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ANALELE STIINTIFICE ALE UNIVERSITATII “AL.I. CUZA” DIN IASI (S.N.)MATEMATICA, Tomul LXI, 2015, f.2

DIVISIBLE AND SEMI-DIVISIBLE RESIDUATEDLATTICES

BY

D. BUSNEAG, D. PICIU and J. PARALESCU

Abstract. The main purpose of this paper is to present new results in divisible andsemi-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, MTL-algebra, BL-algebra, MV -algebra, regular element, dense element, re-flective subcategory, deductive system, maximal deductive system, radical.

1. Introduction

The origin of residuated lattices is in Mathematical Logic without con-traction. They have been investigated by Krull [18], Dilworth [9],Ward and Dilworth [26], Ward [25], Balbes and Dwinger [1] andPavelka [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], FLew- algebras in [19], and integral, residuated,commutative l-monoids in [3].

Apart from their logical interest, residuated lattices have interestingalgebraic properties (see [2], [5], [9], [12], [25], [26]).

MV-algebras are known to be special residuated lattices with a properadditive 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

288 D. BUSNEAG, D. PICIU and J. PARALESCU 2

x∗ = x → 0 with x ∈ L. (MV (L),⊕,∗ , 0) is an MV− algebra iff for allx, 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 residuatedlattice 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. Alsowe present MV -center of a semi-divisible residuated lattice, defined by Tu-runen and Mertanen in [22]. This is a very important construction, whichassociates an MV -algebra to every semi-divisible residuated lattice. In thisway, many properties can be transferred from MV -algebras to residuatedlattices and backwards.

In Section 3 we prove that the category MV of MV -algebras is a re-flective subcategory of the category RLd of divisible residuated lattices andas consequence we obtain some informations relative to injective divisibleresiduated lattices (Corollary 3).

In Section 4, for a residuated lattice L we denote by R(L) the set ofregular 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 theseelements. We prove that in general, B(L) ⊆ R(L) ⊆ MV (L). If L is a semi-divisible MTL-algebra, then B(L) = B(MV (L)).

Theorem 10 characterize the residuated lattices which are Boolean al-gebras.

In Section 5, we present new results relative to lattice of maximal de-ductive systems of a residuated lattice. We introduce the notion of semi-G-algebra and we obtain that if L is a semi-G-algebra, then there is a bijectionbetween Max(L) and RL(L, {0, 1}) = { f : L → {0, 1} | f is a morphism ofresiduated lattices} (Theorem 14). We prove that if L is semi-divisible, thenthere is a bijective correspondence between maximal deductive systems ofL and maximal deductive systems of MV (L).

In Section 6, we characterize Rad(MV (L)), for a semi-divisible residu-ated lattice L.

2. Definitions and preliminaries

We review the basic definitions of residuated lattices. Also we presentMV -center of a semi-divisible residuated lattice, defined by Turunen andMertanen in [22].

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 follow-ing:

(LR1) (L,∨,∧, 0, 1) is a bounded lattice;

(LR2) (L,⊙, 1) is a commutative ordered monoid;

(LR3) ⊙ and → form an adjoint pair, i.e. a ≤ x → y iff a ⊙ x ≤ y for alla, x, y ∈ L.

The relations between the pair of operations ⊙ and → expressed by(LR3), is a particular case of the law of residuation ([2]). The class RL ofresiduated lattices is equational (see [16]).

Example 1 ([23]). Let p be a fixed natural number and I = [0, 1] thereal unit interval. We define for x, y ∈ I, x⊙ y = (max{0, xp + yp − 1})1/p

and x → y = min{1, (1−xp+yp)1/p}, then (I,max,min,⊙,→, 0, 1) becomea residuated lattice called generalized Lukasiewicz structure. For p = 1 weobtain 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 multi-plication of real numbers and for x, y ∈ I, x → y = 1 if x ≤ y and y/xotherwise, then (I,max,min,⊙,→, 0, 1) is a residuated lattice (called Pro-ducts structure or Gaines structure).

Example 3 ([23]). If (B,∨,∧,′ , 0, 1) is a Boolean algebra, then if we de-fine 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:

(BL1) x⊙ (x → y) = x ∧ y (divisibility);

(BL2) (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 (BL1);


(ii) MTL-algebra if L verify (BL2);

(iii) BL-algebra if L verify (BL1) and (BL2) (that is, L is a divisibleMTL− algebra);

(iv) semi-divisible if L verify (BL∗1).

We denote by RL ( RLd,MT L, BL, RLsd) the class of residuatedlattices (divisible residuated lattices, MTL-algebras, BL-algebras, semi-divisible residuated lattices).

Proposition 1 ([22]). For a residuated lattice L, the following condi-tions are equivalent:

(i) L ∈ RLd;

(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 an MV -algebra is an algebra (M,⊕,∗ , 0) oftype (2, 1, 0) such that:

(MV1) (M,⊕, 0) is a commutative monoid;

(MV2) x∗∗ = x, for every x ∈ M ;

(MV3) (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 presen-tation of MV -algebras can be given as BL-algebras plus condition (MV2).

2. Let (L,∨,∧,⊙,→, 0, 1) be a residuated lattice. If for x, y ∈ L wedenote 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− al-gebras; not every residuated lattice is a BL-algebra (see [23], p.16).


Example 4 ([17]). We give an another example of a finite residuatedlattice, 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 tothe following operations:

→ 0 a b c 1

0 1 1 1 1 1a b 1 b 1 1b a a 1 1 1c 0 a b 1 11 0 a b c 1

,

⊙ 0 a b c 1

0 0 0 0 0 0a 0 a 0 a ab 0 0 b b bc 0 a b c c1 0 a b c 1

.

In what follows by L we denote the universe of a residuated lattice; forx ∈ L and a natural number n, we define x∗ = x → 0, x∗∗ = (x∗)∗, x0 = 1and xn = xn−1 ⊙ x for n ≥ 1.

In residuated lattices we have the following rules of calculus:

Theorem 1 ([5], [6], [8], [12], [23]). Let x, x1, x2, y, y1, y2, z ∈ L. Thenwe have:

(c1) 1 → x = x, x → 1 = 1, 0 → x = 1, x⊙ 0 = 0;

(c2) x ≤ y iff x → y = 1;

(c3) x ≤ y → x, x ≤ (x → y) → y, ((x → y) → y) → y = x → y;

(c4) x → y ≤ (z → x) → (z → y);

(c5) x → y ≤ (y → z) → (x → z);

(c6) (x → y) → (x → z) ≤ x → (y → z);

(c7) x → (y ∧ z) = (x → y) ∧ (x → z), (y ∧ z) → x ≥ (y → x) ∨ (z → x);

(c8) (x ∨ y) → z = (x → z) ∧ (y → z);

(c9) x ≤ y implies z → x ≤ z → y and y → z ≤ x → z;

(c10) x1 → y1 ≤ (y2 → x2) → [(y1 → y2) → (x1 → x2)];

(c11) x ≤ y implies z ⊙ x ≤ z ⊙ y;

(c12) x⊙ (x → y) ≤ x ∧ y, x ≤ y → (x⊙ y);


(c13) x⊙ (x → (x⊙ y)) = x⊙ y;

(c14) x → (y → z) = (x⊙ y) → z = y → (x → z);

(c15) x → y ≤ (x⊙ z) → (y ⊙ z) and (x1 → y1) ⊙ (x2 → y2) ≤ (x1 ⊙ x2) →(y1 ⊙ y2);

(c16) (x1 → y1) ⊙ (x2 → y2) ≤ (x1 ∨ x2) → (y1 ∨ y2);

(c17) (x1 → y1) ⊙ (x2 → y2) ≤ (x1 ∧ x2) → (y1 ∧ y2);

(c18) x ∨ y = 1 ⇒ x⊙ y = x ∧ y;

(c19) x∨ (y⊙ z) ≥ (x∨y)⊙ (x∨ z), so, xm∨yn ≥ (x∨y)mn for any naturalnumbers m,n;

(c20) x⊙ (y → z) ≤ y → (x⊙ z) ≤ (x⊙ y) → (x⊙ z);

(c21) x⊙ (y ∨ z) = (x⊙ y) ∨ (x⊙ z);

(c22) (x → y)n ≤ xn → yn for every n ≥ 1;

(c23) (x ∧ y)n ≤ xn ∧ yn, for every n ≥ 1;

(c24) (x ∨ y)n ≥ xn ∨ yn, for every n ≥ 1.

Theorem 2 ([5], [6], [8], [12], [23]). If x, y ∈ L, then:

(c25) 1∗ = 0, 0∗ = 1;

(c26) x → y ≤ y∗ → x∗;

(c27) x ≤ y ⇒ y∗ ≤ x∗;

(c28) x⊙ x∗ = 0 and x⊙ y = 0 iff x ≤ y∗;

(c29) x ≤ x∗∗, x∗∗ ≤ x∗ → x, x∗∗∗ = x∗;

(c30) (x⊙ y)∗ = x → y∗ = y → x∗;

(c31) (x ∨ y)∗ = x∗ ∧ y∗;

(c32) x∗∗ → y∗∗ = y∗ → x∗ = x → y∗∗;

(c33) (x → y∗∗)∗∗ = x → y∗∗;


(c34) (x → y)∗∗ ≤ x∗∗ → y∗∗;

(c35) x∗∗ ⊙ y∗∗ ≤ (x⊙ y)∗∗, so (x∗∗)n ≤ (xn)∗∗ for every natural number n;

(c36) x∗ ⊙ y∗ ≤ (x⊙ y)∗.

Corollary 1. If L is a divisible residuated lattice, then for every x, y ∈ Lwe have:

(c37) (x∗∗ → x)∗ = 0;

(c38) (x → y)∗∗ = x∗∗ → y∗∗;

(c39) (x⊙ y)∗∗ = x∗∗ ⊙ (x∗∗ ∧ y∗)∗, (x ∧ y)∗∗ = x∗∗ ∧ y∗∗;

(c40) y∗ ≤ x ⇒ x → (x⊙ y)∗∗ = y∗∗;

(c41) x⊙ (y ∧ z) = (x⊙ y) ∧ (x⊙ z), x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).

Proof. (c37). From x∗ ≤ x∗∗ → x ⇒ (x∗∗ → x)∗ ≤ x∗∗ ⇒ (x∗∗ → x)∗ =

x∗∗ ∧ (x∗∗ → x)∗(BL1)

= x∗∗ ⊙ [x∗∗ → (x∗∗ → x)∗](c30)= x∗∗ ⊙ [(x∗∗ → x) →

(x∗∗)∗](c30)= x∗∗ ⊙ [(x∗∗ → x) ⊙ x∗∗]∗

(BL1)= x∗∗ ⊙ (x∗∗ ∧ x)∗ = x∗∗ ⊙ x∗ = 0.

(c38). From (c34) we have (x → y)∗∗ ≤ x∗∗ → y∗∗ = x → y∗∗. Onthe another hand we have (x → y∗∗) → (x → y)∗∗ = (x → y∗∗)∗∗ →

(x → y)∗∗(c34)

≥ [(x → y∗∗) → (x → y)]∗∗ = [(x ⊙ (x → y∗∗)) → y]∗∗(BL1)

=

[(x ∧ y∗∗) → y]∗∗ ≥ [(x → y) ∨ (y∗∗ → y)]∗∗ ≥ (y∗∗ → y)∗∗(c37)= 0∗ = 1 ⇒

(x → y∗∗) → (x → y)∗∗ = 1 ⇒ x → y∗∗ ≤ (x → y)∗∗ ⇒ (x → y)∗∗ = x →y∗∗ = x∗∗ → y∗∗.

(c39). 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∗∗.

(c40). From y∗ ≤ x ⇒ x → (x ⊙ y)∗∗ = (x ⊙ y)∗ → x∗ = (x → y∗) →x∗ = [x⊙ (x → y∗)]∗ = (x ∧ y∗)∗ = y∗∗.

(c41). Clearly x⊙ (y ∧ z) ≤ (x⊙ y) ∧ (x⊙ z).


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](c21)= [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(c30), x

∗ → y∗ = (x∗ ⊙ y)∗ ∈ MV (L)), ∧ (by c31) and in particular, withrespect 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 aresiduated lattice L, then L is semi-divisible.

Corollary 2 ([22]). In any semi-divisible residuated lattice L, (MV (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 algebraif it satisfies the following conditions:

(W1) 1 → x = x;

(W2) (x → y) → ((y → z) → (x → z)) = 1;

(W3) (x → y) → y = (y → x) → x;

(W4) (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


(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 iffx → y = 1. Following Theorems 3, 4 and Corollary 2, if L is a semi-divisibleresiduated 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 ofMV− algebra (hence, Wajsberg algebra) and the order relation ≤W on theWajsberg 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 relationon 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, forjoin 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:

(c42) [x∗ ⊙ (x∗ → y∗)]∗ = (y∗∗ → x∗∗) → x∗∗;

(c43) x∗∗ ∨W y∗∗ = (x∗ ∧ y∗)∗(c31)= (x ∨ y)∗∗.

Proof. (c42). We have [x∗⊙(x∗ → y∗)]∗ = [x∗⊙(y∗∗ → x∗∗)]∗ = [(y∗∗ →x∗∗) ⊙ x∗]∗ = (y∗∗ → x∗∗) → (x∗ → 0) = (y∗∗ → x∗∗) → x∗∗.

(c43). From x∗ ∧ y∗ ≤ x∗ ⇒ x∗∗ ≤ (x∗ ∧ y∗)∗ and analogously y∗∗ ≤(x∗ ∧ y∗)∗. To prove x∗∗ ∨W y∗∗ = (x∗ ∧ y∗)∗ consider t ∈ MV (L) such thatx∗∗, 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 areequivalent:

(i) L ∈ RLsd;

(ii) (x∗∗ → y∗∗) → y∗∗ = (y∗∗ → x∗∗) → x∗∗ for every x, y ∈ L.


Proof. (i) ⇒ (ii). If L ∈ RLsd, then for x, y ∈ L, [x∗ ⊙ (x∗ → y∗)]∗ =[y∗⊙ (y∗ → x∗)]∗ (= (x∗∧y∗)∗) and by (c42) 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∗)∗, thatis, L ∈ RLsd. �

Remark 3. 1. 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 latticeLc = ([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 (Lc) = [0, c)∪{1}. Let x = 3

5c, y = 45c. Then x, y ∈

MV (Lc), (y → x) → x = 1, but (x → y) → y = y. Thus, the conditionfrom Theorem 3 does not hold. So Lc is not semi-divisible. Evidently, eachresiduated lattice Lc is linear, therefore is a MTL-algebra. We deduce thatMTL-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 LsD = ([0, 1],∨,∧,⊙,→, 0, 1) such that for all x, y ∈ [0, 1], x ⊙ y = 0 if x, y ∈ [0, 12 ] andmin{x, y} elsewhere x → y = 1 if x ≤ y; 1

2 if y < x ≤ 12 and y if (y < x, 12 <

x). We have MV (LsD) = {0, 12 , 1} and the condition from Theorem 3 holds,whence, LsD is a semi-divisible residuated lattice. LsD is not divisible.Indeed, let x = 1

3 , y = 12 . Then 1

2 ⊙ (12 → 13) = 1

2 ⊙ 12 = 0 = 1

3 = min{13 ,

12}.

We recall that if (Li,∨,∧,⊙,→, 0, 1), i = 1, 2 are two residuated latticesthen, a map f : L1 → L2 is called morphism of residuated lattices if fsatisfies the following conditions, for every x, y ∈ L1 :

(m1) f(0) = 0;

(m2) f(1) = 1;

(m3) f(x ∧ y) = f(x) ∧ f(y);

(m4) f(x ∨ y) = f(x) ∨ f(y);

(m5) f(x → y) = f(x) → f(y);


(m6) f(x⊙ y) = f(x) ⊙ f(y).

We define Ker(f) = {x ∈ L1 : f(x) = 1}.

Remark 4. 1. If L1, L2 are BL-algebras, then f : L1 → L2 is morfismof residuated lattices iff f verifies (m1), (m5) and (m6).

2. If L1, L2 are divisible, then f : L1 → L2 is morfism of residuatedlattices iff f verifies the conditions (m1), (m4), (m5) and (m6).

3. If L1, L2 are MTL-algebras, then f : L1 → L2 is morfism of residuatedlattices iff f verifies (m1) − (m3) and (m5) − (m6).

So, RL becomes in canonical way a category such that RLd, MT L,RLsd 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 mor-phisms.

In what follows, by L we denote the universe of a divisible residuatedlattice. 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 restrictionof 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) isa BL-algebra (hence a residuated lattice).

Proposition 4. If x, y ∈ L, then (x⊙ y)∗∗ = x∗∗ ⊙W y∗∗.

Proof. We have x∗∗ ⊙W y∗∗(c44)= (x∗∗∗ ⊕ y∗∗∗)∗ = (x∗ ⊕ y∗)∗ = (x∗∗ →

y∗)∗ = (y → x∗)∗ = (y → (x → 0))∗ = [(x⊙ y)∗]∗ = (x⊙ y)∗∗. �We denote R(L) = MV (L) and we define ΦR(L) : L → MV (L) by

ΦR(L)(x) = x∗∗, for all x ∈ L.

Proposition 5. If L is a divisible residuated lattice, then ΦR(L) is amorphism of residuated lattices (between the lattices (L,∨,∧,→,⊙, 0, 1) and(MV (L),∨W ,∧W = ∧,→,⊙W , 0, 1)).

Proof. Clearly, ΦR(L)(0) = 0, ΦR(L)(1) = 1,ΦR(L)(x ∨ y) = (x ∨y)∗∗

(c43)= x∗∗ ∨W y∗∗ = ΦR(L)(x) ∨W ΦR(L)(y), ΦR(L)(x → y) = (x →


y)∗∗(c38)= x∗∗ → y∗∗ = ΦR(L)(x) → ΦR(L)(y) and by Proposition 4,

ΦR(L)(x ⊙ y) = (x ⊙ y)∗∗ = x∗∗ ⊙W y∗∗ = ΦR(L)(x) ⊙W ΦR(L)(y) forall x, y ∈ L.

Since L is divisible, we have x∧y = x⊙ (x → y), hence, ΦR(L)(x∧y) =ΦR(L)(x) ∧W ΦR(L)(y), so, ΦR(L) is a morphism of residuated lattices.�

If L,L′ are divisible residuated lattices and f : L → L′ is a morphism ofresiduated 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∗)) andR(f)((x∗)∗) = (f(x∗))∗ = (f(x)∗)∗ = (R(f)(x∗))∗.

So, the assignments L MV (L) = R(L) and f R(f) define a func-tor (covariant) R : RLd → MV from the category of divisible residuatedlattices to the category of MV -algebras.

Theorem 5. The category MV of MV -algebras is a reflective subcate-gory of the category RLd of divisible residuated lattices and the reflector Rpreserves monomorphisms.

Proof. To prove R is a reflector ([1]), we consider the diagram

L

ΦR(L)

��

f // L′

ΦR(L′)��

R(L)R(f) // R(L′)

with L,L′ ∈ RLd.If x ∈ L, then (ΦR(L′) ◦ f)(x) = ΦR(L′)(f(x)) = (f(x))∗∗ and (R(f) ◦

ΦR(L))(x) = R(f)(ΦR(L)(x)) = R(f)(x∗∗) = (f(x∗))∗ = (f(x)∗)∗ =(f(x))∗∗, hence ΦR(L′) ◦ f = R(f) ◦ ΦR(L), that is, the above diagramis commutative.

Let now L ∈ RLd, M ∈ MV and f : L → M a morphism of residuatedlattices

L

f ��???

????

?ΦR(L) // R(L)

f ′||yyyyyyyy

M

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 ⊙


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. Sincefor x ∈ L, (f ′ ◦ ΦR(L))(x) = f ′(ΦR(L)(x)) = f ′(x∗∗) = f(x)∗∗ = f(x), wededuce that f ′ ◦ ΦR(L) = f.

If we have again f ′′ : R(L) → M a morphism in MV such that f ′′ ◦ΦR(L) = f, then for any x ∈ L, (f ′′ ◦ ΦR(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 monomor-phism in RLd 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 great-est 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 numbern ≥ 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 equiv-alent:

(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 isan injective object in the category RLd.

Proof. By Theorem 6, L is an injective object in the category MV.Since MV is reflective subcategory of RLd and the reflector R : RLd → MVpreserves monomorphisms (by Theorem 5), then by Theorem 6 from [1] wededuce that L is injective object in the category RLd. �

Remark 6. Following Corollary 4.4 from [11] we deduce that if L is aninjective object in the category RLd, then L is a complete lattice.

Open problem. If L is an injective object in the category RLd, thenL is divisible?


4. Boolean center, regular and dense elements

Let (L,∨,∧, 0, 1) be a bounded lattice. Recall (see [1], [13]) that anelement a ∈ L is called complemented if there is an element b ∈ L such thata ∨ b = 1 and a ∧ b = 0; if such element b exists it is called a complementof a. We will denote b = a′ and the set of all complemented elements inL by B(L). Complements are generally not unique, unless the lattice isdistributive.

In residuated lattices however, although the underlying lattices need notbe distributive, the complements are unique.

Lemma 2 ([12]). Let L be a residuated lattice and suppose that a ∈ Lhave 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) a2 = a.

Let L be a residuated lattice and B(L) the set of all complementedelements of the lattice (L,∨,∧, 0, 1). So, if e ∈ B(L), then e′ = e∗ ande∗∗ = 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 theBoolean center of L - see [12]).

Proposition 6 ([5]). If L is a residuated lattice, then for e ∈ L thefollowing are equivalent:

(i) e ∈ B(L);

(ii) e ∨ e∗ = 1.

Theorem 7 ([7]). For every element e in an MV -algebra A, the follow-ing conditions are equivalent:

(i) e ∈ B(A);

(ii) e ∨ e∗ = 1;

(iii) e ∧ e∗ = 0;

(iv) e⊕ e = e;


(v) e⊙ e = e.

Proposition 7 ([5]). Let L be a residuated lattice. For e ∈ L we con-sider the following assertions:

(i) e ∈ B(L);

(ii) e2 = e and e = e∗∗;

(iii) e2 = 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 aboveproposition 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:

(c42) x⊙ (x → e) = x ∧ e, e⊙ (e → x) = e ∧ x;

(c43) e ∨ (x⊙ y) = (e ∨ x) ⊙ (e ∨ y);

(c44) e ∧ (x⊙ y) = (e ∧ x) ⊙ (e ∧ y);

(c45) e⊙ (x → y) = e⊙ [(e⊙ x) → (e⊙ y)];

(c46) x⊙ (e → f) = x⊙ [(x⊙ e) → (x⊙ f)];

(c47) e → (x → y) = (e → x) → (e → y);

(c48) e → (e → x) = e → x;

(c49) (e → x) → e = e;

(c50) (e → x) → x ≤ (x → e) → e;

(c51) e∗ → x = (e → x) → x = e ∨ x;


(c52) e ∨ (x → y) = (e ∨ x) → (e ∨ y);

(c53) (e ∨ x) → (f ∨ x) = (e → f) ∨ x;

(c54) x → (e → f) = (x → e) → (x → f);

(c55) e ∧ (x ∨ y) = (e ∧ x) ∨ (e ∧ y);

(c56) x ∧ (e ∨ f) = (x ∧ e) ∨ (x ∧ f);

(c57) If e, f ≤ x, then e⊙ (x → f) = f ⊙ (x → e);

(c58) (e → x) ∨ (x → e) = 1;

(c59) e ∨ x = [(e → x) → x] ∧ [(x → e) → e].

Proof. For (c42) − (c47) see [5].(c48). We have e → (e → x) = e2 → x = e → x.(c49). See [5].

(c50). We have (e → x) → x(c5)

≤ (x → e)→[(e→x) → e] = (x → e) → e.(c51). 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.

(c52). By (c51) we have e ∨ (x → y) = e∗ → (x → y)(c47)= (e∗ → x) →

(e∗ → y) = (e ∨ x) → (e ∨ y).

(c53). We have (e ∨ x) → (f ∨ x)(c51)= (e∗ → x) → (f∗ → x) = f∗ →

[(e∗ → x) → x](c51)= 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.

(c54). (x → e) → (x → f)(c14)= [x ⊙ (x → e)] → f

(c42)= (x ∧ e) → f =

(x⊙ e) → f(c14)= x → (e → f).

(c55). We have e∧(x∨y) = e⊙(x∨y)(c21)= (e⊙x)∨(e⊙y) = (e∧x)∨(e∧y).

(c56). We have x∧(e∨f) = x⊙(e∨f)(c21)= (x⊙e)∨(x⊙f) = (x∧e)∨(x∧f).

(c57). 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);


(c58). If e ∈ B(A) then e ∨ e∗ = 1. Since e∗ ≤ e → x and e ≤ x → e wededuce that 1 = e∗ ∨ e ≤ (e → x) ∨ (x → e), so (e → x) ∨ (x → e) = 1.

(c59). 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], soe ∨ x ≤ t.

We have t = t⊙1(c58)= t⊙[(e → x)∨(x → e)]

(c21)= [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 soe ≤ e∗ → e. We have that e∗ → e = e.

(iii) ⇒ (i). Applying (c59), e ∨ e∗ = 1 ⇔ (e → e∗) → e∗ = 1 and(e∗ → e) → e = 1. By (iii), e∗ → e = e, hence (e∗ → e) → e = 1. Wealso 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), fromProposition 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 = 0we obtain that e∗ ∧ e = 0. Also, from hypothesis (for x = 0) we obtaine∗ → e = e. So, from (c59) we obtain e∨ e∗ = [(e → e∗) → e∗]∧ [(e∗ → e) →e] = [(e → e∗) → e∗] ∧ (e → e)

= [(e → e∗) → e∗]∧1 = (e → e∗) → e∗(c30)= [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 considerthe following assertions:

(i) e ∈ B(L);


(vi) (e → e∗) ∨ (e∗ → e) = 1.

Then (i) ⇒ (vi) but (vi) ; (i).

Proof. (i) ⇒ (vi) If e ∈ B(L) from (c58) for x = e∗ we deduce that(e → e∗) ∨ (e∗ → e) = 1.

(vi) ; (i). Consider the residuated lattice L = {0, a, b, c, 1} from theExample 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 ∈ Lis 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 denseif for every r ∈ R(L) we have x → r = r. We denote by D(L) the set of alldense elements of L.

We give a new characterization for regular elements:

Theorem 8. Let L be a residuated lattice. For x ∈ L the followingassertions are equivalent:

(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 weobtain 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 (c29) and the relationx∗∗ ≤ 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).


Proof. Let x ∈ R(L); by Theorem 8, x∗∗ = x and x∗ → x = x, sox∗∗ ∈ 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)(c7)= [(x∧y)∗ → x]∧[(x∧y)∗ → y] = x∧y ⇒ x∧y ∈ R(L).�

A residuated lattice L will be called G- algebra if x2 = x, for everyx ∈ 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 thatx∗ ∈ R(L). Indeed, x∗ verifies the conditions (iii) from the Theorem 8because 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 = xthen x → x∗ = x∗.

Proof. x → x∗ = x → (x → 0) = (x⊙ x) → 0 = x → 0 = x∗. �We give a new characterization for boolean elements:


(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 MTL− algebra. For x ∈ L the followingassertions are equivalent:

(i) x ∈ B(L);

(ii) x ∈ R(L) and x⊙ x = x.


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 thatB(L) = R(L) we consider the residuated lattice L = {0, a, b, c, 1} fromExample 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, hencea ∈ 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. Lete ∈ 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. Wededuce that B(L) ⊆ B(MV (L)).

Consider now e ∈ B(MV (L)). Then e = e∗∗ and e ∨W e∗ = 1. Hencee ∨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 MTL-algebra and x ⊙ x = x for every x ∈ L,then B(L) = R(L).

Corollary 10. If L is a semi-divisible MTL-algebra, then B(L) =B(MV (L)).

Proof. By Corollary 8, B(L) ⊆ B(MV (L)) for any semi-divisible resi-duated lattice L. Consider now that L is a semi-divisible MTL-algebra andlet 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 MTL-algebra, soe ∈ 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 MTL-algebra, then B(L) =B(MV (L)) = R(L).

Remark 10. 1. We consider the residuated lattice from Remark 3, (2)which is not semi-divisible. Let c = 1

2 . We have 0∗ = 1, x∗ = 12 − x, if

x ∈ (0, 12), x∗ = 0, if x ∈ [12 , 1]. Then, 0∗∗ = 0, x∗∗ = x, if x ∈ (0, 12), x∗∗ = 1,


if x ∈ [12 , 1]. Also, 0∗ → 0 = 0, x∗ → x = x, if x ∈ (0, 14), x∗ → x = 1, ifx ∈ [14 , 1]. We conclude that MV (L) = {x ∈ L : x∗∗ = x} = [0, 12)∪{1} (butMV (L) is not an MV-algebra), R(L) = {x ∈ L : x∗ → x = x} = [0, 14)∪{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 semi-divisible we have 0∗ = 1, x∗ = 1

2 , if x ∈ (0, 12 ], x∗ = 0, if x ∈ (12 , 1]. Then,0∗∗ = 0, x∗∗ = 1

2 , if x ∈ (0, 12 ], x∗∗ = 1, if x ∈ (12 , 1]. Also, 0∗ → 0 =0, x∗ → x = 1

2 , if x ∈ (0, 12), x∗ → x = 1, if x ∈ [12 , 1]. We conclude thatMV (L) = {x ∈ L : x∗∗ = x} = {0, 12 , 1} (in this case MV (L) is an MV-algebra), 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 semi-divisible, see Corollary 8).

We characterize the residuated lattices which are Boolean algebras:

Theorem 10. For a residuated lattice L the following assertions areequivalent:

(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 naturalordering, 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∗ → yfor every x, y ∈ L.

Indeed, x, y ≤ x∗ → y. Let t ∈ L such that x, y ≤ t. From x ≤ t wededuce 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 Lis a Boolean algebra it will suffice to prove that x∨x∗ = 1, for every x ∈ L.

Obviously, x ∨ x∗ = x∗ → x∗ = 1. �


(i) x ∈ D(L);

(ii) x∗ = 0.


Proof. (i) ⇒ (ii). Since (0 → y) → 0 = 1 → 0 = 0 for every y ∈ L, wededuce 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)(c14)= 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, hencet ≤ 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∗∗(c32)= 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 arey ∈ 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 deduc-tive system of L, ds for short, if the following conditions are satisfied:


(Ds1) 1 ∈ D;

(Ds2) 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 nonemptysubset 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 D1, D2 ∈ Ds(L) we put D1∧D2 = D1∩D2 and D1∨D2 =< D1∪D2 > .The lattice (Ds(L),⊆ ) is a complete Brouwerian lattice (hence distribu-tive).

We say ([21]) that D ∈ Ds(L) is prime if D = L and D verify one ofthe equivalent assertions:

(i) If D = D1 ∩D2 with D1, D2 ∈ Ds(L), then D = D1 or D = D2;

(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 dsof L is maximal if it is proper and it is not contained in any other properds. In a nontrivial residuated lattice L, every proper ds can be extended toa 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 de-ductive systems of a residuated lattice. We prove that if L is semi-divisible,then there is a bijective correspondence between maximal deductive sys-tems of L and maximal deductive systems of MV (L). Also, we characterizeRad(MV (L)), for a semi-divisible residuated lattice L.

Theorem 12 ([5]). Let L be a residuated lattice and M a proper ds ofL. Then the following are equivalent:

(i) M ∈ Max(L),


(ii) For any x ∈ L, x /∈ M iff (xn)∗ ∈ M, for some n ≥ 1.

Corollary 13. Let L be a residuated lattice. If M ∈ Max(L) andx, y ∈ L such that (xn)∗ → y ∈ M, for any n ≥ 1, then x ∈ M or y ∈ M.

Proof. Let x, y ∈ L such that (xn)∗ → y ∈ M for any n ≥ 1 andsuppose that x /∈ M and y /∈ M. From Theorem 12 we deduce that there isn0 ≥ 1 such that (xn0)∗ ∈ M. Since (xn0)∗, (xn0)∗ → y ∈ M ⇒ y ∈ M, acontradiction. �

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 xn → y ∈ M, for every y ∈ L.

Proof. (i) ⇒ (ii). If we suppose that x /∈ M then by Theorem 12 thereis n ≥ 1 such that (xn)∗ ∈ M. Because (xn)∗ ≤ xn → y for every y ∈ L wededuce that xn → y ∈ M, for every y ∈ L.

(ii) ⇒ (i). Let D ∈ Ds(L) proper such that M $ D, hence there isx0 ∈ D such that x0 /∈ M. Then there is n ≥ 1 such that xn0 → y ∈ M, forevery y ∈ L. Hence xn0 → y ∈ D. Since D ∈ Ds(L) and x0 ∈ D we deducethat xn0 ∈ D. From xn0 , x

n0 → 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 ∈ Lsuch that x∗∗ ∈ M and suppose by contrary that x /∈ M. Since M ∈Max(L), then there is n ≥ 1 such that (xn)∗ ∈ M. Following (c35), (x∗∗)n ≤(xn)∗∗, so (xn)∗∗ ∈ M. From (xn)∗, (xn)∗∗ ∈ M ⇒ 0 ∈ M ⇒ M = L, acontradiction. �

Remark 12. In a residuated lattice L, the following are equivalent:

(i) x → x∗ = x∗ for every x ∈ L;

(ii) (x2)∗ = x∗ for every x ∈ L;

(iii) (xn)∗ = x∗ for every x ∈ L and n ≥ 2;


(iv) x⊙ (x → x∗) = 0 for every x ∈ L.

Definition 5. L is called semi G-algebra if it verifies one of equivalentconditions of Remark 12.

We recall that a residuated lattice L is called a G-algebra if x2 = x forevery 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 notG-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,(02)∗ = 0∗ = 1 = 0∗ and for x > 0 ⇒ x∗ = x → 0 = 0/x = 0. Sincex2 > 0 ⇒ (x2)∗ = x2 → 0 = 0/x2 = 0, so (x2)∗ = x∗ = 0, hence L is a semi

G-algebra. Since(12

)2= 1

4 = 12 , L is not a G-algebra.

Consider L from Example 1 for p = 1. Then(12

)∗= 1

2 → 0 = min{1, 1−12 + 0} = min{1, 12} = 1

2 and(12

)2= 1

2 ⊙ 12 = max{0, 12 + 1

2 − 1} = 0, so[(12

)2]∗= 1 = 1

2 =(12

)∗, hence L is not a semi G-algebra.

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(c34), (x → y)∗∗ ≤ x∗∗ → y∗∗ = x → y, so (x → y)∗∗ = x → y. By (c15),(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 φD : L → {0, 1} defined for x ∈ L byφD(x) = 1 if x ∈ D and φD(x) = 0 if x /∈ D.

Lemma 6. Let D ∈ Ds(L) proper. Then:

(i) φD(0) = 0, φD(1) = 1, φD(x⊙ y) = φD(x) ⊙ φD(y) and φD(x ∧ y) =φD(x) ∧ φD(y), for every x, y ∈ L;

(ii) If D ∈ Spec(L), then φD(x ∨ y) = φD(x) ∨ φD(y), for every x, y ∈ L;

(iii) If L is a semi G-algebra and D ∈ Max(L) then φD(x → y) =φD(x) → φD(y), for every x, y ∈ L.


Proof. (i). Since 0 /∈ D and 1 ∈ D, then φD(0) = 0 and φD(1) = 1.

Consider x, y ∈ L. If x ⊙ y ∈ D, then x, y ∈ D, so φD(x ⊙ y) = 1 =1⊙ 1 = φD(x)⊙φD(y). If x⊙ y /∈ D, then we deduce that x /∈ D or y /∈ D,so φD(x⊙ y) = φD(x) ⊙ φD(y) = 0.

If x ∧ y ∈ D then x, y ∈ D, so φD(x ∧ y) = φD(x) ∧ φD(y) = 1.

If x ∧ y /∈ D, then x /∈ D or y /∈ D (since if by contrary x ∈ D andy ∈ D ⇒ x⊙ y ∈ D ⇒ x ∧ y ∈ D ), so φD(x ∧ y) = φD(x) ∧ φD(y) = 0.

(ii). If x∨y ∈ D, since D is prime, then x ∈ D or y ∈ D, so φD(x∨y) =φD(x) ∨ φD(y) = 1.

If x∨y /∈ D, then x /∈ D and y /∈ D, so φD(x∨y) = φD(x)∨φD(y) = 0.

(iii). Consider x, y ∈ L such that x → y ∈ D.

If x ∈ D, then y ∈ D, so φD(x → y) = φD(x) → φD(y) = 1.

If x /∈ D and y ∈ D, then φD(x) → φD(y) = 0 → 1 = 1 ⇒ φD(x →y) = φD(x) → φD(y) = 1.

If x /∈ D and y /∈ D, then φD(x) → φD(y) = 0 → 0 = 1 ⇒ φD(x →y) = φD(x) → φD(y) = 1.

If x → y /∈ D, then y /∈ D. To prove the equality φD(x → y) = φD(x) →φD(y) it is necessary to prove that φD(x) = 1, that is x ∈ D.

If by contrary x /∈ D, since D ∈ Max(L), then there is n ≥ 1 such that(xn)∗ ∈ D. Since L is supposed to be a semi G-algebra, then (xn)∗ = x∗, sox∗ ∈ 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 φM :L → {0, 1} is a morphism of residuated lattices and Ker(fM ) = M .

Proof. Using Lemma 6, φM is a morphism of residuated lattices.

Obviously, Ker(φM ) = M , because Ker(φM ) = {x ∈ L : φM (x) =1} = {x ∈ L : x ∈ M} = M . �

Theorem 14. If L is a semi G-algebra, then there is a bijection be-tween Max(L) and RL(L, {0, 1}) = { f : L → {0, 1} | f is a morphism ofresiduated lattices}.

Proof. We define α : Max(L) → RL(L, {0, 1}) by α(M) = fM , forevery 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). ForM ∈ Max(L), (β ◦α)(M) = β(α(M)) = Ker(fM ) = M, from Corollary 14,


that is, β ◦ α = 1Max(L). If f ∈ RL(L, {0, 1}), then (α ◦ β)(f) = α(β(f)) =fKer(f).

We prove that fKer(f) = f. If x ∈ Ker(f), then f(x) = 1, so fKer(f)(x) =1 and if x /∈ Ker(f), then f(x) = 0, so fKer(f)(x) = 0. We deduce thatα ◦ β = 1RL(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. Forx, 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) isan 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), theny ∈ D and y = x∗ (with x ∈ L). We have y = x∗ = x∗∗∗ = y∗∗. �

Suppose L is a semi-divisible residuated lattice. Then MV (L) is anMV− 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 forx, 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, hencey ∈ 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 inMV (L). To prove M ∩MV (L) is maximal in MV (L), suppose by contrarythat there is N ∈ Ds(MV (L)) such that M∩MV (L) ⊂ N ⊆ MV (L). Thenthere is x0 ∈ N such that x0 /∈ M ∩MV (L). Then x0 /∈ M hence there is


n ≥ 1 such that (xn0 )∗ ∈ M ⇒ (xn0 )∗ ∈ M ∩ MV (L) ⊂ N ⇒ (xn0 )∗ ∈ N.Since xn0 ∈ 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)) wedenote 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 isproper in L.

To prove M ∈ Max(L), consider N ∈ Ds(L) such that M ⊂ N ⊆ L.Then there is x0 ∈ N such that x0 /∈ 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 x0 ∈ 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 byLemma 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. �


Theorem 15. If L is a semi-divisible residuated lattice, then there is abijection between Max(L) and Max(MV (L)).

Proof. Following Lemma 9 and Lemma 10 we deduce that we candefine f : Max(L) → Max(MV (L)), by f(M) = M ∩ MV (L), for everyM ∈ Max(L) and g : Max(MV (L)) → Max(L), by g(N) = N, for everyN ∈ Max(MV (L)). By Lemma 11 we deduce that f ◦g = 1Max(MV (L)) andg ◦ f = 1Max(L), hence we deduce that f is bijective and f−1 = g. �

Definition 6 ([23]). A residuated lattice L is called semilocal (local) ifit 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 asemilocal 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 systemsof a residuated lattice L is called the radical of L and will be denoted byRad(L).

For n ≥ 1 and x ∈ L we denote nx = [(x∗)n]∗. Following [11], [12],

Rad(L) = {x ∈ L : for every n ≥ 1 there is kn ≥ 1 such that kn(xn) = 1} ={x ∈ L : for every n ≥ 1 there is kn ≥ 1 such that [(xn)∗]kn = 0}.

If L is a BL-algebra, then Rad(L) = {x ∈ L : (xn)∗ ≤ x, for everyn ≥ 1} (see [23]).

Proposition 12. If L is a chain, then Rad(L) = {x ∈ L : xn > x∗, forevery n ≥ 1}.

Proof. If L is a chain, then there is an unique maximal deductivesystem M of L, so Rad(L) = M .

If x ∈ M then xn ∈ M, for every n ≥ 1, hence xn > (xn)∗ (because ifby contrary xn ≤ (xn)∗ ⇒ (xn)∗ ∈ M ⇒ 0 ∈ M, a contradiction). Sincexn < x ⇒ (xn)∗ > x∗, so xn > x∗.

Conversely, let x ∈ L such that xn > x∗, for every n ≥ 1. Then xn = 0for every n ≥ 1 ⇒< x > is proper⇒ x ∈< x >⊆ M = Rad(L). �

Now, we characterize Rad(MV (L)), for a semi-divisible residuated latti-ce L.


Theorem 17. If L is a semi-divisible residuated lattice, thenRad(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)). Thenthere is M ∈ Max(MV (L)) such that x∗∗ /∈ M. Thus, there is n ≥ 1such 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(MV (L)) =Rad(L) ∩MV (L).


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Received: 25.VI.2012 Faculty of Exacte Sciences,

Revised: 20.VII.2012 Department of Mathematics,

Accepted: 24.VII.2012 University of Craiova,

200585, Craiova,

ROMANIA

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