the claim holds for n = 0. Now suppose it holds for n, i.e., ∆n
is Σ -consistent. ∆n+1 is either ∆n ∪ {An} is that is Σ -consistent,
otherwise it is ∆n ∪ {¬An}. In the first case, ∆n+1 is clearly Σ -
consistent. However, byProposition 3.39(3), either ∆n∪ {An}or
∆n ∪ {¬An} is consistent, so ∆n+1is consistent in the other case
as well. □
Corollary 4.4. Γ ⊢Σ A if and only if A ∈ ∆ for each complete Σ - consistent set ∆ extending Γ (including when Γ = ∅, in which case we get another characterization of the modal system Σ .)
Proof. Suppose Γ ⊢Σ A, and let ∆ be any complete Σ -consistent set extending Γ. If A ∉ ∆ then by maximality ¬A ∈ ∆ and so ∆ ⊢Σ A (by monotony) and ∆ ⊢Σ ¬A (by reflexivity), and so ∆ is
inconsistent. Conversely if Γ ⊬Σ A, then Γ ∪{¬A} is Σ -consistent, and by Lindenbaum’s Lemma there is a complete consistent set ∆ extending Γ ∪ {¬A}. By consistency, A ∉ ∆. □
4.4
Modalities and Complete Consistent
Sets
When we construct a model MΣ whose set of worlds is given by the complete Σ -consistent sets ∆ in some normal modal logic Σ , we will also need to define an accessibility relation RΣ between such “worlds.” We want it to be the case that the accessibility relation (and the assignment VΣ)are defined in such a way that
MΣ, ∆ ⊩ A iff A ∈ ∆. How should we do this?
Once the accessibility relation is defined, the definition of truth at a world ensures that MΣ, ∆ ⊩ □A iff MΣ, ∆′ ⊩ A for all ∆′ such that RΣ∆∆′. The proof that MΣ, ∆ ⊩ A iff A ∈ ∆
requires that this is true in particular for formulas starting with a modal operator, i.e., MΣ, ∆ ⊩ □A iff □A ∈ ∆. Combining this requirement with the definition of truth at a world for □A yields:
68 CHAPTER 4. COMPLETENESS AND CANONICAL MODELS
Consider the left-to-right direction: it says that if □A ∈ ∆, then A ∈ ∆′ for any A and any ∆′ with RΣ∆∆′. If we stipulate that RΣ∆∆′ iff A ∈ ∆′ for all □A ∈ ∆, then this holds. We can write the condition on the right of the “iff” more compactly as: {A : □A ∈ ∆} ⊆ ∆′.
So the question is: does this definition of RΣ in fact guarantee that □A ∈ ∆ iff MΣ, ∆ ⊩ □A? Does it also guarantee that ♢A ∈ ∆ iff MΣ, ∆ ⊩ ♢A? The next few results will establish this.
Definition 4.5. If Γ is a set of formulas, let
□Γ = {□B : B ∈ Γ } ♢Γ = {♢B : B ∈ Γ } and
□−1Γ = {B : □B ∈ Γ} ♢−1Γ = {B : ♢B ∈ Γ}
In other words, □Γ is Γ with □ in front of every formula in Γ; □−1Γ is all the □’ed formulas of Γ with the initial □’s removed. This definition is not terribly important on its own, but will simplify the notation considerably.
Note that □□−1Γ ⊆ Γ:
□□−1Γ = {□B : □B ∈ Γ}
i.e., it’s just the set of all those formulas of Γ that start with □. Lemma 4.6. If Γ ⊢Σ A then □Γ ⊢Σ □A.
Proof. If Γ ⊢Σ A then there are B1, . . . , Bk ∈ Γ such that Σ ⊢
B1 → (B2 → · · · (Bn → A) · · · ). Since Σ is normal, by rule rk,
Σ ⊢ □B1→ (□B2→ · · · (□Bn→ □A) · · · ), where obviously □B1,
69 4.4. MODALITIES AND COMPLETE CONSISTENT SETS
Lemma 4.7. If □−1Γ ⊢Σ
A then Γ ⊢Σ □A.
Proof. Suppose □−1Γ ⊢Σ A; then byLemma 4.6, □□−1Γ ⊢ □A. But since □□−1Γ ⊆ Γ, also Γ ⊢Σ □A by Monotony. □
Proposition 4.8. If Γ is complete Σ -consistent, then □A ∈ Γ if and only if for every complete Σ -consistent ∆ such that □−1Γ ⊆ ∆, it holds that A ∈ ∆.
Proof. Suppose Γ is complete Σ -consistent. The “only if” direc- tion is easy: Suppose □A ∈ Γ and that □−1Γ ⊆ ∆. Since □A ∈ Γ,
A ∈ □−1Γ ⊆ ∆, so A ∈ ∆.
For the “if” direction, we prove the contrapositive: Suppose □A ∉ Γ. Since Γ is complete Σ -consistent, it is deductively closed, and hence Γ ⊬Σ □A. By Lemma 4.7, □−1Γ ⊬Σ A. By
Proposition 3.39(2), □−1Γ ∪ {¬A} is Σ -consistent. By Linden- baum’s Lemma, there is a complete Σ -consistent set ∆ such that □−1Γ ∪ {¬A} ⊆ ∆. By consistency, A ∉ ∆. □ Lemma 4.9. Suppose Γ and ∆ are complete Σ -consistent. Then: □−1Γ ⊆ ∆ if and only if ♢∆ ⊆ Γ.
Proof. “Only if” direction: Assume □−1Γ ⊆ ∆ and suppose ♢A ∈ ♢∆ (i.e., A ∈ ∆). In order to show ♢A ∈ Γ it suffices to show □¬A ∉ Γ for then by maximality ¬□¬A ∈ Γ. Now, if □¬A ∈ Γ then by hypothesis ¬A ∈ ∆, against the consistency of ∆ (since A ∈ ∆). Hence □¬A ∉ Γ , as required.
“If” direction: Assume ♢∆ ⊆ Γ. We argue contrapositively: suppose A ∉ ∆ in order to show □A ∉ Γ. If A ∉ ∆ then by maximality ¬A ∈ ∆ and so by hypothesis ♢¬A ∈ Γ. But in a normal modal logic ♢¬A is equivalent to ¬□A, and if the latter is in Γ, by consistency □A ∉ Γ, as required. □
70 CHAPTER 4. COMPLETENESS AND CANONICAL MODELS
Proposition 4.10. If Γ is complete Σ -consistent, then ♢A ∈ Γ if and only if for some complete Σ -consistent ∆ such that ♢∆ ⊆ Γ , it holds that A ∈ ∆.
Proof. Suppose Γ is complete Σ -consistent. ♢A ∈ Γ iff ¬□¬A ∈ Γ by dual and closure. ¬□¬A ∈ Γ iff □¬A ∉ Γ by Proposi- tion 4.2(4)since Γ is complete Σ -consistent. ByProposition 4.8, □¬A ∉ Γ iff, for some complete Σ -consistent ∆ with □−1Γ ⊆ ∆, ¬A ∉ ∆. Now consider any such ∆. ByLemma 4.9, □−1Γ ⊆ ∆ iff ♢∆ ⊆ Γ. Also, ¬A ∉ ∆ iff A ∈ ∆ byProposition 4.2(4). So ♢A ∈ Γ iff, for some complete Σ -consistent ∆ with ♢∆ ⊆ Γ, A ∈ ∆. □