Properties of tense logisFrank WolterShool of Information SieneJAISTTatsunokuhi, Ishikawa 923-12, Japane-mail: wolter�jaist.a.jpAbstratBased on the results of [11℄ this paper delivers uniform algorithms for deidingwhether a �nitely axiomatizable tense logi has� the �nite model property,� is omplete with respet to Kripke semantis,� is strongly omplete with respet to Kripke semantis,� is d-persistent,� is r-persistent.It is also proved that a tense logi is strongly omplete i� the orresponding variety ofbimodal algebras is omplex and that a tense logi is d-persistent i� it is omplete andits Kripke frames form a �rst order de�nable lass. From this we obtain many naturalnon-d-persistent tense logis whose orresponding varieties of bimodal algebras areomplex.Mathematis Subjet Classi�ation: 03B45, 03B25Keywords: Modal Logi, Tense Logi, strong ompleteness, omplex variety, d-persisteny, Kripke semantis1 IntrodutionTogether with [12℄ this paper shows that for nearly all interesting properties P there existalgorithms whih deide whether a �nitely axiomatizable tense logi enjoys P . It is to beemphasized that we also investigate tense logis with respet to the properties introduedabove in order to get some insight into those properties in general. This is indeed thease sine tense logis turn out to be a lass of polymodal logis in whih we an alreadyobserve most of the interesting phenomena whih o

ur in the more general framework ofpolymodal logi.We use the notation from [11℄. Reall that a tense logi is a normal bimodal logiwhih ontains the bimodal logi Lin:t of the lass of weak orderings (i.e. sets with a1

1 INTRODUCTION 2relation R whih is transitive and onneted). We assume that tense logis are formulatedin the propositional language L! with ! propositional variables and modal operators 2+and 2�. A tense frame is a struture G = hg;R+; R�; Ai suh that R� = (R+)�1, hg;R+iis a weak ordering, and A � 2g is losed under boolean operations and2a = fx 2 gj(8y 2 g)(xSy ) y 2 a)g;for (2; S) = (2+; R+); (2�; R�). (We shall often abbreviate hg;R+; R�; Ai by writinghg;R;Ai, where R = R+). It is well known that eah tense logi is determined by tenseframes. G is a Kripke frame i� A = 2g and we denote it by hg;Ri or simply by g. The lassof Kripke frames is denoted by Fr. (Consult [11℄ for a detailed disussion of the semantisfor tense logis).We shall now introdue those properties of tense logis whih will be investigated inthis paper. A tense logi � has the �nite model property i� it is determined by �nite tenseframes. � is omplete (with respet to Kripke semantis) i� � is determined by Kripkeframes, i.e. i�, for all � 62 �, there exists a Kripke frame g with g j= � suh that g 6j= �.Call a set � � L! of formulas �-onsistent i� ? is not derivable from � [ � by usingmodus ponens. Now � is alled strongly omplete (alias ompat) i� eah �-onsistentset � is satis�able in a Kripke-frame for �. (As usual, � is alled satis�able in a Kripkeframe g i� there exists a valuation V and x 2 g suh that hg; V; xi j= �). Consult e.g. [8℄,[9℄, [3℄ for results on as well as disussion of strong ompleteness. A tense logi � is alledelementary i� it is omplete and the Kripke frames for � form a �rst order de�nable lass(in the language with equality and a relational symbol R). Consult e.g. [2℄ and [1℄ fororrespondene theory.The following onepts will be introdued in the general setting of normal bimodallogis. Thus we shall use the notion of a (bimodal) frame in the usual sense and do notrestrit attention to rooted frames. (Note that the restrition to rooted frames is impliitin our de�nition of tense frames). Reall that an algebra A = hA;\;�;2+;2�; 1i is abimodal algebra if hA;\;�; 1i is a boolean algebra and 21 = 1 and 2(a\b) = 2a\2b, forall a; b 2 A and 2 = 2+;2�. The lass V(�) is the variety (equational lass) of bimodalalgebras A with A j= � = 1, for all � 2 � (f. [4℄). For a frame G = hg;Ai the set Aforms a bimodal algebra G+ = hA;\;�;2+;2�; gi. The underlying Kripke frame of G is�G = g. (In [11℄ �G is denoted by G+). Following Goldblatt [5℄ we all a variety V ofbimodal algebras omplex i�, for eah A 2 V, there exists a frame G with (�G)+ 2 V suhthat G+ ' A. Consult [5℄ also for a motivation of this de�nition. Strongly omplete logis� are losely related to omplex varieties. Call a normal modal logi � �-omplete, � � !a ardinal, i� eah �-onsistent set � in the language L� with � propositional variables issatis�able in a Kripke frame g with g j= �. (All the unde�ned terms here are the anonialextensions of the notation for L!). Correspondingly, all a variety V �-omplex i�, foreah A 2 V with jAj � �, there exists a frame G with (�G)+ 2 V suh that G+ ' A. Thefollowing is proved in [10℄. (Independently, this result was proved by Goldblatt, f. [6℄.)Theorem 1 For all in�nite ardinals � and normal (poly-) modal logis �, � is �-ompletei� V(�) is �-omplex.

1 INTRODUCTION 3In general, it is an open problem whether the ardinality onditions an be dispensedwith, that is, it is not known whether V(�) is omplex whenever � is strongly omplete.However, for tense logis we shall show in the present paper (f. Theorem 13) that thetwo notions are equivalent.A frame G = hg;R+; R�; Ai is re�ned i�[ref1℄ x = y , 8a 2 A(x 2 a, y 2 a);[ref2℄ xSy , 8a 2 A(x 2 2a) y 2 a);for (2; S) = (2+; R+); (2�; R�). Call a logi � r-persistent i� �G j= � whenever G j= �and G is re�ned. [7℄ is an extensive study of re�ned frames and r-persistent logis and[2℄ shows that all r-persistent logis are elementary. (Note that [2℄ alls r-persistent logisnatural). A frame G is alled desriptive if it is re�ned and TU 6= ;, for all ultra�ltersU in hA;\;�; gi. The lass of desriptive frames is denoted by Dfr. It is well knownthat, for eah algebra A, there exists a (uniquely) determined desriptive frame A+ withA ' (A+)+ (f. [4℄). A logi � is alled d-persistent i� �(A+) j= �, for eah A 2 V(�);equivalently, � is d-persistent i� �G j= � whenever G j= � and G 2 Dfr. (We note that,by a result of [7℄, d-persisteny oinides with anoniity. Reall that a normal modallogi � is alled anonial if �((F�(�)+) j= �, for all ardinals �. Here F�(�) denotesthe �-generated free algebra in V(�)). Although, in general, d-persisteny does not implyelementarity (f. [2℄), this turns out to be true for tense logis (f. Theorem 21). Thus weshall show, for all tense logis �, � is r-persistent+� is d-persistent (, � is elementary)+� is strongly omplete (, V(�) is omplex)+� is ompleteNo other impliation holds for all tense logis. The failure of an impliation from thethird line to the seond line is the interesting ase. In [3℄ it is shown that the logi R:t ofthe reals hR;

2 NOTATION AND BASIC RESULTS 42 Notation and basi resultsWe shall briey introdue only the basi notation from Part I. Reall that an interval ina Kripke frame hg;Ri is a non-empty subset I of g suh that fz 2 gjxRzRyg � I, for allx; y 2 g. An interval in a tense frame G is an interval I in �G suh that I 2 G+. Let Ibe an interval in a tense frame G = hg;R;Ai and de�ne a set of basi interval types asfollows.� I is a joker if I is just an interval, we write I 2 Jok.� I 2 C(0) i� I is a degenerate luster.� I 2 [J;!℄ i� all x 2 I have a proper R-su

essor in I.� I 2 [J;C(m)℄, m > 0, i� I has a non degenerate luster of size � m to the right.� I 2 [J;!n℄, n � 1, i� I 2 [J;!℄ or I 2 [J;C(n)℄.The meaning of other terms, for instane I 2 [C(n); J ℄ or I 2 [ n;!℄, are de�ned in theobvious way. If T is a set of interval types then preisely the expressionsf9p~S : ~S = hSij1 � i � ni is a sequene of interval types from Tgbelong to LT. A sequene of intervals ~I = hIij1 � i � ni in a frame G satis�es ~S = hSi :1 � i � ni, in symbols G; ~I j= ~S;i� g = (I1 � : : : � In) and Ii 2 Si, for all 1 � i � n. We say that 9p~S holds in G, insymbols G j= 9p~S, i� there exists a partition ~I = hIij1 � i � ni of G with G; ~I j= ~S. Put:LT = f:P jP 2 LTg. Then G j= :9p~S is an abbrevation for G 6j= 9p~S. For � � :LT or� � L! and K � Fr [ Dfr putModK(�) = fG 2 Kj8P 2 �(G j= P )gand all ModK(�) the lass of frames for � in K. The following result from Part I will bebasi for our investigation. PutM = fJok;C(0)g [ f[J;!m℄jm > 0g [ f[ m; J ℄jm > 0g [ f[ m;!m℄jm > 0g:Theorem 2 There is an e�etive proedure whih gives for � 2 L! a �nite set N(�) ofmodal interval types suh thatModDfr[Fr(�) =ModDfr[Frf:9p(~S)j~S 2 N(�)g:Conversely, there is an e�etive proedure whih gives for eah ~S 2 M a tense formula �with ModDfr[Fr(�) = ModDfr[Fr:9p(~S):

2 NOTATION AND BASIC RESULTS 5With this result at hand we shall work in the language :LM in order to solve the problemsintrodued. Nearly all the notation for tense logis is easily translated from L! to :LM.First, we simply say that a subset � of :LM is a tense logi i�(ModDfr[Fr(�) � ModDfr[Fr(P ))) P 2 �;for all P 2 :LM. The smallest tense logi ontaining a set � � :LM is denoted byLin:t��. For G 2 Fr [Dfr de�ne the theory of G by putting ThG = fP 2 :LMjG j= Pg.Now we are going to introdue the frames we shall work with. The lass of framesintrodued in Part I will not be suÆient for dealing with d-persisteny, r-persisteny andstrong ompleteness. So we add another type of frames. First reall that, for an ordinal�, we denote by �< the frame (or the empty set) h�;i. �> denotes the mirror imageof �. Now de�ne for n > 0 and m 2 !! [f(ni; ni)ji 2 !gi;m [f(ni; ni)jni < mgi:The mirror images of !(n) respetively.Reall that the lass of basi Kripke frames isBK = f!(n) : n 2 !g [ fC(n) : n 2 !g:De�nition 3 Suppose that C(m) = fx0; : : : ; xm�1g.(1) For n 2 ! and m > 0 with m > 1 if n > 0 de�ne the general tense frameG[n;C(m)℄ = h! 1 if n1 + n2 > 0,G[n1; C(m); n2℄ = h!(n2); Ai;where A is the tense algebra generated by fbi : 0 � i � m� 1g withbi = fxig [ f(nj + i; 0) : j 2 !g [ f(n; j + i; 1) : j 2 !g;for i < m. (Here we assume that !(m2) are based on the sets ! � f0g and! � f1g, respetively). The lass of frames de�ned under (3) is denoted by Dl.

2 NOTATION AND BASIC RESULTS 6The lass Df has already been introdued in Part I. In this paper we shall need thefollowing lasses BD0 = Df [ fC(m) : m 2 !g and DS = BD0 [Dl:In Part I we disussed the de�nition of Df and the reason why the ase m = 1 and n > 0is omitted in the de�nition of Df . In a similar way we here omit the ase n1 + n2 6=0 and m = 1 for Dl sine in this ase we would get a non-�nitely generated algebra(G[m1; C(1);m2℄)+. There will be no need for those frames.Proposition 4 [~G℄ is desriptive and [~G℄+ is �nitely generated, for all ~G 2 SDS.Proof. A straightforward proof shows that the [~G℄ are desriptive. Assume that ~G = hGi :1 � i � ni. Then [~G℄ is generated byfgi : 1 � i � ng [ faij : 1 � j � ni; 1 � i � ng;if faij : 1 � j � nig generates G+i , for 1 � i � n. The G+i are �nitely generated, byde�nition. aThe following Lemma plays a role similar to Lemmas 11 and 13 of Part I. De�ne form 2 ! and n > 0, the frame G+[C(n);m℄ by puttingG+[C(n);m℄ = G[C(n);m℄� C(0)m�1 �G[m;C(n);m℄;for m > 1, and G+[C(n);m℄ = G[C(n);m℄�G[m;C(n);m℄;otherwise. G+[m;C(n)℄ is the mirror image of G+[C(n);m℄. PutG2[m1; C(n);m2℄ = G[m1; C(n);m2℄�C(0)m2�1 �G[m2; C(n);m2℄;for m2 > 1, andG2[m1; C(n);m2℄ = G[m1; C(n);m2℄�G[m2; C(n);m2℄;otherwise. Also put(m1;m2)k = (m1(k + 1))(m2);for m1 > 1, and (m1;m2)k = (m1(k + 1))(m2);otherwise. Reall the following de�nitions. For ~S 2 M denote by l(~S) the length of ~S andputCl(~S) = ( 1 : no !m and no m o

urs in ~S;maxfm :!m or m o

urs in ~Sg : otherwise.Now de�ne, for �; � � ! and G 2 Dfr [ Fr,M(�; �) = f~S 2 Mjl(~S) � �;Cl(~S) � �g and Th�;�G = fP 2 :LM(�;�)jG j= Pg:

2 NOTATION AND BASIC RESULTS 7Lemma 5 For all m;m1;m2 2 ! and n > 0,1. Thk;!G[m1; C(n);m2℄ = Thk;!(m1;m2)k, for k > 0.2. ThG[m1; C(n1);m2℄ = ThG[m1; C(n2);m2)℄, for n1; n2 > 0.3. ThG[C(n);m℄ = ThG+[C(n);m℄.4. ThG[m1; C(n);m2℄ = ThG2[m1; C(n);m2℄.Proof. Similar to the proof of Lemma 13 in Part I. aLemma 5 is of basi importane in what follows. It says how we an manipulate frameswithout e�et on the validity of formulas. Certainly this is useful when we are interestedin the representation of algebras by frames. Another easy but important onsequene ofthis Lemma isTheorem 6 Th[~G℄ is deidable, for all ~G 2 SDS, and Th[~g℄ is deidable, for all ~g 2 SBK.Proof. (1) follows from Theorem 14 of Part I and Lemma 5 (1) above (by appliation ofProposition 5 of Part I). (2) is Theorem 18 (4) of Part I. aOften it will be helpful to have the following notation. A gap in a frame hg;Ri is apair hI1; I2i of disjoint intervals with g = I1 [ I2. We shall identify objets hI1; I2i withthe gap = hfx 2 g : (9y 2 I1)xRyg [ I1; I2 [ fx 2 g : (9y 2 I2)yRxgi;if I1 and I2 are disjoint intervals and I1 [ I2 is an interval. For two gaps 0 = hI1; I2i and1 = hI3; I4i in hg;Ri we write 0R1 if I1 � I3. In this ase we de�ne the interval [0; 1℄by putting [0; 1℄ = I2 \ I3:Hene, eah �nite set of gaps G = f0 R : : : R ng gives rise to a partitionpar[G℄ = f[1; 0℄; [0; 1℄; : : : ; [n�1; n℄; [n;1℄g:Conversely, given a partition ~I = hIi : 1 � i � ni of hg;Ri we putgap(~I) = fhIi; Ii+1i : i < ng:For a �nite set of partitions f~Ii : i � ng put_h~Ii : i � ni = par[f[fgap(~Ii) : i � ng℄:We shall need the following paramaters for talking about sequenes ~G 2 SDS. Thede�nitions are anonial extensions of the de�nitions in Part I for BD.� Cl(~G) 2 ! is the ardinality of the maximal luster in [~G℄.

2 NOTATION AND BASIC RESULTS 8� R(~G) 2 ! denotes the number of Gi in ~G whih are not irreexive points.� IR(~G) 2 ! is the length of the maximal in�x subsequene of the form (C(0))n in ~G.� S(~G) 2 ! is the maximal n suh that there is a Gi in ~G with Gi = G[n;C℄ orGi = G[C;n℄ or Gi = G[n;C; k℄ or Gi = G[k;C; n℄.Clearly there exist only �nitely many sequenes ~G 2 SDS where all those parameters arebounded. The following Proposition will be basi for most of the deision proedures.Proposition 7 Suppose that H0 = H1 � G � H2 2 Dfr and that � � :LM(m;n) withH0 j= �. Let H be a set of frames withH1 �H�H2 6j= �;for all H 2 H. Then there exist ~F1, ~F2 2 SBD0 with R( ~F1)+R( ~F2) � 4mjHj, Cl( ~Fj) � n,S( ~Fj) � m+ 2, and IR( ~Fj) � m+ 1, for j 2 f1; 2g, suh that [ ~F1℄� G � [ ~F2℄ j= � and[ ~F1℄�H� [ ~F2℄ 6j= �;for all H 2 H.Proof. For eah H 2 H take a partition ~IH = hIHi : 1 � i � nHi of H1 �H �H2 suhthat H1 �H�H2; ~IH j= ~S;for a ~S with :9p~S 2 �. We onsider the restritions of the ~IH to Hj, de�ned byHj \ ~IH := hhj \ IHi : 1 � i � nHi;for j = 1; 2. Now form, for j = 1; 2,~Ij =_hHj \ ~IH : H 2Hi:Then l(~I1) + l(~I2) � mjHj. Suppose that ~Ij = hIji : 1 � i � nji, for j = 1; 2, and �x anIji . It follows immediately from the proof of Theorem 14 (1) in Part I that there exists~F ji 2 SBD0 with R( ~F ji ) � 4, S( ~F ji ) � m+2, IR( ~F ji ) � m+1, and Cl( ~F ji ) � n, suh thatThm;nHIji � Thm;n[ ~F ji ℄;(8S 2 M(m;n))(HIji 2 S) [ ~F ji ℄ 2 S):Thus, by Proposition 5 of Part I, the sequenes ~Fj = h ~F ij : 1 � i � nji, j = 1; 2, provethe Theorem up to the ondition on IR( ~F j). We only know IR( ~Fj) � 2(m + 1). But, byLemma 13 (2) of Part I, we an redue sequenes C(0)l, l > m + 1, in ~Fj to sequenesC(0)m+1. a

3 COMPLEX VARIETIES AND STRONG COMPLETENESS 93 Complex varieties and strong ompletenessSine :LM is not talking about bimodal algebras and not even about arbitrary frames fortense logis but only about rooted frames we need the following easily proved translationof the onept of omplex varieties. Call a tense logi � (�-) omplex i� V(�) is (�-)omplex.Proposition 8 Let � be an in�nite ardinal. For eah tense logi �, � is (�-) omplexi�, for eah desriptive tense frame G (with jG+j � �) with G j= � there exists a tenseframe H suh that G+ ' H+ and �H j= �.We are going to show that a tense logi � is omplex i� all �-frames of the form [~G℄,~G 2 SDS, an be represented by Kripke frames for �. First we need some more informationon gaps in desriptive tense frames. A gap = hI1; I2i in hg;Ri is� simple i� I1 2 [J;C(n)℄ and I2 2 [C(m); J ℄, for some m;n 2 !. The set of simplegaps in hg;Ri is denoted by SG(hg;Ri).� left-proper i� I1 2 [J;!℄.� right-proper i� I2 2 [ ; J ℄.� proper i� it is both left and right-proper.The set of non-simple gaps in hg;Ri is denoted by PG(hg;Ri). Applied to frames in DSwe see that �G[C(m); n℄ ontains preisely one non-simple gap whih is right-proper. Theframe �G[m1; C(n);m2℄ ontains two non-simple gaps, one is right-proper and the otheris left proper.Lemma 9 Suppose that G is desriptive. Then �G ontains no proper gap.Proof. Assume that = hI0; I1i with I0 2 [J;!℄, I1 2 [ ; J ℄. Certainly there exists a setof intervals H in G with Ii \ I 6= ;, for all I 2 H, suh that Ii \TH = ;, for i = 0; 1. Butthen TH = ; and TH 0 6= ;, for all �nite subsets H 0 of H. Hene G is not desriptive. aThus, for eah non-simple gap = hI1; I2i in �G, G desriptive, there is a luster Csuh that I1 = fx 2 g : xRCg or I2 = fx 2 g : CRxg. We all C the luster orrespondingto . Reall the de�nition of the anonial type assignment for G from Part I. In termsof this notion we have, by Theorem 3 of Part I, that +C = j if C orresponds to aright-proper gap and �C = j i� C orresponds to a left-proper gap. If = hI1; I2i is asimple gap in �G then I1 2 G+. Thus, both I1 and I2 are intervals in G.For an algebra A, we are going to desribe the frames H with H+ ' A. The replae-ment of non-simple gaps is the basi operation. For a set of gaps G in hg;Ri and a set of

3 COMPLEX VARIETIES AND STRONG COMPLETENESS 10(mutually disjoint) objets fh : 2 Gg whih are Kripke frames or empty sets we denoteby hg;Ri[h= : 2 G℄the frame resulting when all gaps ; 2 G, are replaed by h in hg;Ri.Suppose that G = hg;R;Ai is a desriptive frame and G � PG(hg;Ri). Let fh : 2 Gg be a set of Kripke frames. For eah ; 2 G; take an x from the luster C

orresponding to . De�ne a set B by putting b 2 B i�(9a 2 A)(b = a [[fh : x 2 a; 2 Gg):Lemma 10 G+ ' hhg;Ri[h= : 2 G℄; Bi+.Proof. We show that a 7! a[Sfh : x 2 a; 2 Gg, a 2 A, de�nes an isomorphism fromG+ onto hhg;Ri[h= : 2 G℄; Bi+. This is lear for the boolean reduts. The proof for3+ and 3� is symmetri. So it suÆes to show that, for all a 2 A,3+1 (a [[fh : x 2 a; 2 Gg) = 3+2 a [[fh : x 2 3+2 a; 2 Gg;where 3+1 is omputed in hg;Ri[h= : 2 G℄ and 3+2 is omputed in hg;Ri. But thisequation is an immediate onsequene of the desription of non-simple gaps in desriptiveframes. aMostly it will be suÆient to replae gaps by frames from the following setFillm = fn< � C(1)� n< : n � m+ 1g [ fn< : n � m+ 1g:The following lemma explains why this is so.Lemma 11 For eah �nite frame h there exists m 2 ! and g 2 Fillm with Thh � Thg.For eah �nite frame h and m 2 ! there exists g 2 Fillm with Thm;!h � Thm;!g.Proof. Follows from Lemmas 10 and 13 (2) of Part I. aLemma 12 Let � = Lin:t�� with � � :LM(m;n). Suppose that � is not omplex. Thenthere exists a ~G 2 SDS with [~G℄ j= � but(�[~G℄)[h= : 2 PG(�[~G℄)℄ 6j= �;for all fh : 2 PG(�[~G℄)g � Fillm.Proof. By Proposition 8 there exists a desriptive tense frame H = hh; S;Bi with H j= �suh that �F 6j= �, for eah F with F+ ' H+. Hene, by Lemma 10,(�H)[h= : 2 PG(�H)℄ 6j= �;

3 COMPLEX VARIETIES AND STRONG COMPLETENESS 11for all fh : 2 PG(�H)g � Fillm.Some notation. For a partition ~I = hIi : 1 � i � ri of (�H)[h= : 2 PG(�H)℄denote by ~IH the partition ~IH = hIi \ h : 1 � i � ri of �H. Denote by the anonialtype assignment for H.Certainly there exists a �nite set D1 � PG(�H) and a �nite set D2 � SG(�H) suhthat, for all s = fh : 2 D1g � Fillm, there exists ~S with :9p~S 2 � and a partition ~I(s)of �H[h= : 2 D1℄ with gap(~I(s)H) � D1 [D2and (�H)[h= : 2 D1℄; ~I(s) j= ~S:We may assume that for any two gaps 1; 2 2 D1 with 1S2 suh that there exists asimple gap in �H with 0SS2 there already exists a 2 D2 with 0SS2. (Theonly possibility that there is no simple gap with 0SS1 is that 0 is left-proper and1 is right-proper suh that the luster C orresponding to 0 oinides with the lusterC orresponding to 1. In this ase C = (j; j), for this luster C. This situation ausesno speial problems and is overed by the onstrution in ase (4.) and (5.) below).Denote by ~I = hIi : 1 � i � ki the partition par(D2) de�ned by D2 in �[ ~H℄. All gapsin D2 are simple. Hene ~I is also a partition of H, i.e. Ii 2 H+, for 1 � i � k. Consideran interval Ii from ~I. Depending on Ii we shall onstrut a sequene ~Gi 2 SDS with1. Th[ ~Gi℄ � ThHIi,2. HIi 2 S) [ ~Gi℄ 2 S, for all S 2 M(m;n),but so that in �[ ~Gi℄ we still �nd the required non-simple gaps.Case 1. No gap from D1 is in Ii. Similar to the proof of Proposition 7 in Part I we�nd a sequene ~Gi 2 SBD0 satisfying (1.) and (2.).Case 2. A right-proper gap from D1 is in Ii and �C = m, for the luster Corresponding to . Similar to the proof of Proposition 7 in Part I we �nd a~Hi 2 S(BD0 [ fG[C(1); k℄ : k > 0g)suh that a G[C(m1);m2℄ o

urs in ~Hi, for some m2;m1 2 ! with m1 = jCj if C is �niteand m1 = n, otherwise, and suh that ~Hi satis�es (1.) and (2.). Replae all G[C(1); l℄,l > 0, in ~Hi by G[l; C(2); l℄ 2Dl and denote the result by ~Gi. Clearly ~Gi still satis�es (2.).By Lemma 5 (1.), ~Gi satis�es (1.).Case 3. A left-proper gap from D1 is in Ii and +C = m, for the luster C orre-sponding to . Construt ~Gi as the mirror image of ~Gi in ase 2.

3 COMPLEX VARIETIES AND STRONG COMPLETENESS 12Case 4. A right-proper gap from D1 is in Ii and �C = j, for the luster C orre-sponding to . Similar to the proof of Proposition 7 in Part I we �nd a ~Gi 2 SDS suhthat a G[m1; C(n);m2℄ 2 Dl o

urs in ~Gi and ~Gi satis�es (1.) and (2.). (Here Lemma 5(1.) is useful).Case 5. A right-proper gap from D is in Ii and +C = j, for the luster C orre-sponding to j . Construt ~Gi as the mirror image of ~Gi in ase 4.Put ~G = h ~Gi : 1 � i � ki. Then Th[~G℄ � ThH, by Proposition 5 of Part I and (1.)above. Hene [~G℄ j= �. Moreover, by (2.) and the onstrution of the ~Gi,(�[~G℄)[h= : 2 PG(�[~G℄)℄ 6j= �;for all fh : 2 PG(�[~G℄)℄g � Fillm. aFor G 2 DS and k 2 ! de�ne a set of Kripke frames �kG by putting�kG = 8>>>>>: fGg : G �nitef!(n) : h 2 Fillkg : G = G[C(m); n℄;f!(m2) : h 2 Fillkg : G = G[m1; C(m);m2℄;and put, for a sequene ~G = hGi : 1 � i � ri 2 SDS,�k ~G = fhfi : 1 � i � ri : fi 2 �kGi; 1 � i � rg:Note that the set �kG nearly oinides with the set of frames �G[h= : 2 PG(�G)℄,fh : 2 PG(�G)g � Fillk, for G 2 DS. Reall that an algebra A is simple if it has nonon-trivial ongruenes. We are in a position now to haraterize omplex tense logis.Theorem 13 Suppose that � = Lin:t� � with � � :LM(m;n). The following onditionsare equivalent.1. � is omplex.2. � is strongly omplete.3. There exists a tense frame G with �G j= � and G+ ' A, for all �nitely generatedsimple algebras A 2 V(�).4. For all ~G 2 SDS with [~G℄ j= � there exists ~f 2 �m ~G with [~f ℄ j= �.5. For all ~G 2 SDS with [~G℄ j= � there exists a set fh : 2 PG(�[~G℄)g � Fillm with(�[~G℄)[h= : 2 PG(�[~G℄)℄ j= �:

3 COMPLEX VARIETIES AND STRONG COMPLETENESS 13Proof. (5) ) (1) is Lemma 12.(4) ) (5). Suppose that (4) holds and suppose that ~G = hGi : 1 � i � ri 2 SDS with[~G℄ j= �. Put, for 1 � i � r,Fi = 8>>>>>: Gi : Gi �nite;G+[C(k); l℄ : Gi = G[C(k); l℄;G+[l; C(k)℄ : Gi = G[l; C(k)℄;G2[m1; C(k);m2℄ : Gi = G[m1; C(k);m2℄:By Lemma 5, ~F j= �. It is easy (but a bit tedious) to hek that for eah ~f 2 �m ~F thereexists a k 2 ! and fh : 2 PG(�[~G℄)g � Fillk withTh[~f ℄ � Th(�([~G℄)[h= : 2 PG(�[~G℄)℄):But then, by Lemma 11, for eah ~f 2 �m ~F there exists fh : 2 PG(�[~G℄)g � Fillm withThm;![~f ℄ � Thm;!(�([~G℄)[h= : 2 PG(�[~G℄)℄):By (4), there is an ~f 2 �m ~F with [~f ℄ j= �. Hene there is a set fh : 2 PG(�[~G℄)g � Fillmwith �([~G℄)[h= : 2 PG(�[~G℄)℄ j= �.(3) ) (4). Assume that (3) holds and suppose that ~G 2 SDS with [~G℄ j= �. It iseasy to hek that [~G℄+ is simple. By Proposition 4, [~G℄+ is �nitely generated. Hene (3)applies to A = [~G℄+. Thus there exists a H = hh; S;Bi with �H j= � and H+ ' [~G℄+.Certainly we may assume that all the atoms inH+ are of the form fxg for an x 2 h. Thus,we may assume there is a sequene ~h = hhi : 1 � i � ri with hh; Si = [~h℄ and suh that� hi = Gi, whenever Gi is �nite,� hi = !(m2), for an arbitrary (possibly empty) fi, whenever Gi =G[m1; C(k);m2℄.� hi = !(l), for some fi with fi = ; or fi 2 [ k; J ℄, whenever Gi = G[C(k); l℄.Let 1 � i � r. By Theorem 18 (1) of Part I, we have that fi = ; or, for all S 2 Mwith fi 2 S, there exists ~g 2 SBK with Thfi � Th[~g℄ and [~g℄ 2 S. Using this fat it isstraightforward to show that there exists l 2 ! and h1i 2 �lGi suh that Thhi � Thh1i .Hene, by Lemma 11 and Proposition 5 of Part I, there exists a h2i 2 �mGi withThm;!hi � Thm;!h1i = Thm;!h2i :But then, [~h2℄ j= � and ~h2 2 �m ~G, for ~h2 := hh2i : 1 � i � ri.(1) ) (3) is lear. Hene (1), (3), (4), and (5) are equivalent. The equivalene (1) ,(2) follows from (1) , (3) with Theorem 1. a

3 COMPLEX VARIETIES AND STRONG COMPLETENESS 14Theorem 14 f� � :LM �nite : Lin:t�� is strongly omplete g is reursive. Moreover,� = Lin:t��, � � :LM(m;n), is strongly omplete i�, for all ~G 2 SDS with [~G℄ j= � andCl(~G) � n, S(~G) � m+ 2, IR(~G) � m+ 1, and R(~G) � 1 + 4m(1 + (m+1)(m+ 2)), thereexists a ~f 2 �m ~G with [~f ℄ j= �.Proof. By Theorem 6 it suÆes to show the moreover part. The diretion from left toright follows from Theorem 13. Conversely suppose that � is not strongly omplete. ByTheorem 13, there is a ~H = hHi : 1 � j � ri 2 SDS with [ ~H℄ j= � suh that [~f ℄ 6j= �, forall ~f 2 �m ~H. For G 2 DS de�ne a set of frames tnmG by puttingtnmG = 8>>>>>: fGg : G �nite,fG[l; C(n)℄g [ fG[l; C(n)℄ � h� C(k) : h 2 Fillmg : G = G[l; C(k)℄;fG[C(n); l℄g [ fC(k)� h�G[C(n); l℄ : h 2 Fillmg : G = G[C(k); l℄;fG[m1; C(n)℄� h�G[C(n);m2℄ : h 2 Fillmg : G = G[m1; C(k);m2℄By Lemma 17 of Part I there is a bijetion � : �mG ! tnmG suh that1. Th!;nh = Th!;n�(h), for all h 2 �mG.2. Also, ThH � ThG, for all H 2 tnmG.Note that eah frame F 2 tnmG is of the form [ ~F ℄, for some ~F 2 SDS. Thus, by (1.) and(2.) we may assume that there is an in�nite Hi in ~H suh thatH1 �H�H2 6j= �;for H1 = (H1 � : : : �Hi�1) and H2 = (Hi+1 � : : : �Hr), and all H 2 tnmHi. (We brieyjustify this assumption. Assume that there is no ~H 2 SDS with(i) [ ~H℄ j= � and (ii) [~f ℄ 6j= �, for all ~f 2 �m ~H,in whih suh a Hi o

urs. Put ~H(0) = ~H for some ~H with (i) and (ii). Certainlythere is an in�nite Hi in ~H. Also, there is a sequene ~F1 with [ ~F1℄ 2 tnmHi suh thatH1 � [ ~F1℄ �H2 j= �. Denote the sequene resulting by replaing Hi by [ ~F1℄ in ~H(0) by~H(1). Then ~H(1) also satis�es (i) and (ii). We go on with ~H(1) and form ~H(2); ~H(3)et. in the same way. Finally we get a ~H(j) 2 SDS with k � n, for all G[C(k); l℄ andG[l; C(k)℄ in ~H(j), and in whih there does not o

ur a frame from Dl. But then [~f ℄ j= �,for a ~f 2 �m ~H(j), by Lemma 17 of Part I. This ontradits ondition (ii) for ~H(j)).By Proposition 7, there exist ~F1; ~F2 2 SBD0 with R( ~F1) +R( ~F2) � 4mjHj, Cl( ~Fj) �n, S( ~Fj) � m+ 2, IR( ~Fj) � m+ 1, for j 2 f1; 2g, suh that[ ~F1℄�Hi � [ ~F2℄ j= � and [ ~F1℄�H� [ ~F2℄ 6j= �;for all H 2 H = tnmHi. We have jHj � 1+ (m+1)(m+2). Certainly we may assume thatCl(Hi) � n, S(Hi) � m+ 2. Then the sequene ~G = h ~F1;Hi; ~F2i satis�es the ardinalityonditions and, by (1.) and (2.), [~f ℄ 6j= �, for all ~f 2 �m ~G. aExamples. Put, for n > 0, gapn = :9ph[J;!n℄; [ n; J ℄i, and de�ne

3 COMPLEX VARIETIES AND STRONG COMPLETENESS 15� Gapn = Lin:t� gapn,� Gap+n = Gapn � :9pC(0)� :9p[Jok;C(0)℄� :9p[C(0); Jok℄,� �n = Gap+n�Dens2, whereDens2 is the :LM-formula orresponding to the densityaxiom 2+2+p! 2+p.Dens2 and gap = gap1 are disussed in Part I where is it also shown that �1 = ThhR; 0, Gapn, Gap+n and �n are strongly omplete and omplexbut not d-persistent.Proof. Let � be one of those logis and suppose that ~G 2 SDS with [~G℄ j= �. It is readilyheked that �([~G℄)[1 1.There is an m > 1 with gapm 2 �. But then �G[2; C(m); 2℄ 6j= �. Hene, by Proposition4, � is not d-persistent. aIt is well known and easy to hek that d-persisteny transfers under forming joins oflogis, i.e., if �1 and �2 are d-persistent, then so is �1 t �2. We are now going to showthat this does not hold for strong ompleteness.Reall from Part I the axiomatization the logi Lin:t�RDis determined by the lassof weak orderings in whih eah irreexive point has an immediate irreexive su

essor.This logi is d-persistent, hene omplex and strongly omplete.Corollary 16 (1) There are strongly omplete logis �1 and �2 suh that �1 t �2 is notstrongly omplete. (2) There are omplex varieties V1 and V2 suh that V1 \ V2 is notomplex.Proof. (1) It suÆes to show that � = Gap1t (Lin:t�RDis) is not strongly omplete.But G[0; C(2)℄ j= � and (!

4 D-PERSISTENCY AND ELEMENTARITY 164 d-persisteny and elementarityAgain we need an easily proved simpli�ation of the notion of d-persisteny.Proposition 17 A tense logi � is d-persistent i� �G j= �, whenever G is a desriptivetense frame validating �.In order to prove that eah d-persistent tense logi is elementary we introdue anotherlanguage for talking about intervals in Kripke frames. Consider the following intervaltypes. Q = f[C(n);C(m)℄jn;m 2 !g [ fC(0)g[ f[C(n);!℄jn 2 !g[ f[ ;C(n)℄jn 2 !g[ f[ ;!℄gLemma 18 Let P 2 :LM. Then there exist Q1; : : : ; Qn 2 :LQ suh thatg j= P , (8i � n)(g j= Qi);for all Kripke frames g.Proof. Certainly it suÆes to show that for eah S 2 M there exists a set QS of intervaltypes in Q suh that g j= :9pS, (8B 2 QS)(g j= :9pB):For instane, for S = [J;!m℄ the setQS = f[C(0);C(m)℄; [C(0);!℄; [C(1);C(m)℄; [C(1);!℄; [ ;!℄gis as required. All the other ases are similar.aFor an interval type B 2 Q denote by L(B) and R(B) the symbol to the left and thesymbol to the right in B, respetively. PutLNG = f9phBi : 1 � i < ni 2 LQ : :(9i � n)(R(Bi) =! ^L(Bi+1) = )g:Lemma 19 Suppose that Q 2 :LNG. Then ModFrQ is �rst order de�nable.Proof. Obvious.aLemma 20 Suppose that �[~G℄ j= �, for all ~G 2 SDS with [~G℄ j= �. Then ModFr� isde�nable in :LNG.

4 D-PERSISTENCY AND ELEMENTARITY 17Proof. Suppose that ModFr� is not de�nable in :LNG. Put� = f:Q 2 :LNG : (8g)(g j= Q) g 6j= �)g:We have M � ModFr�, for M = ModFr�. Moreover h 2M �ModFr� i�1. there exists :9p~S 2 � suh that h j= 9p~S.2. :9p~S 62 �, for all ~S 2 M suh that there is a partition ~I of h with h; ~I j= ~S suh thatgap(~I) ontains no proper gap.Take an hf;Ri 2 M �ModFr� and a partition ~I = hIi : 1 � i � ri of hf;Ri suh thathf;Ri; ~I j= ~S, for some ~S with :9p~S 2 �. We may assume that the size of lusters in hf;Riis bounded by Cl(~S). We show that, for 1 � i � r; there is a sequene ~hi 2 SBK with(a) fIi 2 B) [~hi℄ 2 B, for all B 2 Q.(b) For all ~S 2 M, if [~hi℄; ~J1 j= ~S and gap( ~J1) ontains no proper gaps then the exists apartition ~J2 of fIi suh that gap( ~J2) ontains no proper gaps and so that fIi; ~J2 j= ~S.() Th[~hi℄ � ThfIi .Note that it follows from (a) that [~hi℄ 2 Si. One has to distiguish the ases Ii 2 B = [C(0)℄,Ii 2 B = [C(m);C(n)℄, where m;n 2 ! are maximal with this property, Ii 2 B = [ ;C(n)℄,where n 2 ! is maximal with this property, Ii 2 B = [C(n);!℄, where n 2 ! is maximalwith this property, and Ii 2 B = [ ;!℄. In all those ases there exists a ~hi satisfying (b)and () with [~hi℄ 2 B. The proof is similar to the proof of Proposition 7 in Part I. Someare has to be taken in order to satisfy ondition (b). Instead of delivering the tediousdetailed proof we give two instrutive examples. AssumefIi = hQ; (1) � !(1)� !

4 D-PERSISTENCY AND ELEMENTARITY 18The frame C(1)� !>(0)� !>>>>: gi : gi �nite,G[l; C(n)℄ : gi = !(l); i 2 D2;G[m1; C(n);m2℄ : gi = !(m2);and put ~G = hGi : i 2 D1 [D2i. It follows from ondition (b) for ~hi, 1 � i � r; andthe onstrution of ~G that [~G℄ j= �. (Case four in the de�nition of Gi is essential). On theother hand, by ondition (a) for ~hi and the onstrution of ~G, we have �[~G℄ 6j= �. Detailsare left to the reader. aTheorem 21 The following onditions are equivalent, for eah tense logi �.1. �[~G℄ j= �, for all ~G 2 SDS with [~G℄ j= �.2. � is omplete and ModFr� is de�nable in :LNG.3. � is elementary.4. � is d-persistent.Proof. (1) ) (2) is Lemma 20. (2) ) (3) is Lemma 19. (3) ) (4) is proved in [2℄ and(4) ) (1) is trivial. aTheorem 22 f� � :LM �nite : Lin:t � � is d-persistent g is reursive. Moreover,� = Lin:t � �, � � :LM(m;n), is d-persistent i�, for all ~G 2 SDS with [~G℄ j= � andCl(~G) � n, S(~G) � m+2, IR(~G) � m+1 and R(~G) � 1 + 4m, we have �[~G℄ j= �, as well.Proof. By Theorem 6, it suÆes to show the moreover part. The diretion from left toright follows from Proposition 4. Now suppose that � is not d-persistent. By Theorem21, there is a ~H = hHi : 1 � i � ri 2 SDS with [ ~H℄ j= � but �[ ~H℄ 6j= �. Put, for G 2 DSand k > 0,skG = 8>>>>>: fGg : G �nite,G[n;C(k)℄� C(m) : G = G[n;C(m)℄;C(m)�G[C(k); n℄ : G = G[C(m); n℄;G[m1; C(k)℄� C(m)�G[C(k);m2℄ : G = G[m1; C(m);m2℄:

5 R-PERSISTENCY 19By Lemma 17 of Part I, Th!;n�G = Th!;nsnG;for all G 2 SDS. Hene we may assume that there exists a Hi suh that H1�snHi�H2 6j=�, for H1 = (H1 � : : :�Hi�1) and H2 = (Hi+1 � : : :�Hr). By Proposition 7 there exist~F1; ~F2 2 SBD0 with R( ~F1)+R( ~F2) � 4m, Cl( ~Fj) � n, S( ~Fj) �m+2, and IR( ~Fj) � m+1,for j 2 f1; 2g, suh that[ ~F1℄� Gi � [ ~F2℄ j= � and [ ~F1℄� tnHi � [ ~F2℄ 6j= �:(Here we put H = fsnHig). We may assume that Cl(Hi) � n, S(Hi) � m+ 2. But thenthe sequene ~G = h ~F1;Hi; ~F2i satis�es the ardinality onditions and �[~G℄ 6j= �. a5 r-persistenyIn [7℄ it is shown that eah re�ned frame H is, in a ertain sense, embeddable into adesriptive frame G with H+ ' G+. (Intuitively, this is ahieved by adding preisely thoseultra�lters U in H+ to h for whih TU = ;). The preise formulation is as follows.Proposition 23 Suppose that H+ ' G+, where H = hh; S;Ai is desriptive and G isre�ned. Then there exists H1 = hh1; S1; A1i, where h1 � h, S1 = S\h21 and A1 = fa\h1 :a 2 Ag, suh that G ' H1 and suh that a 7! a \ h1, a 2 A, de�nes an isomorphism fromH onto H1. Moreover, for the anonial type assignment for H and all lusters C in H,� C � h1 or C \ h1 in�nite, for C = (m;m).� C � h1 or C \ h1 is in�nite or C \ h1 = ;, for C 2 f(m; j); (j;m)g.� C \ h1 is arbitrary, for C = (j; j).Proof. The �rst part is shown in [7℄ and the moreover part is an immediate onsequeneof the �rst part and the desription of lusters in desriptive tense frames. aFor re�ned frames we have to take some are when working with LM instead of L!.Theorem 2 does not hold for re�ned frames. (For example, the theories of G[0; C(1)℄ andG := h!

5 R-PERSISTENCY 20� C � h1, for C = (m;m).� C � h1 or C \ h1 = ;, for C 2 f(m; j); (j;m)g.� C \ h1 arbitrary, for C = (j; j).Proof. Suppose that � is not r-persistent. There exists a re�ned tense frame G with�G 6j= � and G j= �. Certainly there exists a desriptive frame H = hh; S;Ai withH+ ' G+. Take a frame H01 = hh01; S01; A01i whih satie�es the ondition of Proposition 23.Now put, for C a luster in h, C1 = h01 \C if h01 \C is �nite, and C1 = C, otherwise. Puth1 = h01 [[fC1 : C luster in hg;and H1 = hh1; S \ h21; fa \ h1 : a 2 Agi: Only the size of in�nite lusters is manipulated.So it is straightforward to show that H1 is as required. aPut, for G 2DS,sG = 8>>>>>: fGg : G �nite,f!(n)g [ f�Gg : G = G[C(m); n℄;f!(m2)g [ f!(m2)g : G = G[m1; C(m);m2℄;and de�ne, for ~G = hGi : 1 � i � ri 2 SDS,s~G = fhfi : 1 � i � ri : fi 2 sGi; 1 � i � rg:Note that, for ~G 2 SDS and H = [~G℄, the frames in f[~f ℄ : ~f 2 s~Gg basially oinide withthe frames of the form �H1, for H1 satisfying the onditions of Corollary 24 with respetto H.Theorem 25 The following properties are equivalent, for � = Lin:t��, � � :LM(m;n).1. � is r-persistent.2. [~f ℄ j= �, for all ~f 2 s~G with ~G 2 SDS and [~G℄ j= �.3. [~f ℄ j= �, for all ~f 2 s~G with ~G 2 SDS and Cl(~G) � n, S(~G) � m+2, IR(~G) � m+1,R(~G) � 4m and [~G℄ j= �.The set f� � :LM �nite : Lin:t�� is r-persistent g is reursive.Proof. (1) ) (2). Consider hh; Si = [~f ℄, for a ~f 2 s~G with [~G℄ j= �. For [~G℄ = hg;R;Aiwe may assume that h � g and S = R\h2. Put B = fa\h : a 2 Ag and H = hh; S;Bi. It

6 THE FINITE MODEL PROPERTY 21is easy to hek that a 7! a\h, a 2 A, de�nes an isomorphism from G+ onto H+. ClearlyH is re�ned. Hene, by (1), hh; Si j= �.(2) ) (3) is trivial.(3) ) (1). (Sketh) Suppose that � is not r-persistent. Take a desriptive frame Hwith H j= � and a re�ned frame H1 satisfying the onditions of Corollary 24. So we have�H1 j= :9p~S, for a :9p~S 2 �. Similar to a lot of previous examples and espeially to theproof of Proposition 7 of Part I we an redue H1 in suh a way that we get a ~G 2 SDSsatisfying the ardinality onditions of (3) suh that there is an ~f 2 s~G with [~f ℄ j= :9p~S.We leave the onstrution to the reader. aA simple example of a d-persistent tense logi whih is not r-persistent is the logiMk:t de�ned in Part I. It is easily veri�ed that this logi is d-persistent. On the otherhand, Mk:t is not r-persistent sineG[0; C(1)℄ j=Mk:t but !>>>>>>>>: fGg : G �nite,fn(k + 1) 0;f(k + 1)

6 THE FINITE MODEL PROPERTY 22and de�ne, for a sequene ~G = hGi : 1 � i � ri 2 SBD0�lk ~G = fhfi : 1 � i � ri : fi 2 �lkGi; 1 � i � rg and �k ~G =[f�lkG : l 2 !g:Lemma 27 For all ~G 2 SBD0 and all �nite frames h, h j= dgk ~G i� there exists ~f 2 �k ~Gwith Thh � Th[~f ℄.Proof. We have [~f ℄ j= dgk ~G, for all ~f 2 �k ~G. Conversely, suppose that h j= dgk ~G, for a�nite frame h. Assume that dgkGi = 9p~Si, for 1 � i � r, and that ~I = hIi : 1 � i � riis a partition of h suh that hIi j= 9p~Si, 1 � i � r. The Lemma follows if there existfi 2 �kGi with ThhIi � Thfi, for 1 � i � r. This is lear if Gi is �nite. Suppose thatGi = G[n;C(m)℄; n > 0. Certainly we may assume thathIi = n(k + 1)

7 COMPLETENESS 23SBD0 and a k � m suh that [~f ℄ 6j= �, for all ~f 2 �k ~G. By (a), [~f ℄ 6j= �, for all ~f 2 �mm ~G.We may assume, by (b), that there is an in�nite Gi in ~G suh thatH1 � f �H2 6j= �;for H1 = (G1� : : :�Gi�1) and H2 = (Gi+1� : : :�Gr) and all f 2 �mmGi. (The justi�ationof this assumption is similar to the proof of Theorem 14). By Proposition 7, there exist ~F1,~F2 2 SBD0 with R( ~F1)+R( ~F2) � 4mjHj, Cl( ~Fj) � n, S( ~Fj) � m+2, and IR( ~Fj) � m+1,for j 2 f1; 2g, suh that[ ~F1℄� Gi � [ ~F2℄ j= � and [ ~F1℄� f � [ ~F2℄ 6j= �;for all f 2 H := �mmGi. We have jHj � (m + 1)(m + 2). Certainly we may assumethat Cl(Gi) � n, S(Gi) � m + 2. Thus ~H = h ~F1;Gi; ~F2i refutes (3.), i.e. ~H satis�es theardinality onditions of (3.) and there does not exist a ~f 2 �mm ~H suh that [~f ℄ j= �. a7 CompletenessFor G 2 BD0 and l; k 2 ! de�ne a set of frames lkG by puttinglkG = 8>: fGg : G �nite,�lkG [ f!(n) : g 2 Filllg : G = G[C(m); n℄;and put, for a sequene ~G = hGi : 1 � i � ri 2 SBD0;lk ~G = fhfi : 1 � i � r : fi 2 lkGi; 1 � i � rg and k ~G =[flkG : l 2 !g:Lemma 29 For all ~G 2 SBD0 and all Kripke frames h, if h j= dgk ~G then there exists~f 2 k ~G with Thk;!h � Thk;![~f ℄.Proof. Suppose that h j= dgk ~G, for a Kripke frame h. Let dgkGi = 9p~Si, for 1 � i � r,and assume that hIi : 1 � i � ri is a partition of h suh that hIi j= 9p~Si, 1 � i � r. TheLemma follows if there exist fi 2 kGi with Thk;!fi � Thk;!hIi , for all 1 � i � r. This islear if Gi is �nite. Suppose that Gi = G[n;C(m)℄; n > 0. Certainly we may assume that(a) hIi = n(k + 1)

7 COMPLETENESS 24Case 2. f1 is in�nite. By Theorem 18 of Part I, t here exists a sequene ~f = hfi :1 � i � ti 2 SBK with Thf1 � Th[~f ℄. Suppose that there is a minimal j suh thatfj = !>(m), for some m 2 !. ThenThhIi � Th(n(k + 1)(m) sine hfi : 1 � i � j � 1i ontains no!>(m). Now take a maximal j � t suh that fj = !

REFERENCES 25for H1 = (G1 � : : : � Gi�1) and H2 = (Gi+1 � : : : � Gr). By Proposition 7 there exist ~F1,~F2 2 SBD0 with R( ~F1) + R( ~F2) � 4mjHj, Cl( ~Fj) � n, S( ~Fj) � m+ 2, IR( ~Fj) � m+ 1,for j 2 f1; 2g suh that[ ~F1℄� Gi � [ ~F2℄ j= � and [ ~F1℄�H� [ ~F2℄ 6j= �;for all H 2 H := gmnGi. Again we may assume that Cl(Gi) � n and S(Gi) � m+ 2. Thush ~F1;Gi; ~F2i refutes (3). aReferenes[1℄ J. van Benthem. Modal Logi and Classial Logi, Napoli, 1983[2℄ K. Fine. Some onnetions between elementary and modal logi, in Proeedings ofthe Third Sandinavian Logi Symposion, edited by S. Kanger, North-Holland, Am-sterdam, 1975[3℄ D. Gabbay & I. Hodkinson & M. Reynolds, Temporal Logi, Oxford, 1994[4℄ R. Goldblatt. Metamathematis of modal logi, Reports on Mathematial Logi 6:41 - 77, 7: 21 - 52, 1976[5℄ R. Goldblatt. Varieties of Complex Algebras, Annals of Pure and Applied Logi 38:173 - 241, 1989[6℄ R. Goldblatt. Algebrai Polymodal Logi, in Handbook of algebrai logi, forthoming[7℄ G. Sambin & V. Va

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