ON THREE-VALUED MOISIL ALGEBRAS Manuel ARAD and Luiz MONTEIRO In this paper we investigate the properties of the family I(A) of all intervals of the form [x, x] with x x, in a De Morgan algebra A, and we obtain a necessary and sufficient condition for a complete De Morgan algebra to be a Kleene algebra in terms of I(A) (Lemma 11). We prove that I(A) is a complete Boolean algebra if A is a complete Moisi! algebra (Lemma 12). Let A be a distributive lattice which is bounded, that is, has a largest element 1 and a smallest element O. We use x V y for the join, and x Ay fo r the meet of two elements x, y in A. I f is a unary operation defined on A satisfying x = x and (x y ) = x A - A is called a De Morgan algebra. The set o f all complemented elements of A is noted by B (A). A Kleene algebra is a De Morgan algebra satisfying: x A - algebra is a system (L, 1,, A, V, V) such that (L, 1,, A, V) is a De Morgan algebra and V is a unary operation (called possibility operation) defined on A with the properties: xv Vx = 1, x A - - The necessity operation A is defined by Ax = x. L is called a centered three-valued Lukasiewicz algebra, o r a three-valued Post algebra, if it has a center, that is, an element c of L such that c c. The center of L (if it exists) is unique [6], [7], [8], [12], [15]. An axis of a three-valued Lukasiewicz algebra L is an element e of L with the properties: Ae = 0 and V x A x v Ve for all x of L. Again, if the axis of L exists, it is unique. Following A. Monteiro, L is called a t h r e e of three valued Moisil algebras we refer to [12], [13], [15]. Let E be a three-valued Moisil algebra, and e E E the axis of E. In [12] L. Monteiro proved that: (1) x ( A x ye) A Vx, for all x ee, and (2) x = (Ax - Let us introduce the following set:
408 M A N U E L ABAD and LUIZ MONTEIRO S = { t E E : x = (Ax t ) A Vx, for all x EE}. From (I) and (2) it is clear that e and e belong to S. On the other hand, we know that Ae = 0 and this is equivalent to say that e e, thus the interval [e, e LEMMA 1. S = [e, e Proof. If y [ e, e], e y e, then we have x = (Ax e ) A Ve (A x V y ) A V x ( A x (Ax y ) A Vx for ail x ee, whence y If s ES, then x = (A x Vs) A Vx for all x EE. Then e s, for e = (A e Vs ) A V e =- ( 0 Vs ) A V e = s A Ve s, a n d s (A e Vs ) A V e = (A e Vs ) A 1 = A e V s It is well known that S = [e, e] is a distributive lattice, the elements e and e being the least and greastest elements o f S respectively [4]. I t is not difficult to see that (S, e,, A, V) is a Kleene algebra. Furthemore, S is a Boolean algebra. For, if s ES, then e ES and t = s A s ES, then (i) x = (Ax ( s A x EE. Since (ii) At = A(s A s) =- As A A s 0, from (i) and (ii) from definition and uniqueness of the axis, we have (iii) s A whence (iv) s s = e. So S is a Boolean algebra. Now we define the following mapping f: E > [e, el, f(x) = ( e A x) v e, for each x LEMMA 2. The mapping f has the following properties. Fl) f (l) = e ; F2) f(0) = e; F3) f(x A y) = fix) A f(y); F4) f( x) = f(x); F5) f(x y ) = f(x) v f(y); F6) f(s) = s if and only if S [ e, e From Fi to F4, it follows that f is an M-homomorphism (homomorphism between Morgan algebras). From F6, it follows that f is an M-epimorphism. L. Monteiro [12] proved that if E is an algebra with axis e, then E is isomorphic to the direct product of a Boolean algebra B and a centered three-valued Lukasiewicz algebra C. More precisely, B i s t h e principal ideal generated b y Ve, I( ve) = {x EE: x y e }, and C is I(Ve)=-
ON THREE-VALUED MOISIL ALGEBRAS 4 0 9 LEMMA 3. I( Ve) and S are isomorphic Boolean algebras. Proof. The correspondence h(x) = x A Ye sets up an M-epimorphism from E onto l( ye). I t is an easy verification that Ker f = Ker h, which completes the proof. LEMMA 4. A t h r e e bra if S = L : x = (Ax Vs) A Vx, for all x cl }, is a nonvoid set. Proof. It is clear that S is closed under A and V. I f s es, then x ( x Vs) A V x, for all x G L, hence x = (Y x A s) V A x for all x e L a n d then s G S. F ro m th is, s A s S f o r s ES. B u t A(s A s) 0 and x = (Ax ( s A s)) A Vx f or all x EL, therefore e = s A s is the axis of algebra L. From lemma I, S = [e, e We remark that L has a center c if and only if S = tel. For if L has a center c, we know that c is also an axis [12] and then S [ c. cl [c, c] = {e an axis of L and moreover c e S = Icl, thus c = c and so c is the center of L. We now deal with the dual concept of ideal. Recall that a filter is a nonvoid subset F of a lattice L with the properties: a ef, x E L, a imply x EF, and a EF, b e F imply a Ab CF. Giv en an element a in any lattice L, the set F(a) = E L : a LÇ,. x i s evidently a filter; it is called a principal filter of L. LEMMA 5. Let E be a three-valued Moisil algebra, e its axis. Then the ordered sets F(e), I( e) and B(E) are isomorphic. Proof. Define H: F(e) B ( E ) by H(x) = Ax, and G: F(e) ) R e) by G(x) = x. DEFINITION 6. Let (A, I,, A, V) be any De Morgan algebra, x Yo elements of A such that x be Boolean if the system (A, y
410 M A N U E L ABAD and LUTZ MONTEIRO EXAMPLES. - x 0 a 1 d Fig. I - x 0 a 1 d d In the algebra of fig. 1, [a, d] is Boolean, and [0, 1] is not Boolean. In the algebra of fig. 2, [a, e] with its natural ordering is a Boolean algebra, but it is not a Boolean interval because c = c *d. The only Boolean intervals are [c, cl and [d, dl. REMARK. I f S [ x x A -x = x SgS, and conversely, if [x
ON THREE-VALUED MOISIL ALGEBRAS 4 1 1 belong to [x every Boolean interval can be written as [ x However, an interval of this kind need not be Boolean as we can see for [0, o f the examples above. It should pointed out that a De Morgan algebra need not have Boolean intervals, as the following example shows. EXAMPLE. Let T { 0, c, 1}, with 0 < c < 1, and 0 = 1, c = c, 1 = O. T is a Kleene algebra, and if N is the set of all positive integers, it is readily verified that A = TN is a Kleene algebra with pointwise operations. Let K and K2 the set of all f e TN such that there exists a nonvoid finite subset F of N with f(x) = c for all x ef, and f(x) { 0, 1} if x F. Clearly K a proper Kleene subalgebra of A. Suppose [f, I] is a Boolean interval of K. Since f f, then f(x) BO, cl for all x en, so there exists a finite subset F of N such that f(x) c for all x e F and f(y) = 0 if y ÊF. Let i if x EF f i b A b = b w h i c h contradicts that [f, f] is Boolean. LEMMA 7 I f A is a Kleene algebra and there exists a Boolean interval in A, it is unique. Proof. Let S Since A is a Kleene algebra: (1) e = e A e v f = f. Then e v f f Vf = f, hence f e vf f, so e V f 0 S (e Vt) A (e vt) = f. By (I), t - - e then e f V e e v e e, hence e V f S and e = f. Then S Denote the set of intervals of a De Morgan algebra A of the form [x, xl with x x by I(A). I(A) 4-0 seeing that A = [0, I ] ei(a). Notice that I (A) = t x A - x, x - x l, x c A l. LEMMA 8. Any minimal element o f the ordered set (I(A), g ), whenever it exists, is a Boolean interval.
412 M A N U E L ABAD and LUIZ MONTEIRO Proof. Let S = [xo, x01 be minimal element of I(A). Clearly xo t A t x S t A t = xo for every t ES- Thus Sis Boolean. LEMMA 9. I f A is a complete De Morgan algebra, then the ordered set I(A) has the Zom's property for descending chains. Proof. Let C = {S Since A is complete, there exists the element x = x that x x x x [ t x that x COROLLARY 10. Any complete De Morgan algebra has Boolean intervals. LEMMA 11. A complete De Morgan algebra A is a Kleene algebra if and only if I(A) has least element. Proof. By the above lemmas, if A is a Kleene algebra, it has a unique Boolean interval, which is the least element in I(A). For the converse let P = [b, b] be the least element in I(A). Then, Pg [x A x, x V x] for all x EA. Since b EP, then x A x b for all x EA and b y V y fo r y EA. Therefore x A x y V y fo r x, y EA, that is, A is a Kleene algebra. (Note that in the converse, the completeness of A has not been used). Let A be Kleene algebra. Define the following binary operations on I(A): [x, x] v [y, y] ---- [x Ay, x V y Tx, xl A [y, y] = [x Vy, x A From Kleene condition, it follows at once that [x Ay, x y ] and [x v y, x A y] belong to I(A), if x x and y 5.. y.
ON THREE-VALUED MOBIL ALGEBRAS 4 1 3 Moreover, it is easy to check that (I(A), A, A, V) is a distributive lattice with greatest element A. In fact, [x, x] A [y, y] [ x, x] if and only if [x, x] g [y, y] for [x, x] and [y, y] E I(A). Observe that if A is a complete three-valued Moisil algebra, then I(A) has least element [e, e Indeed, we know that [x, x] L e, e] if and only if x e l and x [ e, e [e, e element of I(A) is Boolean, and since [e, el is the unique Boolean interval of A, thus [e, e] is the unique minimal interval of A. Then [e, e LEMMA 12. I f A is a complete Mo isil algebra, then I (A ) is a complete Boolean algebra. Proof. To prove that for all [x, x] I ( A ) there exists [y, y] el(a) such that [x, x] A [y, --y] = [e, e] and [x, x] v [y, y] = [0, I] it is equivalent to prove that for all x e there exists y e such that x Vy = e and x Ay = O. Let us put y = A x Ae. Then ( i) x A(A x A e) 0 because x A A x O. (ii) x V(A x Ae) = e. I n d e e d, A ( x V (A x Ae)) = Ax V (A x A Ae) = A x = 0 = A e a n d 7 ( x v (A x Ac)) Vx V (A x A ye) ( V x V A x) A (Vx V Ve) = (Vx v A x) A Ve = I A Ve = Ve, and by determination principle (see [12] and [141) (ii) holds. The completeness of 1(A) follows from n [xi, -x i] =, Y Observe that the mapping H: F( e) I ( A ) defined by H( y) = [y, y], for all element y E R e), is an order isomorphism. Indeed, H is clearly a bijection and y y ' if and only if H( y) Thus we can state COROLLARY 13. I f A is a three-valued Post algebra such that I(A) is a complete Boolean algebra, then A is complete. Proof. By the above remark, if 1(A) is a complete Boolean algebra,
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