On the Products of Linear Modal Logics

On the products of linear modal logics

MARK REYNOLDS,School of Information Technology, MurdochUniversity, Perth, Western Australia.E-mail: [email protected]

MICHAEL ZAKHARYASCHEV, Department of Computer Science, King’sCollege London, Strand, London WC2R 2LS, UK.E-mail: [email protected]

AbstractWe study two-dimensional Cartesian products of modal logics determined by infinite or arbitrarily long finite linearorders and prove a general theorem showing that in many casesthese products are undecidable, in particular, suchare the squares of standard linear logics likeK4:3, S4:3,GL:3,Grz:3, or the logic determined by the Cartesiansquare of any infinite linear order. This theorem solves a number of open problems of Gabbay and Shehtman [7]. Wealso prove a sufficient condition for such products to be not recursively enumerable and give a simple axiomatisationfor the squareK4:3�K4:3 of the minimal liner logic using non-structural Gabbay-type inference rules.

Keywords: Modal logic, Kripke frame, Cartesian product, decidability.

1 Introduction

The formation of Cartesian products of Kripke frames is probably the most natural way ofintroducing a concept of dimension in modal logic in order toreflect interactions betweenmodal operators representing time, space, knowledge, actions, etc. Products of modal logics(i.e., sets of multi-modal formulas that are valid in the Cartesian products of Kripke frames forthose logics) have been studied in both pure modal logic (seee.g. [23, 24, 18, 7, 15, 19, 31])and applications in computer science and artificial intelligence (see e.g. [20, 3, 1, 21, 4, 27,28, 29]) since the 1970s. It would not be an exaggeration to say that now multi-dimensionallogics in general and Cartesian products in particular are becoming the subject of one ofthe most important and interesting research fields in pure and applied modal logic. (Seee.g. applications of results and techniques developed in multi-dimensional logic to first-orderclassical, modal and temporal logics in [8, 30, 13].)

Unfortunately, as was observed by Gabbay and Shehtman [7], there is no general transfertheorem that could guarantee the preservation of such properties of logics as decidabilityor axiomatizability under the formation of products. If we consider only 2D products of“standard” modal systems, then the results obtained so far can roughly be described as follows(for more details consult [7] and [31]):� logics of the formK�L andS5�L are usually decidable and finitely axiomatisable, in

particular forL 2 fK;T;K4;S4;S5g (see [19] for complexity results concerning someof these logics);� the known undecidable (and not recursively axiomatizable)2D products are the squareof the logic of the framehN; <i (see [25]) and the “compass” logic of Venema [26] (see[17]).

J. Logic Computat., Vol. 11 No. 00-34, pp. 1–23 2001 c Oxford University Press

2 On the products of linear modal logics

As concerns 3D products, Maddux [16] proved (in the algebraic setting) that any logic be-tween[S5;S5;S5℄ andS5�S5�S5 is undecidable.1 And quite recently Hirschet al. [12]have shown that no logic betweenK3 andS53 is decidable and finitely axiomatisable: seealso [15]. (For more details consult [6].)

The main aim of this paper is to study the 2D products of modal logics determined byinfinite or arbitrarily long finite linear frames. Namely, weshow that the logic characterisedby a class of product-frames containing the Cartesian product of two rooted linear orders,both having an infinite ascending chain (or both having an infinite descending chain), isundecidable. Moreover, if two classes of finite linear frames both contain arbitrarily longchains, then the product of these classes determines an undecidable logic as well. It followsthat the Cartesian squares of all standard (non-tabular) “linear” modal logics, sayK4.3, S4.3,GL.3, Grz.3, etc., are undecidable. We show also that some of these logics (say,S4:3Dum�S4:3, GL:32, Grz:32, and the logics ofhN; <i � hN; <i, hR; <i � hR; <i), are not evenrecursively enumerable (and so not recursively axiomatisable). These results give “negative”solutions to a number of open problems of [7], in particular to Problem 12 which asks whetherthere exists an infinite strict linear orderF such that the logic ofF� F is decidable.

We prove our undecidability theorems using a tiling technique (save Theorem 2.2 (iii),where we simulate the behaviour of Turing machines). Reductions to the tiling problem areknown to be rather straightforward when logics have both the“next-time” operator and theuniversal modality (see e.g. [19]). In our case the former isnot available. To simulate it, weadopt the ideas of [17] and [25]. In particular, we enumeratethe whole tiling plane along adefined diagonal and use various propositional variables aspointers to certain diagonal points.

In Section 7 we give a simple axiomatisation for the minimal linear logicK4:3 �K4:3using non-structural Gabbay-type inference rules. Note that the standard techniques usingSahlqvist and negative definability (see [18]) are not appropriate here as the language has nodifference operator, i.e., a modality which captures the concept of “anywhere but here.” Un-fortunately, we have not been able to eliminate the non-structural rules and so, as with manyother multi-modal logics, it is not known whether there is anorthodox finite axiomatisationforK4:3�K4:3.

Acknowledgments.We are grateful to Robin Hirsch, Ian Hodkinson, Agnes Kurucz, Sz-abolcs Mikulas, Valentin Shehtman, and Frank Wolter for stimulating discussions, commentsand suggestions. The work of the first author was partially supported by U.K. EPSRC grantGR/L82441 and an Australian Research Council Small Grant. The work of the second authorwas partially supported by U.K. EPSRC Visiting Fellowship GR/M36748 and by grant no.99–01–0986 from the Russian Foundation for Basic Research.

2 Preliminaries

In this paper we consider bimodal logics determined by the Cartesian products of KripkeframesF = hW;<i that are� transitive, i.e.,8x; y; z 2 W (x < y ^ y < z ! x < z), and� weakly linear, i.e.,8x; y 2 W (x = y _ x < y _ y < x).

1[S5;S5;S5℄ is the fusion (or independent join) of threeS5s extended with the commutativity and Church-Rosser axioms; see [7].

On the products of linear modal logics3

Theproductof two framesF1 = hW1; <1i andF2 = hW2; <2i is the birelational frameF1 � F2 = hW1 �W2; <h; <viin which, for allu1; u2 2 W1, v1; v2 2W2,(u1; v1) <h (u2; v2) iff u1 <1 u2 andv1 = v2;(u1; v1) <v (u2; v2) iff u1 = u2 andv1 <2 v2:(The subscriptsh andv appeal to the geometrical intuition of considering<h as the “hori-zontal” accessibility relation inF1 � F2 and<v as the “vertical” one.)

Given two classes of framesC1 andC2, we define theirproductC1 � C2 by takingC1 � C2 = fF1 � F2 : F1 2 C1;F2 2 C2g:Let ML2 be thebimodal propositional languagewith the boolean connectives (:, ^, _,!) and two diamonds3 and3. We interpretML2-formulas in models of the formM =hF1 � F2;Vi, whereV, avaluationin F1�F2, is a function associating with every proposi-tional variablep a set of worldsV(p) � W1 �W2 in which, according toM, p holds. Thetruth-relation(M; x) j= ' (or simplyx j= ', if M is understood), wherex 2 W1 �W2 and' is anML2-formula, is defined in the standard way for the booleans, andx j= 3 iff 9y >h x y j= ;x j= 3 iff 9y >v x y j= :The following modal operators are defined as abbreviations:3+' = ' _3';3+' = ' _3';2' = :3:';2' = :3:';2�' = ' ^ 2' ^ 2' ^ 22':A formula' is said to bevalid in a frameF if ' is true in all worlds in all models based onF; this fact is denoted byF j= '.

The set ofML2-formulas that are valid in all frames in the classC1�C2 is called thelogicof C1 � C2 and denoted byLog(C1 � C2).

Given two Kripke complete unimodal logicsL1 andL2, we define theirproductL1 � L2asLog (C1 � C2), whereC1 and C2 are the classes of all Kripke frames forL1 andL2,respectively. As in the unimodal case, it is not hard to see thatL1 � L2 = Log(Cr1 � Cr2);whereCri , i = 1; 2, contains allrootedframes inCi (in which9x8y(x 6= y ! x < y)). SaythatL1 �L2 has theproduct finite model propertyif L1 �L2 = Log (Cf1 �Cf2 ), whereCfi isthe set of all finite frames inCi, i = 1; 2.

Here is a list of some standard unimodal logics (in the language with3 and2) deter-mined by transitive weakly linear frames. It begins with theminimal logic, known asK4.3

4 On the products of linear modal logics

and characterised by the class of all such frames. The classes of (rooted transitive weaklylinear) frames of the formhW;<i for other logics in the list are described in brackets. Theoperator�means: add the axiom and take the closure under modus ponens,substitution andnecessitation'=2'.� K4:3 =K4�2(p ^ 2p! q) _ 2(q ^ 2q ! p);� KD4:3 =K4:3�3> (serial, i.e.,8x9yx < y);� K4:3Den =K4:3�22p! 2p (dense, i.e.,8x; z(x < z ! 9y x < y < z));� K4:1:3 = K4:3� 23p! 32p (8x9y(x < y ^ 8z(y < z ! y = z)); this first-order

formula is known as the McKinsey condition);� KD4:3Z = KD4:3 � 2(2p ! p) ! (32p ! 2p) (serial and without an infiniteascending chainx0 < x1 < : : : (of not necessarily distinct points) having an upperbound);� GL:3 = K4:3 � 2(2p ! p) ! 2p (irreflexive and noetherian, i.e., containing noinfinite ascending chain of distinct points);� S4:3 =K4:3�2p! p (reflexive);� S4:1:3 = S4:3�23p! 32p (reflexive and satisfying the McKinsey condition);� Grz:3 = S4:3 � 2(2(p ! 2p) ! p) ! p (reflexive, noetherian and containing noproper cluster, i.e., distinctx, y such thatx < y andy < x);� S4:3Dum = S4:3 � 2(2(p ! 2p) ! p) ! (32p ! p) (reflexive and without aninfinite ascending chainx0 < x1 < : : : which has an upper bound and contains at leasttwo distinct points).

Transitive weakly linearly ordered frames that areantisymmetricin the sense that8x; y(x < y ^ y < x! x = y)will be called in this paperlinear orders. A linear order all points in which are irreflexive isknown also as astrict linear order.

Using the standard bulldozing technique of modal logic (seee.g. [2]) and the obvious factthatF1 � F2 is a p-morphic image ofG1 �G2 wheneverFi is a p-morphic image ofGi, fori = 1; 2, one can prove the following:

PROPOSITION2.1SupposeL1 andL2 are logics from the list:K4.3, KD4.3, K4.3Den, KD4.3Z, K4.1.3, S4.3,S4.1.3, S4.3Dum. Suppose also thatC1 andC2 are the classes of all rooted linear orders forL1 andL2, respectively. ThenL1 � L2 = Log(C1 � C2).

All the unimodal logics in the list above are known to have thefinite model property andto be decidable (consult e.g. [2]). Our main aim in this paperis to show that this is not thecase for their products. Namely, we are going to prove the following general theorem:

THEOREM 2.2(i) Let C1 andC2 be classes of linear orders such that at least one frame in each of these classescontains an infinite ascending chain of distinct points. ThenLog (C1�C2) is undecidable anddoes not have the product finite model property.

(ii) Let C1 andC2 be classes of rooted noetherian linear orders such that at least one frame ineach of these classes contains an infinite descending chain.ThenLog(C1�C2) is undecidableand does not have the product finite model property.

On the products of linear modal logics5

(iii) Let C1 andC2 be classes of finite linear orders both containing arbitrarily long (butfinite) chains. ThenLog(C1 � C2) is undecidable.

As a consequence we shall have, in particular, the followingresults.

THEOREM 2.3If F is an infinite linear order thenLog(F� F) is undecidable.

PROOF. Let Ci = fFg, i = 1; 2. If the Ci satisfy neither (i) nor (ii), thenF is an infinitenoetherian linear order without root. LetC0i be the class of all point-generated (finite) sub-frames ofF. It should be clear thatLog (C1 � C2) = Log (C01 � C02), and so we can applyTheorem 2.2 (iii).

THEOREM 2.4(i) The product of any two unimodal logics from the intervalsK4:3 � L � Log h! + 1;�i or K4:3 � L � S4:3Dum = Log hN;�iis undecidable and does not have the product finite model property.

(ii) The following logics are undecidable and do not have theproduct finite model property:LoghN; <i2, LoghN;�i2, LoghQ; <i2, LoghQ;�i2, Log hR; <i2, Log hR;�i2.(iii) GL:32, Grz:32, andGL:3 �Grz:3 do not have the product finite model property

and are undecidable.(iv) The logicsLog hN; >i2 andLog hN;�i2 are undecidable (and so not recursively enu-

merable).2

REMARK 2.5(1) Note that the transitivity of frames is essential in the proof of Theorem 2.2. For instance,the logicsAlt1 andAlt1 � 3> can be considered as the logics of someintransitivelinearframes containing infinite ascending chains; yet their squares have the product finite modelproperty and are decidable, as was shown in [7].

(2) Although some of the logics mentioned in theorems above do not have the productfinite model property, they may still enjoy the finite model property in the class of all frames,not only the products.

(3) It may be worth noting that the following logics are decidable:� Lin � S5, whereLin is the temporal logic with the diamonds in the future and the paston arbitrary strict linear orders (cf. [21]).� Actually, the products of many other temporal logics withS5 turn out to be decidable, inparticular,S5� LoghN; <i, S5� LoghQ; <i; see [13] for more details.� Lin�K,K� Log hN; <i,K� Log hQ; <i (cf. [31]).� S4�K,K4�K and evenCPDL�K (cf. [31]).

On the other hand, it is not known whether such natural logicsasK4 � K4, S4 � S4,K4� Log hN; <i, S4� Log hN; <i and similar ones are decidable.

2Log hN; >i2 andLog hN;�i2 are different fromGL:32 andGrz:32, althoughLog hN; >i = GL:3 andLog hN;�i = Grz:3. Indeed, bothLog hN; >i2 andLog hN;�i2 have clearly the product finite model property,whileGL:32 andGrz:32 do not enjoy it.

6 On the products of linear modal logics

4 7

6

5

3

2

10������wall

4,33,32,31,3 ���0,3

0,2

0,1

4,03,02,01,00,0floor

��� �� �� ����� ���� floor

��� �� �� ����� ����FIGURE 1. The enumerationpair.

3 Linear orders with infinite ascending chains

In this section we prove part (i) of Theorem 2.2 by reducing the tiling problem forN � N,which is known to be undecidable [22], to the satisfiability problem inC1 � C2.

Let us recall first thattilesare 4-tuples of colourst = hleft(t); right(t); up(t); down(t)i :A finite setT of tiles is said totile N � N if there is a map� : N � N 7! T such that for eachi; j 2 N,� up(�(i; j)) = down(�(i; j + 1)) and� right(�(i; j)) = left(�(i+ 1; j)).If we think of a tile as a physical1� 1-square with colours along its four edges, then a tiling� of N�N is just a way of placing an infinite number of tiles, each of a type fromT , togetherto cover the first quarter of the infinite plane, with no rotation of the tiles allowed and thecolours on adjacent edges of adjacent tiles matching.

The tiling problemfor N � N is formulated as follows: “given a finite setT of tiles, doesT tile N � N?”The reduction we are looking for would be pretty simple if ourframes were intransitive or

the language contained the “next-time” operators (see e.g.[10] or [19]); for then we wouldbe able to refer to the tiles on the right and above directly. As this is not the case, we will usethe idea of [17] to enumerate the pairs of natural numbers andrefer to the right and abovepairs indirectly via special pointers. And a little trick of[25] will make it possible to dealwith both strict and non-strict linear orders.

Let pair : N 7! N � N be the enumeration of the pairs of natural numbers defined recur-sively by:� pair(0) = (0; 0),� if pair(n) = (0; j) thenpair(n+ 1) = (j + 1; 0),� otherwise, ifpair(n) = (i+ 1; j) thenpair(n+ 1) = (i; j + 1);

On the products of linear modal logics7

see Fig. 1. The pairs of the form(n; 0) are said to be on thefloor, while those of the form(0; n) are on thewall. Let right(n) denote the number of the pair to the right ofpair(n) andabove(n) the number of the pair abovepair(n). For instance,right(3) = 6, above(3) = 7.An important property of the enumeration is thatright(n+ 1) = (above(n); if pair(n) is not on the wall;above(n) + 1; if pair(n) is on the wall.

(3.1)

To explain the general idea of the reduction, let us imagine first that we need to show thatLog(hN; <i�hN; <i) is undecidable. Then, given a set of tilesT , we will construct a formula'T with the propositional variables� t, for every tilet 2 T ,� tile (= Wft : t 2 Tg),� next (a pointer to the next tile according to the enumeration),� right (a pointer to the tile on the right),� above (a pointer to the tile above),� wall (a tile is on the wall),� oor (a tile is on the floor)

in such a way that� : N � N ! T is a tiling iff 'T holds at(0; 0) under the valuation inhN; <i2 defined by: (n;m) j= t iff m = n and�(pair(n)) = t; (3.2)(n;m) j= tile iff n = m; (3.3)(n;m) j= next iff m = n+ 1; (3.4)(n;m) j= right iff m = right(n); (3.5)(n;m) j= above iff m = above(n); (3.6)(n;m) j= wall iff m = n andpair(n) is on the wall; (3.7)(n;m) j= oor iff m = n andpair(n) is on the floor. (3.8)

(This situation is depicted in Fig. 2.) It will follow thatLog hN; <i2 is undecidable.Suppose now that we have the productF = hW1 �W2; <h; <vi of two arbitrary rooted

linear ordersF1 = hW1; <1i andF2 = hW2; <2i each of which contains an infinite as-cending chain of distinct points. One important differencefrom hN; <i � hN; <i is that therelations<h and<v are not in general irreflexive (the diamond operators can’t say “here butnot later”), and our first task is to simulate strict orders inboth directions ofF.

To this end, we partitionW1 �W2 into “black” and “white” squares using the followingformulas containing propositional variablesh (horizontal) andv (vertical):2((h! 3:h) ^ (:h! 3h));2((v ! 3:v) ^ (:v ! 3v));2�(h! 2h); 2�(:h! 2:h);2�(v ! 2v); 2�(:v ! 2:v):

8 On the products of linear modal logics

ttt t t t t t t t

t t t t t tt tt t t tt

tilewall oor�(pair(0))nextrightabove tile oor�(pair(1))tilewall�(pair(2))tile oor�(pair(3))tile�(pair(4))tilewall�(pair(5))tile oor�(pair(6))tile�(pair(7))

next next next next next nextright rightabove above right rightabove

1 2 3 4 5 6 7

0

1

2

3

4

5

6

7

FIGURE 2. The case of infinite ascending chains.

The conjunction of these formulas will be denoted byChessboard. Letw = (h ^ v) _ (:h ^ :v); b = (:h ^ v) _ (h ^ :v):Say that a pointx in F (under some valuation) iswhite(black) if x j= w (respectively,x j= b).Let square(x) denote the largest setS of points inF such thatx 2 S, all points inS havethe same colour, andS is continuous in the sense that ify; z 2 S theny <h u <h z ory <v u <v z imply u 2 S. Thus, ifChessboard is true at the rootr of F (under somevaluation) thenF can be viewed as an infinite chessboard: it is divided into infinite columnsand rows of black and white squares in such a way that every square has a horizontal and avertical successors of different colour.

Now we can define new possibility operators3 and3 by taking, for any formula ,3 = (h! 3(:h ^3+ )) ^ (:h! 3(h ^3+ ));3 = (v ! 3(:v ^3+ )) ^ (:v ! 3(v ^3+ )):It should be clear that for everyx on the chessboard we have:x j= 3 iff 9y >h x (y j= & square(x) 6= square(y));x j= 3 iff 9y >v x (y j= & square(x) 6= square(y)):

On the products of linear modal logics9

A squareS on the chessboard is called ap-square, p a propositional variable, if the follow-ing two conditions hold:� 8x 2 S x j= p;� 8x 2 S8y =2 S(x <h y _ x <v y ! y 6j= p).Let p-square be the conjunction of the following formulas (in whichq0 andq00 are auxiliaryvariables different fromp): 2�(p! (:3p ^ :3p));2�((3p ^ :3p)! p);2�((3p ^ :3p)! p);2�(p! (3q0 ^3q00 ^ :33q0 ^ :33q00));2�(q0 ! :3q0);2�(q00 ! :3q00);2�(p! 2(:q0 ^3q0 ! p));2�(p! 2(:q00 ^3q00 ! p)):The reader can readily check that ifp-square andChessboard hold at the root ofF andx j= pthensquare(x) is ap-square: the first conjunct ofp-square ensures that the secondp-squarecondition holds; the second and third conjuncts ensure thatwherep holds in a square it alsoholds to the left and downwards within that square; the last five formulas say that the auxiliaryvariables hold in the immediate right and upwards neighbours of ap-square and use this toensure that wherep holds in a square it also holds to the right and upwards in thatsquare.

To simulate the tiling problem we use the formulaTiling which is the conjunction of thefour formulas: 2�(tile$ _t2T t);2� t6=t0 :(t ^ t0);2� ^up(t)6=down(t0):(t ^3(above ^3t0));2� ^right(t)6=left(t0):(t ^3(right ^3t0)):The intended meaning of these formulas should be clear.

We are in a position now to write down a formula'T which is satisfiable inF iff T tilesN � N. Define'T to be the conjunction ofChessboard, Tiling, t-square, for all t 2 T ,tile-square, next-square, right-square, above-square, wall-square, oor-square,3 and the fol-

3We always take distinct pairs of auxiliary variablesq0 andq00 in thesquare-formulas.

10 On the products of linear modal logics

lowing formulas as well: 2�(tile! 3next); (3.9)2�(next! 3tile); (3.10)2�:(3next ^3tile); (3.11)2�(tile! 3right); (3.12)2�(right! 3tile); (3.13)2�(next! :3(3tile ^3right)); (3.14)2�:(3right ^3right); (3.15)2�(tile! 3above); (3.16)2�(above! 3tile); (3.17):3(next ^ :right); (3.18)2�(right! 3above); (3.19)2�(3above ^3tile! right _3right); (3.20) oor ^ wall ^ :33( oor ^ wall); (3.21)2�( oor _ wall! tile); (3.22)2�(wall! :3(next ^3(tile ^ : oor))); (3.23)2�(wall! :3(above ^3(tile ^ :wall))); (3.24)2�(tile ^ :wall! :3(above ^3wall)); (3.25):3(3(tile ^ :wall) ^ :(3right _ right)); (3.26)2�:(3tile ^3tile); (3.27):3(3wall ^3right): (3.28)

Figure 2 will give the reader some general intuitions about these formulas. In particular,(3.14) says that as we go rightwards from anext-square to atile-square we do not pass belowa right-square. (3.18) says that the onlynext-square above the origin is also aright-square.(3.26) says that everytile-square which is not awall-square is on the same horizontal levelas someright-square. We will see how all of these formulas are used in the proof of the nextlemma.

LEMMA 3.1T tilesN � N iff 'T is satisfiable in a frameF 2 C1 � C2.PROOF. ()) Suppose� : N � N ! T is a tiling andF 2 C1 � C2 is the product of twoframesF1 andF2 with infinite ascending chains. Then the formulaChessboard is satisfiedin F under some valuationV of h andv. Let us select two infinite sequences of squareson the chessboard: one in the first horizontal row and anotherone in the first vertical rowso that the root ofF belongs to the first square in both sequences. We enumerate them bynatural numbers;(n;m) is then the square with the horizontal coordinaten and the verticalcoordinatem according to this enumeration. Now we extendV to the other variables in'T

On the products of linear modal logics11

by taking (3.2)–(3.8) as the definition in which(n;m) j= p means thatp holds at every pointin the square(n;m). It is not hard to check that under this valuation we have(0; 0) j= 'T .(() Suppose'T is satisfied at the rootx0 of someF 2 C1 � C2 under some valuation.Thenx0 belongs to atile-square; let us denote this square by(0; 0). By (3.9) and (3.10), wehave an infinite sequence of squares(0; 0) <v (0; 1) <h (1; 1) <v (1; 2) <h � � � <h (i; i) <v (i; i+ 1) <h : : : ;in which every(i; i), i 2 N, is atile-square and every(i; i + 1) is anext-square ((0; 0) <v(0; 1)means that for everyx 2 (0; 0) there isy 2 (0; 1) such thatx <v y). Thus(n; n) j= tileand(n; n + 1) j= next, for everyn 2 N. Denote by(m;n) the square located in the samecolumn with(m;m) and the same row with(n; n).

Observe that by (3.21),(0; 0) j= oor ^ wall. By (3.23) we then have(1; 1) j= oor andby (3.21)(1; 1) j= :wall.

Consider an arbitrary square(n; n). By (3.12) and (3.13) there is aright-squarern suchthat(n; n) <v rn <h t, for sometile-squaret. We will say thatrn points tot. Notice that by(3.18)r0 = (0; 1), and sor0 points to(1; 1).

By (3.16) and (3.17) there is anabove-squarean such that(n; n) <v an <h t, for sometile-squaret. We will say thatan points tot. Because of the second square condition,rnandan are the uniqueright-square andabove-square above(n; n), respectively. Notice thatby (3.19) and (3.20)a0 = (0; 2) points to(2; 2): if a0 was further above(0; 2) then (3.20)would imply thatr0 = (0; 1) is not a uniqueright-square above(0; 0). By (3.24), we have(2; 2) j= wall.

Now we prove by induction onn � 0 that

(i) if pair(right(n)) is on the floor then(right(n); right(n)) j= oor;(ii) (right(n); right(n)) j= :wall;(iii) rn = (n; right(n)) (i.e.,rn points to(right(n); right(n)));(iv) an = (n; above(n)) (i.e.,an points to(above(n); above(n));(v) if pair(above(n)) is on the wall then(above(n); above(n)) j= wall.

The basis of induction was established above. Consider the induction step forn > 0.(i) If pair(right(n)) is on the floor thenpair(n � 1) is on the wall. By (3.1), we haver(n)� 1 = a(n� 1) and sopair(right(n)� 1) = pair(above(n� 1))

is on the wall too, whence by IH (v),(right(n)�1; right(n)�1) j= wall. It follows directlyby (3.23) that we must have(right(n); right(n)) j= oor.

(ii) If pair(right(n)) is on the floor then(right(n); right(n)) j= :wall by (3.21) and(i). Otherwiseright(n) = above(right(k)) for somek, 0 � k < n. By IH (ii), we have(right(k); right(k)) j= :wall and (n � 1; n � 1) j= :wall, becausen � 1 = right(k).Since IH (iv) tells us thatan�1 points to(above(n� 1); above(n� 1)), by (3.25) we obtain(above(n� 1); above(n� 1)) j= :wall which is(right(n); right(n)) j= :wall as required.

(iii) It follows from (ii) and (3.26) that there is aright-squarer in the same horizontal lineas(right(n); right(n)). According to (3.12) and (3.15), this is possible only when we haver <h (right(n); right(n)). We will show thatr = (n; right(n)).

By (3.14) there is somei such that0 � i < right(n) with (i; i) <v r. It follows thatr = (i; right(i)). If i < n then by IHri = (i; right(i)), which is impossible. Suppose that

12 On the products of linear modal logicsi > n. Clearly,rn cannot be in the same row with(right(n); right(n)). If rn is below thatrow then, by (3.11) and (3.27)rn = (n; j) for some0 < j < right(n). If j = right(k),for somek < n, then by IHrk = (k; j), which is a contradiction. And ifj 6= right(k)for anyk < n thenpair(j) is on the wall, i.e.,j = above(k) for somek < n, from whichby IH (j; j) j= wall, contrary to (3.28). Finally, the case whenrn is above the row with(right(n); right(n)) is impossible by (3.15). Thus,i = n.

(iv) By (iii) rn j= right, and so by (3.19) there is anabove-squarea >v rn. It follows by(3.11) and (3.20) thata = (n; above(n)).

(v) If pair(above(n)) is on the wall thenpair(n) is on the wall as well. Sincen > 0,pair(n) = pair(above(k)) for somek < n, and so by IH(n; n) j= wall. By (iv) we havean = (n; above(n)) and by (3.24),(above(n); above(n)) j= wall.A tiling � of N � N is defined now as follows:�(m;n) is the uniquet 2 T for whichtpair�1(m;n) j= t. Using the formulaTiling it is readily seen that� is a tiling indeed.

Suppose now thatC1 andC2 are classes of linear frames each of which contains a framewith an infinite ascending chain of distinct points. To complete the proof of Theorem 2.2 (i),it remains only to observe that, for any set of tilesT , we clearly have::'T 2 Log (C1 � C2)iff 'T is not satisfiable inC1 � C2 iff, by Lemma 3.1,T does not tileN � N. It follows thatLog(C1 � C2) is undecidable.

4 Linear orders with infinite descending chains

Now let us prove the second claim of Theorem 2.2. SupposeF1 andF2 are rooted noetherianlinear orders containing infinite descending chains, i.e.,infinite sequences of the form� � � <i xn <i � � � <i x1 <i x0;for i = 1; 2. As in Section 3, we partitionF = hW1 �W2; <h; <vi into “black” and “white”squares. This can be done using the formulas:2�((h _3h! 2h) ^ (:h _3:h! 2:h)); (4.1)2�((v _3v ! 2v) ^ (:v _3:v ! 2:v)); (4.2):h ^ :v ^332�(h ^ v); (4.3)2�(2?^ 2?! p); (4.4)23(p ^ 2:p); (4.5)23(:p ^2p); (4.6)2�(p! 2(p ^3p)); (4.7)2�(:p! 2(:p ^3:p)); (4.8)

where3' and3' are defined as in Section 3;2 and2 are the duals of3 and3, respectively.Denote the conjunction of (4.1)–(4.8) byDiagonal. If it holds somewhere inF (under

some valuation) then, in view of (4.3)–(4.4), there is a top-right square inF all points inwhich validateh, v, andp. We denote this square by(0; 0). Let (1; 0) be its left neighbour(its points validate:h andv), (0; 1) the neighbour below (thereh and:v hold), if they exist,etc. The square(m;n) is located in the column with(m; 0) and in the row with(0; n). We

On the products of linear modal logics13

write x �h (m;n) to say thatx =2 (m;n) and there isy 2 (m;n) such thatx <h y;(m;n) j= means thatx j= for everyx 2 (m;n),LEMMA 4.1If Diagonal is true at some pointx in F under some valuation then, for alln;m 2 N,

(a) (m;n) j= p if m � n,

(b) (m;n) j= :p if m > n, and

(c) 9y x �h y �v (n; n).PROOF. Supposex j= Diagonal for somex in F. By induction onn we are going to showthat (i) there is a pointy in F such thatx �h y �v (n; n), and (ii) (m;n) j= p if m � n and(n;m) j= :p wheneverm < n. The basis of induction follows from (4.1)–(4.4).

Assume now thatn > 0 and our claim holds for allk < n. So we havey such thatx �h y �v (n � 1; n � 1), p is true everywhere in the triangle(0; 0), (n � 1; n � 1),(0; n� 1) and is false everywhere above this triangle. By (4.6), thereare pointsz andu suchthatx �v z �h u �h (n� 1; n� 1) andu j= :p. So there is a square(n; n� 1), that is theleft neighbour of(n� 1; n� 1). By (4.7),(n; k) 6j= p, for all k � n� 1. In view of (4.5), wehave somev such thatx �h v �v (n; n). And by (4.8),(k; n) j= p for all k � n.

LEMMA 4.2Diagonal is satisfied inF.

PROOF. SinceF1 andF2 are noetherian, we can define a valuation ofh andv in F so that(0; 0) consists of the point in the top-right corner,(1; 0) of its immediate<h-predecessor,(0; 1) of the immediate<v-predecessor of(0; 0), and so forth. Besides, we putz j= p iffz = (m;n) for somem � n. Let x be the point such that condition (c) of Lemma 4.1 holdsfor all n 2 N and whenevery h x or y v x theny = (m;n) for somem;n 2 N (it exists,sinceFi are rooted and contain no infinite ascending chains). The reader can easily checkthat under this valuationx j= Diagonal.

Note by the way that as a consequence of these two lemmas we obtain

COROLLARY 4.3Let C1 andC2 be as in the formulation of Theorem 2.2 (ii). ThenLog (C1�C2) does not havethe product finite model property.

Given a finite setT of tiles, we construct now a formula'T which is satisfiable inF iff TtilesN � N. As in the previous section, we will use the enumerationpair : N ! N � N andrepresent the tiles onN � N by the squares(n; n) on the diagonal ofF using the variablestile andt, for t 2 T . Let left(n) denote the number of the pair to the left ofpair(n) andbelow(n) the number of the pair belowpair(n). As before,wall and oor will indicate that acertain pair is on the wall or on the floor. But instead of the pointersnext, right andabove wewill use now the variablesprev, left andbelow which are supposed to point to the previouspair in the enumeration, to that on the left and below, respectively (see Fig. 3).

14 On the products of linear modal logics

x j= 'Tt r r r ......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

......

....

t t t t t ttttttttttt p

:pdiagonal012345 tilewall oor1 prevleft2 below345

tile oorprevleftbelowtilewallprevleftbelow

tile oorprevtileprevtilewall��������������

������������

FIGURE 3. The noetherian case.

LetTiling be the conjunction of the following formulas:2�(tile$ _t2T t);2�(t6=t0 :(t ^ t0));^down(t)6=up(t0)2�:(t ^3(below ^3t0));^left(t)6=right(t0)2�:(t ^3(left ^3t0)):Define'T to be the conjunction ofTiling, Diagonal, t-square, for all t 2 T , prev-square,

On the products of linear modal logics15left-square, below-square, wall-square, oor-square,4 and the formulas:2�(tile$ p ^ :3p); (4.9)2�(prev $ 3tile ^ :33tile); (4.10)2�(tile ^ :wall! 3left); (4.11)2�(tile ^ wall! :3left); (4.12)2�(left! 3tile ^ :3left); (4.13)2�(tile ^ : oor! 3below); (4.14)2�(tile ^ oor! :3below); (4.15)2�(below! 3tile ^ :3below); (4.16)2�(tile ^ : oor ^ :wall! 3(left ^3below ^ :33below)); (4.17)2�(2?^2? $ oor ^ wall); (4.18)2�((2?^2?) _ (tile ^3(prev ^3wall))$ oor); (4.19)2�(3> ! (tile ^3(below ^3wall)$ wall)); (4.20)2�(tile ^ : oor! 2(below! 3left ^ :33left)); (4.21)2�(tile ^ oor! :3(3 oor ^3left)); (4.22)2�( oor! 2(left! 3 oor)): (4.23)

LEMMA 4.4T tilesN � N iff 'T is satisfiable inF.

PROOF. ()) Suppose� : N � N ! T is a tiling, and letx be the “limit” of the diagonal inF as in the proof of Lemma 4.2. Define a valuation inF by taking, for anyy 2 W1 �W2,y j= h iff y = (m;n) andm is even;y j= v iff y = (m;n) andn is even;y j= p iff y = (m;n), for somem � n;y j= t iff y = (n; n) and�(pair(n)) = t;y j= tile iff y = (n; n) for somen 2 N;y j= prev iff y = (n� 1; n);y j= left iff y = (m;n) andm = left(n);y j= below iff y = (m;n) andm = below(n);y j= wall iff y = (n; n) andpair(n) is on the wall;y j= oor iff y = (n; n) andpair(n) is on the floor.

It is a matter of routine to check that under the defined valuation'T holds atx.(() Suppose'T is true atx in F under some valuation. Sincex j= Diagonal and in viewof Lemma 4.1,p is true in all squares(m;n), form � n, and false in all(m;n) with m > n,m;n 2 N. Moreover, all squares on the diagonal are accessible fromx via<h and<v.

4These formulas were constructed in Section 3.

16 On the products of linear modal logics

By (4.9), tile holds in (n; n) on the diagonal and is false in(n;m) whenevern 6= m.And by (4.10),prev holds in(m;n) iff n = m + 1, i.e., it holds only in the squares locatedimmediately below the diagonal ones. (Note thatp, tile, prev may be true at some other pointsthat do not belong to the depicted grid, but they are of no concern to us.)

By (4.18),(0; 0) j= oor ^ wall. By (4.19),(1; 1) j= oor and in view of (4.18), we have(1; 1) j= :wall. By (4.11),(0; 1) j= left. And by (4.15),(0; 1) j= :below.Now by induction onn � 0 we show that

(i) pair(n) is on the floor iff(n; n) j= oor;(ii) (m;n) j= below iff m = below(n);(iii) pair(n) is on the wall iff(n; n) j= wall;(iv) (m;n) j= left iff m = left(n).

The casen � 1 has been already considered. So letn > 1. Observe that by (4.13) and (4.16),if (m;n) j= left or (m;n) j= below thenn > m, i.e.,(m;n) is below the diagonal.

Suppose thatpair(n) is on the floor. Thenpair(n � 1) is on the wall, and so by IH(n � 1; n� 1) j= wall. It follows that(n; n) j= tile ^ 3(prev ^ 3wall). By (4.19) we thenhave(n; n) j= oor. The converse implication also follows by IH from (4.19). This proves(i).

Assume now that(m;n) j= below. By (4.15),(n; n) j= : oor. And by (4.21) we have(m;n � 1) j= left, which by IH means thatm = left(n � 1), and sopair(m) is belowpair(n) indeed. Conversely, supposem = below(n). Thenn is not on the floor, and so(n; n) j= : oor. By (4.14),(k; n) j= below for somek < n. But then, as we know,k = m(uniquely by the definition ofbelow-square. This yields (ii).

Supposepair(n) is on the wall andpair(m) is located immediately belowpair(n). Thenby (i) (n; n) j= : oor, and so by (4.14)(k; n) j= below for somek < n. By (ii) k = m. Soby IH (n;m) j= wall and we have(n; n) j= tile^3(below^3wall). And since(n; n) j= 3>,we obtain by (4.20) that(n; n) j= wall. The converse implication follows from IH and (4.20).Thus we have (iii).

Finally, to prove (iv), suppose that(m;n) j= left. By (4.12), we have(n; n) j= :wall. If(n; n) j= : oor then by (4.17) and (ii)pair(m�1) is belowpair(n), whencem = left(n).And if (n; n) j= oor (and sopair(n) is on the floor) then by (4.22)(k; k) j= oor for no ksuch thatm < k < n, which means by IH and (4.23) thatm is the greatest integer< n withpair(m) on the floor. Thuspair(m) is on the floor immediately to the left ofpair(n). Thusagainm = left(n). Conversely, supposem = left(n). Then(n; n) j= :wall. By (4.11),(k; n) j= left for somek < n, from whichk = m, as required.

It follows from (ii), (iv) andx j= Tiling that the map� defined by taking�(n; n) = t iff(n; n) j= t, for t 2 T , is a tiling ofN � N.

Theorem 2.2 (ii) follows immediately from Lemma 4.4.

5 Finite linear orders

In this section we prove the last claim of Theorem 2.2, namelythat if C1 andC2 are classesof finite linear orders (no matter reflexive or irreflexive) both containing arbitrarily long (butfinite) chains, then the logicLog(C1 � C2) is undecidable.

For simplicity we will assume here that bothC1 andC2 contain onlystrict linear orders. Ifthis is not the case, one can use variablesh andv satisfying (4.1) and (4.2) to partition framesinto black and white squares as in the two previous sections.

On the products of linear modal logics17

We are going to reduce the following undecidable problem to the satisfiability problem inC1�C2: given a Turing machine, to determine whether it comes to a stop having started fromthe empty tape. The basic idea of the proof is to represent a run of the Turing machine as asequence of columns, each of which represents a configuration.

LetA be a single-tape right-infinite deterministic Turing machine with state spaceS, initialstates0, halt states1, tape alphabetA (b 2 A stands for blank) and transition functionÆ. Weassume thatA comes to a stop only if it reachess1; if the current state is different froms1then the machine makes another step. The configurations ofA will be represented by finitewords of the form$a0 : : : ai : : : anbm, where$ marks the left side of the tape,m > 0, alla0; : : : ; an save one, sayai, are inA, while ai belongs toS � A and represents the activecell and the current state. The start configuration, for instance, is represented by any word ofthe form:$(s0; b)bm, wherem > 0. LetA0 = A [ f$g [ (S � A). With each� 2 A0 weassociate a propositional variablep�.

We show how to construct a formula'A which is satisfiable in a frame fromC1 � C2 iffA comes to a stop having started from$(s0; b)bm. Define'A to be the conjunction of thefollowing formulas, wherep ranges over all the variables occurring in'A:d; (5.1)2�:(p ^ (3p _3p)); (5.2)2�(d ^3> ! 3next ^ :33next); (5.3)2�(next ^3> ! 3d ^ :33d); (5.4) oor ^ wall ^ :33( oor ^ wall); (5.5)2(next! right); (5.6)2�(right ^3> ! 3above ^ :33above ^ :3above); (5.7)2�(:wall ^ d! 2(above ^3> ! 3right ^ :33right)); (5.8)2�(wall ^ d! 2(above ^3>! 3q ^ :33q ^ :3right)); (5.9)2�(q ^3> ! 3right ^ :33right); (5.10)2�(wall! 2(next! 2(d! oor))); (5.11)2�(wall! 2(above! 2(d! wall))); (5.12)2�(above! :3 oor); (5.13)2�(right! :3wall); (5.14)2�(d$ _�2A0 p�); (5.15)2� �2A0(p� ! : _�6=�2A0 p�); (5.16)2�(l$ d ^3(above ^3s)); (5.17)2�(s$ _(s;a)2A0 p(s;a)); (5.18)2�(s$ 3(above ^3(d ^ r))); (5.19)

18 On the products of linear modal logics2� ^Æ(�;�; )=(�0;�0; 0) �p� ^ l ^3(above ^3(p� ^3(above ^3p )))!3(right ^3(p�0 ^3(above ^3(p�0 ^3(above ^3p 0)))))� ; (5.20)2� �2A0(p� ^ :l ^ :s ^ :r ! 2(right! 2(d! p�))); (5.21)p$ ^3(above ^3(p(s0;b) ^3(above ^3pb))): (5.22)

The intended meaning of this formula is as follows. Suppose'A is true at the root of aframeF 2 C1�C2 under some valuation. Then the variabledmarks the diagonal ofF startingfrom the root. Let0; : : : ; ` be the consecutive points on this diagonal. We will think of themas representing the first`+1 cells in theN�N grid under the enumeration defined in Section3. For anyi; j � `, denote byxij the point located in the column containingi and in the rowcontainingj. Then we havexij j= right andxik j= above, for i; j; k � `, iff right(i) = jandabove(i) = k. The variableswall and oor mark the wall and the floor, respectively. Theformulas (5.1) to (5.14) are used to ensure this in a variation on the argument in Lemma 3.1.

Every point on the diagonal is marked also by a variablep�, for some� 2 A0. Configu-rations are represented by tupleshi0; : : : ; iki such thati0 j= oor andxij ij+1 j= above. Wehaveij j= s iff ij j= p(s;a) for some(s; a) 2 A0, ij�1 j= l andij+1 j= r. The formulas(5.20) and (5.21) describe the transition function: we moveto the column on the right andchange the active cell with its left and right neighbours, leaving other cells intact. Finally,(5.22) says thatA starts working on the empty tape.

With this explanation, it is not hard to show that'A is as required. We leave this to thereader.

6 Dedekind complete linear orders

Harel [11] proved that the following problem is�11-complete:� Given a finite setT of tiles and a tilet 2 T , canT tile N�N such thatt appears infinitelyoften in the first column?

We will use this result to show that logics of certain classesof products of linear frames arenot recursively enumerable, and so not recursively axiomatisable.

Say that a linear orderF = hW;<i is Dedekind completeif every bounded (with respect to<) infinite ascending chain of distinct points inW has a least upper bound inF.

THEOREM 6.1Let C be a class of Cartesian products of linear orders satisfyingthe following conditions:

min there is a framehW1; <1i � hW2; <2i 2 C with eachhWi; <ii, for i = 1; 2, containingan infinite ascending chain;

max if hW1; <1i � hW2; <2i 2 C, thenhW1; <1i is Dedekind complete.

Then the satisfiability problem inC is�11-hard, whenceLogC is not recursively enumerable.

PROOF. Given a setT of tiles containingt, let T be the conjunction of'T defined in Sec-

On the products of linear modal logics19

tion 3 and the following three formulas:2�(tile! 33(t ^ wall)); (6.1)2 �(:3tile ^ :33tile) _3tile _3(3tile ^ :next ^ :3next)� ; (6.2)2(3tile! next _3next): (6.3)

We will show that T is satisfiable inC iff T tilesN � N with t appearing infinitely often upthe wall.

If we have a tiling witht appearing infinitely often along the wall then it is clear that Thas a model based on a frame inC: just use the construction given in Section 3 above on oneof the frames having ascending chains in both dimensions acting as the natural numbers. Thethree new conjuncts can easily seen to be satisfied.

Conversely, if T has a model based on a frame(W1; <1)� (W2; <2) from C then this isalso a model of'T . So in the same way as in Section 3 we can construct a tiling ofN � N.(Recall that we use a sequence of squares(0; 0); (1; 1); : : : at whichtile holds.)

We will show that the only points at whichwall holds are those in squares(i; i) for whichpair(i) is on the wall. (6.1) then tells us thatt appears infinitely often as a tile on the wall inthe tiling, which is precisely what we need.

In fact it is sufficient to show that there are notile-squares apart from(n; n), n 2 N. Thisis because by (3.22)wall is only true intile-squares and in Section 3 we had shown that(n; n)is awall-square iffpair(n) is on the wall (i.e. not right of any otherpair(k)).

For eachi < !, choosexi 2 W1 andyi 2 W2 such that(xi; yi) 2 (i; i). If x0 <1 x1 <1: : : is unbounded inhW1; <1i then we are done as (3.9) and (3.11) can easily be used to showthat there are no othertile-squares.

So consider the situation in whichx0 <1 x1 <1 : : : is bounded inhW1; <1i. As hW1; <1iis Dedekind complete, there is some least upper boundz to the sequence. We will show that“from z on” there are no moretile-squares, i.e.,8u 2 W28z0 2W1(z0 �1 z ! (z0; u) 6j= tile):By (6.2) there are three possibilities for the pair(z; y0):case 1:We may have:3tile ^ :33tile true at(z; y0). But then indeedtile is false fromz

on. Let us show that the other cases cannot occur.

case 2:(z; y0) makes3tile true. Thentile is true at(z; u) for someu �2 y0. By (6.3),nextis true at(z0; u) for somez0 �1 x0. By (3.10) and the definition of atile-square,z0 <1 z.As z is the least upper bound of the sequence ofxis, we know there is somexi such thatz0 <1 xi. There are three cases depending on the ordering ofu andyi. We can not haveu <2 yi by (3.27). We cannot haveu >2 yi by (3.11). And finally, we cannot haveu = yi by the definition of atile-square.

case 3:3(3tile^:next^:3next) is true at(z; y0). This case is similar to case 2 and cannothappen.

So all thetile-squares lie before the least upper boundz and we are done by (3.9) and (3.11).

COROLLARY 6.2The following logics are not recursively enumerable:LoghN; <i2, Log hN;�i2, Log hR; <i2,Log hR;�i2, Log hZ; <i2, andLog hZ;�i2.

20 On the products of linear modal logics

COROLLARY 6.3LetL1 2 fKD4:3Z;S4:3Dumg and letL2 = LogC, whereC is a class of linear orders atleast one of which contains an infinite ascending chain of distinct points (e.g.L2 is any logicmentioned in Theorem 2.4 (i)). ThenL1 � L2 is not recursively enumerable.

By using an equally devious set of extra formulas we can also produce a similar theoremfor classes of frames with infinite dscending chains.

THEOREM 6.4Let C be a class of Cartesian products of linear orders satisfyingthe following conditions:

min there is a framehW1; <1i�hW2; <2i 2 C with eachhWi; <ii, for i = 1; 2, being rootedand containing an infinite descending chain;

max if hW1; <1i�hW2; <2i 2 C, then bothhW1; >1i andhW2; >2i are Dedekind complete.

Then the satisfiability problem inC is�11-hard, whenceLogC is not recursively enumerable.

PROOF. As before, suppose that we have a finite setT of tiles containingt. We extend'T ofSection 4 to T by adding the following conjuncts to'T ::33'T ; (6.4)2�(:p! :r); (6.5)2�(t ^ wall! r ^ 2r); (6.6)2(3r ! 3(t ^ wall)); (6.7)23r: (6.8)

These are designed to ensure that:� T can be true only at the Dedekind limit of the diagonal squares(n; n), n 2 N, (6.4)(for 'T holds at this limit),� r is false above the diagonal squares (6.5),� r is true in everyt-square and to the right it whenever this square validateswall (6.6),� and only there (6.7), and� if one moves even a bit to the right from the square where T holds, one seesr above(6.8).

Thus, if T is satisfied in some square thenr (and sot on the wall) must be infinitely closeto this square.

Conditionmin guarantees that ifT tilesN�N so thatt appears infinitely often on the wall,then T is satisfiable inC.

COROLLARY 6.5GL:32,Grz:32, andGL:3�Grz:3 are not recursively enumerable.

7 An axiomatisation ofK4:3�K4:3The results of the previous section showed that it is often impossible to axiomatize productsof linear modal logics. Even when it is possible it is still hard. Here we give an examplefor K4:32. It is difficult to see how to find a complete axiom system whichonly uses the

On the products of linear modal logics21

traditional structural rules: there is a problem with marshalling maximal consistent sets intoa grid-like pattern. As we will see, we solve the problem by using Gabbay-style irreflexivityrules to allow us to introduce new atoms into the construction which indicate lattitudes andlongitudes of the sets of formulas. Because the connectivesdo not have duals in the lan-guage we rely on a series of irreflexivity rules (rather than just one) for each dimension. Theproof then follows standard techniques for creating canonical structures using such rules (seeSection 6.11 in [9]) and standard techniques for modifying the canonical structures as seenin [26] (for an interval logic), [9] (for Vlach andAqvist’s logic), [5] (for a two-dimensionaluntil-since logic) and [18] (for the compass logic). Thus wedo not present the proof.

So here is a complete axiom system—call itK4:32—for K4:3 � K4:3. The inferencerules ofK4:32 are modus ponens and the rules of necessitation for both2 and2:'; '! ; '2'; '2'along with Gabbay-style irreflexivity rules (cf. [9]) for eachn < !:IRRn : :(�n ^ �n(�n�1 ^ �n�1(� � � ^ �1(�0 ^ (q))) � � � )):(�n ^ �n(�n�1 ^ �n�1(� � � ^ �1(�0)) � � � ))where ; �0; �1; : : : ; �n are any formulas, (q) is a substitution instance of by any atomsq not occurring in the�i, and�1; : : : ;�n are each either3 or3.

The axiom schemas ofK4:32 include the usual ones forK4.3:X0 the schemas of classical logic,X1v 2('! )! (2'! 2 )X2v 2'! 22'X3v 3' ^3 ! 3(' ^ ) _3(' ^3 ) _3( ^3')with horizontal versions (X1h,X2h,X3h) of the latter three and two schemas to describe theinteraction of the two dimensions:X4 33'$ 33' “commutativity ”X5 32'! 23' “Church-Rosser.”

THEOREM 7.1The axiom systemK4:32 is sound and complete forK4:32.8 Questions

As we have seen, products of linear logics aboveK4 are usually undecidable and sometimeseven not recursively enumerable. We did not consider, however, products of frames one ofwhich contains an infinite ascending chain and the other an infinite ascending chain. So thefollowing problem remains open:

(1) AreK4:3�GL:3 and similar logics decidable (recursively enumerable)?

Nothing is known about the behaviour of products of linear and non-linear logics aboveK4.For instance, it would be interesting to answer the following questions:

(2) Are logics of the formK4:3�K4, S4:3� S4, etc. decidable?

22 On the products of linear modal logics

(3) Is the product of a logic with the difference operator anda Priorean temporal logic decid-able?

And of course products of “transitive” non-linear logics remain a challenge:

(4) Are logics of the formK4�K4, S4� S4,GL�GL etc. decidable?

Turning to axiomatizations, we have seen, once again, the benefits of using IRR-style rules.However, as with many other two-dimensional temporal logics, we have the usual open prob-lem:

(5) Is there a finite axiomatization forK4:32 (or for some other product of linear logics)without using IRR-style rules?

Even making use of IRR-style rules we have only presented a complete axiom system forK4:32. Other products may be straightforward modifications of this system but some seemto be more difficult:

(6) Give complete axiomatisations for products such asK4:1:32 andK4:3�GL:3.

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Received 31st August 1999