In this section we give tableau calculi for the modal logics that have im-portant readings as logics of \provability" where2Ais read as \it is prov-able in Peano Arithmetic thatAholds"; see Fitting [Fit83, page 241] and Boolos [Boo79]. These systems are obtained either by adding the axiom G:2(2A!A)!2A, named after Godel-Lob and sometimes calledGL, or adding the axiomGrz:2(2(A!2A)!A)!A, named after Grzegor-czyk, or adding the axiom 4 and the axiomGo:2(2(A!2A)!A)!2A, to
K
.It is known that both Gand Grzimply the transitivity axiom 4 when they are respectively added to
K
[vB78]. But the logicK4G
owhose frames share some of the properties ofG
-frames andGrz
-frames, explicitlycon-(G) 2X;:2P
X;2X;:P;2P (Grz) 2X;:2P X;2X;:P;2(P!2P)
Fig. 9.
Tableau Rules for logics of provability tains 4 as an axiom. It is also known thatGrz implies reexivity.Once again, all the tableau calculi contain the rules ofC
PC
and one ormore logical rules from Figure 9 on page 52 as shown below:
C
L
Static Rules Transitional Rules XCLC
G
CPC
(G) XeC
K4G
o CPC
(Grz) Sf2(Xe !2Xe)C
Grz
CPC
, (T) (Grz) Sf2(Xe !2Xe) The semantic and axiomatic intuitions behind these rules are more en-lightening than any technical proof (of soundness) so we present these as well.Intuitions for
(G):
We know that axiomatically formulated logicG
is characterised by
G
-frames. Therefore, axiom G must be valid on anyG
-frame; hence true in any world of anyG
-model. The axiomGis2(2A!A)!2A:
Its contrapositive is
:2A!:(2(2A!A)) which is the same as
:2A!3(2A^:A):
Thus, if the numerator represents a worldwwhere:2P is true, then there exists another worldw0 where2P is true andP is false, andw0is reachable fromw. The denominator represents this world.
Intuitions for
(Grz):
The axiomGrz is2(2(A!2A)!A)!A:
It is known that 4 and T are theorems of
Grz
[HC84, page 111], henceS4
Grz
. Segerberg [Seg71, page 107], and more recently Gore et al [GHH95], show thatGrz
=S4Grz
=S4G
o whereGois2(2(A!2A)!A)!2A
which gives the following (contraposed formulae) as theorems of
Grz
::2A ! :2(2(A!2A)!A)
:2A ! 3(2(A!2A)^:A):
Thus, if :2P is true at the numerator, then there exists some world where 2(P ! 2P)^:P eventually becomes true. The denominator of (Grz) represents this world.
Theorem 4.16.1 (Soundness).
The calculi CG
,CGrz
andCK4G
oaresound with respect to
G
-frames,Grz
-frames andK4G
o-frames respec-tively.Proof Outline :
For each rule in CL
we have to show that if the numerator of the rule isL
-satisable then so is at least one of the denomi-nators.Proof of
CG :
SupposeM =hW;R;Vi is aG
-model, w0 2W and w0 j= 2X;:2P: Thus there exists some w1 2 W withw0Rw1 and w1 j= X;2X;:Pby the transitivity ofR:SinceRis irreexive,w06=w1:Suppose w1 6j=2P: Then w1 j= :2P and there exists some w2 2W with w1Rw2 andw2j=X;2X;:P by transitivity ofR:SinceR is irreexive,w16=w2: Since R is transitive, w2 = w0 would give w1Rw0Rw1 implying w1Rw1 and contradicting the irreexivity ofR;hencew06=w2:Supposew26j=2P then ::: Continuing in this way, it is possible to obtain an innite path of distinct worlds inMcontradicting theG
-frame condition onM. Thus there must exist somewi2W withw0Rwi andwij=X;2X;:P;2P and we are done.Proof of
(T)for
CGrz :
The (T) rule is sound forGrz
-frames since everyGrz
-frame is reexive.Proof of
(Grz)for
CK4G
o:
Suppose M = hW;R;Vi is aK4G
o -model, then R is transitive, there are no proper clusters, and there are no proper 1-R-chains. Suppose w0 2 W is such that w0 j= 2X;:2P. We have to show that there exists somewn 2W withw0Rwn andwn j= X;2X;:P;2(P !2P). SinceRis transitive,w0j=2Xmeans that8w2 W;w0Rw impliesw j=X;2X. Thus our task is reduced to showing that there exists somewn 2W such thatw0Rwn andwn j=:P;2(P !2P): Suppose for a contradiction that no such world exists in W. That is,(a) 8w2W;w0Rwimpliesw6j=:P;2(P !2P).
Since w0 j=:2P, there exists some w1 2 W with w0Rw1 and w1 j=
:P. By (a), w1 6j= 2(P ! 2P) and hence w1 j= :2(P ! 2P): Thus there exists some w2 2 W with w1Rw2 and w2 j= :(P ! 2P); that is, w2j=P^:2P. Sincew1j=:P,w16=w2and since
K4G
o-models cannot contain proper clusters, w0 6= w2. Since w2 j= :2P there exists some w3 2 W with w3 j= :P. Since w2 j= P, w3 6= w2. And w3 6= w0 and w36=w1 as either would give a proper cluster. By (a),w36j=2(P !2P)and hencew3j=:2(P !2P). Continuing in this way, we either obtain an innite path of distinct points, giving a proper1-R-chain, or we obtain a cycle, giving a proper cluster. Both are forbidden in
K4G
o-frames. Hence (a) cannot hold and 9w 2W;w0Rw and w j=:P;2(P !2P): That is, the desiredwn exists.Proof of
(Grz)for
CGrz :
EveryGrz
-frame is aK4G
o-frame, hence the proof above suces.As we saw in Subsection 4.11, proving completeness boils down to prov-ing the followprov-ing: ifX is a nite set of formulae and noC
L
-tableau for X is closed then there is anL
-model forX on anL
-framehW;Ri.Lemma 4.16.2.
If there is a closedCL
-tableau forX then there is a closedC
L
-tableau for X with all nodes in the nite setXCL .Proof:
Obvious from the fact that all rules forCL
operate with subsets ofXCL only.Lemma 4.16.3.
For each CL
-consistentX there is an eective procedure to construct some niteCL
-saturatedXswith XXsXCL .Theorem 4.16.4 (Completeness).
If X is a nite set of formulae and X isCL
-consistent then there is anL
-model forX on a niteL
-frame.As usual we construct someC
L
-saturatedw0 fromX with X w0XCL .
Proof for
CG:
If no :2P occurs in w0 thenhfw0g;;i is the desired model graph as (i)-(iii) are satised. Otherwise, letQ1;Q2;;Qm be all the formulae such that :2Qi 2 w0. Create a CG
-saturatedQi-successor for eachQi using () and (G) giving the nodes vi of level one. Repeating this construction on the nodes of level one gives the nodes of level two, and so on for other levels. Consider any sequencewiwi+1wi+2. Since wihas a successor, there is some:2Q2wiand2Q2wi+j for allj1 by (G). Thuswi6=wi+jfor anyj1 and each such sequence must terminate sinceXCG is nite. LetRbe the transitive closure of; that is putwRw0 ifww0 and putwRvifww0 v. The resulting tree is a model graphhW0;RiforX which is also a
G
-frame.Proof for
CGrz:
If no:2P occurs inw0thenhfw0g;f(w0;w0)giis the desired model graph as (i)-(iii) are satised. Otherwise, letQ1;Q2;;Qmbe all the formulae such that:2Qi 2w0 and :Qi62w0. Create aC
Grz
-saturatedQi-successor for eachQiusing () and (Grz) giving the nodesvi
of level one, and so on for other levels. Consider any sequencewiwi+1
wi+2. Sincewi has a successor, there is someQ such that:2Q2wi,
:Q 62 wi, and by (Grz), 2(Q ! 2Q) 2 wi+j for all j 1. Suppose wi+j =wi, then2(Q!2Q)2wi and henceQ!2Q2wi by (T). Since Q!2Qis just abbreviation for :(Q^:2Q), we know that :Q2wi or
::2Q2wi. We created a successorwi+1forwiprecisely because:Q62wi
and so the rst case is impossible. And if::2Q2withen2Q2wiby (:),
contradicting the
Grz
-consistency of wi since:2Q 2wi by supposition.Thus each such sequence must terminate (without cycles). Let R be the reexive and transitive closure of to obtain a model graph hW0;Rifor X which is also a
Grz
-frame.Proof for
CK4G
o:
If no :2P occurs in w0 then hfw0g;f;gi is thedesired model graph as (i)-(iii) are satised. Otherwise, letQ1;Q2;;Qm
be all the formulae such that :2Qi 2 w0. A C
K4G
o-saturated set v is reexive i 2A 2 v implies A 2v. If v is non-reexive then there exists some2B2v butB62v.Ifw0 is reexive then create aC
K4G
o-saturatedQi-successor for each:2Qiwith:Qi62w0, otherwise ifw0is non-reexive then create aC
K4G
o -saturated Qi-successor for each :2Qi, 1 i m. This gives the nodes of level one. Continue creating successors in this fashion for these nodes using () and (Grz).Consider any sequence wi wi+1wi+2. Sincewi has a successor, there is some:2Q2wi that gives rise towi+1. Also,2(Q!2Q)2wi+j for allj1.
If wi is reexive then :Q 62 wi, and yet :Q2 wi+1 by (Grz); hence wi 6= wi+1. Suppose wi+j = wi, j 2. That j 2 is crucial! Then
2(Q ! 2Q)2 wi and Q! 2Q 2 wi by (Grz). Since Q ! 2Q is just abbreviation for:(Q^:2Q), we know that:Q2wior::2Q2wi. Since wiis reexive, we created a successorwi+1forwiprecisely because:Q62wi
and so the rst case is impossible. And if::2Q2withen2Q2wiby (:), contradicting the
K4G
o-consistency ofwisince:2Q2wi by supposition.If wi is non-reexive then there is some 2B 2 wi, withB 62 wi, and yet both 2B and B are inwi+j by (Grz), for all j 1; hencewi 6=wi+j, j1.
Thus each such sequence must terminate (without cycles). Let R be the transitive closure of and also putwRw ifwis reexive to obtain a model graphhW0;RiforX which is also a
K4G
o-frame.As Amerbauer [Ame93] points out, this means that
K4G
o is charac-terised by nite transitive trees of non-proper clusters refuting the conjec-ture of Gore [Gor92] thatK4G
o is characterised by nite transitive trees of degenerate non-nal clusters and simple nal clusters.4.16.1 Bibliographic Remarks and Related Systems
The tableau systemC
G
is from Fitting [Fit83] who attributes it to [Boo79], whileCGrz
is from Rautenberg [Rau83]. Rautenberg gives a hint on how to extend these to handle CK4G
o but Gore [Gor92] is unable to give an adequate system forCK4G
o, leaving it as further work. The givenCK4G
ois due to Martin Amerbauer [Ame93] who following suggestions of Rauten-berg and Gore also gives systems for
KG
:2
andKGL
(which Amerbauer callsK4
:3G
).Provability logics have also been studied using Gentzen systems, and
ap-propriate cut-elimination proofs have been given by Avron [Avr84], Bellin [Bel85], Borga [Bor83], Borga and Gentilini [BG86], Sambin and Valentini [SV80, VS83, SV82], and Valentini [Val83, Val86].