branch and that we apply the (l5) rule to some 1:n::2P to get a branch
C containing 1 ::22P. We have to show thatC is also
K5
-satisable.As before there is some
K5
-model M = hW;R;Vi and someK5
-interpretation I in M such that I(1:n) 2 W and I(1:n) j= 2P. Since 1 is used onBand 11:n, there must be someI(1)2W withI(1)RI(1:n).
Now suppose for a contradiction thatI(1)j=:22P; then there is some w2W such thatI(1)Rw andwj=:2P, which in turn implies that there is some w0 2 W such that wRw0 and w0 j= :P. Since R is euclidean, I(1)RI(1:n) andI(1)RwgiveswRI(1:n), and thenwRw0 givesI(1:n)Rw0. But then I(1:n)j=2P impliesw0 j=P; contradiction. HenceI(1)j=22P andC is
K5
-satisable underI in M.Theorem 6.3.3.
If the systematic tableau for X closes then X isL
-unsatisable.Proof:
For a contradiction, suppose the tableau for X is closed and thatX isL
-satisable. The latter means that there is someL
-modelM=hW;R;Vi and some worldw2W such thatw j=X. Our tableau begins with nodes 1 ::Ai, for eachAi 2X so dene an
L
-interpretationI inM such thatI(1) =w. Then the initial tableau comprising the linear sequence of these nodes 1 :: Ai isL
-satisable (under I in M). Since each of our tableau rules is sound, any tableau obtained from this initial tableau by these rules is alsoL
-satisable. Hence our tableau isL
-satisable.SupposeBis some branch of this closed tableau. ThenBitself is closed and hence contains some labelled formula ::P and also contains:::P. Now any
L
-interpretation I0 for B in anyL
-model M0 would entail that I0()j=P and also that I0()j=:P, which is clearly impossible. HenceB is not
L
-satisable. Since Bwas an arbitrary branch this must be true for all branches of this closed tableau. Then, by denition, our tableau is notL
-satisable. Contradiction, hence if the tableau for X closes thenX isL
-unsatisable.Corollary 6.3.4 (soundness).
If the systematic tableau forf:Agis closed thenA isL
-valid.The systematic tableau is a nitely generated tree in that each node has at most two immediate children (since branches are caused only by the (_) rule). By Konigs lemma, an innite but nitely generated tree must contain an innite branch (see Fitting [Fit83, pages 404-407]). Hence there are four ways in which the systematic procedure can go on ad innitum:
1. by constructing an innite branch containing a sequence of distinct labelled formulae :: P1; ::P2; ::P3;; ::Pn; all with the same label;
2. by constructing an innite branch containing a sequence of labelled formulae :1 :: P1;:2 :: P2;:3 :: P3;;:n :: Pn; all simple extensions of some common ;
3. by constructing an innite branch containing a sequence of labelled formulae1::P1;2::P2;3 ::P3;;n ::Pn; all with dierent labels ; and
4. by traversing a set of formulae that repeatedly switch from asleep to awake and vice-versa on the visit sequence.
We show that items (1), (2) and (4) cannot occur.
Lemma 6.4.1.
In any branch of a systematic tableau for the nite set of formulae X, the maximum number of formulae with some given label is nite.Proof:
By induction on the length of . If jj = 1 then = 1 and the only possible formulae with this label are either subformulae of X, negations of a subformula of X, or are obtained from some subformulae of X by the building up rules (l5) and (lD). But no innite sequence of building up rules is possible. Ifjj1 then must have been created by (l) which adds only the negation of a subformula of its numerator. For details see Fitting [Fit83, page 411].Item 1 above is then impossible since any branch has but a nite num-ber of formulae with label and we do not permit the branch to contain repetitions. We leave it to the reader to compute actual bounds noting the presence of the \building up rules" (lD) and (l5); see Massacci [Mas94]
Lemma 6.4.2.
In any branch of a systematic tableau for the nite set of formulae X, the numberNk of dierent labels of lengthkis nite.Proof:
Proof by induction onkand the fact that the systematic tableau construction avoids repetitions. See Fitting [Fit83, pages 410-412] and Massacci [Mas94] for more exact bounds but once again beware that these need to be adjusted for the \building up rules".Thus no branch can contain an innite number of labels all of the same lengthkfor anyk, and item 2 above is also impossible.
We now turn to item 4 in some detail since these details cannot be found elsewhere. First note that although a branch does not contain repetitions,
the visit sequence may do so.
Lemma 6.4.3.
A particular labelled formula occurrence :: 2Q on the visit sequence can be awakened only a nite number of times.Proof:
The only way to awaken a-formula occurrence ::2Q is to visit some-formula occurrence :::2P that appears on the same branch as ::2Q. Since the systematic tableau is nitely branching, the number of such branches is nite. A branch can contain :: :2P at most once, hence the number of occurrences of :::2P on the visit sequence is (also) nite. Sincemust be of nite length, Lemma 6.4.1 guarantees that there are only a nite number of formulae with label on any branch of the tableau. Hence there are a nite number of -formulae occurrences that can awaken::2Q.If none of these-formulae occurrences is visited then ::2Qis never awakened. On the other hand, whenever one of these -formulae occur-rences is visited, it is marked as nished, and-formulae are never reawak-ened, hence::2Qcan be awakened only a nite number of times. Since this formula occurrence was an arbitrary -formula occurrence we know that every -formula occurrence can be awakened only a nite number of times.
Lemma 6.4.4 (fairness).
If a labelled formula occurrence ::A on the visit sequence is awake at the end of Stage n, the systematic procedure is guaranteed to visit it at some later stage.Proof
: By induction on the number of -formulae occurrences that precede :: A in the visit sequence. Clearly, if :: A is the root then it is immediately visited at Stagen+ 1. Similarly, if there are no-formulae occurrences between the root and ::A on the visit sequence then every subsequent stage will visit the next intervening formulae occurrence in the visit sequence and mark it as asleep or nished. The absence of intervening -formulae occurrences means that no formulae occurrences can awaken until after :: A is visited. Hence there must come a stage that visits ::A.Suppose the lemma holds for any labelled formula occurrence with j -formulae occurrences preceding it in the visit sequence.
Consider some::Aoccurrence that is awake at the end of stagenbut that has j+ 1 -formulae occurrences preceding it in the visit sequence.
Let :: :2B be the last -formula occurrence in the visit sequence that precedes::A.
If :::2B is not awake at the end of stagenthen it must be nished, meaning that all -formula occurrences preceding :: A in the visit se-quence must be nished. Each subsequent stage must visit one of the awake -formulae occurrences preceding ::Aand mark each one as asleep. No -formulae occurrences can awaken during this process since there are no
awake -formula occurrences preceding :: A. Hence there must come a stage that visits ::A.
If :::2Bis awake at the end of stagenthen it satises the induction hypothesis, so it will eventually be visited at some later stage, and marked as nished, meaning that no-formula occurrences preceding::Ain the visit sequence are awake. Some -formulae occurrences preceding :: A may be awakened by the visit to :::2B but each of these will be visited in turn and put to sleep in the stages that follow. Again, no formulae occurrences will be awakened in this process. Hence there must come a stage when we visit the formula occurrence immediately after :::2B in the visit sequence. If this is::Athen we are done. Otherwise this stage and subsequent stages must bring us closer and closer to::Asince none of these intervening formulae occurrences is a-formula.
Lemma 6.4.5.
No labelled formula occurrence on the visit sequence can remain awake for ever.Proof:
Suppose the occurrence::Ais awake at stagen. Lemma 6.4.4 guarantees that ::A will be visited at some later stage mwith m > n. If :: A is not a -formula then it will be marked as nished and will remain so hereafter. Else :: A is a -formula and it will be marked as asleep at the end of stagem. If ::Aever awakens at some later stagek then Lemma 6.4.4 again guarantees that it will be visited and put back to sleep. But this can happen only a nite number of times since Lemma 6.4.3 guarantees that :: A can awaken only a nite number of times. Hence there must come a stage when::Ais put to sleep, never to awaken again.Thus the systematic procedure is \fair" in that item 4 is also impossible.
The only way the systematic procedure can go ad innitum is for some branch to have at least one innite sequence of longer and longer labels of the form, :n1, :n1:n2 where each label is a simple extension of its predecessor. In fact, since every label starts with a 1 we can be more precise as below (again following Fitting [Fit83]).
A
chain
is a sequence of labels 1, 1 , 2 where each label in the sequence is a simple extension of its predecessor [Fit83]. A chain of labels 1,1,2from branchBisperiodic
if there exist distinct labelsiand j in the chain (i < j) such that i ::Ais onB ij ::A is onB; that is iffAji ::A on B g=fBjj ::B onB g. A branch isperiodic
if everyinnite chain (of labels) onBis periodic.
Lemma 6.4.6.
If any branch of a systematic tableau for the nite set of formulae X is innite, then it must be periodic [Fit83].Proof:
Basically, given a nite X, there is a limit to the number of dierent (unlabelled) formulae we can play with, even with the building up rules. Thus any innite chain of prexed formulae from any one branchmust repeat formulae at some stage. Since this is true for every chain on an innite branch, the branch must become periodic.
We thus have a handle on the systematic construction since an innite branch is not as bad as it rst seemed. If we could keep track of cycles then we could obtain a decision procedure. We briey return to this point later.