Algebra Universalis, 22 (1986) 215-228

0002-5240/86/030215-14501.50 + 0.20/0 9 1986 Birkh~iuser Verlag, Basel

On "constructions o f p-algebras" P. V. RAMANA MURTY AND SR. TERESA ENGELBERT

1. Introduction The topic of our special interest is the extension of the triple method construction to quasi-modular p-algebras by T. Katrifi~k and P. Mederly in [6]. The main theorem of this paper (Theorem 6) answers problem 1 of [6] ([6], problem 1, page 309). Simultaneously we observe that problem 3 of [6] ([6], page 314) is also solved under the assumption that q0 is a join-homorphism. Moreover condition (iv) of problem 3 is implied in condition (v) and hence it is not necessary. Under this hypothesis it follows that (B, D, q0) is a quasi-modular F-triple. Another important observation is that the dense filter of a quasi-modular p-algebra L is a neutral element of the lattice of all filters of L (Sex Theorem 5 of the present paper). This fact simplifies the proofs of most of the results of section 6 of [6]. In [6] it is proved that a decomposable PCS S is admissible if and only if (i) Every congruence relation a qg(S), a ~ B ( S ) is comonomial and (ii) If c + e denotes the greatest element of [e]cq~(S), then (a v b) + d = a + d(b(p(S)) and (a v b) + d =- b + d(a(p(S)) for every d ~ D(s) and arbitrary a, b c B(S). But we show by means of an example that there are decomposable PCS's satisfying the above two conditions which are not admissible thus showing that 4.1 of [6] is not valid. However modular PCS's satisfying condition (i) are admissible (See Theorem 1). Also we note that the dense filter of a p-algebra L with a modular frame is neutral in the lattice of all filters of L and hence the proofs of some of the results of paper [5] by the same authors become much more simplified.

2. Preliminaries We follow the same notations and definitions of [6]. A pseudocomplemented semilattice (=PCS) is a universal algebra (S; A, Presented by Ralph Freese. Received June 3, 1985. Accepted for publication in final form April 28, 1986. 215

P. V. RAMANA MURTY AND SR. TERESA ENGELBERT

216

ALGEBRA UNIV.

*, 0, 1) of type (2, 1, 0, 0) where (S; ^, 0, 1) is a bounded meet-semilattice and, for every a e S, the element a* is the pseudocomplement of a i.e., x-
d=-e(agp(S))

iff a* A d = a * Ae.

a q~(S) is a semilattice congruence relation on D(S). Also it is a (0, !) isotone map from B(S) into Con (D(s)) a q~(S) is called the structure map. (B(S), D(S), fig(S)) is the triple associated with S. We recall the following definitions and results from [6]. Definition 2([6]). An abstract triple is (B, D, ~) where (i) B = (B; v, A,', 0, 1) is a Boolean algebra (ii) D is a ^-semilattice with 1 (iii) q~ is a (0, 1) isotone map from B into Con (D). D E F I N I T I O N 3([6]). An isomorphism of the triples (B, D, (p) and (B1, D1, qJ1) is a pair ~, g) where f is an isomorphism of B and B~, g is an B~ f

Co~ (o) g

Bl~-------~Con (DI)

isomorphism of D and D~ such that the diagram is commutative. (g stands for the isomorphism of Cond (D) and Con (D 0 induced by g). T H E O R E M 1([6]). Two decomposable PCS's are isomorphic if and only if their associated triples are isomorphic. T H E O R E M 2([6]). Let (B, D, (p) be a triple. Then we can construct a decomposable PCS S such that

(B(S), D(S), (p(S)) ~ (B, D, (p). We also recall the definitions of standard and neutral elements of a lattice

Vol. 22, 1986

On "constructions of p-algebras"

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from [4]. Let L be a lattice. An element s e L is called a standard element of L if and only if x A (S V y ) = (X A S) V (X A y) for all x, y e L. Let L be a lattice. An element n E L is a neutral element of L if and only if (n A X) V y = (n v y) A (X V y)

for all x, y e L and n is standard.

3. Admissible PCS's. A decomposable PCS S is called admissible if there is a map

f: B(S) x D(S)--> D(S) such that for any x e S, x <-f(a, d) if and only i f x A a --
EXAMPLE 1. Consider the semilattice S in Figure 1. First we note that S is decomposable.

B(S) = {a, a*, O, 1}

D(S) = {1, h, s, g, e, d}. Now we shall show that every congruence class [d]a (p (S); a ~ B(S), d ~ D(S) is comonomial, g is the greatest element of [d] a ~ (S) and [g]a r (S). s is the greatest element of [e]a 6p (S) and [s]a ~ (S). 1 is the greatest element of [1]

218

P, V. RAMANA MURTY AND SR. TERESA ENOELBERT

ALGEBRA UNIV.

!

g

a~

0 Figure 1.

a 62 (S) and [h]a 62 (S). Finally i is the greatest element of [d]a* ~ (S) for all d e D(S). Hence the first condition is satisfied. g is the greatest element of [d]a (p (S) and a v a * = 1 and 1 is the greatest element of [d]l q~ (S) and l ~ g ( a * 6 2 (S)). Also 1 is the greatest element of [alla* 62 (S) and 1 ----l(a ~0 (S)). Hence the second condition also is satisfied. But now we show that S is not admissible. If S is admissible there exists a map f : B ( S ) xD(S)---~D(S) such that for any x e S , x<-f(a,d) if and only if x A a - d . Now we claim that f(a*, e) cannot exist. We know that e <-f(a*, e). Hence the only possible values o f f ( a * , e) are e, h, s, 1.

f(a*,e)g=e

since

gAa*=x<--e=~g<-e

f(a*, e)4:h; g A a* =x<-ez~g<-h f(a*, e) vSs; g A a* = x < - - e ~ g <--s f(a*, e) g: 1; a* <-- l = ~ a * Aa* =a*<--e Thus S is not admissible and hence the above result is wrong. But in the following theorem we prove that if S is a modular decomposable PCS satisfying only the first condition of 4.116], then S is admissible, T H E O R E M 1. Let S be a modular PCS. Then S is admissible if every congruence relation a 62 (S), a e B(S) is comonomial.

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On "constructionsof p-algebras"

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Proof. A modular PCS is decomposable and filter decomposable and hence a ~ ( S ) = 0a~([7]). Now we shall prove that S is admissible. S is admissible if there exists a map f : B ( S ) • D(S)---> D(S) such that for any x ~ S, x <-f(a, d) iff xAa<_d. Let f(a, d) = t h e greatest element of [dlO,.~. Then d A a = f ( a , d) A a. x <--f(a, d) ~ x A a <--f(a, d) A a = d ^ a <- d. Next suppose x A a ----x, such that d A a = x l h a since S is modular, a = a * * ^ l = ( a Ad)** = (a AXl)** = a AXe* SO that xl**>--a. L e t x l = x ~ * Ae. a A x l = a A x ~ * A e = a A e. a A e = a a x ~ = a A d ~ ( d , e ) ~ O a . ~ o ~ e e [ d ] O , . ~ . So that e < - f ( a , d ) , e>_ xl -->x and hence we get x <-f(a, d). Thus S is admissible. We recall the result 4.3 of [6]. 4.3[6]: Let L be a decomposable PCL. Then L satisfies the identity x =X**A (X V X*) if and only if (i) For every a ~ B ( L ) and d ~ D ( L ) there exists t e a q0 (L) where a q0 (L) = K e r a q~ (L) = {x c O ( L ) / x ->a*} such that d = t ( a * (p (L)). (ii) For every a ~ B ( L ) , the restriction a ~ (L) = a (g (L)/a*cp(L) is a monomial congruence relation. In this result we note the following two points. First is ( i ) ~ (ii). Second, the identity x = x * * ^ (x v x*) is equivalent to condition (i) above.

4. Filter-decomposable PCS's. A decomposable PCS S is said to be filter-decomposable if the congruence relations a q~ (S) are filter-congruences. That is a ~ ( s ) = O[a cp (S)] where a q0(S) = Ker a q) (S) = {x ~ D ( S ) / x >- a*}. Also [[x]a ~ (S)) = a cp (S) v [x)([6], Page 300) we recall the following result of [6]. 5.116]: Let S be a decomposable PCS. Then the following statements are equivalent. (i) S is filter-decomposable (ii) a A d = a A e for a e B ( S ) and d, e e D ( S ) implies a*qo(S) v [ d ) = a* (s) v [e). (iii) [a ^ d) ^ O(S) = a* q0 (S) v [ d ) for every a ~ B(S) and d e O(S). First we prove a theorem about the dense filter of a decomposable PCS.

220

e. V. R A M A N A

M U R T Y A N D SR. T E R E S A E N G E L B E R T

,/~GEBRA

UNIV.

T H E O R E M 2. In a decomposable PCS S, the dense filter D ( S ) is a standard element o f the lattice o f all filters o f S.

Proof. We have to show that (D v B) A C = (D A C) v (B n C) where D is the dense filter of S and B and C are any filters of S. Evidently (D v B) ^ C _ (D A C) v (B

A

C)

Let x e (D v B) A C. This implies that x e D v B and x e C. x e D v B ~ x -> d ,~, s where d e D and s e B. Since S is decomposable, x = x * * A e; e >--x e C so that e e C and hence e e D A C. T h e r e f o r e e e (D A C) v (B A C) Further x** >-x e C ~ x * * e C. Also x --- d A S SO that x** >- (d A S)** = S** >---S e B. H e n c e x** e B so that x** e B A C and hence x** e (D A C) v (B A C). Thus x = x * * ^ e e (D A C) v ( B A C ) . Thus we get ( D v B ) A C = (D A C) v (B A C). H e n c e D is standard. Thus we see that the dense filter is a standard element of the lattice of all filters of a decomposable PCS. Now in a filter-decomposable P C L we prove the following theorem. T H E O R E M 3. In a filter-decomposable P C L L, the dense filter D ( L ) is a neutral element o f the lattice o f all filters o f L if and only if (a v b) cp = a cp v bcp for every a, b e B ( L ) .

Proof. Let L be a filter-decomposable P C L and D be the dense filter of L. First suppose that D is neutral. (a v b)(p = [(a v b)*) A D

= [a* A b*) A D = ([a*) v [ b * ) ) A D = ([a*) A D ) v ([b*) A D ) since D is neutral = a(p v b(p.

Conversely suppose that (a v b)q9 = aq0 v bep for every a, b e B ( L ) W e shall prove that D is neutral. By T h e o r e m 2, D is a standard element of the lattice of all filters of L as a filter decomposable PCS is by definition a decomposable PCS. H e n c e to show that D is neutral it is enough to show that DA([x) vLv))=(DA[x))v(DA[y))

for

x, y 6 L .

Voi. 22, 1986

On "constructions of p-algebras"

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Let x = a A d and y = b A e be a decomposition of x and y D ^

(Ix) v [y)) = D ^ ([a ^ d) v [b ^ e)) = D ^ ([a) v [b) v [d) v [e))

= O ^ ([a ^ b) v [d ^ e)) = (a ^ b)* cp (L) v [d ^ e) by 5.1 (iii) of [6] = (a* v b*) q0 (L) v [d A e) = a* q9 (L) v b* q9 (L) v [d ^ e) by assumption = (a* q9 (L) v [ d ) ) v (b* tp (L) v [e)) = (D ^ [a A d)) v (D ^ [b ^ e)) by 5.1 (iii) of [6] = (D ^ [x)) v (O ^ [y)) Hence D is neutral Here we give an example of a filter-decomposable PCL in which the dense filter is not neutral. E X A M P L E 2. Consider the PCL in Figure 2. D = {d, e, f, g, 1} is the dense filter. D = [d) and the sublattice of L generated by {d, t, z} is not distributive. Hence D is not neutral. 1

|

g

a

el.

0 Figure 2.

222

P, V. RAMANA MURTY AND SR. TERESA ENGELBERT

ALGEBRA UNIV.

5. Quasi-modularp-algebras A p-algebra L is said to be quasi-modular if [(xAy) vz**]Ax=(x^y)v(z**^x)

for

x,y, zeL([6])

Proposition 6.116] states that any quasi-modular p-algebra satisfies the identity x =x** ^ (x v x*) and is filter decomposable. In a quasi-modular p-algebra we prove the following result. T H E O R E M 4. In a quasi-modular p-algebra L, (B v [a)) ^ C =/3 v ([a) A C) where B and C are filters of L such that B ~ C and a is a closed element of L.

Proof. Clearly (B v [a)) A C D B v ([a) ^ C). Let t e (B v [a)) ^ C. This impliesteCandt---b^aforsomebeBc_CsothattAbeC. Sincet__bAawe have t ^ b --- b A a. Since L is quasi-modular, ((t n b) v a) A b = ( t A b ) v ( a A b ) = t n b < - t ((t ^ b) v a) ^ b ~ B v ([a) ^ C) and hence t e B v ([a) ^ C) Hence (B v [a)) ^ C _ B v ([a) ^ C) so that (B v [a)) A C = B v ([a) ^ C) Now we prove the following important theorem. T H E O R E M 5. In a quasi-modular p-algebra L, the dense filter D is a neutral element in the lattice of all filters of L.

Proof. By ([6]; 6.1) L is filter decomposable and by ([6], Theorem 3) (a v b)q0 = aq9 v bqo for every a, b e B(L) so that by theorem 3 of this paper D is a neutral element in the lattice of all filters of L. Now 6.3. [6] becomes a corollary to theorem 4 of this paper. Moreover the condition a ^ b-> c is not a, necessary condition for the equality in 6.3: 6.3[6] reads as follows. 6.3[6]. Let L be a quasi-modular p-algebra and let a, b, c, g ~ B ( L ) and x, y, z e D(L). Let a ^ b -> c and let (a q0 (L) v ix)) ^ (b q0 (L) v [y)) ~ c q0 (L) v [z), then (a q0 (L) v Ix)) ^ (b q9 (L) v [y)) A (g q9 (L) v c q~ (L) v [z)) = [(a q~ (L) v [x)) A (b q~ (L) v [y)) ^ g q0 (L)] v c q~ (L) v [z).

VoL 22, 1986

On "constructions of p-algebras"

223

C O R O L L A R Y 1. Let L be a quasi-modular p-algebra. Let a, b, c, g ~ B(L) and x, y, z ~ D(L). Then (a q~ (L) v Ix)) A (b q0 (L) v [y)) A (g q~ (L) v c q~ (L) v [z)) = [(a q~ (L) v Ix)) A (b q0 (L) v [y)) A g Cp (L)] v c q~ (L) v [z)

whenever (a qo (L) v Ix)) A (b q9 (L) v [y)) _~ c qv (L) v [z) Proof. a q9 (L) v Ix) = ([a*) A O(L)) v Ix) = ([a*) v [x)) A D(L) since D (L) is neutral and x is dense = [a* A X) A D(L). By the given condition we get [a* AX) A [b* A y ) A D(L) ~_ [c* A Z) A D(L) Now (a q~ (L) v [x)) A (b q9 (L) v [y)) A (g q~ (L) v c q0 (L) v [z)) = [a* A X) A [b* A y) A D(L) A ([g*) v ([C* A Z) A D(L)) A D(L) since D(L) is neutral = ([a* A x) A [b* A y) A D(L) A [g*)) V ([C* A Z) A D(L)) by the given condition = [(a q9 (L) v Ix)) A (b q9 (L) v [y)) A g q0 (L)] v (c qo (L) v [z)) H e n c e the result. H e n c e we can redefine a quasi-modular F-triple following the definition 5 of [6]. A n F-triple (B, D, qg) is said to be quasi-modular if (i) D is a lattice with 1 (ii) q0 :B---~F(D) is a (0, 1, v ) h o m o m o r p h i s m , such that for every a E B we have da ~ D with a q0 A a'q0 = [da)

224

~'. V. RAMANA MURTY AND SR. TERESA ENGELBERT

ALGEBRA UNIV.

(iii) If (a qo v [x)) A (b qo v [y)_~c q9 v [z) where a, b, c, g e B; x, y, z e D ; then (a q) v [x)) A (b qo v [y)) A (g qo V (C q9 V [Z)) = [(a qo v Ix)) A (b qJ v ly)) ^ g

v c

v [z)

Now we prove the following important theorem which shows that tile condition (iii) of definition 5 in [6] can be replaced by the condition stated in problem 1 of [6]; thus solving problem 1 of [6]. In fact we prove the following. T H E O R E M 6. Let (B, D, q)) be an F-triple where D is a lattice with 1:. Let a, b, c, g e B(L) and x, y, z c D(L). Then the following five conditions are equivalent. (i) [ d A e ) A ( a q ) v [ d ) ) = ( [ d A e ) A a q o ) v [ d ) ; d , eeD(L) (ii) (a r v [x)) A (C r V [Z)) = [(a r v [x)) A C qO] V [Z) ifzeacpv[x). (iii) (a qo v [x)) A (g q) V C q) V [Z)) = [(a q) v Ix)) A g q0] V C ~ V [Z) if a q9 v Ix) ~ c q9 v [z) (weaker condition stated in problem 1 of [6]). (iv) (a qo v Ix)) A (b go v [y)) A (C qO V [Z)) = [(a q) v [x)) A (b qo v [y)) A C q0] V [Z) i f z e (a q9 v [x)) A (b q) v [y)) (v) (a q9 v [x)) A (b qo v [y)) A (g qo v c q9 v [z)) = [(a q) v Ix)) A (b q9 v [y)) A g q0] V C qO V [Z)) if (a q9 v Ix)) A (b qo v [y)) ~ c qo v [z) (Strong condition)

Proof. (i) 9 (ii). Assume (i) clearly (a qo v [x)) A (C q9 V [Z)) _~ [(a cp v Ix)) A Cq)]V[Z)) Let d e ( a q o v [ x ) ) A ( C q O V [ Z ) ) so that d e a q o v [ x ) and d e c q o v [z)). H e n c e d ~ [d A z) A (C q) V [Z)) = ([d A Z) A C Cp) V [Z) (By (i) since z ~ [d A z)) d e a qo v [x), z e a q9 v [x) implies [d A Z) c__a qo v [x) and hence d e ([d A z) A C Cp) V [Z) __ [(a go v [x)) A C q)] V [Z). Thus (a q9 v [x)) A (Cq0 V [Z)) c__ [(a q9 v [x)) A C 4O] V [Z) SO that (a cp v [x)) A (C 4O V [Z)) = [(a q9 v [x)) A C Cp] v [Z). Thus (i) ~ (ii). Next we prove that ( i i ) ~ (iii). Assume (ii). Let L(R) denote the left-hand (right-hand) side term of equality (iii), Then for every q e c q9 v [z) we have L(q) = ([q] v g qo) A (a q9 v [x)) = [q) v [g qo A (a q9 v [x)) l

= R(q) by (ii). Clearly, L = v (L(q): q e ccp v [z)) = v (R(q): q s c q9 v [z)) = R. Hence (ii) (iii).

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Now we shall prove that ( i i i ) ~ (ii). Assume (iii). Putting g = c, c = 0 in (iii) we get (ii). Thus ( i i i ) ~ (ii). Now assume (ii). Putting z = d, a = 0, x = d ^ e, c = a in (ii) we get (i). Thus (ii) ~ (i). Now we shall prove that (i) ~ (iv). Assume (i) and let z e (a 97 v Ix) ^ (b 97 v [y)). Clearly (a 99 v Ix)) A (b q) v [y)) ^ (c 97 v [z)) _~ [(a 97 v Ix)) ^ (b 97 v [y)) ^ c 97] v [z). Let d e (a 97 v [x)) ^ (b 97 v [y)) ^ (c 97 v [z)) so that d e [a ^ z) ^ (c 97 v [z)) = ([d A z ) a c 97) v [z) by (i) ___((a 97 v Ix)) ^ (b 97 v [y)) ^ c 97) v [z)

since [d ^ z) __ (a 97 v [x) a (b tp v [y)) so that (a 97 v Ix)) ^ (b 97 v [y)) A (C 99 v Z)) c_ [(a 97 v [x)) ^ (b 97 v [y)) ^ c 97] v [z) and hence the required inequality. Thus ( i ) ~ (iv). Now we shall prove that ( i v ) ~ (v). Assume (iv). Let (a 97 v [ x ) ) ^ (b 97 v [y)) _Dc 97 v [z). Clearly (a 97 v Ix)) ^ (b 97 ,, [ y ) ) ^ (g ~o v c 97 ,, [z)) _= [(a 97 v Ix)) ^ (b ~ v [y)) ^ g ~1 v c 97 v [z)

Let L(R) denote the left-hand (right-hand) side term of equality (v). For every q e c 97 v [z) there is

L(q) = ([q] v g qo) ^ (a 97 v Ix)) ^ (b 97 v [y)) = [q) v (g 97 ^ (a 97 v [x)) ^ (b 97 v [y)) = R(q) by iv. Clearly L = v (L(q): q ~ c 97 v [z)) = v (R(q): q ~ c 97 v [z)) = R. Thus (iv) ~ (v).

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ALGEBRA UNIV.

Finally we shall prove that ( v ) ~ (i). Assume (v). Putting a = 0, x = d .~, e, b = l, y = l, g = a, c = O and z = d we get (i). Thus all the five conditions are equivalent. C O R O L L A R Y 2. In a quasi-modular p-algebra the above five conditions are equivalent. Proof. Obvious by the above theorem. The above corollary shows that the weaker condition in problem 1 of [6] is not really weak and in fact is equivalent to (iii) of definition 5 of [6], thus solving problem 1 of [6]. Therefore the definition of a quasi-modular F-triple can certainly be reworded again as follows:

An F-triple (B, D, qJ) is said to be quasi-modular if (i) D is a lattice with 1 (ii) cp : B ~ F ( D ) is a (0, 1, v ) homomorphism such that for every a ~ B we have da e D with a go ^ a'q9 = [da) and (iii) If a q0 v [x) _~ c q) v [z)), then (a go v [x)) ^ (g go v c go v [z)) = [(a go v Ix)) ^g~p] v cqo v [z). For a, c, g e B a n d x , y, z e D . Problem 3 of [6] is also solved under the additional assumption that q0 is a join-homomorphism. Moreover condition (iv) of problem (3) is implied in condition (v) by Theorem 6. Thus we get a solution to problem 3 under the new hypothesis in the following corollary to Theorem 6. C O R O L L A R Y 3. An F-triple (B, D, cp) is quasi-modular if and only if (i) D is a lattice with 1 (ii) qD: B ~ F ( D ) is a (0, 1, v ) h o m o m o r p h i s m such that for every a c B we have da ~ D with a cp ^ a' cp = [do) and (iii) [d ^ e) ^ (a tp v [d)) = ([d ^ e) ^ a go) v [d) Proof. Obvious by Theorem 6.

6. p-algebras with a modular frame

A p-algebra with a modular frame is quasi-modular by Lemma 1 of [5]. ([5], page 68). Hence the dense filter is a neutral element of the lattice of all filters of L so that results like Lemma 3, Lemma 4 and Theorem 4 of [5] become quite evident.

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On "constructions of p-algebras"

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L E M M A 3[5]. Let L be a p-algebra with a modular frame. Let a, b ~ B(L). Then, (i) ([a) A D(L)) v ([b) A D(L)) = [([a) ^ D(L)) v [b)l ^ (ii) ([b) A [a)) v (D(L) A [a)) = ([b v a) v D(L)) ^ [a) (iii) (D(L) v [b)) A [a A b) = [ ( D ( L ) v [b)) A [a)] v [b).

D(L)

L E M M A 4[5]. Let L be a p-algebra with a modular frame. Let a ~ B(L) and x 6 D ( L ) . Then [ a A x ) A D ( L ) = a O ( L ) v [ x ) where aO(L)=a*cp(L)=[a)A D(L). Now we shall give a direct proof of T h e o r e m 4[5] using the neutrality of D(L). T H E O R E M 4[5]. Let L be a p-algebra with a modular frame. Let a, b, c ~ B(L) and x, y, z ~ D(L). Let a <-c and aO(L) v [x)_~c0(L) v [z). Let a ~ b or b ~ c. Then

[(aO(L) v [x)) A (bO(L) v [y))] v cO(L) v [z) = (aO(L) v [x)) ^ [(bO(L) v [y)) v cO(L) v [z)l Proof. Let s y=bAy, g = Z A C , aO(L) v [ x ) ~ c O ( L ) v [ z ) i.e., [a A X) A D(L) ~ [c A Z) ^ D(L) which implies [a A X) _~ [C A Z) i.e., ~ --<2. By L e m m a 1 [5] (~ v )7) ^ z = x v 07 A 2) i.e., [~) i (D7) v [2)) = ([s i [.9)) v [2) so that [[~) t ([y) v [2))] A D(L) = [([~) A DT)) v [2)] A D(L). By the neutrality of D(L) we get ([s

i

D(L))

A [([y) A

D(L))

v ([2) A D ( L ) ) ]

= [(s A D ( L ) ) A ([.9) i D ( L ) ) ] v ([2) i D ( L ) ) i.e., ([a ^ x) A D ( L ) ) n [([b A y) A D ( L ) ) v ([c ^ z) A D(L))] = [[a ^ x) ^ D ( L ) ^ [b ^ y) ^ D ( L ) ] v ([c ^ z) ^ D ( L ) ) i.e.,

(aO(L) v

= [(aO(L) v

[x)) ^

[x)) A

[(bO(L) v

(bO(L) v

[y)) v

[y))] v

(cO(L) v

cO(L) v

[z))]

[z).

In conclusion, we highly thank the Referee for his valuable suggestions and comments which helped us to shape this paper into the present form.

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P. v. RAMANAMURTYAND SR. TERESAENGELBERT

ALGEISRAUNiV.

[3] G. GRATZER, Lattice Theory, First Concepts and Distributive lattices, W. H. Freeman and Co., 1971. [4] G. GRATZER and E. T. SCHMIDT, Standard ideals in lattices, Acta Math. Acad, Sci. Hungar !2 (1961) 17-86. [5] -F. KATRII~.~Kand P. MEDERLY, Construction of p-algebras with a modular frame, Houston J. Math, 4 (1978), 67-79. [6] T. KATRIr~AKand P. MEDERLY, Construction of p-algebras, Algebra Univ., 17 (1983) 288-316. [7] P. MEDERLY,A characterization of modularpseudocomplemented semilattices, Colloq. Math~ Soc. J. Bolagi, Szeged 14 (1974) 231-248. [8] P. V. R. MURTY and V. V. R. RAO, Characterization of certain pseudocomplemented simi-lattices~ Algebra Uni. 4 (1974) 289-300. [9] W. C. NEMITZ, Implicative semilattices, Trans. Ameri. Math. Soc. 117 (1965) 128-I42.

Department of Mathematics Andhra University Waltair India