Brock University. A Discrete Representation of Dicomplemented Lattices. Department of Computer Science. Ivo Düntsch, Léonard Kwuida, Ewa Orloska,

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1 Brock University Department of Computer Science A Discrete Representation of Dicomplemented Lattices Ivo Düntsch, Léonard Kwuida, Ewa Orloska, Brock University Department of Computer Science St. Catharines Ontario Canada LS 3A Institute of Telecommunications Szachowa, Warszawa, Poland Berner Fachhochschule Fachbereich Wirtschaft CH-3005, Bern, Switzerland Technical Report # CS-7-0 March 07

2 A discrete representation of dicomplemented lattices Ivo Düntsch Dept of Computer Science Brock University St Catharines, ON, LS 3A, Canada Ewa Orłowska Institute of Telecommunications Szachowa , Warszawa, Poland Léonard Kwuida Berner Fachhochschule Fachbereich Wirtschaft CH-3005, Bern, Switzerland Abstract Dicomplemented lattices were introduced as an abstraction of Wille s concept algebras which provided negations to a concept lattice. We prove a discrete representation theorem for the class of dicomplemented lattices. The theorem is based on a topology free version of Urquhart s representation of bounded general lattices. Introduction In philosophical logic a concept is characterized by its extent and intent. The extent of a concept C meaningful in a universe of discourse U is a subset e(c) of U consisting of those objects which are the instances of the concept. The intent of a concept is the set i(c) of properties which qualify the objects of e(c) as instances of the concept. In [3] the notion of Dedekind cuts used in the definition of real numbers was generalized with the intuition that, when considered on a lattice, the cuts represent concepts. More precisely, a cut is a pair ha,b)i of subsets of a lattice such that there is a concept C with e(c) =A and i(c) =B. Concept lattices were defined endowed with a negation and a representation theorem was proved. In [6] an abstract algebraic representation of concepts and their negations is presented within the framework of dicomplemented lattices. The lattices are endowed with two negations; details can be found in Section 5. When considered on a distributive lattice, they are the counterparts to the Heyting and Brouwer negation, respectively, see []. Ivo Düntsch gratefully acknowledges support by the Natural Sciences and Engineering Research Council of Canada Discovery Grant 5053 and by the Bulgarian National Fund of Science, contract DN0/5/ Ewa Orłowska gratefully acknowledges support from the National Science Centre project DEC-0/0/A/HS/00395.

3 Since the early 0th century a variety of logical systems whose languages included the operators turning propositions into the opposite propositions have been a subject of extensive studies. Helena Rasiowa contributed substantially to that line of research. In her book An Algebraic Approach to non-classical Logics [0] a survey of classes of systems with negations and their classification is provided. Dicomplemented lattices are a valuable enlargement of that classification. In the present paper we prove a discrete representation theorem for the class of dicomplemented lattices. The definition of representation algebra does not involve any topology and therefore it is referred to as a discrete representation. The theorem is based on a discrete version of Urquhart s representation of bounded general lattices []. It is well known that in case of distributive lattices there are discrete representation theorems both for the lattices and the lattice frames, e.g. discrete versions of the Stone and Priestley dualities. In this paper we show that there are Urquhart general lattice frames (doubly ordered sets) which do not admit a representation theorem for frames. First definitions and notation A two sorted frame is a triple hx,y,ri where X,Y are nonempty sets and R X Y is a binary relation among elements of X and Y. Two sorted frames are called polarities in [] and formal contexts in [3]. If X = Y, we usually just write hx,ri and speak of a frame. We let R be the relational converse of R, i.e. R = {hy,xiy X : xry}. If x X, then R(x)={y Y : xry}; R (y) is defined analogously. Common operators on the powerset of frames are hri(z) ={x X : R(x) \ Z 6= /0 } [R](Z)={x X : R(x) Z} Possibility operator, Necessity operator. For useful facts about these operators we invite the reader to consult [9, Section.8]. In particular, we shall use hri(z)= [R]( Z), (.) hr;si(z)=hrihsi(z) and [R;S](Z)=[R][S](Z). (.) If f : X! Y is a function, ran f = { f (x) : x X} is the range of f. Throughout, hl,, +, 0, i is a bounded lattice, possibly with additional operators. The set of all proper filters, respectively, ideals, of L is denoted by F, respectively, by I. Furthermore, " M = {a L : (9b)[b M and b apple a]}; the set # M is defined dually. If M = {a} we usually just write " a instead of "{a}. A subset M of L is called join dense if every nonzero element of L is a join of elements of M; meet density is defined dually. The filter of L generated by M is denoted by [M] filt. A closure operator on a partially ordered sethp,applei is a mapping f : P! P such that for all a,b P. a apple f (a), (Extensive). a apple b implies f (a) apple f (b), (Isotone)

4 3. f ( f (a)) = f (a)). (Idempotent) a P is called closed (with respect to f ) if a = f (a). An interior operator on hp,applei is defined dually. In the case of h X, i the smallest closed element is f (/0 ) and the largest closed element is X. Let hp,applei and hq, i be partially ordered sets. A Galois connection is a pair of functions f : P! Q, g : Q! P such that for all a P,b Q It is well known that both f and g are antitone (order reversing). b f (a) () a apple g(b). (.3) There are close connections among closure operators, Galois connections and complete lattices. In particular, each closure operator on some X induces a complete lattice; G. Birkhoff [, p. 49] credits this result to E.H. Moore [7, pp 53 80]: Theorem.. [, Ch IV, Theorem ] Let f be a closure operator on the ordered powerset h X, i of some nonempty set X, and let C be the collection of closed subsets of X. Then, C can be made into a complete lattice with the operations [ ÂA i = f A i!, (.4) ii ii A i = \ A i. (.5) ii ii Theorem... [4, Theorem 4] Let hp,applei be a partially ordered set and c be a closure operator on P. Then, there are an ordered set hq, i and a Galois connection h f,gi such that c(a)=g( f (a)) for all a P.. [8, Theorem ] If h f,gi is a Galois connection between the partially ordered sets hp,applei and hq, i, then g f : P! P is a closure operator. 3 Polarities and concept lattices A sufficiency operator on L is a function f : L! L for which for all a,b L, f (0)=, Co normal (3.) f (a + b)= f (a) f (b) Co additive. (3.) A dual sufficiency operator is a function g : L! L for which for all a,b L, g()=0, (3.3) g(a b)=g(a)+g(b). (3.4) These operators were first considered in a logical setting in [5]. It is easy to see that sufficiency operators and dual sufficiency operators are antitone. 3

5 Define functions [[R ]] : X! Y, [[R]] : Y! X by [[R ]](V )={y Y : V R (y)]}, (3.5) = {y Y : (8x X)[x V ) xry]} [[R]](W)={x X : W R(x)}, (3.6) = {x X : (8y Y )[y W ) xry]}. These mappings are complete sufficiency operators. They are called the polars of hx,y, Ri by Birkhoff [, p. 56], and derivation operators by Wille [3]. Theorem 3.. [, p. 54] Let A X,B Y.. The pair h[[r ]],[[R]]i forms a Galois connection.. The correspondences A 7! [[R ]](A) and B 7! [[R]](B) are complete dual isomorphisms between the complete lattices of [[R]][[R ]] closed subsets of X and [[R ]][[R]] closed subsets of Y. To simplify notation we shall write f R for [[R]][[R ]] and g R for [[R ]][[R]]. We call the complete lattice of f R closed subsets of X the B complex algebra of the polarity hx,y,ri, and denote it by Cm B hx,y,ri. Birkhoff s construction is the basis of formal concept analysis introduced by Wille [3]. A formal concept of hx,y,ri is a pair ha,bi X Y such that [[R ]](A)=B and [[R]](B)=A. The set of all formal concepts of hx,y,ri is denoted by Con(X,Y,R). Theorem 3.. [3] Let hx,y,ri be a frame. Con(X,Y,R) is a complete lattice, called the concept lattice of hx,y,ri under the lattice ordering ha,biappleha 0,B 0 i if and only if A A 0. Conversely, a complete lattice M is isomorphic to Con(X,Y,R) if and only if there are mappings g : X! M, µ : Y! M such that rang is join dense, ran µ is meet dense, and xry if and only if g(x) apple µ(y) for all x X,y Y. It is not hard to see that the lattice Con(X,Y,R) is isomorphic to the lattice of f R closed sets of X and dually isomorphic to the g R closed sets of Y. Thus, the concept lattice of a frame is (isomorphic to) the complex algebra of the polarity hx,y,ri in Birkhoff s sense. 4 Urquhart s representation of lattices In this section we briefly review the lattice representation of []. A doubly ordered set is a structure hx,apple,apple i such that apple,apple are quasiorders on X and x apple y and x apple y imply x = y. (4.) If the relations are clear from the context we shall name a doubly ordered set just by its base set. If Y X and i {,}, we let # i Y and " i Y be the downset, respectively, the upset of Y with respect to apple i. Define two mappings l,r : X! X by l(y ) df = {x :" x \Y = /0 }, (4.) r(y ) df = {x :" x \Y = /0 }. (4.3) 4

6 l and r can be viewed as intuitionistic negations; for example, l(y ) is the largest apple increasing subset of X disjoint from Y. These can be written in modal form as l(y )=[apple ]( Y ), (4.4) r(y )=[apple ]( Y ). (4.5) Y is called a stable set, if Y = l(r(y )). The collection of stable sets is denoted by L X. Observe that l(r(y )) = l([apple ]( Y )) = [apple ]( [apple ]( Y )) = [apple ]happle i(y ). (4.6) Lemma 4.. [] Let (X,apple,apple ) be a doubly ordered set.. The mappings l and r form a Galois connection between the lattice of apple increasing subsets of X and the lattice of apple increasing subsets of X.. If Y is apple increasing, then l(y ) is a stable set. Thus, if Y is apple increasing and Z is apple increasing, then Y l(z) if and only if Z r(y ). For Y,Z L X let Theorem 4.. [] The structure hl X,_ X,^X, /0,Xi is a complete lattice. Y _ X Z df =[apple ]happle i(y [ Z), (4.7) Y ^X Z df = Y \ Z. (4.8) We call this structure the Urquhart complex algebra of X, and denote it by Cm U (X). Next, we go from lattices to frames. A filter ideal pair is a pair hf,ii, where F F, I I, and F \I = /0. Define the component wise quasiorders on the set of all filter ideal pairs by hf,i i hf,i i df hf,i i () F F, (4.9) df hf,i i () I I, (4.0) and let be the intersection of and. A filter ideal pair is called a maximal pair, if it is maximal with respect to. Lemma 4.3. Each filter ideal pair can be extended to a maximal pair. The collection of all maximal pairs is denoted by X L. To facilitate notation, if x = hf,iix L, we let x = F and x = I. Furthermore, if x,y X L, we let x i y if and only if x i y i. With these definitions hx L,, i is a doubly ordered set, which we call the Urquhart canonical frame of L, denoted by Cf U (L). Theorem 4.4. [] Let L be a lattice and define h : L! X L by h(a) df = {x X L : a x }. Then h is a lattice embedding into Cm U Cf U (L). 5

7 Whereas every lattice can be embedded into the complex algebra of its Urquhart canonical frame, an analogous embedding for doubly ordered frames is not always possible, even on the set level: Suppose that X, apple is the identity, and apple is the universal relation. Clearly, hx,apple,apple i is a doubly ordered frame. Since apple is the identity, [apple ](Y )=Y for every Y X. If Y 6= /0 and y Y, then x apple y for all x X, since apple is the universal relation. Hence, happle i(y )=X for every nonempty Y X. It follows that L X = {X, /0 }, and therefore, Cf U Cm U (X) ={h{x},{/0 }i}, which has only one element. Since X, there is no injective mapping X! Cf U Cm U (X). Thus, we obtain the following theorem: Theorem 4.5. If X, there is a doubly ordered frame X which cannot be embedded into Cf U Cm U (X). 5 Concept algebras and dicomplemented lattices Formal concept analysis is based on the formalization of the notion of concept. Concepts are considered as basic unit of thought. They are determined by their extent and their intent. The extent consists of all objects belonging to the concept while the intent is the set of all attributes shared by all objects of the concept. Each of these sets (extents and intents) should uniquely define the corresponding concept. To develop a logic based on concepts as units of thought, the logical operators need to be formalized appropriately. The universe of discourse is a binary relation involving objects and attributes of interest. The set of concepts of this context forms a complete lattice [3]. Towards developing a Boolean concept logic, the conjunction was encoded by the meet and the disjunction by the join in the concept lattice. > corresponds to the top element and? to the bottom element of the concept lattice. To formalize negation we face the problem that the complement of extents are not always extents, and idem for intents. Two options for negation were considered: The first approach relaxes the definition of concept to accommodate the complement of extents/intents. This leads to the notion of semi-, pre-, protoconcepts and double Boolean algebras [4]. The second approach obtains the negations as concepts generated by the complement of the extents or complement of the intents. This leads to concept algebras and dicomplemented lattices [4, 6]. Here we consider the second approach. If L = ConhX,Y,Ri, and ha,bil, we define mappings 4, 5 : L! L by ha,bi 4 = h f R ( A),[[R ]]( A)i, (5.) ha,bi 5 = h[[r]]( B),g R ( B)i. (5.) A concept algebra is a structure hl,+,, 4, 5,0,i of type h,,,,0,0i which is isomorphic to a concept lattice with the additional operations defined by (5.) and (5.). Generalizing these structures, Kwuida [6] has introduced the following class of algebras: A dicomplemented lattice (DL) is a structure hl,+,,4,5,0,i such that 4, 5 : L! L are operators such that for all a,b L, These are called weakly dicomplemented lattices in [6]; here we follow the original notation [4]. 6

8 a 44 apple a, (5.3) a apple b ) b 4 apple a 4, (5.4) a apple (a b)+(a b 4 ). (5.5) a apple a 55, (5.6) a apple b ) b 5 apple a 5, (5.7) (a + b) (a + b 5 ) apple a. (5.8) Theorem 5.. [4] Each concept algebra is a dicomplemented lattice. Lemma 5.. [3] The following are equivalent:. (5.6) and (5.7),. a apple b 5 () b apple a 5. Similarly, Lemma 5.3. The following are equivalent:. (5.3) and (5.4),. a 4 apple b () b 4 apple a. Every bounded lattice can be made into a DL by defining ( a 4, if a 6=, = 0, if a =, ( a 5 0, if a 6= 0, =, if a = 0. (5.9) (5.0) These are called trivial complementations. Lemma 5.4. [6, 4] Let L be dicomplemented and a,b L. Then, 44 is an interior operator and 55 is a closure operator. Furthermore, for all a,b L, a + a 4 =, a a 5 = 0, (5.) 0 4 = 0 5 =, 4 = 5 = 0, (5.) (a b) 4 = a 4 + b 4, (a + b) 5 = a 5 b 5, (5.3) a 444 = a 4 a 5 = a 555, (5.4) a 45 apple a 44 apple a apple a 55 apple a 54, a 5 apple a 4, (5.5) a b 4 apple b ) a apple b, a apple b + a 5 ) a apple b, (5.6) a b = 0 ) a apple b 4 a + b = ) a 5 apple b, (5.7) a a 4 = 0 ) a = a 44 a + a 5 = ) a = a 55. (5.8) 7

9 (5.) and (5.3) show that 5 is a sufficiency operator and 4 is a dual sufficiency operator. Therefore, we can replace (5.4) and (5.7), by the equations in (5.) and (5.3). In particular, the class of dicomplemented lattices is equational, and we denote it by D. Let L be a DL. A proper filter F of L is called primary if a F or a 4 F for all a L. A proper ideal I is called primary if a I or a 5 I for all a L. The set of all primary filters, respectively, primary ideals, is denoted by F P, respectively, I P. The following separation lemma and its corollary are decisive for the representation results based on Urquhart s representation: Lemma 5.5. [6, Lemma..] If F F and I I with F \ I = /0, then there are a primary filter F 0 such that F F 0, and a primary ideal I 0 such that I I 0 and F 0 \ I 0 = /0. Proof. Let F F and I I such that F \ I = /0. Let K = {G F : F G and G \ I = /0 }. Since F K and every chain in K has an upper bound in K, it contains a maximal element F 0. Similarly, let I 0 be an ideal containing I and maximal with respect to I 0 \ F = /0; then F 0 \ I 0 = /0. Assume that F 0 is not primary; then there is some a L such that a 6 F 0 and a 4 6 F 0. Since F 0 is maximal disjoint to I, [F 0 [{a}] filt \ I 6= /0 and [F 0 [{a 4 }] filt \ I 6= /0. Thus, there exist b,c F 0 such that b a,c a 4 I. Since F 0 is a filter, b c F 0, and since I is an ideal we may suppose that b = c. Therefore, b a + b a 4 I. (5.5) now implies b I, contradicting that F 0 \ I = /0. Dually, it can be shown that I 0 is primary. Corollary 5.6. If hf,ii is a maximal pair (in Urquhart s sense), then both F and I are primary. Proof. Let hf,ii be a maximal pair. By Lemma 5.5 there is a filter ideal pair hf 0,I 0 i such that F F 0 F P and I I 0 I P. The maximality of hf,ii implies that F = F 0 and I = I 0. For later use, we introduce the following convention: If M L, we let M 5 = {a : a 5 M} and M 4 = {a : a 4 M}. Lemma Let F be a proper filter. Then, F 5 is a proper ideal of L disjoint from F.. Let I be a proper ideal. Then, I 4 is a proper filter of L disjoint from I. Proof.. If F 5, then 0 = 5 F, contradicting that F is proper. If a F 5 and b apple a, then a 5 F and a 5 apple b 5, since 5 is antitone. If a,b F 5, then a 5,b 5 F, and thus, a 5 b 5 F since F is a filter. By (5.3), a 5 b 5 =(a + b) 5, showing that a + b F 5. If a F and a F 5, i.e. a 5 F, then a a 5 = 0 F, contradicting that F is proper.. This can be shown dually. 8

10 6 A discrete representation for DLs We are now going to establish the discrete representation theorem for DLs, and consider the reducts hl, 5 i and hl, 4 i separately. A 5 frame is a structure hx,apple,apple,ci such that hx,apple,apple i is a doubly ordered set, and C is a binary relation on X satisfying the following conditions: FC. (8x,y,z)[xCy and y apple z ) xcz], C ; apple C, FC. (8x,y)[xCy ) (9z)(x apple z and ycz)] C apple ; C. FC3. (8x,y,z)[x apple y and x apple z ) zcy)] ; apple C. The class of all 5 frames is denoted by Frm 5. Theorem 6.. Let X = hx,apple,apple,ci be a 5 frame, and hl X,_ X,^X, /0,Xi be its Urquhart complex algebra. Furthermore, for Y L X set Y 5 X Then, Y 5 X L X and hl X,_ X,^X, 5 X, /0,Xi satisfies (5.6) (5.8). df = {x X : (8y)[xCy ) y 6 Y ]} =[C]( Y ). (6.) Proof. The frame conditions and the definition of 5 X were presented in [3]; here we give a simpler proof. Y 5 X L X : Since [apple ]happle i is a closure operator, we have Y 5 X [apple ]happle i(y 5 X ), and only the reverse inclusion needs to be proved. Let x [apple ]happle i([c]( Y )), and assume that C(x) 6 Y, i.e. that there is some y Y and xcy. By FC, there is some s such that x apple s and ycs. Now, x [apple ]happle i([c]( Y )) implies that " x happle i[c]( Y ), and thus there is some t such that s apple t and C(t) \Y = /0. From ycs and s apple t we obtain y(c ; apple )t, and now FC implies that yct. Using FC again, it follows that y(apple ; C )t, and thus, there is some z such that y apple z and tcz. Since Y L X it is apple closed, and thus, z Y. This contradicts C(t) \Y = /0. (5.6): First, note that Y 5 X 5 X =[C]( Y 5 X )=[C]( [C]( Y )) = [C]hCi(Y )={x : (8y)[xCy ) (9z)(z Y and zcy}. Now, suppose that x Y, and xcy. By FC, there is some z such x apple zc y. Since Y L X, it is " apple increasing, and thus, x Y and x apple z imply z Y. The claim now follows from ycz. (5.7): If Y Z X, then Z Y. Since [C] is isotone, we obtain [C]( Z) [C]( Y ). (5.8): Let Y,Z L X ; we need to show that (Y _ X Z) \ (Y _ X Z 5 X ) Y. Set M =(Y _ X Z) \ (Y _ X Z 5 X )= [apple ]happle i(y [ Z) \ [apple ]happle i(y [ [C]( Z)). Now, M =[apple ]happle i(y [ Z) \ [apple ]happle i(y [ [C]( Z)), =[apple ](happle i(y [ Z) \happle i(y [ [C]( Z))), [apple ] distributes over \, =[apple ][(happle iy [happle iz) \ (happle i(y ) [happle i[c]( Z))] happle i distributes over [, =[apple ][happle iy [ (happle iz \happle i[c]( Z))], [ distributes over \ = Y [ (happle iz \happle i[c]( Z))], Y =[apple ]happle i(y ). We are done if we can show that happle iz \happle i[c]( Z)) = /0, since Y =[apple ]happle iy. Assume that x happle iz and x happle i[c]( Z)). Then, there is some y Z such that x apple y; also, there is some z such that x apple z and C(z) \ Z = /0. By FC3 we have zcy, contradicting C(z) \ Z = /0. 9

11 If X is a 5 frame, its complex algebra is the structure hl X,_ X,^X, 5 X, /0,Xi which we denote by Cm 5 (X). Conversely, let L be a dicomplemented lattice and X L the set of all maximal pairs. Define a relation C L on X L by xc L y if and only if x 5 y. Theorem 6.. Let L be a DL, and C L the relation on its Urquhart canonical frame defined by xc L y if and only if x 5 y. Then, C L satisfies FC FC3. Proof. It was shown in [3] that C L satisfies FC FC. For FC3, suppose that x y and x z. We need to show that zc L y, i.e. a 5 z implies a y for all a L; thus, suppose that a 5 z. Then, a 5 6 z, and therefore, a 5 6 x, since x z. It follows that a x, since x is primary, and x y implies zcy. A 4 frame is a structure hx,apple,apple,qi such that hx,apple,apple i is a doubly ordered set, and Q is a binary relation on X satisfying the following conditions: FC. (8x,y,z)[xQy and y apple z ) xqz], Q ; apple Q, FC. (8x,y,z)[z apple x and xqy ) zqy], apple ; Q Q, FC3. 8x,y)[xQy ) (9z)(x apple z and yqz)] FC4. (8x,y,z)[x apple y and x apple z ) yqz and zqy)] Q apple ; Q. ; apple Q \ Q. The class of all 4 frames is denoted by Frm 4. The following lemma lists some properties of Q which we shall use later on. Lemma 6.3. Suppose Q satisfies FC FC.. Q = Q ; apple.. [Q](Y )=" [Q](Y ) for all Y L X. Proof.. Since apple is reflexive, xqy implies x apple xqy. The other direction is just FC.. Since apple is reflexive, we have [Q](Y ) " [Q](Y ). For the other direction, let x " [Q](Y ). Then, there is some y X such that y apple x and Q(y) Y. Let xqz; then, y apple xqz, and FC implies that yqz. Therefore, z Y since Q(y) Y. Theorem 6.4. Let X = hx,apple,apple,qi be a 4 frame, and hl X,_ X,^X, /0,Xi be its Urquhart complex algebra. Furthermore, for Y L X set Y 4 X Then, Y 4 X L X and hl X,_ X,^X, 4 X, /0,Xi satisfies (5.3) (5.5). 4 X is the operator defined in [9, p. 69] for dual sufficiency. df =[apple ]hqi( Y ). (6.) 0

12 Proof. Y 4 X L X : We need to show that [apple ]happle i(y 4 X )=Y 4 X for Y L X. Since [apple ]happle i is a closure operator, we only consider the part. Let Y L X. By Lemma 6.3(3), [Q](Y ) is apple increasing, in other words, [apple ][Q](Y )=[Q](Y ). Thus, [apple ]happle i[q](y )=[apple ]( [Q](Y )) = [apple ]hqi( Y )=Y 4 X is stable. 5.3: Let Y L X ; then Thus, Y 4 X 4 X =[apple ]hqi( Y 4 X ) (6.3) =[apple ]hqi( [apple ]hqi( Y )) (6.4) =[apple ]hqihapple i[q](y ) (6.5) =[apple ]hq ; apple i[q](y ) by (.) (6.6) =[apple ]hqi[q](y ) by Lemma 6.3(). (6.7) hqi[q](y ) happle i(y ) ) Y 4 X 4 X =[apple ]hqi[q](y ) [apple ]happle i(y )=Y. (6.8) Thus, it suffices to show that hqi[q](y ) happle i(y ). Let x hqi[q](y ); then, there is some y X such that xqy and Q(y) Y. By FC3, there is some z X such that x apple z and yqz. Now, Q(y) Y implies that z Y, and it follows that " x \Y 6= /0, i.e. x happle i(y ). 5.4: Suppose that Y Z X. Then, Z Y and the fact that both [apple ] and hqi are isotone with respect to implies that Z 4 X =[apple ]hqi( Z) [apple ]hqi( Y )=Y 4 X. 5.5: We need to show that Y (Y ^X Z) _ X (Y ^X Z 4 X ) for all Y,Z L X. By definition of the lattice operations (Y ^X Z) _ X (Y ^X Z 4 X )=[apple ]happle i((y \ Z) [ (Y \ [apple ]hqi( Z))), =[apple ]happle i(y \ (Z [ [apple ]hqi( Since Y =[apple ]happle i(y ) it is enough to show that X = Z [ [apple ]hqi( Z). Thus, let x X, and assume that x 6 Z [ [apple ]hqi( Z)); then, x 6 Z and x 6 [apple ]hqi( Z). Thus, there is some y such that x apple y, and y 6 hqi( Z), i.e. y [Q](Z). By FC4 we have yqx, and thus, x Z, a contradiction. Z))). If L is a DL, define a relation Q L on X L by xq L y if and only if x 4 y. Theorem 6.5. Q L satisfies FC FC4. Proof. FC : Let xqy and y apple z. Then, x 4 y z, showing xqz. FC : Let x apple z and zqy. Then, x z and z 4 y. Now, x z implies x 4 z 4, and thus, x 4 y. It follows that xqy. FC3 : Let xqy, i.e. x 4 y. First, assume that a y 4 \ x. Then, a 4 y and a 44 x, since a 44 apple a and x is an ideal. Hence, a 4 x 4 y, contradicting a 4 y. It follows that hy 4,x i is a filter ideal pair which can be extended to a maximal pair z. Then, x z and y 4 z, showing z apple zq y. FC4 : Let x,y,z X L such that x y and x z. Let a y 4, i.e. a 4 y. Then, a 4 6 y, and x y implies that a 4 6 x. Since x is a primary filter by Corollary 5.6, it follows that a x. Hence, a z, which implies yqz. Reversing y and z shows that zqy.

13 We can now state our representation result. A DL frame is a structure X = hx,apple,apple,c,qi such that hx,apple,apple i is a doubly ordered set, and C and Q are binary relations on X which satisfy, respectively, FC FC3 and FC FC4. Its complex algebra hl X,_ X,^X, 4 X, 5 X, /0,Xi is denoted by Cm DL (X). Theorem 6.6. Each DL can be embedded into the complex algebra of its Urquhart canonical frame. Proof. Let h : L! CmCf(L) be defined by h(a)={x X L : a x }. It was shown in [] that h is a lattice embedding, so all that is left to show is that h preserves 4 and 5. First, recall that for all a L, h(a) 4 X L =[apple ]hqi(x \ h(a)), Definition of 4 X L (6.9) = {x : (8y)[x apple y ) y hqi(x \ h(a))]}, Definition of [apple ] (6.0) = {x : (8y)[x apple y ) Q(y) \ (X \ h(a)) 6= /0 ]}, Definition of hqi (6.) = {x : (8y)[x apple y ) (9z)(y z and a 6 z )]}, Definition of Q (6.) h(a) 5 X L =[C](X \ h(a)), Definition of 5 X L (6.3) = {x : (8y)[xCy ) a 6 y }, (6.4) = {x : (8y)[x 5 y ) a 6 y }. (6.5) h(a 4 ) h(a) 4 X L : Let x h(a 4 ), i.e. a 4 x, and x y. Then, a 4 y, and thus, a 4 6 y. It follows that a 6 y 4, and therefore, y 4 \#a = /0. By Lemma 4.3 there is a maximal pair z such that y 4 z and a z. This shows that yqz and z 6 h(a). h(a) 4 X L h(a 4 ): We prove the contraposition; thus, let x 6 h(a 4 ), i.e. a 4 6 x. Then, x \#a 4 = /0, and by Lemma 4.3 there is some maximal pair y such that x y and a 4 y. By (6.) it suffices to show that a z whenever y 4 z. Thus, let y 4 z. From a 4 y we obtain a y 4, and therefore, a z. Next, we show that h preserves 5 as well. h(a 5 ) h(a) 5 X : Let x h(a 5 ), i.e. a 5 x. Suppose that xcy; then x 5 y by definition of C. Now, a 5 x implies a x 5, and x 5 y implies a y. Since y \ y = /0 it follows that a 6 y. h(a) 5 X h(a 5 ): Let x h(a) 5 X, and assume that a 5 6 x. Then, a 6 x 5 which is a proper ideal by Lemma 5.7. Therefore, we can extend h" a,x 5 i to a maximal pair y by Lemma 4.3. Since x h(a) 5 X, and x 5 y, we obtain a 6 y by (6.5), contradicting " a y. 7 Conclusion and outlook We have shown that each dicomplemented lattice L can be embedded into the Urquhart complex algebra Cm U Cf U (L) of its canonical frame. Since its lattice reduct is a canonical extension of L [] this also shows that the variety of dicomplemented lattices is canonical. In further work we intend to investigate the representation problem raised in [6]: Is every dicomplemented lattice embeddable into a concept algebra? We shall also endeavour to extend the logic based on doubly ordered frames presented in [9, Chapter.9] to the dicomplemented case. In view of Theorem 4.5 it would be interesting to know whether there is an axiomatic extension of doubly ordered frames which would guarantee that hx,apple,apple i can be embedded into Cf U Cm U (X). We suspect that the class of such frames is not first order axiomatizable.

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