374 K.ORIUCI Figure 1. one dimensional irreversile positive or negative diretions. Figure 2. one dimensional reversile Y > ^ Figure 3. two dimensional

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1 Sientiae Mathematiae Vol. 2, No. 3(1999), 373{ TRICE AND TWO DELEGATES OPERATION KIYOMITSU ORIUCI Reeived Otoer 1, 1999 Astrat. The onept of \trie " was introdued in [5] and [] for the purpose of using truth value of fuzzy sets. In this paper, we introdue the notion of trie and some of its mathematial properties. Moreover, we shall show some typial onrete examples of trie inluding those having two delegates operations. 1. Introdution. A semilattie (S ) is a set S with a single inary, idempotent, ommutative and assoiative operation. (1) (2) (3) a a = a a = a a ( ) =(a ) (idempotent) (ommutative) The following Proposition is wellknown (see [1] p.10). Proposition 1. (assoiative) Let (S ) e a semilattie. Under the relation dened y a () a = any semilattie (S ) is a partially ordered set (S ). A lattie has two operations. Let L e a lattie with two operations _ (join) and ^ (meet). Then, (L _) and (L ^) are semilatties. We an onstrut two ordered sets (L _ ) and (L ^), respetively, y using the Proposition 1 (a _, a _ = and a ^, a ^ = ). The dual of (L _ )is(l ^). That is, the order _ is nothing ut the reverse of the order ^. The reason why _ and ^ introdue the reverse order is that latties satisfy the asorption law: (4) a _ (a ^ ) =a a ^ (a _ )=a (asorption) So, when a _ (that is, a _ = ), we have a ^ = a ^ (a _ ) =a y (4). ene ^ a. Suppose that there is an ojet with a rope on a line, and that we are ale to pull the rope from the right (see Fig. 1). Then, we anmove the ojet to the right diretion, ut not to the left diretion. This situation is onsidered to e irreversile. Next, see Fig. 2. By pulling the rope from either right or left diretions, we an move the ojet to any point on the line. This situation is onsidered to e reversile. From Proposition 1, a semilattie is a set having one order, i.e. one diretion, like in Fig. 1. A lattie is a set having two orders, i.e. two diretions, like in Fig. 2. Any ojet on the line an e moved to another aritrary point on the line y pulling the ojet to the 1991 Mathematis Sujet Classiation. 0A12, 20N10, 20M10. Key words and phrases. semilattie, trie, roundaoutasorption law, triangular situation, two delegates operation.

2 374 K.ORIUCI Figure 1. one dimensional irreversile positive or negative diretions. Figure 2. one dimensional reversile Y > ^ Figure 3. two dimensional reversile Suppose that there is an ojet not on a line ut in a plane (two dimensional Eulidian spae). If we want tomove the ojet to an aritrary point in the plane, it is suient to have three \suitale" diretions to pull the ojet, as shown in Fig. 3. If the three diretions are not \suitale", as shown in Fig. 4, we annot move the original ojet to an aritrarily hosen target point. Movements in three diretions are not neessarily restrited to the two dimensional plane (see Fig. 5 right). Consider that a lattie orresponds to the one dimensional reversile ase (Fig. 2). What systems with three inary operations (semilattie) orrespond to the two dimensional reversile ases (Fig. 3)? 2. RoundaoutAsorption Law. For A a nonempty set and n a positive integer, let (A 1 2 ::: n ) e an algera with n inary operations, and (A i ) e a semilattie for every i 2f1 2 ::: ng. Then, (A 1 2 ::: n )isalledansemilattie (See [7]). In this paper, we will deal mainly with triplesemilattie, (i.e. n = 3). We denote eah order on A y (5) respetively. a i () a i = Denition 1. Let S n e the symmetri group on f1 2 :::ng. An algera (A 1 2 ::: n ) has the nroundaoutasorption law if it satises the following n! identities: () ((((a (1) ) (2) ) (3) )::: (n) )=:

3 TRICE 375 ; ; f;?? CO C Cf > j Y f > = ; ; ; ; ; Figure 4. two dimensional irreversile for all a 2 A and for all 2 S n. f j f ; A ; A j A A A AU Figure 5. three dimensional ases We remark that \nroundaoutasorption law" have dierent name \asorptive law." And arlgeai researhes yielded interesting results. (See [7]). owever, we adopt n roundaoutasorption law from the image of Proposition 2. Of ause, the 2roundaoutasorption law is the asorption law. An algera (A 1 2 ) whih satises the 2roundaoutasorption law is a lattie. The operations 1 and 2 are denoted y _ and ^. Denition 2. An algera (A )whih satises the 3roundaoutasorption law issaidtoeatrie. To simplify explanation, we often omit \3" of \3roundaoutasorption law." The operations 1, 2 and 3 will e denoted y % 1, 2 and # 3. The \trie" is a notion to orrespond to \lattie." Let T e a set. We introdue three orders into T, under the ondition that all two elements of T have a least upper ound for eah order. Then, we an onstrut the set into a triplesemilattie. Example 1. Let T e a set whih onsists of six points. We introdue three orders in the set y arrows of Fig.. Suppose that arrowhead is larger than the other end (e.g. d 2 e). Then, T is a trie. That is, T has the 3roundaoutasorption law ] ] e d ] Figure. Example 1 ^ ^ ^

4 37 K.ORIUCI Example 2. Denote y R the set of all real numers. We introdue three orders in the two dimensional Eulidian spae R 2 as follows : x 1 y () x 1 y 1 and 0 y 2 ; x 2 p 3(y 1 ; x 1 ) x 2 y () x 1 y 1 and 0 y 2 ; x 2 p 3(x 1 ; y 1 ) x 3 y () x 2 ; y 2 p 3 jy 1 ; x 1 j: for x =(x 1 x 2 ) y =(y 1 y 2 ) 2 R 2. This (R 2 % 1 2 # 3 )isatrie. Needless to say, we anmake another trie on R 2. Denition 3. of T is a suset S of T suh that (7) Let (T % 1 2 # 3 ) e a triplesemilattie. A sutriplesemilattie a 2 S imply a % 1 a 2 a # 3 2 S: If T is a trie, then a sutriplesemilattie S is a trie. We say that S is a sutrie of T. Let T e a trie. The empty susetoft is a sutrie. For any a 2 T,onepoint set fag is a sutrie of T.Inaove Example 1, the set f, d, eg isasutrieoft.weanemed Example 1 in Example 2, as a sutrie. We an easily hek that a lattie L has following properties: Proposition 2. a 1 2 d 3. 8a 2 L 9 2 Ls:t:a _ ^ 8a 2 L 9 2 L s:t: a ^ _ : Let T e a trie. For every a 2 T, there exist d 2 T suh that Proof Oviously, a 1 a % 1 2 (a % 1 ) 2. From ((a % 1 ) 2 ) # 3 = (the roundaoutasorption law), (a % 1 ) 2 3. Let = a % 1 and d =(a % 1 ) 2. Then, we have a 1 2 d 3. This ompletes the proof. One an prove the next Proposition in a similary way. Proposition 3. Let an algera (A 1 2 ::: n )have the nroundaoutasorption law. For every a 2 A and for evry 2 S n, there exists 1 ::: n;1 2 A suh that (8) a (1) 1 (2) 2 (3) ::: (n;1) n;1 (n) : Denition 4. Wesay that (A 1 2 ::: n )isattainale if (8) is true for every a 2 A. The \attainality" of lattie orresponds to the notion of reversility in the ase of one dimension (Fig. 2). We an onsider that the \attainality" of trie orresponds to the ase of Fig. 3, that is, the two dimensional reversile ases * * Y ] Y d ] a Figure 7. Example 3? ^ ^?

5 Example 3. TRICE 377 Let T e a set whih onsists of seven points. We introdue three orders in the set y arrows of Fig. 7. Then, T has not the roundaoutasorption law. But T is attainale. From the point of algerai view, the onept of tries are availale. Now, we will take one example of many. The following proposition is well knwon. (See [8]). Proposition 4. near unanimity identities, then V is ongruene distriutive. If V is a variety whih admits a kary polynomial m that satises the Proposition 5. The variety of tries is ongruenedistriutive. 1 Proof Consider the following 4ary term: M(x y z t) = ((x % 1 y) 2 (y % 1 z) 2 (x % 1 z)) # 3 ((x % 1 y) 2 (y % 1 t) 2 (x % 1 t)) # 3 ((x % 1 z) 2 (z % 1 t) 2 (x % 1 t)) # 3 ((y % 1 z) 2 (y % 1 t) 2 (z % 1 t)): We have that: M(a a a ) = ((a % 1 a) 2 (a % 1 a) 2 (a % 1 a)) # 3 ((a % 1 a) 2 (a % 1 ) 2 (a % 1 )) # 3 ((a % 1 a) 2 (a % 1 ) 2 (a % 1 )) # 3 ((a % 1 a) 2 (a % 1 ) 2 (a % 1 )) = a # 3 (a 2 (a % 1 )) # 3 (a 2 (a % 1 )) # 3 (a 2 (a % 1 )) = a # 3 (a 2 (a % 1 )) = a: Similarly, M(a a a) =a, M(a a a) =a and M( a a a) =a. From Proposition 4, variety of tries is ongruene distriutive. Similary, one an prove that a variety of algeras (A 1 ::: n ), whihhave nroundaoutasorption law, is ongruenedistriutive. 3. Triangular Situation. In this setion, we disuss the onept to distinguish etween trie and lattie. Let (T % 1 2 # 3 ) e a trie and a 2 T (a = = = a). Denition 5. We say that an ordered triple (a ) isinatriangular situation if (a ) have the following properties: (9) a # 3 = and a 2 = and % 1 = a: Stritly speaking, a orrespond to % 1, orrespond to 2 and orrespond to # 3. Therefore, if neessary, wemust say that (a ) isina % 1 2 # 3 triangular situation. If there is a triple in the triangular situation on T,wesay that T has a triangle. 1 After I gave a presentation at the onferene on latties and universal algera in Szeged 1998, I reeived an message from Professor A. Tepaveviof University of Novi Sad saying that she proved Proposition 5.

6 378 K.ORIUCI If we add another operation to a lattie, we an onstrut a trie. owever, this trie has no triangle. ene, triangular situation is an important onept in the theory of tries. Let T e a trie. If (a ) is in a triangular situation in T, then the set fa g is a sutrie and ( 2 a) % 1 = % 1 = a = (a # 3 ) 2 a = 2 a = = a ( % 1 ) # 3 = a # 3 = = : ene, two operations in T do not satisfy the asorption law. That is, eah of(t % 1 2 ), (T 2 # 3 ) and (T 2 # 3 ) is not a lattie. If we desire to onsider the ase that tries orresponding to the two dimensional reversile ases (Fig. 3), we should suppose that T is not a lattie. Of ourse, there is an example in whih \(T % 1 2 # 3 ) is a trie and (T % 1 2 ) is a lattie". Let I e the unit interval in R, then (I _ ^) is a lattie. Denote _ y % 1 and ^ y 2. Suppose # 3 is a \modetype operator" in [4]. Then (I % 1 2 # 3 ) is a trie. The triangular situation is easy to understand. Let n 4. When an algera (A 1 2 ::: n ) have the nroundaoutasorption law, it is diult to represent a onept orresponding with triangular situation on trie. Thus, the "triangular situation" would e a key word for tries. From the algerai point of view, distriutivity is important and essential. owever, if T has a triangle, then T is not distriutive. Eah of the Example 1, 2 and 3 has a triangle. This shows that we must deal with \nondistriutive" algera, and the aim of this paper is to investigate harateristi poperties of nonalgerai algera. Example 4. Let T e a set of four points. We introdue three orders in the set y 1 j * 2 Y a d Figure 8. Example 4 3 j? arrows of Fig. 8. This T is a trie, ut T does not have a triangle. Let S e the set fa d g 2T. Then, S is a sutrie of T, ut again S does not have a triangle. Note that (S, % 1, 2 ) is a lattie. There are some properties related to the triangular situation. We dene \the triangle onstrutive law" and \the triangle natural law". Denition. Let T e a triplesemilattie. We say that T has the triangle onstrutive law if T has the following properties: (10) (11) (12) for all d e 2 T. (d % 1 e) # 3 (d 2 e)=d # 3 e (d % 1 e) 2 (d # 3 e)=d 2 e (d # 3 e) % 1 (d 2 e)=d % 1 e

7 TRICE 379 Suppose that a triplesemilattie T satises the identities ((10) (12)) for d e 2 T. Let a = d % 1 e, = d 2 e, = d # 3 e. If a = = = a, then (a ) is in a triangular situation. That is why we named the identities \the triangle onstrutive law". Denition 7. Let T e a triplesemilattie. We say that T has the triangle natural law if T has the following properties: (13) (14) (15) for all x y z 2 T. if x % 1 y = zandx 2 z = y if x # 3 y = z and x 2 z = y if x % 1 y = zandx# 3 z = y then y # 3 z = x then y % 1 z = x then y 2 z = x Example 1 and Example 2 have oth the triangle onstrutive law and the triangle natural law. Example 3 has neither the triangle onstrutive law nor the triangle natural law. Now, we give other examples. Example 5 has the triangle onstrutive law, ut it has neither the roundaoutasorption law nor the triangle natural law. Example has the roundaoutasorption law, ut it has neither the triangle onstrutive law nor the triangle natural law. Example 7 has the triangle natural law, ut it has neither the roundaoutasorption law nor the triangle onstrutive law. Example 5. Let T e a ve points set fa d eg. See Fig. 9. We introdue three 1 * ; ; ; 2 e a 3 ^? Figure 9. Example 5 orders in T y : (1) (17) (18) 1 a 1 a d 1 a e 1 a a 2 2 d 2 e 2 a 3 3 d 3 e 3 : From (( % 1 e) 2 e) # 3 e =, thist does not satisfy the roundaoutasorption law. From e % 1 = a, e 2 a = and # 3 a =, T does not have the triangle natural law. But T has the triangle onstrutive law. Example. Let T e a set of four points. We introdue three orders in the set y arrows of Fig. 10. We an easily hek T has the roundaoutasorption law, and the (a d) is a in a triangular situation on T. But, from ( % 1 ) # 3 ( 2 )=a # 3 = d and # 3 =, T does not have the triangle onstrutivelaw. From % 1 = a, a 2 = and # 3 a = d =, T does not have the triangle natural law, either.

8 380 K.ORIUCI 1 * 2 Y d a 3 j? Figure 10. Example Example 7. Let T e a set fx n y n z n j n 2 N [f0gg. Weintrodue three orders in T y : (19) (20) (21) z 0 1 x 0 y 0 1 x 0 x k 1 x k+1 y k 1 x k+1 z k 1 x k+1 x 0 2 y 0 z 0 2 y 0 x k 2 y k+1 y k 2 y k+1 z k 2 y k+1 y 0 3 z 0 x 0 3 z 0 x k 3 z k+1 y k 3 z k+1 z k 3 z k+1 for k 2 N [f0g. In this triplesemilattie, the only (x 0 y 0 z 0 ) is in a triangular situation. In ase of another omination, the assumption of statements (13) (14) (15) does not hold. Therefore, T has the triangle natural law. Let a = x 2 = x 1,then((a % 1 ) 2 ) # 3 = ((x 2 % 1 x 1 ) 2 x 1 ) # 3 x 1 =(x 2 2 x 1 ) # 3 x 1 = y 3 # 3 x 1 = z 4 =. ene, T does not have the roundaoutasorption law. From a % 1 = x 2, a 2 = y 3 and a # 3 = z 3, T does not have the triangle onstrutive law. Proposition. (22) Every trie T has the following property: if x % 1 y = z and x 2 z = y then y # 3 z 3 x: Proof From x % 1 y = z and x 2 z = y, x 2 (x % 1 y)=y. By the roundaoutasorption law, x # 3 (x 2 (x % 1 y)) = x. ene, we have x # 3 y = x. This implies y 3 x. Similarly, wehave x # 3 z = x, that is, z 3 x. Therefore, we see that y # 3 z 3 x. Proposition 7. triangle natural law. Let T e a trie. If T has the triangle onstrutive law, then T has the Proof By the Proposition, if x % 1 y = z and x 2 z = y, theny # 3 z 3 x for x y z 2 T. Let d e y # 3 z. And suppose that d = x. From x 2 y, x 2 y = y. ene (x % 1 y) # 3 (x 2 y)=z # 3 y = d. But, from y 3 y # 3 z = d 3 x, x # 3 y = x. This implies T does not have the triangle onstrutive law. It is in ontradition with the assumption. Therefore, d = x. ene, T has the triangle natural law. Question 1. Does every trie whih has the triangle natural law have the triangle onstrutive law? We answer this y Example 9 in the next setion. Let L e a lattie and a 2 L. If a, then the set fa g is a sulattie of L. Moreover, the interval [a ] is a sulattie of L. The interval an e onsidered the set fx 2 Ljx ^ a x _ g. Similarly, lett e a trie and a 2 T (a = = = a). If (a ) is in a triangular situation on T, then the set fa g is a sutrie of T. Now, we onsider the set fx 2 T jx 1 a x 2 x 3 g. We will denote it y [a ], and all it hyperinterval. It seems the onept orresponding with the interval in a lattie.

9 TRICE 381 Example 8. in T y : (23) (24) (25) Let T e a set of seven points fa d e f gg. Weintrodue three orders 1 a 1 a d 1 a e 1 a a 1 g f 1 g a 2 2 d 2 e 2 f 2 g 2 a 3 3 d 3 f e 3 f f 3 g 3 : This T is a triplesemilattie. The (a ) is in a triangular situation on T. And the set [a ] isfa d eg. From d # 3 e = f =2 [a ], the hyperinterval [a ] is not a sutriplesemilattie of T. Proposition 8. Let T e a triplesemilattie. If T has the triangle onstrutive law, the hyperinterval [a ] isasutrieoft for every a 2 T. Proof To prove this proposition, it sues to show that x % 1 y 2 [a ] (that is, x % 1 y 1 a, x % 1 y 2 and x % 1 y 3 ) for x y 2 [a ]. From x 1 a and y 1 a, it is lear that x % 1 y 1 a. By the triangle onstrutive law, (x % 1 y) 2 (x # 3 y)=x 2 y. ene x % 1 y 2 x 2 y. From x 2 and y 2, x 2 y 2. Therefore x % 1 y 2. Similarly, we an prove x % 1 y 3. This ompletes the proof. Question 2. Let T e a trie and (a ) e in a triangular situation on T.Isittrue that the hyperinterval [a ] isasutrieoft? 4. Two Delegates Operation On linearly ordered sets. Suppose that X is a linearly ordered set and jxj 2. Let X(2) e the set of all two points suset of X, thatis, X(2) = ffx 1 x 2 gj x 1 x 2 2 X s:t: x 1 <x 2 g: Suppose that fa 1 a 2 g f 1 2 g2x(2). Let 2 e a 2 _ 2.This 2 is the largest memer of fa 1 a g. And take 1 = 8 < : a 2 _ 1 if a 2 < 2 a 1 _ 2 if 2 <a 2 a 1 _ 1 if a 2 = 2 Then 1 is the seond largest memer of fa 1 a g. We dene the inary operation t on X(2) y fa 1 a 2 gtf 1 2 g = f 1 2 g: Clearly, (X(2) t) is a semilattie. For example, let X e a set of students, and we want to selet two students as deledats to attend a onvention. Some memers of ommittee reommend a 1 and a 2. The other memers of ommittee reommend 1 and 2. Then, we may selet 1 and 2 from a 1, a 2, 1 and 2 aording to this method. Next, we dene the dual operation of t. Let d 1 e the a 1 ^ 1. This d 1 e the smallest memer of fa 1 a g. And take d 2 = 8 < : a 2 ^ 1 if a 1 < 1 a 1 ^ 2 if 1 <a 1 a 2 ^ 2 if a 1 = 1

10 382 K.ORIUCI Then, d 2 e the seond smallest memer of fa 1 a g.we dene the inary operation u on X(2) y fa 1 a 2 guf 1 2 g = fd 1 d 2 g: This (X(2) u) is also a semilattie. But, (X(2) t u) is not a lattie! And the operations t and u do not satisfy the distriutive law. We have tosay \this (X(2) t u) isnogood algera"? Now, we dene another operation on X(2) y fa 1 a 2 gf 1 2 g = fd 1 2 g: This is the operation whih hose the smallest memer and the largest memer among fa 1 a g. Clearly, (X(2) ) is a semilattie. The algera (X(2) t u ) is a trie. That is, it satises the roundaout Proposition 9. asorption law. Proof It sues to oserve the ase of four points, that is, next Example. Example 9. Let X e the four points linearly ordered set f g. We denote the two points suset f1 2g, f1 3g, f1 4g, f2 3g, f2 4g and f3 4g y a,,, d, e and f. Then, the set X(2) is fa d e fg (see Fig. 11). This (X(2) t u ) is a trie. We denote t, u and y % 1, 2 and # 3. From (a % 1 d) 2 d = a, the algera (X(2) % 1 2 )is not a lattie. From ( % 1 e) 2 (d % 1 e)= and ( 2 d) % 1 e = e, the operations % 1 and 2 do not satisfy the distriutive. From (e 2 d) % 1 e = e and e 2 (d % 1 e)=, that is, the operations % 1 and 2 do not satisfy Birkho system j a ] ] d e ] f Figure 11. Example 9 ^ ^ ^ The X(2) of Example 9 has the triangle natural law. But, from (d # 3 e) % 1 (d 2 e)=e and d % 1 e =, thex(2) does not have the triangle onstrutive law. This is an answer of Question 1. Note that we an onsider replaing "two delegates" with "three delegates". Then, we add two operations, we an onstrut an algera with 4roundaoutasorption law On partially ordered sets. Next, we will try to expand our operations t and u on a partially ordered set. Suppose that P is a partially ordered set. And let P (2) e the set ffx 1 x 2 gj x 1 x 2 2 P s:t: x 1 <x 2 g. We use the notation a to denote a is overed y, and the notation ak to denote a and are inomparale. We an not expand operations on general partially ordered sets. To denition of operation t, we add next two onditions on P : (I _ ) P is a _semilattie,

11 TRICE 383 (II _ )if hki for h i 2 P, then there exists a unique j 2 P suh that h _ i j. Let fa 1 a 2 g, f 1 2 g2p (2). And let j 2 P suh that a 2 _ 2 j. Now, set 8 fa 2 _ 2 jg if a 2 k 2 fa 2 jg if a 2 > 2 and a 1 _ 2 = a 2 >< f 2 jg if a 2 < 2 and a 2 _ 1 = 2 f 1 2 g = fa 1 _ 2 a 2 g if a 2 > 2 and a 1 _ 2 <a 2 fa 2 _ 1 2 g if a 2 < 2 and a 2 _ 1 < 2 fa >: 2 jg if a 2 = 2 and a 1 _ 1 = a 2 fa 1 _ 1 a 2 g if a 2 = 2 and a 1 _ 1 <a 2 : If P is a linearly ordered set, then the aove f 1 2 g is fa 1 a 2 gtf 1 2 g (this t is the opration on X(2)). ene, we dene the inary operation t on P (2) y fa 1 a 2 gtf 1 2 g = f 1 2 g: We an dene the operation u in a similar way. We add next two onditions on P : (I^) P is a ^semilattie, (II^) if hki for h i 2 P, then there exists a unique j 2 P suh that j h ^ i. Let fa 1 a 2 g, f 1 2 g2p (2). And let j 2 P suh that j a 1 ^ 1. Set 8 fj a 1 ^ 1 g if a 1 k 1 fj a 1 g if a 1 < 1 and a 2 ^ 1 = a 1 >< fj 1 g if a 1 > 1 and a 1 ^ 2 = 1 fd 1 d 2 g = fa 1 a 2 ^ 1 g if a 1 < 1 and a 2 ^ 1 >a 1 f 1 a 1 ^ 2 g if a 1 > 1 and a 1 ^ 2 > 1 fj a >: 1 g if a 1 = 1 and a 2 ^ 2 = a 1 fa 1 2 ^ a 2 g if a 1 = 1 and a 2 ^ 2 >a 1 : We dene the inary operation u on P (2) y fa 1 a 2 gtf 1 2 g = fd 1 d 2 g: If P satisfy the aove onditions (I _ )and(i^) (i.e. P is a lattie), then we an dene operation on X(2) y fa 1 a 2 gf 1 2 g = fa 1 ^ 1 a 2 _ 2 g: Consequently, assume that a lattie L satisfy the aove onditions (II _ ) and (II^). Aove(P (2) t), (P (2) u)and(p (2) ) are semilatties. Then, The algera (L(2) t u ) is a trie. Instead of this, we an use next operation on P (2) y The algera (L(2) t u ) is also a trie. fa 1 a 2 gf 1 2 g = fd 1 2 g:

12 384 K.ORIUCI 1 a \ \ \ \ 0 L 1 L L L L L L L L L L L 2 \ \ x \ \ C Figure 12 The lattie L 1 at Fig. 12 is the most simple nonlinear example whih satisfy the onditions (II _ )and(ii^). owever, we feel the onditions are partiular. Even the lattie L 2 at Fig. 12 an not satisfy the onditions. Moreover, if there exist a point whihover two points and whih isovered y two points (as an example x of C at Fig. 12), it an not satisfy the ondition. Question 3. Is there etter expansion of the onept of X(2)? Referenes 1. G. Birkho. Lattie Theory (third ed.) Amer. Math. So. Colloq. Pul., S. Burris,.P. Sankappanavar, A Course in Universal Algera SpringerVerlag, New York, G. Gratzer. General Lattie Theory (Seond ed.) Birkhauser, K. oriuhi. ModeType operators on fuzzy sets. Fuzzy Sets and Systems, 272:131{139, K. oriuhi, Fuzzy Mathematis (In apanese) Osaka Kyoikutosyo, Osaka apan, K. oriuhi. Triefuzzy sets. to appear. 7. A. Knoeel and A. Romanowska, Distriutive multisemilatties. DissertationesMathematiae (Rozprawy Matematyzne), Polska Akademia Nauk Instytut Matematyzny. 309 (1991), 42 pp. 8. A. Mitshke, Near unanimity identities and ongruene distriutivity in equational lasses. Algera Universalis 8 (1978) Faulty of Siene, Konan University, Okamoto, igashinada, Koe , apan