fusion is order preserving on both coordinates. Now from(∗)and(∗∗)we have
θ◦ψ0∈∆ θ◦ψ”∈∆ because ∆ is
a downset. Since∆is up-directed, then there is someθ0∈∆ with(∗ ∗ ∗)
θ◦ψ0 `θ0 θ◦ψ”`θ0 . Now θ→θ0∈∆2
because θ◦(θ→θ0) =θ0 ∈∆ andθ∈Σ. Now
ψ0 `θ→θ0 ψ”`θ→θ0 by applying residuation to (∗ ∗ ∗), so∆2 is
updirected with θ→θ0 as witness.
The following lemma is the key to the truth-lemma and thus to the completeness result:
Lemma 118. (Existence lemma) LetΣbe an optimal theory and ∆ an optimal co-theory of our language
• if ♦ψ /∈∆, then
{ψ},♦−1[∆]can be extended to a maximal pair
hΣ0,∆0i. • if ♦ψ /∈Σ, thenhΣ,{♦ψ}i can be extended to a maximal pairhΣ0,∆0i. • if ψ /∈Σ, then
−1[Σ],{ψ}can be extended to a maximal pair
hΣ0,∆0i. • if ψ /∈∆, thenh{ψ},∆ican be extended to a maximal pair hΣ0,∆0i. • if /ψ /∈∆ then
/−1[∆],{ψ}can be extended to a maximal pair
hΣ0,∆0i. • if /ψ /∈Σ, thenhΣ,{/ψ}i can be extended to a maximal pairhΣ0,∆0i. • if .ψ /∈Σ, then{ψ}, .−1[Σ] can be extended to a maximal pairhΣ0,∆0i. • if .ψ /∈∆, thenh{.ψ},∆i can be extended to a maximal pairhΣ0,∆0i. • if χ◦ψ /∈∆, then
h{χ},∆1iwith ∆1={χ0|χ0◦ψ∈∆}can be extended to a maximal pair hΣ0,∆0i.
h{ψ},∆2iwith∆2={ψ0|∃χ0(χ0 ∈Σ0&χ0◦ψ0 ∈∆)}can be extended to a maximal pairhΣ”,∆”i.
Proof. Given the case per case assumptions (regarding modal formulas not belonging toΣor∆), all the
pairs to be extended are indeed disjoint. Now given the global assumption aboutΣand∆being respectively
a theory and cotheory, these are in fact a lter and an ideal. Now a straightforward application of corollary
102 will suce.
5.2.1. The truth lemma. Let(c,c)be the satisfaction and co-satisfaction relations associated with Vc (or more precisely, with its unique homomorphic extension), then:
Lemma 119. (Truth lemma) For every ϕ∈F m, everyΣ∈X and every ∆∈Y:
• Σcϕ i ϕ∈Σ;
• ∆c ϕ i ϕ∈∆.
Proof. By induction on the complexity ofϕ.
Base case: Ifϕ=p∈AtProp, then: Σcp i Σ≤Vc(p) =V {∆|∆∈Y andp∈∆} i ∀∆[(∆∈Y andp∈∆) ⇒ Σ≤∆] i ∀∆[(∆∈Y andΣ∆) ⇒ p /∈∆] i ∀∆[(∆∈Y andΣ∩∆ =∅) ⇒ p /∈∆].
So suppose thatΣcpand assume towards a contradiction thatp /∈Σ. Since Σ∈X, then it is an optimal
lter, so in particular Σis maximal w.r.t. some ∆0 ∈ Y. Since∆0 ∈Y and Σ∩∆0 =∅then we conclude
5.2. PROPOSITIONAL SUBSTRUCTURAL LOGIC COMPLETENESS ON NON-DISTRIBUTIVE SETTING 86
Σ0∩∆0 6=∅. Since ∆0 is a co-theory, this implies thatΣ0 6`∆0, against the maximality ofΣw.r.t.∆0. This
shows that p∈Σ. The proof that∆c p i p∈∆is analogous. Therefore, for everyp∈AtProp:
• Σcp i p∈Σ;
• ∆c p i p∈∆.
Inductive step:
As for the inductive step, we need to consider various cases:
The primary cases do the hard job of breaking down the complexity of the formula, reason for which they need the existence lemma, and the secondary cases rely on the primary ones.
♦-case
Primary subcase.
Assume thatϕ=♦ψand that for everyΣ∈X and every∆∈Y,Σcψ i ψ∈Σand∆cψ i ψ∈∆.
Let us x∆∈Y and let us show that:
∆c♦ψ i ♦ψ∈∆.
(⇐) Assume that ♦ψ ∈ ∆. By denition of ∆ c
♦ψ in (5.1.11), we need to show that if Σ0 ∈ X and Σ0cψ, then
♦[Σ0]∩∆6=∅. By induction hypothesis,Σ0 c ψmeans thatψ∈Σ0, so♦ψ∈♦[Σ0], and since
by assumption♦ψ∈∆, then indeed♦[Σ0]∩∆6=∅.
(⇒) Conversely, assume that ♦ψ /∈∆. We need to show that there exists someΣ0 ∈X such thatψ ∈ Σ0
and♦[Σ0]∩∆ =∅, i.e.Σ0∩♦−1[∆] =
∅. Since♦ψ /∈∆, by the Existence Lemma (118){ψ},♦−1[∆]can
be extended to a maximal pair hΣ0,∆0i. Then Σ0 ∈X, ∆0 ∈ Y andψ ∈ Σ0. Moreover, ♦−1[∆]⊆∆0 and Σ0∩∆0 =∅implies thatΣ0∩♦−1[∆] =∅. So result is proven.
Secondary subcase.
Now let us xΣ∈X and show that:
Σc♦ψ i ♦ψ∈Σ.
(⇐)Assume that♦ψ∈Σ. By denition ofΣc
♦ψin (5.1.12), we need to show that if∆∈Y and∆c
♦ψ
then Σ≤∆, that is Σ∩∆ 6=∅. So suppose ∆ ∈Y and∆ c ♦ψ. By the previous case this means that
♦ψ∈∆. But then clearly Σ∩∆6=∅since we assumed♦ψ∈Σ.
(⇒)Assume that♦ψ /∈Σ. We need to show that there is some∆∈Ysuch that∆c
♦ψandΣ∆, that
is Σ∩∆ =∅. Since♦ψ /∈Σ, by the Existence Lemma (118) hΣ,{♦ψ}ican be extended to a maximal pair
hΣ0,∆0i. ThenΣ0 ∈X, ∆0 ∈Y and♦ψ∈∆0, that is∆0c
♦ψ. Moreover,Σ⊆Σ0 andΣ0∩∆0=∅implies
5.2. PROPOSITIONAL SUBSTRUCTURAL LOGIC COMPLETENESS ON NON-DISTRIBUTIVE SETTING 87