Completeness for non-distributive propositional case
5.2. PROPOSITIONAL SUBSTRUCTURAL LOGIC COMPLETENESS ON NON-DISTRIBUTIVE SETTING 87 case
Primary subcase.
LetΣ∈X. We shall show that: 1
Σcψ i ψ∈Σ.
(⇐) For the easy direction, from right to left, assume that ψ ∈ Σ. Then by denition of Σ c
ψ in
(5.1.13), we have to show∀∆[∆V ψ ⇒ ΣRc∆]. So let ∆∈Y and suppose that∆V ψ. Then by IH ψ∈∆and thus[∆]∩Σ6=∅so ΣRc∆by denition ofRc. Since∆∈Y was arbitrary, the implication is
shown and thereforeΣc
ψ.
(⇒) For the other direction, assume thatψ /∈Σ. To show thatΣ1c
ψ we have to nd a ∆ ∈Y such
that ∆ V ψ and such thatΣRc∆ doesn't hold, that is [∆]∩Σ =∅. Since ψ /∈ Σ, by the Existence
Lemma (118)
−1[Σ],{ψ}can be extended to a maximal pair
hΣ0,∆0i. ThenΣ0∈X,∆0 ∈Y andψ∈∆0.
Thus by IH ∆0
V ψ. Moreover,−1[Σ]⊆Σ0 so Σ0∩∆0 =∅ implies that−1[Σ]∩∆0 =∅. Hence ∆0 is
the desired counterexample andΣc
ψdoesn't hold.
Secondary subcase.
Let∆∈Y. We shall show that:
∆c
ψ i ψ∈∆.
(⇐) For the easy direction, from right to left, assume that ψ ∈ ∆. Then by denition of ∆ c
ψ in
(5.1.14), we have to show ∀Σ[Σcψ⇒ Σ≤∆]. So letΣ∈X and supposeΣcψ. Then ψ∈Σas
we just have shown above and therefore∆∩Σ6=∅soΣ≤∆by denition of≤. SinceΣ∈X was arbitrary,
the implication is shown and therefore∆c
ψ.
(⇒) For the other direction, assume that ψ /∈ ∆. To show that ∆ ψ we have to nd a Σ∈ X such
that Σc
ψ but Σ∆. Since ψ /∈∆, by the Existence Lemma (118) h{ψ},∆ican be extended to a
maximal pairhΣ0,∆0i. Then Σ0∈X, ∆0 ∈Y andψ∈Σ0. Moreover,∆⊆∆0 so Σ0∩∆0 =∅implies that ∆∩Σ0=∅which meansΣ∆, as desired.
/-case
Primary subcase.
Let∆∈Y. We shall show that:
∆c/ψ i / ψ∈∆.
(⇐)Let/ψ ∈∆. By denition of co-satisfaction (5.1.15) we have to show ∀∆0[∆0cψ⇒∆Rc
/∆0]. So let ∆0cψfor some∆0∈Y, then by IHψ∈∆0and thus∆∩/[∆0]6=
∅which by denition leads to the desired
result : ∆Rc
/∆0. Thus∆c/ψ.
(⇒)Let/ψ /∈∆. By the denitions of co-satisfaction and ofR/we have to show∃∆0[∆0 cψ& ∆∩/[∆0] =
∅],
with∆0 c ψrewritten asψ∈∆0 by IH. Since/ψ /∈∆then
/−1[∆],{ψ}
is a pair, which by the Existence Lemma (118) can be extended to a maximal pair hΣ0,∆0i. Then Σ0 ∈ X, ∆0 ∈ Y and ψ ∈∆0. Moreover, /−1[∆]⊆Σ0 soΣ0∩∆0 =
∅implies that/−1[∆]∩∆0=∅which means∆∩/[∆0] =∅, as desired.
Secondary subcase.
LetΣ∈X. We shall show that:
5.2. PROPOSITIONAL SUBSTRUCTURAL LOGIC COMPLETENESS ON NON-DISTRIBUTIVE SETTING 88
(⇐) Let /ψ ∈ Σ. By (5.1.16) we have to show ∀∆ [∆c/ψ⇒Σ≤∆]. Let us x a ∆ ∈ Y and assume ∆c/ψthen by the previous proof/ψ∈∆ and thusΣ∩∆6=
∅. ThereforeΣ≤∆as desired. Since∆was
arbitrary, this shows the implication.
(⇒) Let/ψ /∈Σ. We have to show that there exists ∆ c /ψ and Σ
∆, that is Σ∩∆ = ∅. SinceΣ is
a theory, it is clear thatΣ0c /ψ, hence we can try and extendhΣ,{/ψ}i to a maximal pairhΣ0,∆0iby the
Existence lemma (118). ThenΣ0 ∈X, ∆0 ∈Y and/ψ ∈∆0, thus by the previous results∆0 c /ψ. Since Σ⊆Σ0 thenΣ0∩∆0=∅impliesΣ∩∆0 =∅and thusΣ∆0 as desired.
.-case
Primary subcase.
LetΣ∈X. We shall show that:
Σc.ψ i . ψ∈Σ.
(⇐)For the easy direction, from right to left, assume that.ψ∈Σ. Then by denition ofΣc.ψin (5.1.17),
we have to show ∀Σ0[Σ0 c ψ ⇒ ΣRc
.Σ0]. So let Σ0 ∈ X and suppose that Σ0 c ψ, which by IH means ψ∈Σ0 . But then .[Σ0]∩Σ6=∅, as desired.
(⇒)For the other direction, assume that.ψ /∈Σ. To show thatΣ1c .ψwe have to nd aΣ0 ∈X such that Σ0 c ψ(by IH ψ∈Σ0) butΣRc.Σ0 doesn't hold, that is .[Σ0]∩Σ =∅. Since .ψ /∈Σthen
{ψ}, .−1[Σ]
is a pair and by the Existence lemma (118), it can be extended to a maximal pair hΣ0,∆0i. ThenΣ0 ∈X, ∆0∈Y andψ∈Σ0. Since.−1[Σ]⊆∆0 thenΣ0∩∆0 =
∅impliesΣ0∩.−1[Σ] =∅, that is .[Σ0]∩Σ =∅as
desired.
Secondary subcase.
Let∆∈Y. We shall show that:
∆c.ψ i . ψ∈∆.
(⇐) Assume .ψ ∈∆. Given the denition in (5.1.18) we have to show∀Σ [Σc.ψ⇒Σ≤∆]. So x an Σ∈X and letΣc .ψ, then.ψ∈Σby previous proof, and thenΣ∩∆6=
∅which givesΣ≤∆by denition.
SinceΣwas arbitrary, the implication is proven.
(⇒)Assume .ψ /∈∆. We have to show∃Σ [Σc.ψ& Σ∆]. By the previous resultΣc .ψ amounts to
.ψ ∈Σ. Since .ψ /∈ ∆ then h{.ψ},∆i is disjoint and can be extended to a maximal pair hΣ0,∆0iby the Existence Lemma (118). Then Σ0 ∈ X, ∆0 ∈ Y and .ψ ∈ Σ0. Since ∆ ⊆ ∆0, then Σ0 ∩∆0 =
∅ implies
Σ0∩∆ =∅, that isΣ0 ∆ as desired.
◦-case
Primary subcase.
Assume thatϕ=χ◦ψand that for everyΣ∈X and every∆∈Y:
• Σcχ i χ∈Σand∆cχ i χ∈∆.
• Σcψ i ψ∈Σand∆cψ i ψ∈∆.
Let us x∆∈Y and let us show that:
∆c χ◦ψ i χ◦ψ∈∆.
(⇐) Assume thatχ◦ψ∈ ∆. By denition of ∆ c χ◦ψ in (5.1.19), we need to show that if (a) Σ ∈X
5.2. PROPOSITIONAL SUBSTRUCTURAL LOGIC COMPLETENESS ON NON-DISTRIBUTIVE SETTING 89 induction hypothesis, (a) means thatχ∈Σ, and likewise (b) means thatψ∈Σ0, soχ◦ψ∈Σ◦Σ0 and since
by assumptionχ◦ψ∈∆, then indeed Σ◦Σ0∩∆6=∅.
(⇒)Conversely, assume thatχ◦ψ /∈∆. We need to show that there exists someΣ,Σ0 ∈X such thatϕ∈Σ
and ψ∈Σ0 and Σ◦Σ0∩∆ =∅. Let ∆1 ={χ0|χ0◦ψ∈∆}, thenχ /∈∆1. Sinceχ /∈∆1 thenh{χ},∆1iis
disjoint pair and by lemma 118 it can be extended to a maximal pairhΣ1,∆01i, our rst witness point. Now
let∆2={ψ0|∃χ0(χ0 ∈Σ1&χ0◦ψ0∈∆)}be our starting base for our 2nd witness. Notice thatψ /∈∆2. For
suppose otherwise, ifψ∈∆2 then there existsβ∈Σ1 such thatβ◦ψ∈∆. But thenβ ∈∆1 and therefore
β ∈ Σ1∩∆1. However hΣ1,∆01i is a maximal pair with ∆1 ⊆∆01 and thus Σ1∩∆1 =∅, contradiction!
Therefore h{ψ},∆2i is a disjoint pair which can be extended to a maximal pair hΣ2,∆02i by lemma 118.
Now we have hΣ1,∆01i with∆1⊆∆01&χ∈Σ1 hΣ2,∆02i with∆2⊆∆02&ψ∈Σ2
. Let δ ∈Σ1. If there is some δ0 such that δ◦δ0 ∈∆
then δ0 ∈ ∆2 ⊆ ∆02. Therefore δ0 ∈/ Σ2 because hΣ2,∆02i is a disjoint pair. As δ, δ0 were arbitrary then
Σ1◦Σ2∩∆ =∅, as desired.
Secondary subcase.
Now let us xΣ∈X and show that:
Σcχ◦ψ i χ◦ψ∈Σ.
By denition of Σ c χ◦ ψ in (5.1.20), Σ
c χ◦ ψ if and only if ∀∆ [∆V ϕ◦ψ⇒Σ≤∆], that is
∀∆ [ϕ◦ψ∈∆⇒Σ≤∆] by the previous result. But clearly (ϕ◦ψ∈∆) ⇒ Σ ≤ ∆ i (ϕ◦ψ∈∆) ⇒
Σ∩∆6=∅iϕ◦ψ∈Σ. SoΣcχ◦ψ i χ◦ψ∈Σ, as desired.
The Truth lemma is proven.
5.2.2. Completeness theorem and proof. The moment arrived for us to present the completeness theorem.
Theorem 120. (completeness) Given a pairΣ0Λ∆ and a SML logic Λ there is a model Mbased on some
SM L-frame F such that Σ1M ∆ (i.e. there exists a two-sorted point hx, yi which is a maximal pair with
x∈X andy∈Y and such that M, xΣandM, y∆).
Proof. Assume Σ0 ∆. Then hΣ,∆i is a disjoint pair which via corollary (102) can be extended to a maximal pair hΣ0,∆0i. ConsequentlyΣ0 ∈ X and ∆0 ∈ Y in the DM L-canonical frame as described in
denition 112, and the canonical valuation guarantees thatΣ0 Mc Σand∆0Mc ∆and thereforeΣ1Mc∆.
CHAPTER 6