DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES

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1 ANALELE ŞTIINŢIFICE ALE UNIVERSITĂŢII AL.I. CUZA DIN IAŞI (S.N.) MATEMATICĂ, Tomul LXI, 2015, f.2 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES BY D. BUŞNEAG, D. PICIU and J. PARALESCU Abstract. The main purpose of this paper is to present new results in divisible and semi-divisible residuated lattices. It also presents a new characterization for boolean elements of a residuated lattice (Theorem 9). Mathematics Subject Classification 2010: 03G10, 06D35, 06D72, 03G05, 06B20. Key words: divisible and semi-divisible residuated lattices, Boolean algebra, semi- G-algebra, MT L-algebra, BL-algebra, MV -algebra, regular element, dense element, reflective subcategory, deductive system, maximal deductive system, radical. 1. Introduction The origin of residuated lattices is in Mathematical Logic without contraction. They have been investigated by Krull [18], Dilworth [9], Ward and Dilworth [26], Ward [25], Balbes and Dwinger [1] and Pavelka [20]. In [16], Idziak prove that the class of residuated lattices is equational. These lattices have been known under many names: BCK- latices in [15], full BCK- algebras in [18], FL ew - algebras in [19], and integral, residuated, commutative l-monoids in [3]. Apart from their logical interest, residuated lattices have interesting algebraic properties (see [2], [5], [9], [12], [25], [26]). MV-algebras are known to be special residuated lattices with a proper additive operation. MV-algebras fulfill a double negation law x = x, too. For a residuated lattice L we denote by MV (L) the set of all elements

2 288 D. BUŞNEAG, D. PICIU and J. PARALESCU 2 x = x 0 with x L. (MV (L),,, 0) is an MV algebra iff for all x, y L, (x y ) y = (y x ) x (see [22]), where for x, y MV (L), x y = x y. In particular, for any semi-divisible residuated lattice L, the subset (MV (L),,, 0) is an MV -algebra (see [22]). The paper is organized as follows. In Section 2 we recall the basic definitions of residuated lattices. Also we present M V -center of a semi-divisible residuated lattice, defined by Turunen and Mertanen in [22]. This is a very important construction, which associates an M V -algebra to every semi-divisible residuated lattice. In this way, many properties can be transferred from M V -algebras to residuated lattices and backwards. In Section 3 we prove that the category MV of MV -algebras is a reflective subcategory of the category RL d of divisible residuated lattices and as consequence we obtain some informations relative to injective divisible residuated lattices (Corollary 3). In Section 4, for a residuated lattice L we denote by R(L) the set of regular elements of L, by D(L) the set of dense elements of L and by B(L) the set of boolean elements of L. We present new characterizations for these elements. We prove that in general, B(L) R(L) MV (L). If L is a semidivisible MT L-algebra, then B(L) = B(MV (L)). Theorem 10 characterize the residuated lattices which are Boolean algebras. In Section 5, we present new results relative to lattice of maximal deductive systems of a residuated lattice. We introduce the notion of semi-galgebra and we obtain that if L is a semi-g-algebra, then there is a bijection between Max(L) and RL(L, {0, 1}) = { f : L {0, 1} f is a morphism of residuated lattices} (Theorem 14). We prove that if L is semi-divisible, then there is a bijective correspondence between maximal deductive systems of L and maximal deductive systems of MV (L). In Section 6, we characterize Rad(M V (L)), for a semi-divisible residuated lattice L. 2. Definitions and preliminaries We review the basic definitions of residuated lattices. Also we present M V -center of a semi-divisible residuated lattice, defined by Turunen and Mertanen in [22].

3 3 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 289 Definition 1. A residuated lattice ([2], [23]) is an algebra (L,,,,, 0, 1) of type (2, 2, 2, 2, 0, 0) equipped with an order satisfying the following: (LR 1 ) (L,,, 0, 1) is a bounded lattice; (LR 2 ) (L,, 1) is a commutative ordered monoid; (LR 3 ) and form an adjoint pair, i.e. a x y iff a x y for all a, x, y L. The relations between the pair of operations and expressed by (LR 3 ), is a particular case of the law of residuation ([2]). The class RL of residuated lattices is equational (see [16]). Example 1 ([23]). Let p be a fixed natural number and I = [0, 1] the real unit interval. We define for x, y I, x y = (max{0, x p + y p 1}) 1/p and x y = min{1, (1 x p +y p ) 1/p }, then (I, max, min,,, 0, 1) become a residuated lattice called generalized Lukasiewicz structure. For p = 1 we obtain the notion of Lukasiewicz structure ( x y = max{0, x + y 1}, x y = min{1, 1 x + y}). Example 2 ([23]). If consider on I = [0, 1], to be the usual multiplication of real numbers and for x, y I, x y = 1 if x y and y/x otherwise, then (I, max, min,,, 0, 1) is a residuated lattice (called Products structure or Gaines structure). Example 3 ([23]). If (B,,,, 0, 1) is a Boolean algebra, then if we define for every x, y B, x y = x y and x y = x y, then (B,,,,, 0, 1) becomes a residuated lattice. In L we consider the following identities: (BL 1 ) x (x y) = x y (divisibility); (BL 2 ) (x y) (y x) = 1 (preliniarity); (BL 1 ) [x (x y )] = (x y ) (semi-divisibility). Definition 2. The residuated lattice L is called: (i) divisible if L verify (BL 1 );

4 290 D. BUŞNEAG, D. PICIU and J. PARALESCU 4 (ii) MT L-algebra if L verify (BL 2 ); (iii) BL-algebra if L verify (BL 1 ) and (BL 2 ) (that is, L is a divisible MT L algebra); (iv) semi-divisible if L verify (BL 1 ). We denote by RL ( RL d, MT L, BL, RL sd ) the class of residuated lattices (divisible residuated lattices, M T L-algebras, BL-algebras, semidivisible residuated lattices). Proposition 1 ([22]). For a residuated lattice L, the following conditions are equivalent: (i) L RL d ; (ii) For every x, y L with x y there exists z L such that x = y z; (iii) For every x, y, z L, x (y z) = (x y) [(x y) z]. We recall ([7], [23]) that type (2, 1, 0) such that: an MV -algebra is an algebra (M,,, 0) of (MV 1 ) (M,, 0) is a commutative monoid; (MV 2 ) x = x, for every x M; (MV 3 ) (x y) y = (y x) x, for every x, y M (where x y = x y). We denote by MV the class of MV -algebras. Remark 1 ([14], [23]). 1. It is not hard to see that an equivalent presentation of MV -algebras can be given as BL-algebras plus condition (MV 2 ). 2. Let (L,,,,, 0, 1) be a residuated lattice. If for x, y L we denote x y = x y, then (L,,, 0) is an MV -algebra iff (x y) y = (y x) x, for every x, y L. Remark 2. Lukasiewicz structures and Boolean algebras are BL algebras; not every residuated lattice is a BL-algebra (see [23], p.16).

5 5 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 291 Example 4 ([17]). We give an another example of a finite residuated lattice, which is not a BL-algebra. Let L = {0, a, b, c, 1} with 0 < a, b < c < 1, but a, b are incomparable. L become a residuated lattice relative to the following operations: 0 a b c a b 1 b 1 1 b a a c 0 a b a b c 1, 0 a b c a 0 a 0 a a b 0 0 b b b c 0 a b c c 1 0 a b c 1. In what follows by L we denote the universe of a residuated lattice; for x L and a natural number n, we define x = x 0, x = (x ), x 0 = 1 and x n = x n 1 x for n 1. In residuated lattices we have the following rules of calculus: Theorem 1 ([5], [6], [8], [12], [23]). Let x, x 1, x 2, y, y 1, y 2, z L. Then we have: (c 1 ) 1 x = x, x 1 = 1, 0 x = 1, x 0 = 0; (c 2 ) x y iff x y = 1; (c 3 ) x y x, x (x y) y, ((x y) y) y = x y; (c 4 ) x y (z x) (z y); (c 5 ) x y (y z) (x z); (c 6 ) (x y) (x z) x (y z); (c 7 ) x (y z) = (x y) (x z), (y z) x (y x) (z x); (c 8 ) (x y) z = (x z) (y z); (c 9 ) x y implies z x z y and y z x z; (c 10 ) x 1 y 1 (y 2 x 2 ) [(y 1 y 2 ) (x 1 x 2 )]; (c 11 ) x y implies z x z y; (c 12 ) x (x y) x y, x y (x y);

6 292 D. BUŞNEAG, D. PICIU and J. PARALESCU 6 (c 13 ) x (x (x y)) = x y; (c 14 ) x (y z) = (x y) z = y (x z); (c 15 ) x y (x z) (y z) and (x 1 y 1 ) (x 2 y 2 ) (x 1 x 2 ) (y 1 y 2 ); (c 16 ) (x 1 y 1 ) (x 2 y 2 ) (x 1 x 2 ) (y 1 y 2 ); (c 17 ) (x 1 y 1 ) (x 2 y 2 ) (x 1 x 2 ) (y 1 y 2 ); (c 18 ) x y = 1 x y = x y; (c 19 ) x (y z) (x y) (x z), so, x m y n (x y) mn for any natural numbers m, n; (c 20 ) x (y z) y (x z) (x y) (x z); (c 21 ) x (y z) = (x y) (x z); (c 22 ) (x y) n x n y n for every n 1; (c 23 ) (x y) n x n y n, for every n 1; (c 24 ) (x y) n x n y n, for every n 1. Theorem 2 ([5], [6], [8], [12], [23]). If x, y L, then: (c 25 ) 1 = 0, 0 = 1; (c 26 ) x y y x ; (c 27 ) x y y x ; (c 28 ) x x = 0 and x y = 0 iff x y ; (c 29 ) x x, x x x, x = x ; (c 30 ) (x y) = x y = y x ; (c 31 ) (x y) = x y ; (c 32 ) x y = y x = x y ; (c 33 ) (x y ) = x y ;

7 7 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 293 (c 34 ) (x y) x y ; (c 35 ) x y (x y), so (x ) n (x n ) for every natural number n; (c 36 ) x y (x y). Corollary 1. If L is a divisible residuated lattice, then for every x, y L we have: (c 37 ) (x x) = 0; (c 38 ) (x y) = x y ; (c 39 ) (x y) = x (x y ), (x y) = x y ; (c 40 ) y x x (x y) = y ; (c 41 ) x (y z) = (x y) (x z), x (y z) = (x y) (x z). Proof. (c 37 ). From x x x (x x) x (x x) = x (x x) (BL 1) = x [x (x x) ] (c 30) = x [(x x) (x ) ] (c 30) = x [(x x) x ] (BL 1) = x (x x) = x x = 0. (c 38 ). From (c 34 ) we have (x y) x y = x y. On the another hand we have (x y ) (x y) = (x y ) (x y) (c 34) [(x y ) (x y)] = [(x (x y )) y] (BL 1) = [(x y ) y] [(x y) (y y)] (y y) (c 37) = 0 = 1 (x y ) (x y) = 1 x y (x y) (x y) = x y = x y. (c 39 ). From x y x (x y) x (x y) = (x y) x = [(x y) ] x = (y x ) x = x [x (y x ) ] = x [x (x y ) ] = x [x (y x ) ] = x [(y x ) x ] = x [(x y ) x ] = x [(x y ) x ] = x [(x y ) x ] = x (x y ). Clearly, (x y) x, y. Consider t L such that t x, y. Then x, y t x y t t (x y ). But t t and (x y ) = x y t x y (x y) = x y. (c 40 ). From y x x (x y) = (x y) x = (x y ) x = [x (x y )] = (x y ) = y. (c 41 ). Clearly x (y z) (x y) (x z).

8 294 D. BUŞNEAG, D. PICIU and J. PARALESCU 8 On the other hand, (x y) (x z) = (x y) [(x y) (x z)] = (x y) [x (y (x z))] = (x y) [y (x (x z))] = x [y (y (x (x z)))] = x [y (x (x z))] = x [(x (x z)) ((x (x z)) y)]. But z x (x z) (x (x z)) y z y, so (x y) (x z) x [x (x z)] (z y) = x z (z y) = x (y z) (x y) (x z) = x (y z). Clearly, x (y z) (x y) (x z). On the another hand we have x (y z) = (y z) [(y z) x] (c 21) = [y ((y z) x)] [z ((y z) x)] [y (y x)] [z (z x)] = (y x) (z x) = (x y) (x z). 3. MV-center of a divisible residuated lattice For a residuated lattice L we consider the subset MV (L) = {x : x L} = {x L : x = x} of L. MV (L) is non-void, since 1 = 0 MV (L). Moreover, MV (L) is closed with respect to the operations (because by (c 30 ), x y = (x y) MV (L)), (by c 31 ) and in particular, with respect to the operation. For x, y L we define x y = x y. Theorem 3 ([22]). (MV (L),,, 0) is an MV -algebra iff for all x, y L, (x y ) y = (y x ) x. Theorem 4 ([22]). If (x y ) y = (y x ) x holds in a residuated lattice L, then L is semi-divisible. Corollary 2 ([22]). In any semi-divisible residuated lattice L, (M V (L),,, 0) is an MV algebra (called MV center of L). Recall that an algebra (W,,, 1) of type (2, 1, 0) is a Wajsberg algebra if it satisfies the following conditions: (W 1 ) 1 x = x; (W 2 ) (x y) ((y z) (x z)) = 1; (W 3 ) (x y) y = (y x) x; (W 4 ) (x y ) (y x) = 1. Following [23], if (W,,, 1) is a Wajsberg algebra, then (W,,, 0) is an MV algebra, where for x, y W, x y = x y. Conversely, if

9 9 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 295 (W,,, 0) is an MV algebra, then (W,,, 1) is a Wajsberg algebra, where for x, y W, x y = x y. The order relation on a Wajsberg algebra W is given by: x y iff x y = 1. Following Theorems 3, 4 and Corollary 2, if L is a semi-divisible residuated lattice and for x, y L we define x y = (x ) y = x y = y x = y x = x y, then MV (L) has a structure of MV algebra (hence, Wajsberg algebra) and the order relation W on the Wajsberg algebra MV (L) is defined by x W y iff x y = 1 x y (in L). Hence the order relation on MV (L) coincides with the order relation on L. According to [22], the lattice operations least upper bound {x, y }, denoted by x W y, and greatest lower bound {x, y }, denoted by x W y, on the Wajsberg algebra MV (L) are defined via x W y = (x y ) y and x W y = (x W y ). In general, the meet operation W on MV (L) may not be the restriction of the meet operation on L and, similarly, for join operation. Proposition 2 ([22]). If L is a semi-divisible residuated lattice, then W and coincide on MV (L). Lemma 1. If x, y L, then: (c 42 ) [x (x y )] = (y x ) x ; (c 43 ) x W y = (x y ) (c 31) = (x y). Proof. (c 42 ). We have [x (x y )] = [x (y x )] = [(y x ) x ] = (y x ) (x 0) = (y x ) x. (c 43 ). From x y x x (x y ) and analogously y (x y ). To prove x W y = (x y ) consider t MV (L) such that x, y t. Then t x, y t x y (x y ) t = t. We present a new characterization for semi-divisible residuated lattices: Proposition 3. For a residuated lattice L, the following conditions are equivalent: (i) L RL sd ; (ii) (x y ) y = (y x ) x for every x, y L.

10 296 D. BUŞNEAG, D. PICIU and J. PARALESCU 10 Proof. (i) (ii). If L RL sd, then for x, y L, [x (x y )] = [y (y x )] (= (x y ) ) and by (c 42 ) we deduce that (x y ) y = (y x ) x. (ii) (i). If x, y L and (x y ) y = (y x ) x, since (x y ) = x W y = (x y ) y and [x (x y )] = (y x ) x, then we obtain that [x (x y )] = (x y ), that is, L RL sd. Remark Every divisible residuated lattice is semi-divisible, so, every BL-algebra is semi-divisible. 2. Not every residuated lattice is semi-divisible. Consider, for example (see [22]) a fixed real number c, 0 < c < 1, and define the residuated lattice L c = ([0, 1],,,,, 0, 1) such that for all x, y [0, 1], such that x y = 0 if x+y c and min{x, y} elsewhere, x y = 1 if x y and max{c x, y} elsewhere. We have MV (L c ) = [0, c) {1}. Let x = 3 5 c, y = 4 5c. Then x, y MV (L c ), (y x) x = 1, but (x y) y = y. Thus, the condition from Theorem 3 does not hold. So L c is not semi-divisible. Evidently, each residuated lattice L c is linear, therefore is a MT L-algebra. We deduce that M T L-algebras are not, in general semi-divisible. 3. There are residuated lattices that are semi-divisible but not divisible. Consider, for example (see [22]) the residuated lattice L sd = ([0, 1],,,,, 0, 1) such that for all x, y [0, 1], x y = 0 if x, y [0, 1 2 ] and min{x, y} elsewhere x y = 1 if x y; 1 2 if y < x 1 2 and y if (y < x, 1 2 < x). We have MV (L sd ) = {0, 1 2, 1} and the condition from Theorem 3 holds, whence, L sd is a semi-divisible residuated lattice. L sd is not divisible. Indeed, let x = 1 3, y = 1 2. Then 1 2 ( ) = = = min{ 1 3, 1 2 }. We recall that if (L i,,,,, 0, 1), i = 1, 2 are two residuated lattices then, a map f : L 1 L 2 is called morphism of residuated lattices if f satisfies the following conditions, for every x, y L 1 : (m 1 ) f(0) = 0; (m 2 ) f(1) = 1; (m 3 ) f(x y) = f(x) f(y); (m 4 ) f(x y) = f(x) f(y); (m 5 ) f(x y) = f(x) f(y);

11 11 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 297 (m 6 ) f(x y) = f(x) f(y). We define Ker(f) = {x L 1 : f(x) = 1}. Remark If L 1, L 2 are BL-algebras, then f : L 1 L 2 is morfism of residuated lattices iff f verifies (m 1 ), (m 5 ) and (m 6 ). 2. If L 1, L 2 are divisible, then f : L 1 L 2 is morfism of residuated lattices iff f verifies the conditions (m 1 ), (m 4 ), (m 5 ) and (m 6 ). 3. If L 1, L 2 are MT L-algebras, then f : L 1 L 2 is morfism of residuated lattices iff f verifies (m 1 ) (m 3 ) and (m 5 ) (m 6 ). So, RL becomes in canonical way a category such that RL d, MT L, RL sd and BL are subcategories of RL. Remark 5 ([1, p.31]). Since the categories MV and RL are equational, then in these categories the monomorphisms are exactly the one-one morphisms. In what follows, by L we denote the universe of a divisible residuated lattice. Since L is semi-divisible, by Corollary 2, (MV (L),,, 0) is a MV - algebra (hence BL algebra), where for x, y MV (L), x = x 0 MV (L) and x y = x y = x y = (x y ) MV (L). The order on MV (L) is defined for x, y MV (L) by x W y iff x y = 1 x y = 1 x y = 1, so the order on MV (L) is the restriction of the order on L to MV (L). Also, if for x, y L we define x W y = [(x ) (y ) ] = (x y ), then (MV (L), W, W =,, W, 0, 1) is a BL-algebra (hence a residuated lattice). Proposition 4. If x, y L, then (x y) = x W y. Proof. We have x W y (c 44) = (x y ) = (x y ) = (x y ) = (y x ) = (y (x 0)) = [(x y) ] = (x y). We denote R(L) = MV (L) and we define Φ R (L) : L MV (L) by Φ R (L)(x) = x, for all x L. Proposition 5. If L is a divisible residuated lattice, then Φ R (L) is a morphism of residuated lattices (between the lattices (L,,,,, 0, 1) and (MV (L), W, W =,, W, 0, 1)). Proof. Clearly, Φ R (L)(0) = 0, Φ R (L)(1) = 1, Φ R (L)(x y) = (x y) (c 43) = x W y = Φ R (L)(x) W Φ R (L)(y), Φ R (L)(x y) = (x

12 298 D. BUŞNEAG, D. PICIU and J. PARALESCU 12 y) (c 38) = x y = Φ R (L)(x) Φ R (L)(y) and by Proposition 4, Φ R (L)(x y) = (x y) = x W y = Φ R (L)(x) W Φ R (L)(y) for all x, y L. Since L is divisible, we have x y = x (x y), hence, Φ R (L)(x y) = Φ R (L)(x) W Φ R (L)(y), so, Φ R (L) is a morphism of residuated lattices. If L, L are divisible residuated lattices and f : L L is a morphism of residuated lattices, then R(f) : MV (L) MV (L ) defined by R(f)(x ) = f(x ) = (f(x)) for every x L is a morphism in MV. Indeed, if x, y L, then R(f)(x y ) = R(f)((x y) ) = (f(x y)) = (f(x) f(y)) = f(x) f(y) = (R(f)(x )) (R(f)(y )) and R(f)((x ) ) = (f(x )) = (f(x) ) = (R(f)(x )). So, the assignments L MV (L) = R(L) and f R(f) define a functor (covariant) R : RL d MV from the category of divisible residuated lattices to the category of MV -algebras. Theorem 5. The category MV of MV -algebras is a reflective subcategory of the category RL d of divisible residuated lattices and the reflector R preserves monomorphisms. Proof. To prove R is a reflector ([1]), we consider the diagram L f L R(L) Φ R(L) R(f) R(L ) Φ R(L ) with L, L RL d. If x L, then (Φ R (L ) f)(x) = Φ R (L )(f(x)) = (f(x)) and (R(f) Φ R (L))(x) = R(f)(Φ R (L)(x)) = R(f)(x ) = (f(x )) = (f(x) ) = (f(x)), hence Φ R (L ) f = R(f) Φ R (L), that is, the above diagram is commutative. Let now L RL d, M MV and f : L M a morphism of residuated lattices R(L) L R(L) Φ f M f For x L, we define f : R(L) M by f (x ) = f(x ) = f(x) (that is, f = f MV(L) ). For x, y L, we have f (x y ) = f ((x y) ) = (f(x

13 13 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 299 y)) = (f(x) f(y)) = f(x) f(y), f((x ) ) = f(x) = f(x) = (f(x )) and f (0) = f (1 ) = f(1) = 1 = 0, hence f is a morphism in MV. Since for x L, (f Φ R (L))(x) = f (Φ R (L)(x)) = f (x ) = f(x) = f(x), we deduce that f Φ R (L) = f. If we have again f : R(L) M a morphism in MV such that f Φ R (L) = f, then for any x L, (f Φ R (L))(x ) = f(x ), hence f (x ) = f(x ) = f (x ), so f = f, so R is a reflector. To prove that R preserves monomorphisms, let f : L L a monomorphism in RL d and x, y L such that R(f)(x ) = R(f)(y ). Then f(x ) = f(y ), hence x = y, that is, R(f) is a monomorphism in MV. We recall that an MV -algebra is called complete if it contains the greatest lower bound and the lowest upper bound of any subset. Also, an MV - algebra L is called divisible if for any a L and for any natural number n 1 there is x L such that nx = a and a [(n 1)x] = x. In [23], p.66, it is proved: Theorem 6. For any MV -algebra L the following assertions are equivalent: (i) L is injective object in the category MV, (ii) L is complete and divisible MV -algebra. So, we obtain the following: Corollary 3. If L is a complete and divisible MV -algebra, then L is an injective object in the category RL d. Proof. By Theorem 6, L is an injective object in the category MV. Since MV is reflective subcategory of RL d and the reflector R : RL d MV preserves monomorphisms (by Theorem 5), then by Theorem 6 from [1] we deduce that L is injective object in the category RL d. Remark 6. Following Corollary 4.4 from [11] we deduce that if L is an injective object in the category RL d, then L is a complete lattice. Open problem. If L is an injective object in the category RL d, then L is divisible?

14 300 D. BUŞNEAG, D. PICIU and J. PARALESCU Boolean center, regular and dense elements Let (L,,, 0, 1) be a bounded lattice. Recall (see [1], [13]) that an element a L is called complemented if there is an element b L such that a b = 1 and a b = 0; if such element b exists it is called a complement of a. We will denote b = a and the set of all complemented elements in L by B(L). Complements are generally not unique, unless the lattice is distributive. In residuated lattices however, although the underlying lattices need not be distributive, the complements are unique. Lemma 2 ([12]). Let L be a residuated lattice and suppose that a L have a complement b L. Then, the following hold: (i) If c is another complement of a in L, then c = b; (ii) a = b and b = a; (iii) a 2 = a. Let L be a residuated lattice and B(L) the set of all complemented elements of the lattice (L,,, 0, 1). So, if e B(L), then e = e and e = e. Also, e x = e x, for every x L, see ([12]). The set B(L) is the universe of a Boolean subalgebra of L (called the Boolean center of L - see [12]). Proposition 6 ([5]). If L is a residuated lattice, then for e L the following are equivalent: (i) e B(L); (ii) e e = 1. Theorem 7 ([7]). For every element e in an MV -algebra A, the following conditions are equivalent: (i) e B(A); (ii) e e = 1; (iii) e e = 0; (iv) e e = e;

15 15 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 301 (v) e e = e. Proposition 7 ([5]). Let L be a residuated lattice. For e L we consider the following assertions: (i) e B(L); (ii) e 2 = e and e = e ; (iii) e 2 = e and e e = e; (iv) (e x) e = e, for every x L; (v) e e = 0. Then (i) (ii), (iii), (iv) and (v) but (ii) (i), (iii) (i), (iv) (i), (v) (i). Remark 7 ([4]). If L is a BL-algebra, then all assertions from above proposition are equivalent. Remark 8. If L is a residuated lattice, then B(L) MV (L). We put in evidence new rule of calculus with boolean elements: Lemma 3. Let x, y L and e, f B(L). Then: (c 42 ) x (x e) = x e, e (e x) = e x; (c 43 ) e (x y) = (e x) (e y); (c 44 ) e (x y) = (e x) (e y); (c 45 ) e (x y) = e [(e x) (e y)]; (c 46 ) x (e f) = x [(x e) (x f)]; (c 47 ) e (x y) = (e x) (e y); (c 48 ) e (e x) = e x; (c 49 ) (e x) e = e; (c 50 ) (e x) x (x e) e; (c 51 ) e x = (e x) x = e x;

16 302 D. BUŞNEAG, D. PICIU and J. PARALESCU 16 (c 52 ) e (x y) = (e x) (e y); (c 53 ) (e x) (f x) = (e f) x; (c 54 ) x (e f) = (x e) (x f); (c 55 ) e (x y) = (e x) (e y); (c 56 ) x (e f) = (x e) (x f); (c 57 ) If e, f x, then e (x f) = f (x e); (c 58 ) (e x) (x e) = 1; (c 59 ) e x = [(e x) x] [(x e) e]. Proof. For (c 42 ) (c 47 ) see [5]. (c 48 ). We have e (e x) = e 2 x = e x. (c 49 ). See [5]. (c 50 ). We have (e x) x (c 5) (x e) [(e x) e] = (x e) e. (c 51 ). From e e x and x e x e x e x. Also from e (e x) x (e x) (e x) and e (e x) e (e x) (e x), since e e = 1 (e x) (e x) = 1 e x e x e x = e x e x = e x. Since e, x (e x) x e x (e x) x. Since e e x (e x) x e x = e x (e x) x=e x. (c 52 ). By (c 51 ) we have e (x y) = e (x y) (c 47) = (e x) (e y) = (e x) (e y). (c 53 ). We have (e x) (f x) (c 51) = (e x) (f x) = f [(e x) x] (c 51) = f (e x) = f (e x) = (f e) x = (f e) x and (e f) x = (e f) x = (e f) x = (e f) x = (e f ) x = (e f ) x, so (e x) (f x) = (e f) x. (c 54 ). (x e) (x f) (c 14) = [x (x e)] f (c 42) = (x e) f = (x e) f (c 14) = x (e f). (c 55 ). We have e (x y) = e (x y) (c 21) = (e x) (e y) = (e x) (e y). (c 56 ). We have x (e f) = x (e f) (c 21) = (x e) (x f) = (x e) (x f). (c 57 ). We have e (x f) = (e x) (x f) = [x (x e)] (x f) = (x e) [x (x f)] = (x e) (x f) = f (x e);

17 17 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 303 (c 58 ). If e B(A) then e e = 1. Since e e x and e x e we deduce that 1 = e e (e x) (x e), so (e x) (x e) = 1. (c 59 ). Denote t = [(e x) x] [(x e) e]. We have e (e x) x and x (e x) x; it follows that e x (e x) x. Analogous, e x (x e) e. Hence e x [(e x) x] [(x e) e], so e x t. We have t = t 1 (c 58) = t [(e x) (x e)] (c 21) = [t (e x)] [t (x e)]; but t (e x) = [((e x) x) ((x e) e)] (e x) [(e x) x] (e x) (e x) x x; similarly, t (x e) e. Hence, t = [t (e x)] [t (x e)] x e. It follows that e x = t. Proposition 8. If L is a divisible residuated lattice, then the assertions (i), (ii), (iii) and (iv) from Proposition 7 are equivalent. Proof. (i) (ii). See Proposition 7. (ii) (iii). We have that e e = e (e 0) = (e e) 0 = e 0 = e. Hence, e e = e (e e ) = e e = 0. Since e e = e e = e (e e) = 0, we obtain that e e e = e. But, e e e so e e e. We have that e e = e. (iii) (i). Applying (c 59 ), e e = 1 (e e ) e = 1 and (e e) e = 1. By (iii), e e = e, hence (e e) e = 1. We also have that e e = e (e 0) = (e e) 0 = e 0 = e. So, (e e ) e = 1. We deduce that e e = 1, that is, e B(A), from Proposition 6. (i) (iv). See Proposition 7. (iv) (i). If x L, then from (e x) e = e we deduce (e x) [(e x) e] = (e x) e, hence (e x) e = e x. For x = 0 we obtain that e e = 0. Also, from hypothesis (for x = 0) we obtain e e = e. So, from (c 59 ) we obtain e e = [(e e ) e ] [(e e) e] = [(e e ) e ] (e e) = [(e e ) e ] 1 = (e e ) e (c 30) = [e (e e )] = (e e ) = 0 = 1, hence e B(L), from Proposition 6. Proposition 9. Let L be a residuated lattice. For e L we consider the following assertions: (i) e B(L);

18 304 D. BUŞNEAG, D. PICIU and J. PARALESCU 18 (vi) (e e ) (e e) = 1. Then (i) (vi) but (vi) (i). Proof. (i) (vi) If e B(L) from (c 58 ) for x = e we deduce that (e e ) (e e) = 1. (vi) (i). Consider the residuated lattice L = {0, a, b, c, 1} from the Example 4; it is easy to verify that B(L) = {0, 1}. We have (c c ) (c c) = (c 0) (0 c) = 1, but c / B(L). Definition 3. If L is a residuated lattice, we say that an element x L is regular if for every y L we have (x y) x = x. We denote by R(L) the set of all regular elements of L. We say that an element x L is dense if for every r R(L) we have x r = r. We denote by D(L) the set of all dense elements of L. We give a new characterization for regular elements: Theorem 8. Let L be a residuated lattice. assertions are equivalent: For x L the following (i) x R(L); (ii) x x = x. (iii) x = x and x (x x) = 0. Proof. (i) (ii). If x R(L), then (x y) x = x, so for y = 0 we obtain x x = x. (ii) (i). Suppose that x x = x and let y L. Then 0 y x 0 x y x x y (x y) x x x = x. Since x (x y) x we deduce that (x y) x = x, so x R(L). (ii) (iii). Let x L such that x x = x. Then x (x x) = x x = 0. To prove that x = x we use (c 29 ) and the relation x x x = x. (iii) (ii). Since x (x x) = 0 we deduce that x x x = x. But, x x x, so x x = x. From this theorem we obtain that: Corollary 4. If L is a residuated lattice, then R(L) MV (L). Corollary 5. If x, y R(L), then x R(L) and x y R(L).

19 19 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 305 Proof. Let x R(L); by Theorem 8, x = x and x x = x, so x R(L). From x y x x (x y) (x y) x x x = x (x y) x = x. Analogously we deduce that (x y) y = y. Then (x y) (x y) (c 7) = [(x y) x] [(x y) y] = x y x y R(L). A residuated lattice L will be called G- algebra if x 2 = x, for every x L. L is a G-algebra iff x (x y) = x y = x y for every x, y L (see [21]). Remark 9. If L is a G-algebra, then for every x L we deduce that x R(L). Indeed, x verifies the conditions (iii) from the Theorem 8 because x = (x ) and (x ) [(x ) x ] = x (x x ) = x x = 0. Lemma 4. Let L be a residuated lattice. If x L such that x x = x then x x = x. Proof. x x = x (x 0) = (x x) 0 = x 0 = x. We give a new characterization for boolean elements: Theorem 9. Let L be a residuated lattice. assertions are equivalent: For x L the following (i) x B(L); (ii) x R(L), x x = x and (x x ) (x x) = 1. Proof. (i) (ii). If x B(L), then by Proposition 7, x x = x, x = x and x x = 0. Since x (x x) x x = 0 x (x x) = 0, so x R(L) (by Theorem 8, (iii) (i)). By Lemma 4 and Proposition 7, (iii), since x B(L) we deduce that (x x ) (x x) = x x = 1. (ii) (i). Since x x = x, by Lemma 4, x x = x. Since x R(L), then by Theorem 8, (ii), x x = x. Then x x = (x x) (x x ) = 1. Using Proposition 6, we conclude that x B(L). Corollary 6. Let L be a MT L algebra. assertions are equivalent: For x L the following (i) x B(L); (ii) x R(L) and x x = x.

20 306 D. BUŞNEAG, D. PICIU and J. PARALESCU 20 Corollary 7. In general, for a residuated lattice L, B(L) R(L). Proof. By Theorem 9 we deduce that B(L) R(L). To prove that B(L) R(L) we consider the residuated lattice L = {0, a, b, c, 1} from Example 4; it is easy to verify that B(L) = {0, 1}. We have a = b, b = a, hence a = b = a and a (a a) = b (b a) = b a = 0, hence a R(L) but a / B(L). Corollary 8. If L is a semi-divisible residuated lattice, then B(L) B(MV (L)) R(L). Proof. Since L is semi-divisible then MV (L) is an MV -algebra. Let e B(L). Then by Proposition 7, e = e = (e ) MV (L) and e e = 0. Since e e = 0 e W e = 0 e B(MV (L)), by Theorem 7. We deduce that B(L) B(MV (L)). Consider now e B(MV (L)). Then e = e and e W e = 1. Hence e W e = 1 e W e = 1 (e e ) e = 1 (e e) e = 1 e e = e, hence by Theorem 8, (ii) (i), we deduce that e R(L), so B(L) B(MV (L)) R(L). Corollary 9. If L is a MT L-algebra and x x = x for every x L, then B(L) = R(L). Corollary 10. If L is a semi-divisible M T L-algebra, then B(L) = B(MV (L)). Proof. By Corollary 8, B(L) B(MV (L)) for any semi-divisible residuated lattice L. Consider now that L is a semi-divisible MT L-algebra and let e B(MV (L)). Then e W e = 1 e W e = 1 (e e ) e = 1 (e e ) e = 1 e e = e. Also e W e = 1 e W e = 1 (e e ) e = 1 (e e) e = 1 e e = e. Then e e = (e e ) (e e) = 1, since L is a MT L-algebra, so e B(L) and B(MV (L)) B(L). We deduce that B(L) = B(MV (L)). Corollary 11. If L is a G semi-divisible MT L-algebra, then B(L) = B(MV (L)) = R(L). Remark We consider the residuated lattice from Remark 3, (2) which is not semi-divisible. Let c = 1 2. We have 0 = 1, x = 1 2 x, if x (0, 1 2 ), x = 0, if x [ 1 2, 1]. Then, 0 = 0, x = x, if x (0, 1 2 ), x = 1,

21 21 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 307 if x [ 1 2, 1]. Also, 0 0 = 0, x x = x, if x (0, 1 4 ), x x = 1, if x [ 1 4, 1]. We conclude that MV (L) = {x L : x = x} = [0, 1 2 ) {1} (but MV (L) is not an MV-algebra), R(L) = {x L : x x = x} = [0, 1 4 ) {1} and B(L) = {x L : x x = 1} = {0, 1}. So, B(L) R(L) MV (L). 2. If consider the residuated lattice from Remark 3, (3) which is semidivisible we have 0 = 1, x = 1 2, if x (0, 1 2 ], x = 0, if x ( 1 2, 1]. Then, 0 = 0, x = 1 2, if x (0, 1 2 ], x = 1, if x ( 1 2, 1]. Also, 0 0 = 0, x x = 1 2, if x (0, 1 2 ), x x = 1, if x [ 1 2, 1]. We conclude that MV (L) = {x L : x = x} = {0, 1 2, 1} (in this case MV (L) is an MValgebra), R(L) = {x L : x x = x} = {0, 1} and B(L) = {x L : x x = 1} = {0, 1}. Since B(MV (L)) = {x MV (L) : x x = 0} = {0, 1} we have B(L) = R(L) MV (L) and B(L) = B(MV (L)) (since L is semidivisible, see Corollary 8). We characterize the residuated lattices which are Boolean algebras: Theorem 10. For a residuated lattice L the following assertions are equivalent: (i) L is a Boolean algebra relative to the natural ordering; (ii) L is a G - algebra and x = x, for every x L. Proof. (i) (ii). If L is a Boolean algebra relative to the natural ordering, then L becomes a residuated lattice (see Example 3) and x x = x x = x, x = x, for every x L. (ii) (i). Let L be a G - algebra such that x = x, for every x L. Then x y = x y for every x, y L. First we shall prove that x y = x y for every x, y L. Indeed, x, y x y. Let t L such that x, y t. From x t we deduce that t x, hence x y t y t t = t t = (t t ) 0 = t 0 = t = t. Following Proposition 6, to prove that L is a Boolean algebra it will suffice to prove that x x = 1, for every x L. Obviously, x x = x x = 1. Theorem 11. Let L be a residuated lattice. For x L the following assertions are equivalent: (i) x D(L); (ii) x = 0.

22 308 D. BUŞNEAG, D. PICIU and J. PARALESCU 22 Proof. (i) (ii). Since (0 y) 0 = 1 0 = 0 for every y L, we deduce that 0 R(L). Let x D(L); since 0 R(L), we obtain x 0 = 0, hence x = 0. (ii) (i). Let now x L such that x = 0 and r R(L) (hence r = r, by Theorem 8). Then x r = x r = x (r 0) (c 14) = r (x 0) = r x = r 0 = r = r, hence x D(L). Proposition 10. If L is a G-algebra, then: (i) For every x L, x x D(L); (ii) x D(L) iff x = y y for some y L; (iii) For every x L, x (x x) = x and (x x) x = x. Proof. (i). Since x, x x x, we deduce that (x x) x x x, hence (x x) x x so (x x) (x x) (x x) (x x) (x x) 0 (x x) = 0 x x D(L). (ii). By (i), if x = y y for some y L, we deduce that x D(L). Conversely, let x D(L). Then for y = x we obtain x x = 0 x = 1 x = x. (iii). First we prove that x (x x) = x. Clearly, x x, x x. Let t A such that t x, x x. We deduce that x t x, hence t x t x, so t t = t x, that is, x (x x) = x. We prove now that (x x) x = x. Since x (x x) = x and L is a G- algebra we deduce that x (x x) = x, so x (x x) x. Hence x (x x) x. Since x x (x x) x (x x) x (c 32) = x (x x). From (i), x x D(L) for every x L, so (x x) x x 0 = x. We deduce that (x x) x = x. Corollary 12. If L is a G-algebra, then for every x L there are y R(L) and z D(L) such that x = y z. Proof. We have y = x R(L) and z = x x D(L). 5. Semi-G-algebras Definition 4 ([12], [23]). A non empty subset D L is called a deductive system of L, ds for short, if the following conditions are satisfied:

23 23 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 309 (Ds 1 ) 1 D; (Ds 2 ) If x, x y D, then y D. A ds D of L is called proper if D L. Remark 11 ([12], [23]). A nonempty subset D L is a ds of L iff: (Ds 1 ) If x, y D, then x y D; (Ds 2 ) If x D, y L, x y, then y D. We denote by Ds(L) the set of all deductive systems of L. For a nonempty subset S L, the smallest ds of L which contains S, i.e. {D Ds(L) : S D}, is said to be the ds of L generated by S and will be denoted by < S >. For D 1, D 2 Ds(L) we put D 1 D 2 = D 1 D 2 and D 1 D 2 =< D 1 D 2 >. The lattice (Ds(L), ) is a complete Brouwerian lattice (hence distributive). We say ([21]) that D Ds(L) is prime if D L and D verify one of the equivalent assertions: (i) If D = D 1 D 2 with D 1, D 2 Ds(L), then D = D 1 or D = D 2 ; (ii) For a, b L, if a b D, then a D or b D. We denote by Spec(L) the set of all prime deductive systems of L. A ds of L is maximal if it is proper and it is not contained in any other proper ds. In a nontrivial residuated lattice L, every proper ds can be extended to a maximal ds. We shall denote by Max(L) the set of all maximal ds of L. Obviously, Max(L) Spec(L). In what follow, we present new results relative to lattice of maximal deductive systems of a residuated lattice. We prove that if L is semi-divisible, then there is a bijective correspondence between maximal deductive systems of L and maximal deductive systems of MV (L). Also, we characterize Rad(M V (L)), for a semi-divisible residuated lattice L. Theorem 12 ([5]). Let L be a residuated lattice and M a proper ds of L. Then the following are equivalent: (i) M Max(L),

24 310 D. BUŞNEAG, D. PICIU and J. PARALESCU 24 (ii) For any x L, x / M iff (x n ) M, for some n 1. Corollary 13. Let L be a residuated lattice. If M M ax(l) and x, y L such that (x n ) y M, for any n 1, then x M or y M. Proof. Let x, y L such that (x n ) y M for any n 1 and suppose that x / M and y / M. From Theorem 12 we deduce that there is n 0 1 such that (x n 0 ) M. Since (x n 0 ), (x n 0 ) y M y M, a contradiction. Theorem 13. Let L be a residuated lattice. If M is a proper ds of L, then the following are equivalent: (i) M Max(L), (ii) If x / M, then there is n 1 such that x n y M, for every y L. Proof. (i) (ii). If we suppose that x / M then by Theorem 12 there is n 1 such that (x n ) M. Because (x n ) x n y for every y L we deduce that x n y M, for every y L. (ii) (i). Let D Ds(L) proper such that M D, hence there is x 0 D such that x 0 / M. Then there is n 1 such that x n 0 y M, for every y L. Hence x n 0 y D. Since D Ds(L) and x 0 D we deduce that x n 0 D. From xn 0, xn 0 y D y D, for every y L. Hence D = L, a contradiction. Proposition 11. Let M Max(L) and x L. Then x M iff x M. Proof. If x M since x x, then x M. Conversely, let x L such that x M and suppose by contrary that x / M. Since M Max(L), then there is n 1 such that (x n ) M. Following (c 35 ), (x ) n (x n ), so (x n ) M. From (x n ), (x n ) M 0 M M = L, a contradiction. Remark 12. In a residuated lattice L, the following are equivalent: (i) x x = x for every x L; (ii) (x 2 ) = x for every x L; (iii) (x n ) = x for every x L and n 2;

25 25 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 311 (iv) x (x x ) = 0 for every x L. Definition 5. L is called semi G-algebra if it verifies one of equivalent conditions of Remark 12. We recall that a residuated lattice L is called a G-algebra if x 2 = x for every x L. We will show that not every residuated lattice is semi G-algebra and, moreover, there are residuated lattices that are semi G-algebras but not G-algebras. Indeed, consider L = [0, 1] such that (L, max, min,,, 0, 1) is a Product structure or Gaines structure, see Example 2. For x = 0, (0 2 ) = 0 = 1 = 0 and for x > 0 x = x 0 = 0/x = 0. Since x 2 > 0 (x 2 ) = x 2 0 = 0/x 2 = 0, so (x 2 ) = x = 0, hence L is a semi ) 2 = G-algebra. Since ( 1 2, L is not a G-algebra. Consider L from Example 1 for p = 1. Then ( 1 ) 2 = = max{0, } = min{1, 1 2 } = 1 2 and ( 1 [ ( 2 1 ) ] 2 2 = = ( 1 2), hence L is not a semi G-algebra. 2 ) = = min{1, 1 1} = 0, so Lemma 5. If L is a semi G-algebra and x, y R(L), then x y R(L). In particular x R(L) x R(L). Proof. If x, y R(L), following Theorem 8, x = x and y = y. By (c 34 ), (x y) x y = x y, so (x y) = x y. By (c 15 ), (x y) (x y) [(x y) (x y) ] [(x y) (x y) ] = [(x y) ] 2 0 = ([(x y) ] 2 ) = [(x y) ] = (x y) = x y, hence (x y) (x y) x y (x y) (x y) = x y x y R(L). For D Ds(L), we consider φ D : L {0, 1} defined for x L by φ D (x) = 1 if x D and φ D (x) = 0 if x / D. Lemma 6. Let D Ds(L) proper. Then: (i) φ D (0) = 0, φ D (1) = 1, φ D (x y) = φ D (x) φ D (y) and φ D (x y) = φ D (x) φ D (y), for every x, y L; (ii) If D Spec(L), then φ D (x y) = φ D (x) φ D (y), for every x, y L; (iii) If L is a semi G-algebra and D Max(L) then φ D (x y) = φ D (x) φ D (y), for every x, y L.

26 312 D. BUŞNEAG, D. PICIU and J. PARALESCU 26 Proof. (i). Since 0 / D and 1 D, then φ D (0) = 0 and φ D (1) = 1. Consider x, y L. If x y D, then x, y D, so φ D (x y) = 1 = 1 1 = φ D (x) φ D (y). If x y / D, then we deduce that x / D or y / D, so φ D (x y) = φ D (x) φ D (y) = 0. If x y D then x, y D, so φ D (x y) = φ D (x) φ D (y) = 1. If x y / D, then x / D or y / D (since if by contrary x D and y D x y D x y D ), so φ D (x y) = φ D (x) φ D (y) = 0. (ii). If x y D, since D is prime, then x D or y D, so φ D (x y) = φ D (x) φ D (y) = 1. If x y / D, then x / D and y / D, so φ D (x y) = φ D (x) φ D (y) = 0. (iii). Consider x, y L such that x y D. If x D, then y D, so φ D (x y) = φ D (x) φ D (y) = 1. If x / D and y D, then φ D (x) φ D (y) = 0 1 = 1 φ D (x y) = φ D (x) φ D (y) = 1. If x / D and y / D, then φ D (x) φ D (y) = 0 0 = 1 φ D (x y) = φ D (x) φ D (y) = 1. If x y / D, then y / D. To prove the equality φ D (x y) = φ D (x) φ D (y) it is necessary to prove that φ D (x) = 1, that is x D. If by contrary x / D, since D Max(L), then there is n 1 such that (x n ) D. Since L is supposed to be a semi G-algebra, then (x n ) = x, so x D. Since x x y, then x y D, a contradiction. Corollary 14. If L is a semi G-algebra and M Max(L), then φ M : L {0, 1} is a morphism of residuated lattices and Ker(f M ) = M. Proof. Using Lemma 6, φ M is a morphism of residuated lattices. Obviously, Ker(φ M ) = M, because Ker(φ M ) = {x L : φ M (x) = 1} = {x L : x M} = M. Theorem 14. If L is a semi G-algebra, then there is a bijection between Max(L) and RL(L, {0, 1}) = { f : L {0, 1} f is a morphism of residuated lattices}. Proof. We define α : Max(L) RL(L, {0, 1}) by α(m) = f M, for every M Max(L) and β : RL(L, {0, 1}) Max(L) by β(f) = Ker(f), for every f RL(L, {0, 1}). Ker(f) Max(L) because if x / Ker(f) then f(x) = 0. Since f(x ) = (f(x)) = 0 = 1, we deduce that x Ker(f), so Ker(f) Max(L). For M Max(L), (β α)(m) = β(α(m)) = Ker(f M ) = M, from Corollary 14,

27 27 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 313 that is, β α = 1 Max(L). If f RL(L, {0, 1}), then (α β)(f) = α(β(f)) = f Ker(f). We prove that f Ker(f) = f. If x Ker(f), then f(x) = 1, so f Ker(f) (x) = 1 and if x / Ker(f), then f(x) = 0, so f Ker(f) (x) = 0. We deduce that α β = 1 RL(L,{0,1}), so α, β are bijections. We recall that an MV -algebra is an algebra (M,,, 0) of type (2, 1, 0) such that (M,, 0) is a commutative monoid, x = x for every x M and (x y) y = (y x) x, for every x, y M. We denote 0 = 1. For x, y M we denote x y = (x y ) and x y = x y. Then ([23]), (M,,,,, 0, 1) is a BL-algebra, where for x, y M, x y = (x y) y and x y = (x y ). Conversely, a BL-algebra (L,,,,, 0, 1) is an MV -algebra iff x = x for every x L, where x = x 0. Lemma 7. Let L be a residuated lattice. If D Ds(L), then D MV (L) = {y MV (L) : y = x for some x D}. Proof. If y MV (L) such that y = x with x D, since x x, then y D, hence y D MV (L). Conversely, if y D MV (L), then y D and y = x (with x L). We have y = x = x = y. Suppose L is a semi-divisible residuated lattice. Then M V (L) is an MV algebra (by Corollary 2), hence a BL algebra and by Ds(MV (L)) we denote the set of all deductive systems of MV (L). We recall that for x, y MV (L), x y = x y (so x y = x y). Lemma 8. If L is a semi-divisible residuated lattice and D Ds(L), then D MV (L) Ds(MV (L)). Proof. Clearly 1 D MV (L). Let x, y MV (L) such that x, x y D MV (L). Since D Ds(L) and x, x y D, then y D, hence y D MV (L), so D MV (L) Ds(MV (L)). Lemma 9. If L is a semi-divisible residuated lattice and M Max(L), then M MV (L) Max(MV (L)). Proof. We have M L. If M MV (L) = MV (L), then MV (L) M 0 M M = L, a contradiction. So, M MV (L) is proper in MV (L). To prove M MV (L) is maximal in MV (L), suppose by contrary that there is N Ds(MV (L)) such that M MV (L) N MV (L). Then there is x 0 N such that x 0 / M MV (L). Then x 0 / M hence there is

28 314 D. BUŞNEAG, D. PICIU and J. PARALESCU 28 n 1 such that (x n 0 ) M (x n 0 ) M MV (L) N (x n 0 ) N. Since x n 0 N 0 N N = MV (L), so M MV (L) Max(MV (L)). Let L be a semi-divisible residuated lattice. For D Ds(MV (L)) we denote by D = {x L : x D} (since x x we deduce that D D). Lemma 10. Let L be a semi-divisible residuated lattice. (i) If D Ds(MV (L)), then D Ds(L); (ii) If M Max(MV (L)), then M Max(L). Proof. (i). Since 1 D and 1 = 1 1 D. If x, y L, x y and x D then x D and from x y y D y D. If x, y D, then x, y D x y D MV (L) x y = (x y ) = [x (y 0)] = (x y ) = (x y ) = (y x ) = (y x ) = (x y ) = [(x y) ] = (x y) x y D. (ii). Suppose M Max(MV (L)). Then M MV (L). If M = L, then 0 M 0 = 0 M M = MV (L), a contradiction. Hence M is proper in L. To prove M Max(L), consider N Ds(L) such that M N L. Then there is x 0 N such that x 0 / M; hence x 0 / M. Since M Max(MV (L)), there is n 1 such that [(x 0 )n ] M M [(x 0 )n ] M N [(x 0 )n ] N. But x 0 N x 0 N (x 0 )n N 0 N N = L. Lemma 11. Let L be a semi-divisible residuated lattice. Then: (i) If M Max(L), then M = M MV (L); (ii) If N Max(MV (L)), then N = N MV (L). Proof. (i). If M Max(L), by Lemma 9, M MV (L) Max(MV (L)) and from Lemma 10, (ii), we deduce that M MV (L) Max(L). Since for x M x M MV (L), then x M MV (L) M M MV (L) M = M MV (L). (ii). Since N Max(MV (L)), by Lemma 10, (ii), N Max(L), so by Lemma 9, N MV (L) Max(MV (L)). If y N MV (L), then y = x for some x N x N N MV (L) N N MV (L) = N.

29 29 DIVISIBLE AND SEMI-DIVISIBLE RESIDUATED LATTICES 315 Theorem 15. If L is a semi-divisible residuated lattice, then there is a bijection between Max(L) and Max(MV (L)). Proof. Following Lemma 9 and Lemma 10 we deduce that we can define f : Max(L) Max(MV (L)), by f(m) = M MV (L), for every M Max(L) and g : Max(MV (L)) Max(L), by g(n) = N, for every N Max(MV (L)). By Lemma 11 we deduce that f g = 1 Max(MV (L)) and g f = 1 Max(L), hence we deduce that f is bijective and f 1 = g. Definition 6 ([23]). A residuated lattice L is called semilocal (local) if it contains only a finite number (only one) of maximal deductive systems. Clearly, if L is a chain, then L is local. Following Theorem 15 we have: Theorem 16. If L is a semi-divisible residuated lattice, then L is a semilocal residuated lattice iff MV (L) is a semilocal MV -algebra. 6. The radical of a semi-divisible residuated lattice Definition 7 ([12]). The intersection of the maximal deductive systems of a residuated lattice L is called the radical of L and will be denoted by Rad(L). For n 1 and x L we denote ñx = [(x ) n ]. Following [11], [12], Rad(L) = {x L : for every n 1 there is k n 1 such that k n (x n ) = 1} = {x L : for every n 1 there is k n 1 such that [(x n ) ] kn = 0}. If L is a BL-algebra, then Rad(L) = {x L : (x n ) x, for every n 1} (see [23]). Proposition 12. If L is a chain, then Rad(L) = {x L : x n > x, for every n 1}. Proof. If L is a chain, then there is an unique maximal deductive system M of L, so Rad(L) = M. If x M then x n M, for every n 1, hence x n > (x n ) (because if by contrary x n (x n ) (x n ) M 0 M, a contradiction). Since x n < x (x n ) > x, so x n > x. Conversely, let x L such that x n > x, for every n 1. Then x n 0 for every n 1 < x > is proper x < x > M = Rad(L). Now, we characterize Rad(MV (L)), for a semi-divisible residuated lattice L.

30 316 D. BUŞNEAG, D. PICIU and J. PARALESCU 30 Theorem 17. If L is a semi-divisible residuated lattice, then Rad(MV (L)) Rad(L) MV (L). Proof. Clearly, Rad(MV (L)) MV (L). We prove now that Rad(MV (L)) Rad(L). Suppose that there is an element x Rad(MV (L)) such that x / Rad(L). Then there is M Max(L) such that x / M. Thus, there is n 1 such that [(x ) n ] M. Because [(x ) n ] = ([(x ) n ] ) we deduce that [(x ) n ] M MV (L), which is maximal in MV (L), see Lemma 9. Since x Rad(MV (L)) M MV (L) we obtain that (x ) n M MV (L), so 0 = (x ) n [(x ) n ] M MV (L), a contradiction because M MV (L) is proper. So, Rad(MV (L)) Rad(L) MV (L). Theorem 18. If L is a semi-divisible residuated lattice and x Rad(L), then x Rad(MV (L)). Proof. Let x Rad(L) and suppose that x / Rad(MV (L)). Then there is M Max(MV (L)) such that x / M. Thus, there is n 1 such that [(x ) n ] M. Because [(x ) n ] = ([(x ) n ] ) we deduce that [(x ) n ] M, which is maximal in L. Since x Rad(L) M we obtain that (x ) n M, so 0 = (x ) n [(x ) n ] M, a contradiction because M is proper. So, if x Rad(L), then x Rad(MV (L)). Theorem 19. Let L be a semi-divisible residuated lattice and x L. Then, x Rad(L) x Rad(MV (L)). Proof. Let x Rad(L). From Theorem 18, (x ) = x Rad(MV (L)). Conversely, let x Rad(MV (L)). By Theorem 17, we deduce that x Rad(L). Corollary 15. Rad(L) MV (L) Rad(MV (L)). From Theorem 17 and Corollary 15 we deduce that: Corollary 16. For a semi-divisible residuated lattice L, Rad(M V (L)) = Rad(L) MV (L).

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