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 noncontingent, a nontrivial 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.
Notes
Note that Δ¬Δ⊥ will still be true if we countenance more falsifiers of Δstatements, as long as these are all parts of an incompletion of a verifier of the argument. For in the present case, this will mean that these added falsifiers are still parts of @, and hence have @ as their completion.
Note that we do not require the assumption that only δstates falsify ΔA to obtain the result; it is enough that some state t with δt = @ falsifies ΔA. So as long as @, or at least its positive part, is included among the falsifiers of any given false ΔA, ΔNFix will be sound.
This qualification will often remain tacit in what follows.
References
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Acknowledgements
For helpful comments on earlier drafts of this paper and/or discussion of the ideas here developed, I would like to thank the Relevance crew: Stefan Roski, Martin Glazier, and Singa Behrens, as well as Stephan Leuenberger, Bruno Whittle, Josh Schechter, Johannes Korbmacher, and especially Kit Fine. I am also grateful to two anonymous reviewers for this journal whose comments led to highly significant extensions of and improvements to the paper. My work on this paper was done as part of the Emmy Noether project Relevance, funded by the Deutsche Forschungsgemeinschaft (Grant number KR 4516/21); I am very grateful for that support.
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Appendices
Appendix A: Soundness
Theorem 4
Let \({\Phi } \subseteq {\mathscr{L}}_{\Delta }\) and \(\varphi \in {\mathscr{L}}_{\Delta }\). Then Φ ⊩_{C}φ implies Φ⊧_{C}φ.
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 ΔEquivC 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 (nonempty) fusions and the fact that the truthvalue of ΔA and ΔB depends only on the closure under fusion of the verifiersets of A and B.
For ΔFact and ΔPFix, assume ΔA is true in a model, so some part s of @ is the sole verifier of ΔA. By the clause for Δ, some δstate is part of s. But the only δstate 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 ΔDisjC, 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 ΔAbsP, 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 δIdentity(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 ΔNE1, 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 δstate other than @ must be the completion of every verifier of A, so there can only be two δstates. Since \(\blacksquare \) is a δstate distinct from @, it follows that @ and \(\blacksquare \) are the only δstates. By δIdentity(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.^{Footnote 1}
For ΔNE2, suppose Δ¬Δ⊥ is true in a model. By the same reasoning as before, the model can only have two δstates, 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 ⊩_{D}φ implies Φ⊧_{D}φ.
Proof
The argument for the soundness of classical propositional logic remains the same. The soundness of ΔEquivD 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 truthvalue of ΔA and ΔB depends only on the verifiers of A and B. ΔDisjD is immediate by the clause for Δ. ΔFact and ΔPFix are immediate given the fact that always δs ⊒ s and δδs = δs.
For ΔNFix, suppose ¬ΔA is true, and so verified by some part s of @. Since falsifiers of ΔA are all δstates, s = @. Since @ is a world, δ@ = @, and so @ verifies Δ¬ΔA.^{Footnote 2}
For ΔAbsT, 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 nonlogical 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 Δliterals. A Δ − DNF is any disjunction of conjunctions in which each conjunct is either a literal or a Δliteral.
We will show that under either logic, every formula of \({\mathscr{L}}_{\Delta }\) is provably equivalent to a ΔDNF. 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 natoms, nliterals, and nformulas. The 0atoms are the atoms, the 0literals are the 0atoms and their negations, and the 0formulas are all the formulas built up from 0atoms without the use of Δ. The (n + 1)atoms are the natoms as well as all expressions of the form ΔA, where A is an nformula. The (n + 1)literals are the n + 1atoms 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 nliterals will be called an nCF, and a disjunction of nCFs will be called an nDNF. The degree of a formula A is the smallest n for which A is an nformula.
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 nformula is provably exactly equivalent to an nDNF.

2.
For A an nformula, ΔA is provably equivalent to a formula \({\Delta } A^{\prime }\), where \(A^{\prime }\) is an nDNF.

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 nCF of degree n > 0, then ΔA is provably equivalent to some nformula.

6.
For n > 0, if A is a formula of degree n, then ΔA is provably equivalent to some nformula.
Proof
(1): This is a straightforward consequence of Theorem 15 in [3, p. 214].
(2): By (1) and ΔEquivC.
(3): By ΔFact, ΔPFix, and ΔAbsP.
(4): Lefttoright: By ΔNE1, from Δ(¬ΔA ∧ B) we infer Δ¬Δ⊥, from which we obtain, using ΔNE1, Δ⊤. Using ⊩_{eC}A ≈ A ∧⊤ and ΔAbsP, 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 ΔNE2 again, we may infer Δ¬A. Righttoleft: Using ΔNE2, we may infer Δ¬ΔA from Δ¬Δ⊥ and Δ¬A. Using ΔAbsP and ΔB, we obtain Δ(¬ΔA ∧ B).
(5): Suppose A is an nCF of degree n > 0. So A is a conjunction of nliterals A_{1},…,A_{m}, at least one of which is of degree n. Let k be the number of degree n literals among A_{1},…,A_{m}. We may then use an ‘extraction’ procedure to obtain a formula provably equivalent to ΔA, which is a conjunction of an nformula and a sentence \({\Delta } A^{\prime }\) with \(A^{\prime }\) an nCF 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 nformulas, and hence itself an nformula.
The extraction procedure is as follows. First, we pick one of the degree n literals among A_{1},…,A_{m}, call it B_{1}, and turn A into a provably exactly equivalent nCF of degree nB = B_{1} ∧… ∧ B_{m}. Then ΔA is provably equivalent to ΔB. B_{1} is either of the form ΔC or of the form ¬ΔC. In the former case, ΔB is Δ(ΔC ∧ B_{2} ∧… ∧ B_{m}), and hence by (3) provably equivalent to ΔC ∧Δ(C ∧ B_{2} ∧… ∧ B_{m}). Since C is an (n − 1)formula, ΔC is an nformula and Δ(C ∧ B_{2} ∧… ∧ B_{m}) has only k − 1 degree n literals, as desired. In the latter case, ΔB is Δ(¬ΔC ∧ B_{2} ∧… ∧ B_{m}), and hence by (4) provably equivalent to Δ¬Δ⊥∧Δ¬C ∧Δ(B_{2} ∧… ∧ B_{m}). Since C is an (n − 1)formula, Δ¬Δ⊥∧Δ□C is an nformula, and Δ(B_{2} ∧… ∧ B_{m}) has only k − 1 degree n literals, as desired.
(5): By (2), ΔA is provably equivalent to ΔB for some nDNF B = B_{1} ∨… ∨ B_{k}. Then by ΔDisjC, ΔA is provably equivalent to ΔB_{1} ∧… ∧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 nformula. In this way we obtain an nformula provably equivalent to ΔA, as desired. □
Theorem 6
Any formula is provably equivalent within ⊩_{C} to a ΔDNF.
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 0formula. By Lemma 6 (2), there is a 0DNF C with ΔB provably equivalent to ΔC. By ΔDisjC, ΔC in turn is provably equivalent to a conjunction of Δliterals. So we may replace any degree 1 atom in A by a provably equivalent conjunction of Δliterals. The result is provably equivalent to A, and, by propositional logic, provably equivalent to a ΔDNF.
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 nformula. Then by Lemma 6 (6), ΔB is provably equivalent to some nformula. We may thus replace each degree n + 1 atom in A by a provably equivalent nformula. The result will itself be an nformula provably equivalent to A. By IH, that formula, and therefore A, is provably equivalent to a ΔDNF. □
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 nformula is provably exactly equivalent to an nDNF.

2.
For A an nformula, ΔA is provably equivalent to a formula \({\Delta } A^{\prime }\), where \(A^{\prime }\) is an nDNF.

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 nCF 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 nformula.
Proof
(1) and (2) are as before. (3) and (4) may be established using ΔFact, ΔAbsT, and ΔPFix or ΔNFix, respectively.
(5): A has at least one conjunct of degree n > 0, i.e. an nliteral 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 ΔEquivD, 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 ΔEquivD, 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 nDNF B = B_{1} ∨… ∨ B_{k}. Then by ΔDisjD, ΔA is provably equivalent to ΔB_{1} ∨… ∨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 nCF, by (5), ΔB_{i} is provably equivalent to B_{i}. So through these replacements we obtain an nformula provably equivalent to ΔA, as desired. □
Theorem 7
Any formula is provably equivalent within ⊩_{D} to a ΔDNF.
Proof
The reasoning is as for Theorem 7, using ΔDisjD in place of ΔDisjC 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 0CF 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 nonlogical literals occurring as a conjunct in A

φ is actual iff φ ∈Φ

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

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

if φ is nonactual: 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 nonlogical 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 0CF, but no minimally complete 0CF. We denote the set of all (nonlogical)^{Footnote 3} literals by P (since we now regard them as protostates). 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 0CFs are verified by @^{p}.)
Definition 2
The canonical statespace for Φ is \((S, \sqsubseteq )\) with \(\sqsubseteq = \subseteq \restriction S\)
In preparation of the proof that \((S, \sqsubseteq )\) is indeed a statespace, 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 statespace for Φ, \((S, \sqsubseteq )\), is a statespace.
Proof
From the fact that the subsetorder 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 Cspace
Proof
It is clear that \(S^{\lozenge }\) is nonempty 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, δContainment 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 Wspace are satisfied—, and that δCompleteness holds.
δRedundancy(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.
δRedundancy(2): Immediate from the fact that by construction, δs decides every literal.
δIdentity(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 δContainment. 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.
δIdentity(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 Cspace, 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 Δliterals 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 0CF. 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 nonlogical 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 nonempty if ¬Δ⊤∈Φ, so @ is an incompletion of \(\Box \) and hence verifies ¬Δ⊤.
(2): Suppose ΔA ∈Φ. Then A is complete, so every nonlogical 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 nonlogical 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 ΔFact 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 ΔFact 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 nonactual verifier. So every nonlogical literal φ ∈ m(A) is pure, and hence occurs in some complete 0CF 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 nonactual 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 nonlogical literals, which must then be pure. But then using the ΔB_{φ} for all nonlogical literals φ in A, using ΔAbsP 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 ΔEquiv and ΔAbsP, from ΔB and ΔB_{φ} for each nonlogical 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 Cspace. 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 0CF. 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 nonlogical 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 nonlogical 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 ΔDNF \(\varphi ^{\prime } \in {\Phi }\). Since Φ is maximal consistent, some disjunct φ^{∗} of \(\varphi ^{\prime }\) is also in Φ. Since \(\varphi ^{\prime }\) is a ΔDNF, φ^{∗} is a conjunction each conjunct of which is a literal or a Δliteral 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⊧_{C}φ implies Φ ⊩_{C}φ.
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 0CF 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 (nonlogical) literal \(\varphi \in {\mathscr{L}}_{\Delta }\), we associate countably many indexed literals (short: iliterals) 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 nonlogical literals occur in minimally complete 0CFs, but tφ is not one of them. (Note that the actual world may be nonempty while no nonlogical literals occur in minimally complete 0CFs, because it would take an infinite 0CF to exhaust the positive part of the world.) Then nφ is nonactual 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: apositive) ones by @^{p}. Note that @^{p} is empty iff Δ⊤∈Φ.
Next, with any 0CF \(A \in {\mathscr{L}}_{\Delta }\) including only nonlogical 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 multiset 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 0CF B is included in a 0CF 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 iliterals minimally complete iff s = m(A) for some minimally complete conjunction of nonlogical 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 nonlogical conjuncts.
We construct the set of states similarly as before, but using as the set of protostates P not the set of literals, but the set of the iliterals together with the shadows of the apositive iliterals. 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 statespace 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 statespace for Φ, \((S, \sqsubseteq )\), is a statespace.
Proof
From the fact that the subsetorder on any set is a partial order together with Lemma 19 (9). □
To extend \((S, \sqsubseteq )\) to a canonical Cspace, we need to define a set of consistent states and the δfunction, 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 coliterals iff φ = ψ, or φ = ¬ψ, or ψ = ¬φ

φ_{i} and ψ_{j} are incompatible iff they are (a) coliterals and (b) not both actual

s is consistent iff (a) no two members of s are incompatible iliterals, 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 coliteral 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 apositive iliteral 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 cfunction 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 nonpositive iliteral, and no shadow is a positive iliteral, 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 apositive iliterals have shadows, and the only apositive iliterals in δs are already in s. Since s is consistent, it contains no shadows of its apositive members, and beyond the states in s, by definition δs contains only shadows of iliterals not in s. Secondly, no pair of iliterals in δs is incompatible, for only coliterals are incompatible, and by construction, any pair of coliterals 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 iliteral in t. Suppose first that s does not decide φ. Then nφ ∈ δs. If nφ≠φ_{i}, then nφ and φ_{i} are distinct coliterals, and since nφ is nonpositive, 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 coliteral 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 apositive. 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 lefttoright 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 apositive 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 righttoleft 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 apositive, 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 apositive. Since φ_{i} is actual, it follows that i = 0, and that tφ = φ, so _{i} = (tφ)_{0}. Since φ_{i} is not apositive, 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 coliteral 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 apositive. So s does not decide φ, and hence φ_{i} = nφ ∈ δs. □
Lemma 24
\((S, S^{\lozenge }, \sqsubseteq , \delta )\) is a Cspace
Proof
From Lemmas 20 (9) and 22 it follows that \((S, S^{\lozenge }, \sqsubseteq )\) is a Wspace,
δContainment: Immediate from the definitions of δs.
δCompleteness: Immediate from the stronger Lemma 22.
δRedundancy(1): Suppose s is a world. δs = s ∪{nφ : s is φneutral\(\} \cup \{\overline {\varphi _{i}}: \varphi _{i} \in @^{p}\!\setminus \! s\}\). Since s is consistent, δs is a world, and since s is a world and \(s \sqsubseteq \delta s\), we have s = δs.
δRedundancy(2): It is easily verified that δs decides every literal and contains every positive iliteral contained in s. From that observation, it is immediate that δδs = δs.
δIdentity(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 apositive iliteral not in s. Suppose (a). Note that nφ is the only coliteral 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 coliteral of φ. But \(t \sqsubseteq \delta s\) and nφ is the only coliteral 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 \(x = \overline {\varphi _{i}} \in \delta t\).
δIdentity(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 apositive iliteral 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 coliteral of φ is a member of both s and t. Let ψ_{j} be a coliteral 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 coliterals, 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 φ_{0} or ¬φ_{0} is nonactual. Suppose φ_{0} is nonactual. Then f{φ_{0}} = {φ_{0}}, so φ_{0}∉w. Since w is a world, w is incompatible with {φ_{0}}. Since φ_{0} is nonactual, it does not have a shadow, so it follows that w has some coliteral of φ_{0} as a member. Any coliteral of φ_{0} 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 ¬φ_{0} nonactual 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 Δliterals in Φ are verified by parts of @.
Lemma 27
In the canonical model \({\mathscr{M}}\), for φ any literal and A any 0CF:

1.
\({\mathscr{M}} \models \varphi \) if φ ∈Φ

2.
If A contains only nonlogical 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 nonlogical literal. Then φ_{0} is an actual iliteral, so \(\{\varphi _{0}\} \subseteq @\). By Lemma 19(4), \(f\{\varphi _{0}\} \sqsubseteq @\). By definition of [⋅], f{φ_{0}} verifies φ. So \({\mathscr{M}} \models \varphi \).
(2): We may set up a oneone 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 ΔFact, 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 nonlogical 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 nonlogical 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 ΔAbsT, if suffices to show that ΔB ∈Φ for some 0CF 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 ⊩_{C}Δ¬Δ⊥, we obtain ΔA ∈Φ. If ¬Δ⊤ is a conjunct in A, then by ΔNFix, 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 nonlogical conjuncts in A is also verified by s. So suppose let \(A^{\prime } = \varphi ^{0} \wedge {\ldots } \varphi ^{n}\) and let s = s_{1} ⊔… ⊔ s_{n} with every s_{i} verifying φ^{i}. 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 0CF 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⊧_{D}φ implies Φ ⊩_{D}φ.
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Krämer, S. That’s It! Hyperintensional Total Logic. J Philos Logic (2023). https://doi.org/10.1007/s1099202209695z
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DOI: https://doi.org/10.1007/s1099202209695z
Keywords
 Truthmaker semantics
 Hyperintensionality
 Totality operator
 Total logic
 Completeness
 The whole truth