ELIMINATION OF PROBLEMS AND DEFICIENCIES

6. Elimination of Problems and Deficiencies

In this section, we explain how to eliminate problems and deficiencies in an adequate

ILM–frame Fn. This is done by adding new nodes and relations to Fn. We show

that appropriate labels can be found for the new nodes, and that the resulting frameGis a quasi–frame.

Recall that a problem inFnis a a pairha,¬(AB)is.t. ¬(AB)∈ν(a)∩ D, but

there is no y ∈ CB

a s.t.A ∈ν(y). In order to eliminate this problem, we will add

a new worldb to Fn, withb ∈ CaB andA ∈ν(b). Asb ∈ C B

a, and as we want the

resulting frame to be adequate, we have to make sure thatν(a)≺B ν(b). Corollary

88 of Theorem 87 shows that an appropriate label can be found forb.

Theorem 87. LetΓ be a maximalILM–consistent set s.t.¬(AB)∈Γ. LetC be s.t. CB∈Γ. Then the set {¬C,2¬C, A,2¬A} is consistent.

Proof. Suppose for contradiction that (¬C∧2¬C)∧(A∧2¬A)` ⊥.Then

A∧2¬A`C∨3C `A∧2¬A→C∨3C `2(A∧2¬A→C∨3C) (necessitation) `A∧2¬AC∨3C (J1) `A∧2¬AC (J5,J3) `AC (lemma 2.1)

ThusAC∈Γ. Since CB in Γ, byJ2alsoAB∈Γ, contradiction. 2

Corollary 88. Let Γ be a maximalILM–consistent set s.t. ¬(AB)∈Γ. Then there exists a maximal ILM–consistent set∆ s.t. Γ≺B∆ andA,2¬A∈∆.

Proof. We want ∆ to be a maximal consistent extension of

S:={¬C,2¬C|CB ∈Γ} ∪ {A,2¬A}.

The first set guarantees the first half of the definition for Γ≺B ∆. For the other

half, note that 2¬A /∈Γ. By Lemma 2.1,2¬A∈Γ would imply A⊥ ∈Γ, and thus also (as ⊥B ∈ Γ) AB ∈ Γ by J2. Suppose for contradiction that S is inconsistent. By compactness, there isks.t.

(¬C1∧. . .∧ ¬Ck)∧(2¬C1∧. . .∧2¬Ck)∧(A∧2¬A)` ⊥.

Since2commutes with∧, we get

(¬C1∧. . .∧ ¬Ck)∧2(¬C1∧. . .∧ ¬Ck)∧(A∧2¬A)` ⊥,

and thus

(A∧2¬A)`(C1∨. . .∨Ck)∨3(C1∨. . .∨Ck).

LetC=:C1∨. . .∨Ck. Since for alli≤k,CiB ∈Γ, byJ3CB∈Γ . But then

(¬C∧₂¬C)∧(A∧₂¬A)` ⊥

6. ELIMINATION OF PROBLEMS AND DEFICIENCIES 82

Recall that a deficiency in Fn is a tripleha, b, CDis.t aRb, CD∈ν(a)∩ D, C∈ν(b)∩ D, but for nozs.t. bSazwe haveD∈ν(z). To eliminate this deficiency,

we will add a new nodecto Fn withD∈ν(c) andbSac. As we want the resulting

frame to be adequate, we need to make sure thatν(b)⊆₂ν(c), and that ifb∈ CB x

for someBandx, thenν(x)≺B ν(c). Corollary 90 of the following theorem shows

that an appropriate label can be found forc.

Theorem 89. Let Γ be a maximal ILM–consistent set s.t. CD ∈Γ, and let ∆ be s.t. Γ ≺B ∆ and C ∈ ∆. There exists a maximal ILM–consistent set ∆0 with

Γ≺B ∆0 andD,2¬D∈∆0.

Proof. If¬(DB)∈Γ, then by Corollary 88 there is a ∆0s.t. Γ≺B ∆0 and D,2¬D∈∆0. If¬(DB)∈/Γ, then by maximal consistency DB∈Γ, whence alsoCB∈Γ byJ2. But then Γ≺B∆ implies¬C,2¬C∈∆, contradiction. 2

Corollary90. LetΓand∆be maximalILM–consistent sets s.t.Γ≺B ∆,CD∈

Γ, and C ∈ ∆. There exists a maximal ILM–consistent set ∆0 s.t. Γ ≺B ∆0, D,2¬D∈∆0 and∆⊆₂∆0.

Proof. We show that for any 2E ∈ ∆, there is a ∆0 with Γ ≺B ∆0 and D,2¬D,2E∈∆0. The desired result follows by compactness, and by commutation of boxes and conjunctions (as the proof of Corollary 88.). As CD ∈ Γ and Γ is maximal ILM–consistent, also C∧2ED∧2E ∈ Γ. Clearly, we have that

C∧2E∈∆. By Corollary 88, we find a ∆0 s.t. Γ≺B ∆0,D,2E,2(¬D∨ ¬2E).

But already in GL, we have that (D∧2E∧2(¬D∨ ¬2E))→2¬D (using that

2E→22E). Hence also2¬D∈∆0 as required. 2 We will now show how a problem or a deficiency inFn can be eliminated in such a

way as to yield a quasi–ILM–frame G. As said before, we will always eliminate the smallest element ofP which is indeed a problem or a deficiency in F.

Problems. Suppose that the least element ofP which is indeed a problem or deficiency inFn is a problemha,¬(AB)i. Using Corollary 88, we find a maximal

consistent set ∆ s.t.ν(a)≺B∆ andA,2¬A∈∆. Fix some b /∈W and define G=hW∪ {b}, R∪ {ha, bi}, S, ν∪ {hb,∆ihha, bi, Bi}i.

We now have to check thatG is a quasi–ILM–frame, i.e. that it is adequate, and satisfies the conditions of Definition 86.

We will prove here the only case where some work has to be done. For adequacy, we have to show that G y ∈ CE

x ⇒ G ν(x) ≺E ν(y). So suppose that Gy∈ CE

x. We only need to consider the case wherey =b, as in all other cases Fn y ∈ CxE ⇔ Gy ∈ CxE. In case x=aand E =B, we get the property by

the choice ofν(b). So suppose thatx6=a. If a∈ CE

x, then we haveν(x)≺E ν(a)

by adequacy ofFn. Sinceν(a)≺ν(b), this impliesν(x)≺Eν(b). In case a /∈ CxE,

there must be somew∈ CE

x s.t.wStra. By adequacy

6 _ofF

n,ν(w)⊆₂ν(a). Thus ν(x)≺E ν(w)⊆2 ν(a)≺ν(b). It is easy to see that then alsoν(x)≺Eν(b).

6_{Note that the adequacy conditionyS}