That’s It! Hyperintensional Total Logic

Abstract

Call a truth complete with respect to a subject matter if it entails every truth about that subject matter. One attractive way to formulate a complete truth is to state all the relevant positive truths, and then add: and that’s it. When the subject matters under consideration are non-contingent, a non-trivial conception of completeness must invoke a hyperintensional conception of entailment, and of the completion operation denoted by ‘that’s it’. This paper develops two complementary hyperintensional conceptions of completion using the framework of truthmaker semantics and determines the resulting logics of totality.

Appendices

Appendix A: Soundness

Theorem 4

Let ${\Phi } \subseteq {\mathscr{L}}_{\Delta }$ and $\varphi \in {\mathscr{L}}_{\Delta }$. Then Φ ⊩_Cφ implies Φ⊧_Cφ.

Proof

The soundness of classical propositional logic is easily established using the fact that @ is a world, and that any formula of ${\mathscr{L}}_{\Delta }$ is assigned an exclusive and exhaustive proposition. It remains to show that our additional axioms are true in every model.

The soundness of Δ-EquivC follows from the soundness for ⊩_eC for the interpretation of A ≈ B as true iff the sets of verifiers of A and B have the same closure under (non-empty) fusions and the fact that the truth-value of ΔA and ΔB depends only on the closure under fusion of the verifier-sets of A and B.

For Δ-Fact and Δ-PFix, assume ΔA is true in a model, so some part s of @ is the sole verifier of ΔA. By the clause for Δ, some δ-state is part of s. But the only δ-state contained in @ is @, so @ is the sole verifier of ΔA. By the clause for Δ again, A has a verifier and every verifier of A is part of @, so A is true in the model, and @ is also the sole verifier of ΔΔA, so ΔΔA is also true in the model.

For Δ-DisjC, note that Δ(A ∨ B) is true in a model iff @ is the sole verifier of Δ(A ∨ B), which holds iff @ is the fusion of all δs where s verifies A ∨ B. This is so just in case @ = δs for all s verifying A ∨ B and hence iff @ = δs for all s verifying A and for all s verifying B, so iff @ is the sole verifier of both ΔA and ΔB, and hence of their conjunction, so iff ΔA ∧ΔB is true in the model.

For Δ-AbsP, suppose ΔA ∧Δ(B ∧ C) is true in a model and hence verified solely by @. Then the same holds for both ΔA and Δ(B ∧ C). It follows that every verifier of A, and every verifier of B ∧ C has @ as its completion, and so that every verifier of B is part of @. But then using δ-Identity(1), we may infer that every verifier of A ∧ B has @ as its completion, and thus that Δ(A ∧ B) is verified by @ and hence true in our model.

For Δ-NE1, suppose Δ(¬ΔA ∧ B) is true in a model. Then ¬ΔA is verified only by parts of @. Since it is verified by any incompletion of any verifier of A, @ must be the only incompletion of a verifier of A. So every δ-state other than @ must be the completion of every verifier of A, so there can only be two δ-states. Since $\blacksquare$ is a δ-state distinct from @, it follows that @ and $\blacksquare$ are the only δ-states. By δ-Identity(1), δ□ cannot be $\blacksquare$, and so must be @. Then Δ⊥ is falsified by the one incompletion of $\blacksquare$, which will be @, and by □, the falsifier of ⊥. Both have @ as their completion, so Δ¬Δ⊥ is verified by @, and hence true in the model.¹

For Δ-NE2, suppose Δ¬Δ⊥ is true in a model. By the same reasoning as before, the model can only have two δ-states, with $@ = \delta \Box$. Now suppose Δ¬A is true in the model, so every falsifier of A is part of @. Then every verifier of A has $\blacksquare$ as its completion, so @ is the only incompletion of any verifier of A. It follows that ¬ΔA is verified only by parts of @, and thus by states that have @ as their completion, so Δ¬ΔA is true in the model. Conversely, suppose Δ¬ΔA is true in the model. Then all verifiers of ¬ΔA, and hence all falsifiers of A, must be parts of @. But then they all have @ as their completion, so Δ¬A is true in the model. □

Theorem 5

Let ${\Phi } \subseteq {\mathscr{L}}_{\Delta }$ and $\varphi \in {\mathscr{L}}_{\Delta }$. Then Φ ⊩_Dφ implies Φ⊧_Dφ.

Proof

The argument for the soundness of classical propositional logic remains the same. The soundness of Δ-EquivD is immediate from soundness for ⊩_eD for the interpretation of A ≈ B as true iff the sets of verifiers of A and B are the same and the fact that the truth-value of ΔA and ΔB depends only on the verifiers of A and B. Δ-DisjD is immediate by the clause for Δ. Δ-Fact and Δ-PFix are immediate given the fact that always δs ⊒ s and δδs = δs.

For Δ-NFix, suppose ¬ΔA is true, and so verified by some part s of @. Since falsifiers of ΔA are all δ-states, s = @. Since @ is a world, δ@ = @, and so @ verifies Δ¬ΔA.²

For Δ-AbsT, suppose part s of @ verifies ΔA ∧ B. Then s = δt ⊔ u for some t and u verifying A and B, respectively. Since @ is consistent, s is consistent, and so δt is consistent, and therefore a world. And since $\delta t \sqsubseteq @$, it follows that δt = @, und $u \sqsubseteq @$. Then also $t \sqsubseteq @$ and hence $t \sqcup u \sqsubseteq @$. So $t \sqsubseteq t \sqcup u \sqsubseteq @ = \delta t$, and hence δ(t ⊔ u) = @. □

Appendix B: Normal Forms

We establish completeness via normal forms. For each logic, we prove a reduction theorem to the effect that any formula of ${\mathscr{L}}_{\Delta }$ is provably equivalent to a formula in a kind of disjunctive normal form.

For these purposes, we count as atoms the members of At, ⊤, as well as—somewhat unusually—the formulas Δ⊤ and Δ⊥. Any atom and any negation of an atom is a literal; the members of At and their negations are the non-logical literals and their set is denoted by Lit. We use the variables φ,ψ,χ as well as variously adorned versions of them to range over the literals of Δ. The converse $\varphi ^{\prime }$ of a literal φ is ¬φ if is an atom, and ψ if φ = ¬ψ. Formulas of the form Δ(₁ ∧… ∧ φ_n) and their negations will be called Δ-literals. A Δ − DNF is any disjunction of conjunctions in which each conjunct is either a literal or a Δ-literal.

We will show that under either logic, every formula of ${\mathscr{L}}_{\Delta }$ is provably equivalent to a Δ-DNF. To do this, we classify formulas of ${\mathscr{L}}_{\Delta }$ in accordance with their maximum nesting depth of Δ. For each n, we define a set of n-atoms, n-literals, and n-formulas. The 0-atoms are the atoms, the 0-literals are the 0-atoms and their negations, and the 0-formulas are all the formulas built up from 0-atoms without the use of Δ. The (n + 1)-atoms are the n-atoms as well as all expressions of the form ΔA, where A is an n-formula. The (n + 1)-literals are the n + 1-atoms and their negations, and the (n + 1)-formulas are all the formulas built up from (n + 1)-atoms without the use of Δ. A conjunction of n-literals will be called an n-CF, and a disjunction of n-CFs will be called an n-DNF. The degree of a formula A is the smallest n for which A is an n-formula.

1.1 B.1 Conjunctive Δ

Throughout this subsection, reference to provability is to be understood in terms of ⊩_C, and reference to provable exact equivalence in terms of ⊩_eC. We first establish some useful facts in preparation of the reduction theorem.

Lemma 1

• 1. Any n-formula is provably exactly equivalent to an n-DNF.

• 2. For A an n-formula, ΔA is provably equivalent to a formula ${\Delta } A^{\prime }$, where $A^{\prime }$ is an n-DNF.

• 3. Δ(ΔA ∧ B) is provably equivalent to ΔA ∧Δ(A ∧ B)

• 4. Δ(¬ΔA ∧ B) is provably equivalent to Δ¬Δ⊥∧Δ¬A ∧ΔB.

• 5. If A is an n-CF of degree n > 0, then ΔA is provably equivalent to some n-formula.

• 6. For n > 0, if A is a formula of degree n, then ΔA is provably equivalent to some n-formula.

Proof

(1): This is a straightforward consequence of Theorem 15 in [3, p. 214].

(2): By (1) and Δ-EquivC.

(3): By Δ-Fact, Δ-PFix, and Δ-AbsP.

(4): Left-to-right: By Δ-NE1, from Δ(¬ΔA ∧ B) we infer Δ¬Δ⊥, from which we obtain, using Δ-NE1, Δ⊤. Using ⊩_eCA ≈ A ∧⊤ and Δ-AbsP, we can derive every instance of ${\Delta } \top \rightarrow ({\Delta } (A \wedge B) \rightarrow {\Delta } A))$ which allows us to infer both Δ¬ΔA and ΔB from Δ(¬ΔA ∧ B). Using Δ-NE2 again, we may infer Δ¬A. Right-to-left: Using Δ-NE2, we may infer Δ¬ΔA from Δ¬Δ⊥ and Δ¬A. Using Δ-AbsP and ΔB, we obtain Δ(¬ΔA ∧ B).

(5): Suppose A is an n-CF of degree n > 0. So A is a conjunction of n-literals A₁,…,A_m, at least one of which is of degree n. Let k be the number of degree n literals among A₁,…,A_m. We may then use an ‘extraction’ procedure to obtain a formula provably equivalent to ΔA, which is a conjunction of an n-formula and a sentence ${\Delta } A^{\prime }$ with $A^{\prime }$ an n-CF with k − 1 degree n literals among its conjuncts. So by repeated application of that procedure, we obtain a sentence provably equivalent to ΔA which is a conjunction of n-formulas, and hence itself an n-formula.

The extraction procedure is as follows. First, we pick one of the degree n literals among A₁,…,A_m, call it B₁, and turn A into a provably exactly equivalent n-CF of degree nB = B₁ ∧… ∧ B_m. Then ΔA is provably equivalent to ΔB. B₁ is either of the form ΔC or of the form ¬ΔC. In the former case, ΔB is Δ(ΔC ∧ B₂ ∧… ∧ B_m), and hence by (3) provably equivalent to ΔC ∧Δ(C ∧ B₂ ∧… ∧ B_m). Since C is an (n − 1)-formula, ΔC is an n-formula and Δ(C ∧ B₂ ∧… ∧ B_m) has only k − 1 degree n literals, as desired. In the latter case, ΔB is Δ(¬ΔC ∧ B₂ ∧… ∧ B_m), and hence by (4) provably equivalent to Δ¬Δ⊥∧Δ¬C ∧Δ(B₂ ∧… ∧ B_m). Since C is an (n − 1)-formula, Δ¬Δ⊥∧Δ□C is an n-formula, and Δ(B₂ ∧… ∧ B_m) has only k − 1 degree n literals, as desired.

(5): By (2), ΔA is provably equivalent to ΔB for some n-DNF B = B₁ ∨… ∨ B_k. Then by Δ-DisjC, ΔA is provably equivalent to ΔB₁ ∧… ∧ΔB_k. Now consider each conjunct ΔB_i. If it is of degree < n, we do nothing. If it is of degree n, then by (3) it may be replaced by a provably equivalent n-formula. In this way we obtain an n-formula provably equivalent to ΔA, as desired. □

Theorem 6

Any formula is provably equivalent within ⊩_C to a Δ-DNF.

Proof

The case in which A is of degree 0 is trivial. Suppose that A is degree 1, and consider any degree 1 atom in A. This will be of the form ΔB, with B a 0-formula. By Lemma 6 (2), there is a 0-DNF C with ΔB provably equivalent to ΔC. By Δ-DisjC, ΔC in turn is provably equivalent to a conjunction of Δ-literals. So we may replace any degree 1 atom in A by a provably equivalent conjunction of Δ-literals. The result is provably equivalent to A, and, by propositional logic, provably equivalent to a Δ-DNF.

Now suppose A is degree n + 1 with n > 0, and suppose the claim holds for all formulas up to degree n (IH). Consider any degree n + 1 atom in A, which will be of the form ΔB, with B a degree n-formula. Then by Lemma 6 (6), ΔB is provably equivalent to some n-formula. We may thus replace each degree n + 1 atom in A by a provably equivalent n-formula. The result will itself be an n-formula provably equivalent to A. By IH, that formula, and therefore A, is provably equivalent to a Δ-DNF. □

1.2 B.2 Disjunctive Δ

Throughout this subsection, reference to provability is to be understood in terms of ⊩_D, and reference to provable exact equivalence in terms of ⊩_eD.

Lemma 2

• 1. Any n-formula is provably exactly equivalent to an n-DNF.

• 2. For A an n-formula, ΔA is provably equivalent to a formula ${\Delta } A^{\prime }$, where $A^{\prime }$ is an n-DNF.

• 3. Δ(ΔA ∧ B)) is provably equivalent to ΔA ∧ B.

• 4. Δ(¬ΔA ∧ B)) is provably equivalent to ¬ΔA ∧ B.

• 5. If A is an n-CF of degree n > 0, then ΔA is provably equivalent to A.

• 6. For n > 0, if A is a formula of degree n, then ΔA is provably equivalent to some n-formula.

Proof

(1) and (2) are as before. (3) and (4) may be established using Δ-Fact, Δ-AbsT, and Δ-PFix or Δ-NFix, respectively.

(5): A has at least one conjunct of degree n > 0, i.e. an n-literal of the form ΔB or ¬ΔB. In the former case, A is provably exactly equivalent to some conjunction of the form ${\Delta } B \wedge B^{\prime }$. So by Δ-EquivD, ΔA is provably equivalent to ${\Delta } ({\Delta } B \wedge B^{\prime })$, which by (3) is provably equivalent to ${\Delta } B \wedge B^{\prime }$, which is provably equivalent to A. In the latter case, A is provably exactly equivalent to some conjunction of the form $\neg {\Delta } B \wedge B^{\prime }$. So by Δ-EquivD, ΔA is provably equivalent to ${\Delta } (\neg {\Delta } B \wedge B^{\prime })$, which by (4) is provably equivalent to $\neg {\Delta } B \wedge B^{\prime }$, which is provably equivalent to A.

(6): By (2), ΔA is provably equivalent to ΔB for some n-DNF B = B₁ ∨… ∨ B_k. Then by Δ-DisjD, ΔA is provably equivalent to ΔB₁ ∨… ∨ΔB_k. Now consider each disjunct ΔB_i. If it is of degree < n, we do nothing. If it is of degree n, then we replace it by B_i. Since B_i is an n-CF, by (5), ΔB_i is provably equivalent to B_i. So through these replacements we obtain an n-formula provably equivalent to ΔA, as desired. □

Theorem 7

Any formula is provably equivalent within ⊩_D to a Δ-DNF.

Proof

The reasoning is as for Theorem 7, using Δ-DisjD in place of Δ-DisjC and Lemma 8 (6) in place of Lemma 6 (6). □

Appendix C: Completeness

Since our logics extend classical propositional logic, strong completeness—that Φ ⊩ φ whenever Φ⊧φ—follows from the claim that every consistent subset of ${\mathscr{L}}_{\Delta }$ has a model. As usual, any consistent subset of ${\mathscr{L}}_{\Delta }$ can be extended to a maximal consistent subset of ${\mathscr{L}}_{\Delta }$, so it suffices to show for each logic that every maximal consistent (within that logic) subset of ${\mathscr{L}}_{\Delta }$ has a model. So in this section we show how to construct appropriate canonical models. We start with the strong, conjunctive interpretation of Δ, since the construction is somewhat simpler in this case.

1.1 C.1 Conjunctive Δ

Let Φ be a subset of ${\mathscr{L}}_{\Delta }$ that is maximal consistent with respect to ⊩_C. We begin by defining some important notions in terms of Φ.

Definition 1

Let φ ∈ Lit and A a 0-CF in ${\mathscr{L}}_{\Delta }$. Then

• A is complete iff ΔA ∈Φ

• A is minimally complete iff A is complete, and the result of replacing any conjunct of A by ⊤ is not complete

• m(A) is the set of non-logical literals occurring as a conjunct in A

• φ is actual iff φ ∈Φ

• φ is pure iff φ is a conjunct in some complete 0-CF

• if φ is actual: φ is positive iff either there is no complete 0-CF, or φ is a conjunct in some minimally complete 0-CF; φ is negative otherwise

• if φ is non-actual: φ is positive iff its actual converse $\varphi ^{\prime }$ is negative; φ is negative otherwise

• nφ is the negative member of $\{\varphi , \varphi ^{\prime }\}$

We write @ for the set of actual non-logical literals and @^p for the set of positive members of @. Note that @^p is empty just in case Δ⊤∈Φ, in which case there is a complete 0-CF, but no minimally complete 0-CF. We denote the set of all (non-logical)³ literals by P (since we now regard them as proto-states). For $s \subseteq P$, we say that s is minimally complete iff some conjunction of all its members is, and we let f(s) be s if no subset of s is minimally complete, and s ∪ @^p otherwise. Let S—the set of states—be $\{f(s): s \subseteq P\}$. (The point of excluding from the set of states those sets of literals which have a minimally complete subset but do not include @^p is that this will make it easy to ensure that all minimally complete 0-CFs are verified by @^p.)

Definition 2

The canonical state-space for Φ is $(S, \sqsubseteq )$ with $\sqsubseteq = \subseteq \restriction S$

In preparation of the proof that $(S, \sqsubseteq )$ is indeed a state-space, we note some useful facts about f and S.

Lemma 3

Let $s, t \subseteq P$.

• 1. @ ∈ S

• 2. @^p ∈ S

• 3. $f(s) \subseteq @^{p}$ if $s \subseteq @^{p}$

• 4. $f(s) \subseteq @$ if $s \subseteq @$

• 5. $S = \{s \subseteq P$: $@^{p} \subseteq s$ or s has no minimally complete subset}

• 6. S is closed under intersection

• 7. If s≠t are minimally complete and s has at least two members, then S is not closed under union

• 8. $f(s) \subseteq f(t)$ if $s \subseteq f(t)$

• 9. For any $T \subseteq S$, $f\bigcup T$ is $\bigsqcup T$, the least upper bound of T in S

• 10. If $T \subseteq \text {wp}(P)$, then $\bigsqcup \{f(t): t \in T\} = f\bigcup T$

Proof

(1) and (2) are obvious from the definition of f.

(3) and (4) are immediate from the fact that f(s) extends s at most by members of @^p, and $@^{p} \subseteq @$.

(5): It is immediate from the definition of f that f(s) always includes @^p or has no minimally complete subset, and that $s \subseteq P$ is equal to s and hence a member of S whenever s includes @^p or has no minimally complete subset.

(6): Let $T \subseteq S$. Either some member of T has no minimally complete subsets, in which case the same is true of $\bigcap T$, or all members of t have @^p as a subset, in which case again the same is true of $\bigcap T$.

(7): Let φ_i ∈ s. Then {φ_i} and s ∖ {φ_i} have no minimally complete subset, but their union s does. Since t is also minimally complete and distinct from s, $s \not \supseteq @^{p}$. So {φ_i} and s ∖ {φ_i} are members of S but their union s is not.

(8): Suppose $s \subseteq f(t)$. If s has no minimally complete subset, then f(s) = s, so $f(s) \subseteq f(t)$. So suppose that s does have a minimally complete subset. Then so does f(t), and hence $@^{p} \subseteq f(t)$. It follows that $f(s) = s \cup @^{p} \subseteq f(t)$.

(9): Let $T \subseteq S$. $f\bigcup T$ is clearly a member of S and an upper bound of T. So let f(x) ∈ S be any upper bound of T in S. Then $\bigcup T \subseteq f(x)$, and so by (8), $f\bigcup T \subseteq f(x)$.

(10): By (9), $\bigsqcup \{f(t): t \in T\}$ is $f \bigcup \{f(t): t \in T\}$. Suppose first that no t ∈ T has a minimally complete subset. Then f(t) = t for all t ∈ T, and hence $\bigcup \{f(t): t \in T\} = \bigcup T$, so $f\bigcup \{f(t): t \in T\} = f\bigcup T$. Suppose now that at least one t ∈ T has a minimally complete subset. Then $f\bigcup \{f(t): t \in T\} = f(\bigcup T \cup @^{p}) = \bigcup T \cup @^{p}$. But then also $\bigcup T$ has a minimally complete subset, so also $f\bigcup T = \bigcup T \cup @^{p}$. □

Lemma 4

The canonical state-space for Φ, $(S, \sqsubseteq )$, is a state-space.

Proof

From the fact that the subset-order on any set is a partial order together with Lemma 10 (9). □

Next, say that a member s of S is consistent iff there is no literal φ such that $\{\varphi , \varphi ^{\prime }\} \subseteq s$ and let $S^{\lozenge } = \{s \in S: s$ is consistent}. Say that s ∈ S decides a given literal φ iff either φ or $\varphi ^{\prime }$ is a member of s, and let δs = s ∪{nφ : φ ∈ P and s does not decide φ}.

Lemma 5

$(S, S^{\lozenge }, \sqsubseteq , \delta )$ is a C-space

Proof

It is clear that $S^{\lozenge }$ is non-empty and closed under part. Next, we show that δs is a world if s is consistent. So suppose s is consistent. Then clearly δs is also consistent. Now let t be any state not contained in δs. Let φ be a literal which is in t but not in δs. Suppose for reductio that $\varphi ^{\prime } \notin \delta s$. Then s does not decide φ, and so nφ ∈ δs. But nφ is either φ or $\varphi ^{\prime }$. Contradiction. So $\varphi ^{\prime } \in \delta s$ and hence δs is incompatible with t. It follows that δs is a world. Moreover, δ-Containment is immediate from the definition of δs. So we may conclude that every consistent state is part of a world—and hence the conditions on a W-space are satisfied—, and that δ-Completeness holds.

δ-Redundancy(1): Suppose s is a world. Suppose for reductio that s does not decide φ. Then at least one of s ∪{φ} and $s \cup \{\varphi ^{\prime }\}$ is a state, and hence a consistent state, contrary to s being a world. So s decides every literal, and hence by definition, δs = s.

δ-Redundancy(2): Immediate from the fact that by construction, δs decides every literal.

δ-Identity(1): Suppose $s \sqsubseteq t \sqsubseteq \delta s$. Suppose that φ ∈ δt. If φ ∈ t then φ ∈ δs is immediate. If φ∉t, then φ = nφ with t not deciding φ. But then s does not decide φ, and hence φ ∈ δs. Suppose φ ∈ δs. If φ ∈ t, then φ ∈ δt by δ-Containment. If φ∉t, then φ = nφ with s not deciding φ. We need to show that t does not decide φ. Suppose otherwise. Then since φ∉t, it follows that $\varphi ^{\prime } \in t$. Then $\varphi ^{\prime } \in \delta s$. Since $\varphi ^{\prime } \neq n\varphi$, it follows that $\varphi ^{\prime } \in s$. But then s decides φ. Contradiction. So t does not decide φ, and hence φ = nφ ∈ δt. So δs = δt, as desired.

δ-Identity(2): Suppose δs = δt. Suppose φ ∈ δs. Suppose first that φ ∈ s. If also φ ∈ t, then φ ∈ s ⊓ t and so φ ∈ δ(s ⊓ t). If φ∉t, then φ = nφ and t does not decide φ. But then s ⊓ t also does not decide φ, and so again φ ∈ δ(s ⊓ t). Suppose φ ∈ δ(s ⊓ t). Clearly if φ ∈ s ⊓ t, then φ ∈ δs. If φ∉s ⊓ t, then φ = nφ and s ⊓ t does not decide φ. If s also does not decide φ, then φ ∈ δs. If s decides φ by having φ as member, then again φ ∈ δs. So suppose s decides φ by having $\varphi ^{\prime }$ as a member. Then $\varphi ^{\prime } \in \delta s$ and hence $\varphi ^{\prime } \in \delta t$. Since φ = nφ, it follows that $\varphi ^{\prime } \in t$, and hence that $\varphi ^{\prime } \in s \sqcap t$. But then s ⊓ t decides φ. Contradiction. So s does not have $\varphi ^{\prime }$ as a member, and hence φ ∈ δs. □

It is straightforward to verify that @ is a world in this C-space, that δs = @ iff $@^{p} \sqsubseteq s \sqsubseteq @$ and that @^p is the positive part of @ under the definition proposed in Section ??. Moreover, a state is wholly positive under the definition from that section just in case it does not contain nφ for any φ, so the set of wholly positive states is closed under fusion, as one would expect.

We now define the interpretation function for the atoms; it is straightforward that it satisfies the conditions of exclusivity and exhaustivity.

Definition 3

For φ ∈ At, let

• [φ]⁺ = {f{φ}} if φ is pure and $\{f\{\varphi \}, \blacksquare \}$ otherwise

• [φ]⁻ = {f{¬φ}} if ¬φ is pure and $\{f\{\neg \varphi \}, \blacksquare \}$ otherwise

Lemma 6

For all φ ∈ At,

• 1. s and t are incompatible whenever s ∈ [φ]⁺ and t ∈ [φ]⁻

• 2. for every world w ∈ S, there is some s ∈ [φ]⁺ ∪ [φ]⁻ with $w \sqsupseteq s$

Definition 4

The canonical model ${\mathscr{M}}$ of Φ is $(S, S^{\lozenge }, \sqsubseteq , \delta , @, [\cdot ])$

Lemma 7

${\mathscr{M}}$ is a model.

Proof

By Lemmas 12 and 13. □

We now show that in the canonical model, the literals and Δ-literals in Φ are verified by parts of @. Note that every formula of ${\mathscr{L}}_{\Delta }$ has a verifier in the canonical model.

Lemma 8

Let ${\mathscr{M}}$ be the canonical model, φ any literal and A any 0-CF. If Δ¬Δ⊥∉Φ, then

• 1. ${\mathscr{M}} \models \varphi$ if φ ∈Φ.

• 2. ${\mathscr{M}} \models {\Delta } A$ if ΔA ∈Φ.

• 3. ${\mathscr{M}} \models \neg {\Delta } A$ if ¬ΔA ∈Φ.

Proof

(1): Let φ ∈Φ. Suppose first that φ is a non-logical literal. Then φ ∈ @. By definition of [⋅], φ is verified by f{φ}. By definition of f, either f{φ} = {φ} or f{φ} = @^p. Either way, $f\{\varphi \} \sqsubseteq @$, and so ${\mathscr{M}} \models \varphi$. Suppose now that φ is a logical literal. ⊥ and Δ⊥ are inconsistent, so there are four logical literals that may occur in Φ. (i) ⊤: $\Box$ verifies ⊤ by definition of [⋅]. (ii) ¬Δ⊥: Since $\Box$ falsifies ⊥, and every falsifier of ⊥ falsifies Δ⊥, $\Box$ verifies ¬Δ⊥. (iii) Δ⊤: By definition, @^p is empty if Δ⊤∈Φ, so then $@^{p} = \Box$ and hence Δ⊤ is verified by $@ = \delta \Box$. (iv): ¬Δ⊤: By definition, @^p is non-empty if ¬Δ⊤∈Φ, so @ is an incompletion of $\Box$ and hence verifies ¬Δ⊤.

(2): Suppose ΔA ∈Φ. Then A is complete, so every non-logical literal φ occurring as a conjunct in A is pure, and therefore is verified only by f{φ}. By Lemma 10 (10) and the clause for conjunction, the sole verifier of the conjunction of non-logical conjuncts of A is fm(A). Note that since Δ¬Δ⊥∉Φ, the only logical literals that can occur in A are ⊤ and Δ⊤. Suppose first that Δ⊤∈Φ, so @^p = □, and Δ⊤ is verified only by @. Then fm(A) = m(A). By Δ-Fact and conjunction elimination, every literal in m(A) is in Φ, so $m(A) \sqsubseteq @$. Since ⊤ is verified only by □ and Δ⊤ only by @, it follows that the sole verifier of A is part of @, and hence ΔA is verified by @. Suppose now that Δ⊤∉Φ. Then Δ⊤ also does not occur as a conjunct in A, and so the sole verifier of A is fm(A). Since A is complete but ⊤ is not complete, some subset of m(A) is minimally complete, so fm(A) = m(A) ∪ @^p. By Δ-Fact and conjunction elimination, every literal in m(A) is in Φ, so $m(A) \subseteq @$. It follows that $@^{p} \sqsubseteq fm(A) \sqsubseteq @$, and since fm(A) is the sole verifier of A, that @ verifies ΔA. So ${\mathscr{M}} \models {\Delta } A$.

(3): Suppose ¬ΔA ∈Φ. By the falsifier clause for Δ, @ will be a verifier of ¬ΔA provided that @ is not the sole verifier of ΔA. So suppose for contradiction that @ is the sole verifier of ΔA, and hence that every verifier of A is between @^p and @. It follows that no literal that occurs as a conjunct in A has a non-actual verifier. So every non-logical literal φ ∈ m(A) is pure, and hence occurs in some complete 0-CF B_φ, so that ΔB_φ ∈Φ. Any such φ is then verified only by f{φ}, so by Lemma 10 (10), the sole verifier of their conjunction is fm(A). Moreover, the only logical literals possibly occurring in A are ⊤ and Δ⊤, and the latter can occur in A only if it is in Φ, since it has non-actual verifiers otherwise.

So suppose first that Δ⊤∈Φ. Then ΔB ∈Φ whenever B is a conjunction solely of ⊤ and Δ⊤. So A must include some non-logical literals, which must then be pure. But then using the ΔB_φ for all non-logical literals φ in A, using Δ-AbsP and the assumption Δ⊤, we can derive ΔA, contrary to the fact that Φ is consistent and ¬ΔA ∈Φ.

Suppose now that Δ⊤∉Φ. Then fm(A) is the sole verifier of A, and hence $@^{p} \subseteq fm(A)$, so some subset M of m(A) is minimally complete. Then for some conjunction B with m(B) = M, ΔB ∈Φ. By Δ-Equiv and Δ-AbsP, from ΔB and ΔB_φ for each non-logical φ ∈ m(A) we can again derive ΔA, contrary to the consistency of Φ. So @ is not the sole verifier of ΔA and hence @ verifies ¬ΔA, and so ${\mathscr{M}} \models \neg {\Delta } A$. □

We now construct a very simple, ‘empty’ model $\mathcal {E}$ for the case in which Φ does include Δ¬Δ⊥. Let S_e = {0, 1} and $S^{\lozenge }_{e} = \{0\}$, @ = 0, δ_e(0) = 0 and δ_e(1) = 1, and let $\sqsubseteq _{e}$ be the partial order on S_e in which $0 \sqsubseteq _{e} 1$. It is readily verified that $(S_{e}, S^{\lozenge }_{e}, \sqsubseteq _{e}, \delta _{e})$ is a C-space. For φ ∈ At, we let $[\varphi ]_{e}^{+} = \{0\}$ if φ is pure, {1} if φ∉Φ, and {0, 1} otherwise, and $[\varphi ]_{e}^{-} = \{0\}$ if ¬φ is pure, {1} if ¬φ∉Φ, and {0, 1} otherwise. Again, exclusivity and exhaustivity are straightforward.

Lemma 9

Let $\mathcal {E}$ be the canonical empty model, φ any literal, and A any 0-CF. If Δ¬Δ⊥∈Φ, then

• 1. $\mathcal {E} \models \varphi$ if φ ∈Φ.

• 2. $\mathcal {E} \models {\Delta } A$ if ΔA ∈Φ.

• 3. $\mathcal {E} \models \neg {\Delta } A$ if ¬ΔA ∈Φ.

Proof

(1): Let φ ∈Φ. If φ is a non-logical literal, then by definition of [⋅]_e, φ is verified by 0 = @, so $\mathcal {E} \models \varphi$. If φ is a logical literal then φ is one of ⊤, Δ⊤, and ¬Δ⊥. It is readily verified that in either case, φ is again verified by 0, and indeed only by 0. So again, $\mathcal {E} \models \varphi$.

(2): Suppose ΔA ∈Φ. Then every non-logical literal in A is pure, and hence by definition of [⋅]_e verified only by 0. Every logical literal in A is in Φ, and hence by (1) also verified only by 0. So ΔA, too, is verified by 0, and so $\mathcal {E} \models {\Delta } A$.

(3): Suppose ¬ΔA ∈Φ. Then since Δ⊤∈Φ, it follows that some literal in A is not in Φ or not pure. Either way, it is verified by 1, and hence so is A. Since 0 is an incompletion of 1, it follows that ¬ΔA is verified by 0, and hence $\mathcal {E} \models \neg {\Delta } A$. □

Lemma 10

• 1. If Δ¬Δ⊥∉Φ, then ${\mathscr{M}} \models \varphi$ for all φ ∈Φ.

• 2. If Δ¬Δ⊥∈Φ, then $\mathcal {E} \models \varphi$ for all φ ∈Φ.

Proof

Suppose φ ∈Φ. By Theorem 7, φ is provably equivalent to some Δ-DNF $\varphi ^{\prime } \in {\Phi }$. Since Φ is maximal consistent, some disjunct φ^∗ of $\varphi ^{\prime }$ is also in Φ. Since $\varphi ^{\prime }$ is a Δ-DNF, φ^∗ is a conjunction each conjunct of which is a literal or a Δ-literal and also a member of Φ. By Lemmas 15 and 16, every such conjunct, and hence their conjunction φ^∗, and thus also $\varphi ^{\prime }$, is true in the relevant model. By soundness, it follows that so is φ. □

From this lemma, completeness follows straightforwardly.

Theorem 8

Let ${\Phi } \subseteq {\mathscr{L}}_{\Delta }$ and $\varphi \in {\mathscr{L}}_{\Delta }$. Then Φ⊧_Cφ implies Φ ⊩_Cφ.

1.2 C.2 Disjunctive Δ

Let Φ be a subset of ${\mathscr{L}}_{\Delta }$ that is maximal consistent with respect to ⊩_D. Again, we begin by defining some important notions in terms of Φ.

Definition 5

Let φ ∈ Lit and A a 0-CF in ${\mathscr{L}}_{\Delta }$. Then

• A is complete iff ΔA ∈Φ

• A is minimally complete iff A is complete, and the result of replacing any conjunct of A by ⊤ is not complete

• o(φ,A) is the number of times φ occurs as a conjunct in A

• p(φ) is ${\max \limits } \{o(\varphi , A): A$ is minimally complete}

With each (non-logical) literal $\varphi \in {\mathscr{L}}_{\Delta }$, we associate countably many indexed literals (short: i-literals) φ_i ($i \in \mathbb {N}$), and with each indexed literal φ_i we associate a unique shadow $\overline {\varphi _{i}}$. Roughly, the various φ_i will correspond to different ways for φ to be true, and $\overline {\varphi _{i}}$ acts as a negation of φ_i. We countenance an infinite number of ways for each literal to be true in order to prevent the sets of verifiers and falsifiers from being closed under fusion. The need for negations of (some of) these will become clear later on.

Definition 6

Let φ ∈ Lit and $i \in \mathbb {N}$. Then

• bφ = b(φ_i) is the sentence letter on which φ is based

• $\varphi ^{\prime }$ is ¬φ if φ = b and bφ otherwise

• tφ = t(_i) is the unique member of $\{\varphi , \varphi ^{\prime }\} \cap {\Phi }$

• φ_i is actual iff φ ∈Φ and either i = 0 or i < p(φ)

• We let nφ and n(φ_i) be: (tφ)₀ if Δ⊤∈Φ; (tφ)₀ if p(tφ) = 0 and p(ψ) > 0 for some ψ ∈ Lit; (tφ)0′ otherwise

• φ_i is negative iff φ_i = nφ and positive otherwise

Intuitively, we may think of tφ as the true, or obtaining state, and of nφ as the negative state w.r.t. φ. We are taking (tφ)₀ to be negative iff the actual world is empty according to Φ, or some non-logical literals occur in minimally complete 0-CFs, but tφ is not one of them. (Note that the actual world may be non-empty while no non-logical literals occur in minimally complete 0-CFs, because it would take an infinite 0-CF to exhaust the positive part of the world.) Then nφ is non-actual if the truth w.r.t. φ is positive, and actual otherwise. We denote the set of actual indexed literals by @, and the set of actual and positive (short: a-positive) ones by @^p. Note that @^p is empty iff Δ⊤∈Φ.

Next, with any 0-CF $A \in {\mathscr{L}}_{\Delta }$ including only non-logical conjuncts, we associate a unique matching set of indexed literals m(A) = {φ_i : i < o(φ,A)}. Note that m(A) = m(B) iff A and B correspond to the same multi-set of literals, i.e. A and B contain the same conjuncts, and they contain them the same number of times, though possibly in a different order. Moreover, say that a 0-CF B is included in a 0-CF A iff each literal occurring in B as a conjunct occurs at least as many times as a conjunct in A. Then $m(B) \subseteq m(A)$ implies that B is included in A.

We call a set s of i-literals minimally complete iff s = m(A) for some minimally complete conjunction of non-logical literals A. Now suppose A is such a conjunction. Suppose φ_i ∈ m(A), so φ occurs at least i + 1 times in A. Then i < p(φ), so φ_i ∈ @^p. So $m(A) \subseteq @^{p}$ whenever A is a minimally complete conjunction of non-logical conjuncts.

We construct the set of states similarly as before, but using as the set of proto-states P not the set of literals, but the set of the i-literals together with the shadows of the a-positive i-literals. Then for $s \subseteq P$, let f(s) = s if no subset of is minimally complete, and s ∪ @^p otherwise, and let S be $\{f(s): s \subseteq P\}$.

Definition 7

The canonical state-space for Φ is $(S, \sqsubseteq )$ with $\sqsubseteq = \subseteq \restriction S$

Since f is defined from P just as before, the Lemma 10 and its proof carry over unchanged.

Lemma 11

Let $s, t \subseteq P$.

• 1. @ ∈ S

• 2. @^p ∈ S

• 3. $f(s) \subseteq @^{p}$ if $s \subseteq @^{p}$

• 4. $f(s) \subseteq @$ if $s \subseteq @$

• 5. $S = \{s \subseteq P$: $@^{p} \subseteq s$ or s has no minimally complete subset}

• 6. S is closed under intersection

• 7. If s≠t are minimally complete and s has at least two members, then S is not closed under union

• 8. $f(s) \subseteq f(t)$ if $s \subseteq f(t)$

• 9. For any $T \subseteq S$, $f\bigcup T$ is $\bigsqcup T$, the least upper bound of T in S

• 10. If $T \subseteq \text {wp}(P)$, then $\bigsqcup \{f(t): t \in T\} = f\bigcup T$

Lemma 12

The canonical state-space for Φ, $(S, \sqsubseteq )$, is a state-space.

Proof

From the fact that the subset-order on any set is a partial order together with Lemma 19 (9). □

To extend $(S, \sqsubseteq )$ to a canonical C-space, we need to define a set of consistent states and the δ-function, both of which is slightly more complicated than before.

Definition 8

Let φ_i,ψ_j ∈ ILit, and set s ∈ S. Then

• φ_i and ψ_j are co-literals iff φ = ψ, or φ = ¬ψ, or ψ = ¬φ

• φ_i and ψ_j are incompatible iff they are (a) co-literals and (b) not both actual

• s is consistent iff (a) no two members of s are incompatible i-literals, and (b) no member of s is the shadow of a member of s

• $S^{\lozenge } = \{s \in S: S$ is consistent}

Next, if φ is a literal, say that a state s decides φ iff some co-literal of is a member of s. Then to form the completion of any given state, we extend it by the negative state w.r.t. any literal it does not decide, as well as the shadows of any a-positive i-literal that are not members of the state.

Definition 9

For s ∈ S, let

• c(s) = {nφ : s does not decide $\varphi \} \cup \{\overline {\varphi _{i}}: \varphi _{i} \in @^{p}\!\setminus \! s\}$

• δs = s ∪ c(s)

We take note of a useful fact about the c-function used in defining δ:

Lemma 13

For s,t ∈ S, if $s \sqsubseteq t$ then $c(t) \subseteq c(s)$

Proof

Since s decides no literal not decided by t, and contains no member of @^p not contained in t, if $s \sqsubseteq t$. □

Lemma 14

Let $s \in S^{\lozenge }$. Then δs is a world in $(S, S^{\lozenge }, \sqsubseteq , \delta )$.

Proof

δs is a state: Since nφ is always a non-positive i-literal, and no shadow is a positive i-literal, δs has a minimally complete subset iff s does, and consequently δs is a state since s is.

δs is consistent: Firstly, no shadow of any member of δs is a member of s. For only a-positive i-literals have shadows, and the only a-positive i-literals in δs are already in s. Since s is consistent, it contains no shadows of its a-positive members, and beyond the states in s, by definition δs contains only shadows of i-literals not in s. Secondly, no pair of i-literals in δs is incompatible, for only co-literals are incompatible, and by construction, any pair of co-literals in δs is already in s, which was assumed to be consistent.

δs contains every state it is compatible with: Suppose that δs is compatible with t. First, suppose $\overline {\varphi _{i}}$ is a shadow in t. Since t is compatible with δs, φ_i∉δs, and hence φ_i∉s, so $\overline {\varphi _{i}} \in \delta s$.

Now suppose φ_i is an an i-literal in t. Suppose first that s does not decide φ. Then nφ ∈ δs. If nφ≠φ_i, then nφ and φ_i are distinct co-literals, and since nφ is non-positive, they are incompatible, contrary to our assumption that δs is compatible with t. So nφ = φ_i and hence _i ∈ δs. So suppose instead s does decide φ, so some member ψ_j of s is a co-literal of φ_i. If ψ_j = φ_i then φ_i ∈ δs. If ψ_j≠φ_i, then since δs and t are compatible, φ_i and ψ_k are both a-positive. But then since t is compatible with δs, $\overline {\varphi _{i}} \notin \delta s$, so by construction of δs we may infer that φ_i ∈ s and hence φ_i ∈ δs. It follows that $t \sqsubseteq \delta s$, as desired. □

Lemma 23

For all s ∈ S, δs = @ iff $@^{p} \sqsubseteq s \sqsubseteq @$

Proof

For the left-to-right direction, suppose δs = @. Then clearly $s \sqsubseteq @$, so it suffices to show that $@^{p} \sqsubseteq s$. So suppose otherwise. Then since s is a state, it follows that s has no minimally complete subset. So let φ_i be an a-positive literal which is not a member of s. Then by construction, $\overline {\varphi _{i}} \in \delta s$, contrary to the assumption that δs = @.

For the right-to-left direction, assume $@^{p} \sqsubseteq s \sqsubseteq @$. We show first that $\delta s \sqsubseteq @$. Since $@^{p} \sqsubseteq s$, δs = s ∪{nφ : s does not decide φ}. So it suffices to show that nφ ∈ @ whenever s does not decide φ. But if s does not decide φ, then (tφ)₀ is not in s, and hence not in @^p, so is not a-positive, so nφ = (tφ)₀. But tφ ∈Φ, so (tφ)₀ = nφ ∈ @, as desired.

Finally, we show that $@ \sqsubseteq \delta s$. Suppose φ_i ∈ @. If φ_i ∈ s then clearly φ_i ∈ δs. So suppose φ_i∉s. It follows that φ_i is not a-positive. Since φ_i is actual, it follows that i = 0, and that tφ = φ, so _i = (tφ)₀. Since φ_i is not a-positive, nφ = (tφ)₀. So to show that φ_i ∈ δs, it suffices to show that s does not decide φ. Suppose otherwise, so s contains some co-literal ψ_j of φ_i. Then ψ_j≠φ_i, since by assumption _i∉s. Since $s \sqsubseteq @$, ψ_j is actual, so ψ ∈Φ. By consistency of Φ, ψ = φ, and so j≠ 0. But by the definition of actuality it then follows that j < p(φ), and hence that i < p(φ), contrary φ_i not being a-positive. So s does not decide φ, and hence φ_i = nφ ∈ δs. □

Lemma 24

$(S, S^{\lozenge }, \sqsubseteq , \delta )$ is a C-space

Proof

From Lemmas 20 (9) and 22 it follows that $(S, S^{\lozenge }, \sqsubseteq )$ is a W-space,

δ-Containment: Immediate from the definitions of δs.

δ-Completeness: Immediate from the stronger Lemma 22.

δ-Redundancy(1): Suppose s is a world. δs = s ∪{nφ : s is φ-neutral$\} \cup \{\overline {\varphi _{i}}: \varphi _{i} \in @^{p}\!\setminus \! s\}$. Since s is consistent, δs is a world, and since s is a world and $s \sqsubseteq \delta s$, we have s = δs.

δ-Redundancy(2): It is easily verified that δs decides every literal and contains every positive i-literal contained in s. From that observation, it is immediate that δδs = δs.

δ-Identity(1): Suppose $s \sqsubseteq t \sqsubseteq \delta s$. We show first that $\delta t \sqsubseteq \delta s$. By assumption, $t \sqsubseteq \delta s$. By Lemma 21, $c(t) \subseteq c(s)$, so $c(t) \subseteq \delta s$, and hence $\delta t \sqsubseteq \delta s$. We now show that $\delta s \sqsubseteq \delta t$. By assumption, $s \sqsubseteq t \sqsubseteq \delta t$. So let x ∈ c(s). Then either (a) x = nφ with φ not decided by s, or (b) $x = \overline {\varphi _{i}}$ with φ_i an a-positive i-literal not in s. Suppose (a). Note that nφ is the only co-literal of φ in δs. If φ is not decided by t, then x = nφ is also in δt. If φ is decided by t, t contains a co-literal of φ. But $t \sqsubseteq \delta s$ and nφ is the only co-literal of φ in δs, so nφ ∈ t and hence x = nφ ∈ δt. Suppose (b). Then φ_i∉δs, and by $t \sqsubseteq \delta s$ also φ_i∉t, so again $x = \overline {\varphi _{i}} \in \delta t$.

δ-Identity(2): Suppose δs = δt. We show first that $\delta (s \sqcap t) \sqsubseteq \delta s$. Any member x of δ(s ⊓ t) is either (a) a member of s ⊓ t, or (b) nφ with φ not decided by s ⊓ t, or (c) $\overline {\varphi _{i}}$ with φ_i an a-positive i-literal not in s ⊓ t. Suppose (a). Then x ∈ s and hence x ∈ δs. Suppose (b). If either s or t also do not decide φ, then clearly nφ ∈ δs = δt. So suppose both s and t decide φ. Since s ⊓ t does not, no co-literal of φ is a member of both s and t. Let ψ_j be a co-literal of φ that is a member of s but not t. Since δs = δt, ψ_j ∈ δt. Since ψ_j is neither in t nor a shadow, it follows that ψ_j = nφ, and hence that nφ ∈ δs. And if (c), then at least one of s and t also does not have φ_i as a member, so again x ∈ δs = δt.

We show finally that $\delta s \sqsubseteq \delta (s \sqcap t)$. By Lemma 21, since $s \sqcap t \sqsubseteq s$, $c(s) \subseteq c(s \sqcap t)$. It remains to show that $s \subseteq \delta (s \sqcap t)$. Since δs = δt, any member of s is a member of t or a member of c(t). In the first case, it is a member of s ⊓ t and hence of δ(s ⊓ t). In the second case, since $s \sqcap t \sqsubseteq t$, it is a member of c(s ⊓ t) and so a member of δ(s ⊓ t). □

Again @^p is the positive part of @ and the set of wholly positive states is closed under fusion under the definitions from Section ??; here, positivity is equivalent to not having any members which are either shadows or equal to nφ for some φ.

We now define the interpretation function for the atoms and prove that it satisfies the conditions of exclusivity and exhaustivity.

Definition 10

For φ ∈ At, let $[\varphi ]^{+} = \{f\{\varphi _{i}\}: i \in \mathbb {N}\}$, and $[\varphi ]^{-} = \{f\{\neg \varphi _{i}\}: i \in \mathbb {N}\}$

Lemma 25

For all φ ∈ At,

• 1. s and t are incompatible whenever s ∈ [φ]⁺ and t ∈ [φ]⁻

• 2. for every world w ∈ S, there is some s ∈ [φ]⁺ ∪ [φ]⁻ with $w \sqsupseteq s$

Proof

(1): Suppose s ∈ [φ]⁺ and t ∈ [φ]⁻, so for some i,j, s = f{φ_i} and t = f{¬φ_j}. By consistency of Φ, φ_i and ¬φ_j are not both actual. Since they are co-literals, they are incompatible. Since _i ∈ s and ¬φ_j ∈ t, s and t are incompatible.

(2): Let w be a world. By consistency of Φ, either φ₀ or ¬φ₀ is non-actual. Suppose φ₀ is non-actual. Then f{φ₀} = {φ₀}, so φ₀∉w. Since w is a world, w is incompatible with {φ₀}. Since φ₀ is non-actual, it does not have a shadow, so it follows that w has some co-literal of φ₀ as a member. Any co-literal of φ₀ is identical with either some φ_n or some ¬φ_n, so for some n, w contains either f{φ_n} or f{¬φ_n}. The case of ¬φ₀ non-actual is analogous. □

Definition 11

The canonical model ${\mathscr{M}}$ of Φ is $(S, S^{\lozenge }, \sqsubseteq , \delta , @, [\cdot ])$

Lemma 26

${\mathscr{M}}$ is a model.

Proof

From Lemmas 24 and 25. □

In the canonical model, the literals and Δ-literals in Φ are verified by parts of @.

Lemma 27

In the canonical model ${\mathscr{M}}$, for φ any literal and A any 0-CF:

• 1. ${\mathscr{M}} \models \varphi$ if φ ∈Φ

• 2. If A contains only non-logical conjuncts, then fm(A) verifies A

• 3. ${\mathscr{M}} \models {\Delta } A$ if ΔA ∈Φ

• 4. ${\mathscr{M}} \models {\Delta } A$ if ¬ΔA ∈Φ

Proof

(1): Let φ ∈Φ and suppose first that φ is a logical literal. If φ is ⊤ or ¬Δ⊥, then it is verified by $\Box$ and hence by part of @. Since Φ is consistent, φ cannot be ⊥ or Δ⊥. If φ is Δ⊤, then by definition, @^p is empty, and hence @ = δ□ verifies φ. If φ is ¬Δ⊤, then by definition, @^p is not empty, so @ is an incompletion of □ and hence verifies ¬Δ⊤. Suppose next that φ is a non-logical literal. Then φ₀ is an actual i-literal, so $\{\varphi _{0}\} \subseteq @$. By Lemma 19(4), $f\{\varphi _{0}\} \sqsubseteq @$. By definition of [⋅], f{φ₀} verifies φ. So ${\mathscr{M}} \models \varphi$.

(2): We may set up a one-one correspondence between occurrences of literals in A and members of m(A) by pairing each φ_i ∈ m(A) with the (i + 1)^th occurrence of φ in A. Since always f{φ_i}∈ [φ]⁺, by the clause for conjunction it follows that $\bigsqcup \{f(\varphi _{i}): \varphi _{i} \in m(A)\} \in [A]^{+}$. By Lemma 19(10), $\bigsqcup \{f(\varphi _{i}): \varphi _{i} \in m(A)\} = f\bigcup \{\varphi _{i}: \varphi _{i} \in m(A)\} = fm(A)$.

(3): Suppose ΔA ∈Φ. By Δ-Fact, every conjunct of A is in Φ, and so by (1) is verified by some part of @, so some verifier s of A is part of @. Suppose first that Δ⊤∈Φ, so $@^{p} = \Box$. Then s is between @^p and @, and so δs = @, so @ verifies ΔA. Suppose now that Δ⊤∉Φ. There are three non-logical literals that can occur as conjuncts in A, namely ⊤, ¬Δ⊤, and ¬Δ⊥. The latter two are verified by @, so if either is a conjunct of A, then @ verifies A and hence ΔA. If at most ⊤ is a logical conjunct of A, then the result of removing ⊤ from A as a conjunct is still complete and extends some minimally complete conjunction $A^{\prime }$ of only non-logical literals. By (2), $fm(A^{\prime })$ verifies $A^{\prime }$, and since $A^{\prime }$ is minimally complete, $fm(A^{\prime }) = @^{p}$. It follows that some verifier s of A is between @^p and @ and hence that δs = @ verifies ΔA, so ${\mathscr{M}} \models {\Delta } A$.

(4): Suppose ¬ΔA ∈Φ. Given the clause for the falsifiers of ΔA, it suffices to show that @ is not a verifier of ΔA, and hence that no state between @^p and @ verifies A. So suppose for contradiction that s verifies A and $@^{p} \sqsubseteq s \sqsubseteq @$. Then ${\mathscr{M}} \models \varphi$ for every conjunct φ of A, and so by (1) and Φ being maximal consistent, A ∈Φ. We show that then ΔA ∈Φ, contrary to the consistency of Φ. Note that by Δ-AbsT, if suffices to show that ΔB ∈Φ for some 0-CF B included in A. If Δ⊤∈Φ, then ΔA ∧⊤∈Φ and hence ΔA ∈Φ. So suppose Δ⊤∉Φ. If ¬Δ⊥ is a conjunct in A, then since ⊩_CΔ¬Δ⊥, we obtain ΔA ∈Φ. If ¬Δ⊤ is a conjunct in A, then by Δ-NFix, Δ¬Δ⊤∈Φ and hence again ΔA ∈Φ. In every other case, there can be no logical conjuncts in A except possibly ⊤. It follows that the conjunction $A^{\prime }$ of the non-logical conjuncts in A is also verified by s. So suppose let $A^{\prime } = \varphi ^{0} \wedge {\ldots } \varphi ^{n}$ and let s = s₁ ⊔… ⊔ s_n with every s_i verifying φⁱ. Note that for all i, there is a j(i) such that $s_{i} = f\{\varphi ^{i}_{j(i)}\}$, and by Lemma 19(10), $s = f\{\varphi ^{i}_{j(i))}: 0 \leq i \leq n\}$. Since by assumption $s \sqsupseteq @^{p}$, by definition of f it follows that some subset x of $\{\varphi ^{i}_{j(i))}: 0 \leq i \leq n\}$ is minimally complete, so x = m(B) for some minimally complete 0-CF B. It follows that ΔB ∈Φ, and that B is included in A, so again ΔA ∈Φ. □

Lemma 28

${\mathscr{M}} \models \varphi$ for all φ ∈Φ.

Proof

As before, now using Theorem 9 and Lemma 27. □

From this lemma, completeness follows straightforwardly.

Theorem 29

Let ${\Phi } \subseteq \mathcal {L}_{\Delta }$ and $\varphi \in {\mathscr{L}}_{\Delta }$. Then Φ⊧_Dφ implies Φ ⊩_Dφ.