QUASI–FRAMES

For this, we will first show that even thoughGis not an adequateILM–frame, it is “close enough”. The meaning of “close enough” here is captured in the notion of a quasi–ILM–frame, which will be defined in the next section. Second, we will show that every frame which is “close enough” to being an adequateILM–frame can in fact be extended to an adequateILM–frame. This extension ofGwill be theFn+1 that we are looking for.

5. Quasi–Frames

Definition86. An adequate frameG=hW, R, S, νiis aquasi–ILM–frame (or just

quasi–frame) if the following properties hold:

i. Ris converse well-founded

ii. ySxz⇒xRy andxRz

iii. Rtr_◦Str _{is converse well-founded}

Lemma 5.1. Let G0 =hW, R, S, νi be a quasi–frame. There is an adequate frame

Fn+1 extendingG.

Proof. We will construct Fn+1 as the union of an infinite chain of quasi– frames {Gj}_j∈ω. We define an imperfection on a quasi frameGj to be a tupleγ

having one of the following forms:

i. γ=h0, a, b, ciwithGjaRbRcbut Gj 2aRc

ii. γ=h1, a, b,iwithGj aRbbutGj2bSab

iii. γ=h2, a, b, c, diwithGj bSacSadbut Gj2bSad

iv. γ=h3, a, b, ciwithGjaRbRcbut Gj 2bSac

v. γ=h4, a, b, c, diwithGj bSacRdbutGj2bRd

Thus an imperfection onGj is just a violation of an ILM–frame condition on Gj.

Imperfections will be eliminated step by step. Each Gj in the chain will have

at least one imperfection less than its predecessor, and the union Fn+1 will have no imperfections at all. Fix some ordering of the imperfections4. To go from

Gj =hWj, Rj, Sj, νito Gj+1, we choose the least imperfection γ in the ordering.

Depending on the form ofγ,Gj+1 will be defined as:

i. hWj, Rj∪ {a, c}, Sj, νi

ii. hWj, Rj, Sj∪ {a, b, b}, νi

iii. hWj, Rj, Sj∪ {a, b, d}, νi

iv. hWj, Rj∪ {a, c}, Sj∪ {a, b, c}, νi

v. hWj, Rj∪ {b, d}, Sj, νi

4_{New imperfections will arise during the process. For example, when we setaRcin case i, we} will have a new imperfection sinceGj+12cSac. We will also add the new imperfections to the

5. QUASI–FRAMES 80

The proof is by induction. It has to be established that in all cases, Gj+1 will remain a quasi–frame. Thus we have to check thatGj+1 is adequate, and satisfies the conditions of Definition 86. In total, 35 cases have to be checked, but most of them are straightforward. We will illustrate the general strategy behind the proofs with a few examples.

Suppose thatGj+1 is defined by case v, i.e.Gj+1=hWj, Rj∪ {b, d}, Sj, νi.

We will show that R is converse well-founded. Suppose that there is an infinite sequence s.t.Gj+1 z1Rz2R . . .. Replace every occurrence of bRdin the sequence bybSacRdand leave the rest unchanged. If there are infinitely manySa transitions

in the new sequence, we get a contradiction to the assumption that Rtr_◦Str _is

converse well-founded onGj. If not, we get a contradiction to the assumption that Ris converse well-founded on Gj.

For adequacy, we have to check (among other things) thaty∈ CB

x ⇒ν(x)≺Bν(y),

and that if A 6= B, then CB x 6= C

A

x. We will instead prove that the critical and

generalized cones ofGj and Gj+1 are the same5. Suppose Gj+1 y ∈ CxB. Then

there arez1. . . , zms.t.Gj+1 xRBz1(Sx∪R∪(R◦Str))z2. . . zm(Sx∪R∪(R◦Str))y.

We transformz1. . . , zminto a new sequencew1. . . , wnby replacing each occurrence

ofbRdbybSacRd. Thusb(R◦Str)d. We leave the rest of the sequence unchanged.

Clearly, Gj xRBw1(Sx∪R∪(R◦Str))w2. . . wn(Sx∪R∪(R◦Str))y, whence Gjy ∈ CxB. A similar proof strategy works for showing that the generalized cones

remain unchanged.

In the end, we also have to make sure thatFn+1:=Sj∈ωGj is an adequateILM–

frame.

By construction,Fn+1 has no imperfections, so it satisfies the correspondingILM– frame conditions. Furthermore, it is clear that the conditions on adequacy are preserved under unions of chains. The only non-obvious thing is the converse well- foundedness ofR. To prove that, we show that for allj,

Gj xRy⇒G0x(Rtr◦Str,refl)try.

We will first show how this gives us the desired result. Suppose for contradiction that Fn+1 x1Rx2Rx3. . .. Each of the R relations in the chain was defined at some Gj. By the above claim, we can reduce each of them to an (Rtr◦Str,refl)tr

relation inG0. So we get an infinite chain{yi}i∈ωs.t.

G0y0Str,refly1Rtry2Str,refly3Rtry4Str,refl. . .

If there are infinitely manyStr _{transitions in this sequence, we get a contradiction}

to the converse well-foundedness ofRtr_◦Str _onG

0. If not, we get a contradiction to the converse well-foundedness ofR onG0.

The claim Gj xRy ⇒ G0 x(Rtr◦Str,refl)try is proven by induction onj. The proof is straightforward, once we note that we have to prove the stronger claim:

GjxRy∨xSzy⇒G0x(Rtr◦Str,refl)try.

This finishes the proof of Lemma 5.1. 2 5_{In fact, the critical and generalized cones will remain unchanged throughout the whole process} of buildingFn+1. In all cases, the process might at most shorten the path that a node has to take