${\mathscr{M}}$ be a model in which s ∈ S^♢ and s = u ⊔ t such that and . This is a counterexample to . By Proposition 22, it is also a counterexample to . Since Γ₀ ∩Δ₀ = ∅, such a model isn’t ruled out by Exclusivity, which is the only principle that could rule out such a model. So, ${\mathscr{M}}$ shows that . □

1.2 A.2 Relation of C L ^∖[cut] to S T, L P, K 3, and T S

I will use a slightly adjusted version of the propositional fragment of Ripley’s [7] sequent calculus presentation of ST, namely the following:

We write iff the sequent is derivable in ST. Ripley [7] uses single-line rules and an additive right-rule for disjunction, and he includes the material conditional. Given [weakening-ST], the definability of the conditional as $A\supset B=_{\textrm {def.}}\neg A\lor B$, and the admissibility of the bottom-to-top rules in the single-line ST calculus, these differences don’t change which sequents are provable.¹ Ripley treats ∧ as defined in the usual way, i.e., A ∧ B =_def.¬(¬A ∨¬B). Using this definition, we can show:

Proposition 11

.

Proof

We can transform any proof-tree for into a proof-tree for and vice versa. Left-to-right: Applications of [ID] in CL^+∖[cut] can be translated into ST by [ID-ST] followed by [weakening-ST]. Applications of [weakening] are merely relabeled as applications of [weakening-ST]. Similarly for applications of the negation rules, truth rules, and [R∨] (all in both directions). Top-to-bottom applications of [L∨] are translated into applications of [L∨-ST], leaving out the third top-sequent. Bottom-to-top applications of [L∨] are translated into similar applications of [L∨-ST] with an application of [weakening-ST] if the desired sequent is the third top-sequent. Top-to-bottom applications of [L∧] are translated by putting negations of both subaltern sentences on the right by [R¬-ST], then using [R∨-ST] to get their disjunction on the right, and finally using [L¬-ST] to get the negation of the disjunction on the left. This negated disjunction is, by definition, the same as the desired conjunction. Bottom-to-top applications of [L∧] are translated by the same route in reverse. Top-to-bottom applications of [R∧] are translated by applying [L¬-ST] to the first two top-sequents, omitting the third, then disjoining the resulting negations on the left via [L∨-ST], and finally putting the negated disjunction on the right via [R¬-ST]. This negated disjunction is, by definition, the same as the desired conjunction. Bottom-to-top applications of [R∧] are translated by the same route in reverse with an addition of [weakening-ST] if the desired sequent is the third top-sequent.

Right-to-left: Applications of [ID-ST], the ST rules for negation, truth, weakening, and [R∨-ST] are translated by merely relabeling them appropriately. This leaves only [L∨-ST]. Top-to-bottom applications are translated by [weakening] to get the required additional top-sequent followed by [L∨]. Bottom-to-top applications can be merely relabeled. □

Proposition 25

If we add a truth-predicate to our language, the clauses (tr + ) and (tr−) to our semantics, and a sentence λ = $\neg Tr(\bar {\lambda )}$, then is trivial, i.e., .

Proof

The clauses for truth imply that a state that verifies the liar sentence also falsifies it, and vice versa. For suppose . Since λ = $\neg Tr(\bar {\lambda )}$, it follows that . By (neg + ), . And by (tr−), . The same reasoning works in reverse. Now, let s be an arbitrary verifier of λ. Hence, and . By Exclusivity, ∀u(s ⊔ s ⊔ u∉S^♢). So, ∀u(s ⊔ u∉S^♢). Now suppose for reductio that and let u be a fusion of verifiers for each element in Γ and falsifiers for each element in Δ. By truth-maker bilateralism, u ∈ S^♢. By Exhaustivity, there is either a verifier or a falsifier of λ that we can fuse with u into a possible state. Without loss of generality, let that state be s. Hence, u ⊔ s ∈ S^♢, which contradicts the earlier result. So, by reductio, . □

In order to prove the completeness of CL^+∖[cut] with respect to , we cannot use the technique of proof-searches from above because proof-searches are no longer guaranteed to terminate. Hence, I follow Ripley [7, 162-63] in using the technique of (possibly infinite) reduction trees from Takeuti [41].²

Definition 14 (Reduction tree)

The reduction tree for a sequent, , is the possibly infinite tree that results from starting with as the root of the tree and then extending at each stage each top-most sequent of the tree as follows, until all branches are closed or else extending the tree ω-many times: (i) If the sequent is an axiom, i.e., is such that the left and the right side share an atomic sentence, then the branch remains unchanged and is closed. (ii) If the sequent has the form or and no reduction has been applied to ¬A in previous stages, they reduce to and respectively. (iii) If the sequent has the form and no reduction has been applied to A ∧ B in previous stages, it reduces to ; and if it has the form and no reduction has been applied to A ∧ B in previous stages, the reduction tree branches into and and . (iv) Similarly, reduces to ; and reduces to and and . (v) reduces to ; and reduces to .

Lemma 3

The set of states deemed impossible by a sequent is the union of the states deemed impossible by the sequents to which it reduces in a reduction tree.

Proof

We look at each clause in the reduction procedure. The lemma holds trivially for clause (i). It holds for (ii) because the truth-makers of ¬A are exactly the falsity-makers of A, and vice versa. The other cases, in particular those for (v), are analogous except for when the reduction tree branches out, such as in the case of . In this case, the lemma holds because the falsity-makers of A ∧ B are the union of the falsity-makers of A, the falsity-makers of B, and any fusion of such falsity-makers, which corresponds to the three sequents that result from the reduction. □

Definition 15 (Sequents resulting from an open branch of a reduction tree)

If an open branch of a reduction tree terminates, the resulting sequent is the leaf of that branch. If the open branch does not terminate, then the resulting sequent is the sequent , where Γ_ω is the union of all the sets on the left side of sequents in this open branch and and Δ_ω is the union of the sets on the right side of sequents in the branch.

Lemma 4

Let be a sequent resulting from an open branch, let Γ^at be the set of atomic sentences in Γ, and let Δ^at be the set of atomic sentences in Δ. Then a state that is deemed impossible by includes as a part a truth-maker for every sentence in Γ and a falsity-maker for every sentence in Δ.

Proof

We argue by induction on the complexity of sentences in Γ ∪Δ. The states deemed impossible by trivially include truth-makers for every atomic sentence in Γ and falsity-makers for every atomic sentence in Δ. Suppose our lemma holds for sentence up to complexity n, and let’s consider sentences of complexity n + 1. Note that since we have an open branch, we know that all possible reduction procedures have been applied. For negations in Γ, we know that the negatum, which is of complexity n, is in Δ. So, by our induction hypothesis states deemed impossible by include a falsity-maker for the negatum, which is a truth-maker for our negation. Similarly for all other connectives where the reduction tree does not branch. For disjunctions on the left, we know that Γ contains also one or both of the disjunctions, which are of complexity n. So by our hypothesis, contains truth-makers for one or both disjuncts, and any of these options ensures that it includes a truth-maker for the disjunction. Similarly for conjunctions on the right. □

Proposition 26

iff .

Proof

Left-to-right: We leave out [cut] and Exhaustivity in Propositions 20, and the proof still shows the validity of [ID] and [weakening]. Since the operational rules haven’t changed, the validity proof for the operational rules from Proposition 21 still applies. Clauses (tr + ) and (tr−) ensure that [Lt] and [Rt] are valid for .

Right-to-left: Suppose that there is no proof of . Hence, a reduction tree for has an open branch. Let be the sequent that results from that branch, and let be the sequent that results from by omitting all complex sentences. We can use as our desired counter-model any model that makes possible one of the states deemed impossible by . For by Lemma 4, any state that is deemed impossible by is also deemed impossible by . And by Lemma 3, any state deemed impossible by is deemed impossible by . So our model is a counter-model to . We know that there is such a model because any model will work that makes only those states impossible that are required to be impossible by Exclusivity, and ${\Gamma }_{\omega }^{at}\cap {\Delta }_{\omega }^{at}=\emptyset$. □

Note that the models that make possible all or some of the states deemed impossible by can respect our semantic clauses for the transparent truth-predicate, even though the atomic sentences in contain truth-predicates applied to complex sentences. For if, e.g., $Tr(\overline {A\wedge B})\in {\Gamma }_{\omega }^{at}$, then A ∧ B ∈Γ_ω. So the atomic part of the reduction of A ∧ B is also in ${\Gamma }_{\omega }^{at}$. And if A or B include truth-predications, this will apply to those again.

For the relation to LP, we will use the following known fact [24, 25]:

Fact 2

iff is provable if we add for all γ ∈Γ as axioms to the sequent calculus for ST.

Proposition 27

iff is provable if we add for all γ ∈Γ as axioms to CL^∖[cut].

Proof

We have seen in the proof for Proposition 24 that we can transform every ST proof-tree into a CL^+∖[cut] proof-tree and vice versa. □

Proposition 28

iff no fusion of falsifiers of everything in Δ is possible if no falsifier of anything in Γ is possible, i.e., $\forall d\in \vert \bigvee {\Delta }\vert ^{-}\thinspace (d\not \in S^{\Diamond })$ in all models in which ∀γ ∈Γ (∀g ∈|γ|⁻(g∉S^♢)).

Proof

Left-to-right: Suppose that . By Proposition 27, is provable if we add for all γ ∈Γ as axioms to CL^+∖[cut]. To add for all γ ∈Γ as axioms to CL^+∖[cut] corresponds in our semantics to the elimination of all models in which any falsifier of any member of Γ is a possible state. Thus, we only look at models in which ∀γ ∈Γ (∀g ∈|γ|⁻(g∉S^♢)). Since is provable and, by Proposition 26, our sequent rules are sound for , it follows that in all such models the fusion of any falsifiers of the members of Δ is impossible, i.e., $\forall d\in \vert \bigvee {\Delta }\vert ^{-}\thinspace (d\not \in S^{\Diamond })$.

Right-to-left: Suppose that $\forall d\in \vert \bigvee {\Delta }\vert ^{-}\thinspace (d\not \in S^{\Diamond })$ in all models in which ∀γ ∈Γ (∀g ∈|γ|⁻(g∉S^♢)). By the completeness of CL^+∖[cut] for , this means that is provable if we add for all γ ∈Γ as axioms to CL^+∖[cut]. So, by Proposition 27, . □

Proposition 28

iff is provable if we add for all δ ∈Δ as axioms to CL^+∖[cut].

Proof

Suppose that . By the duality of K3 and LP, it follows that ¬Δ⊧_LP¬Γ, where ¬X is the set of the negations of the members of X. By Proposition 27, it follows that is provable if we add for all δ ∈Δ as an axiom to CL^+∖[cut]. Given the negation rules of CL^+∖[cut] this means that is provable if we add for all δ ∈Δ as an axiom to CL^+∖[cut]. For the other direction, the same reasoning works in reverse. □

Proposition 29

iff $\forall g\in \vert \bigwedge {\Gamma }\vert ^{+}\thinspace (g\not \in S^{\Diamond })$ in all models in which ∀δ ∈Δ (∀d ∈|δ|⁺(d∉S^♢)).

Proof

The proof is analogous to the one for LP above. □

Proposition 30

iff iff .

Proof

It is well known that is empty. No sequent is derivable in CL^∖[ID] because there is no way to start proofs without [ID]. And is empty because we can let all states be possible. □

Proposition 31

Let $\mathfrak {I}$ be a set of pairs, $\left \langle {\Gamma },{\Delta }\right \rangle$, where Γ and Δ are sets of sentences. Regardless of whether we close $\mathfrak {I}$ under the rules of CL^∖[ID], of TS, or of , the resulting consequence relation is the same.

Proof

Since the correspondences are familiar by now, I give merely some quick pointers. As we have seen in the proofs of Lemma 2 and 24 the rules governing the logical connectives in all three systems are equivalent, given [weakening]. Exhaustivity in corresponds to [cut] in CL^∖[ID] and TS, and Downward-Closure corresponds to [weakening] (see Proposition 20 above). □

1.3 A.3 Relation of C L to Correia’s Logic, Containment, and Entailment

Proposition 32

The leaves of a proof-tree that result from a proof-search on are such that the union of the states that they deem impossible is exactly the set of truth-makers of A.

Proof

A tree that results from a proof-search uses only top-to-bottom applications of operational rules. By Lemma 2, the union of the states deemed impossible by the top-sequents of such a rule-application is the set of states deemed impossible by the bottom sequent. Hence, for any proof-tree that results from a proof-search, the union of the states deemed impossible by the leaves of the tree is the set of states deemed impossible by the root. The set of states deemed impossible by are exactly the truth-makers of A. □

As Correia [29] shows, the following fact holds:

Fact 3

$A\thickapprox B$ is a theorem of Correia’s logic iff A and B have the same truth-makers in every truth-maker model.

Proposition 21

$A\thickapprox B$ iff proof-searches on and yield the same result.

Proof

$A\thickapprox B$ holds iff A and B have the same truth-makers in all models. By Proposition 33, this holds just in case proof-searches in CL on and yield the same result. □

Since proof-searches use only the operational rules of CL, this last result immediately implies our target:

Proposition 33

$A\thickapprox B$ is a theorem of Correia’s logic iff the operational rules of CL suffice to show that the moves from to and vice versa are admissible.

Fact 3

As Correia [29, 117] shows, $A\thickapprox B$ is provable in his dual system iff $\neg A\thickapprox \neg B$ is provable in his original system.

Proposition 34

$A\thickapprox B$ is a theorem of Correia’s dual logic iff the operational rules of CL suffice to show that the moves from to and vice versa are admissible iff A and B have the same falsity-makers in all models.

Proof

Suppose that $A\thickapprox B$ is a theorem of Correia’s dual logic. By Fact 4 and Proposition 35 and the negation rules of CL, this happens iff the moves from to and vice versa are admissible. And this happens just in case proof-searches for and yield the same atomic sequents. By reasoning that is parallel to the proof of Proposition 20, this happens iff A and B have the same falsity-makers in all models. □

Proposition 35

A contains B in virtue of logical form iff, for some Θ, the operational rules of CL suffice to show that the moves from to and vice versa are admissible.

Proof

Left-to-right: Suppose that A contains B. By definition, there is a proposition R such that |A|⁺ = {b ⊔ r : b ∈|B|⁺ and r ∈|R|⁺}. In accordance with Assumption 1, let Θ be a set of sentences such that the set of fusions of verifiers for each of the elements is |R|⁺. Hence, A and $B\wedge \bigwedge {\Theta }$ have the same verifiers. Therefore, by Proposition 33, a proof-search on and on yield the same result. This ensures that the moves from to and vice versa are admissible.

Right-to-left: Suppose the moves from to and vice versa are admissible. This happens only if proof-searches on and on yield the same result. By Proposition 33, it follows that {t ⊔ p : $t\in \vert \bigwedge {\Theta }\vert ^{+}$ and b ∈|B|⁺} = |A|⁺. So every verifier of A includes a verifier of B as a part, and every verifier of B is included as a part in a verifier of A. Therefore, A contains B in virtue of logical form. □

Proposition 36

A entails B in virtue of logical form iff the operational rules of CL suffice to show that the move from to is admissible.

Proof

Left-to-right: Suppose that A entails B in virtue of logical form and, hence, in all models. Then the verifiers of A are a subset of the verifiers of B. So, by Proposition 33, the union of the states deemed impossible by the leaves of the proof-tree for is a subset of the union of the states deemed impossible by the leaves of the proof-tree for . And this holds in all models. Suppose for reductio that there is a leaf, , in the proof-tree for that is not also a leaf in the tree for . We take a model in which s is the unique state that is deemed impossible by , and we ensure that s is not deemed impossible by any of the leaves in the tree for . Then s is a verifier of A but not of B, contradicting our assumption that A entails B. But if the leaves of the proof-tree for is a subset of those for , then the move from to is admissible.

Right-to-left: Suppose that the move from to is admissible. Then proof-searches on and on yield proof-trees such that the leaves of the tree for is a subset of the leaves of the tree for . This must hold in virtue of logical form. The union of the states deemed impossible by the leaves of the tree for is a subset of the corresponding states for . By Proposition 33, every verifier of A is a verifier of B. So, A entails B in virtue of logical form. □