On co-annihilators in residuated lattices

1 Introduction

The concept of a commutative residuated lattice was firstly introduced by M. Ward and R.P. Dilworth [13] as generalization of ideal lattices of rings. H. Ono considered residuated lattices as an algebraic structure of substructural logics in [10]. P. Hájek introduced the notion of BL-algebra as a residuated lattice with two more conditions, namely divisibility and prelinearity to prove the completeness of Łukasiewicz logic as a many valued logic [4]. He showed that these algebras are the best algebraic counterparts of fuzzy logics generated by continuous t-norms [4]. Leuştean introduced the concept of co-annihilator of an element a relative to a filter F in a BL-algebra L as (F, a) = {x ∈ L|x ∨ a ∈ F} [8].

Here, we extend this notion to a residuated lattice L and arbitrary subsets T and Y of L.

This paper is organized as follows: in the next section we give some preliminaries including the basic definition and theorems that are needed in the sequel. In Section 3, we extend co-annihilators to arbitrary subsets T and Y of residuated lattices called co-annihilator of T relative to Y and denoted by (T, Y). We show that if T is a filter, then (T, Y) is a filter but it does not depend on Y, as expected. So we show that (T, Y) = (T, [Y)), where [Y) is the filter generated by subset Y. Also we investigate many properties of (T, Y) corresponding to constraints on T or Y. Examples are given for showing that some properties hold under some necessary condition. In Section 4, we prove some theorems which determine the relationship between this special subset and the other types of filters. Finally, we give conditions on subsets T and Y of residuated lattices L containing elements such as atoms and molecules, for which (T, Y) is a special subset of L.

2 Preliminaries

Definition 2.1. [11] A residuated lattice is an algebra (L, ∧ , ∨ , ∗ , → , 0, 1) of type (2, 2, 2, 2, 0, 0) such that:

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

(ii) (L, ∗ , 1) is a commutative monoid,and

(iii) the operation ∗ and → form an adjoint pair,i.e., x ∗ y ≤ z if and only if x ≤ y → z for all x, y, z ∈ L.

From now on by (L, ∨ , ∧ , ∗ , → , 0, 1) or simply L we mean a residuated lattice, unless otherwise specified.

Theorem 2.2. [11] The following properties hold for all x, y, z ∈ L:

(1)x ∨ y = 1 implies  x ∗ y = x ∧ y,

(2)x ∨ (y ∗ z) ≥ (x ∨ y) ∗ (x ∨ z).

Definition 2.3. [11] A filter of L is a non-empty subset F of L satisfying:

for all x, y in L: if x ∈ F and x ≤ y, then y ∈ F.

Definition 2.4. [1, 14] A filter F of L is called:

Boolean filter (BF) iff for all x ∈ L, x ∨ ¬ x ∈ F.

A proper filter F of L is said to be a maximal filter (MF) iff it is not included in any other proper filter of L.

We collect and summarize the connections between the above filters from [1, 6] into Fig. 1. In Fig. 1, by A ⟶ B we mean a filter of type A is a filter of type B and A/B means that filters of type A and B are equivalent.

Proposition 2.5. [6] Let F be a filter of L. Then F is an implicative filter iff x → x²∈ F, for all x ∈ L.

M. Kondo claimed that fantastic filters and involution filters are equivalent [7]. In the following example we show that it is not true.

Example 2.6. [9] Consider the residuated lattice (L, ∧ , ∨ , ∗ , → , 0, 1) with the operations defined in Table 1, where L = {0, n, a, b, c, d, e, f, m, 1}, 0 < n < a < c < e < m < 1 and 0 < n < b < d < f < m < 1.

It is clear that F = {1} is an involution filter of L but it is not a fantastic filter of L. Since for m and n, we have n → m = 1 ∈ F, but ((m → n) → n) → m = m ∉ F.

3 Relative co-annihilators

In this section, we give definition of relative co-annihilators in a residuated lattice L and investigate some basic properties of these co-annihilators.

Definition 3.1. Let T and Y be subsets of L. The co-annihilator of T relative to Y is the set ( T , Y ) = { x ∈ L | ( ∀ y ∈ Y ) x ∨ y ∈ T }

If Y = {a}, then we denote (T, {a}) by (T, a). For example, by choosing T = {f, m, 1} and Y = {0, a, 1} in Example 2, we have (T, Y) = {f, m, 1}.

In the following proposition, we consider (T, Y) for some special cases of T and Y.

Proposition 3.2. Let T and Y be subsets of L. Then

(i) If Y =∅, then (T, Y) = L.

(ii) If T =∅ , Y ≠ ∅, then (T, Y) =∅.

(iii) (T, {0}) = T, (T, {1}) = L::iff::1 ∈ T.

(iv) ({0} , Y) = {0} ::iff::Y = {0}, ({1} , Y) ⊆ {x ∈ L| (∀ y ∈ Y) :x ∗ y = x ∧ y}.

(v) ({1} , {0}) = {1} , (L, {1}) = L, ({0} , {0}) = {0} , ({1} , {1}) = L.

(vi) (L, {0}) = L.

(vii) 1∉ T iff  (T, L) = ∅.

(viii) (T, L) ⊆ T.

Proof. We only prove (i) , (iv) , (vii), the others are similar.

(i) We write (T, Y) in a more logical form as {x ∈ L| (∀ y)   (y ∈ Y ⇒ x ∨ y ∈ T)}. So (T, ∅) = {x ∈ L| (∀ y)   (y ∈ ∅ ⇒ x ∨ y ∈ T)} = L, since the formula (∀ y) (y ∈ ∅ ⇒ x ∨ y ∈ T) is always true for all x ∈ L.

(iv) We know ({1} , Y) = {x ∈ L| (∀ y ∈ Y) :x ∨ y = 1}, by Theorem 2.2, (1), we have ({1} , Y) ⊆ {x ∈ L| (∀ y ∈ Y) :x ∗ y = x ∧ y}. Also if ({0} , Y) = {x ∈ L| (∀ y ∈ Y) :x ∨ y = 0} = {0}, then we have x = y = 0, for all y ∈ Y, so Y = {0}. The converse is obvious.

(vii) By contrary, assume that (T, L)≠ ∅. So there exist x ∈ L such that x ∨ a ∈ T, for all a ∈ L. Since 1 ∈ L, therefore x ∨ 1 =1 ∈ T, which is a contradiction. Converse is easy to see.□

Since one of our main goals is to extend the notion of co-annihilator to filters, from now on we assume that all subsets under consideration are non-empty.

Proposition 3.3. Let T₁, T₂, Y and Z be non-empty subsets of L. Then

(i) T₁ ⊆ T₂ implies (T₁, Y) ⊆ (T₂, Y).

(ii) Y₁ ⊆ Y₂ implies (T, Y₂) ⊆ (T, Y₁).

(iii) (T₁, Y) ∩ (T₂, Z) ⊆ (T₁ ∩ T₂, Y ∩ Z).

(iv) (T, (T, Y ∩ Z)) ⊆ (T, (T, Y)) ∩ (T, (T, Z)).

(v) (T, ⋃ _i∈IY _i ) = ⋂ _i∈I (T, Y _i ) ⊆ ⋃ _i∈I (T, Y _i ) ⊆ (T, ⋂ _i∈IY _i ).

(vi) ⋂_i∈I (T _i , Y) = (  ⋂ _i∈IT _i , Y) and   ⋃_i∈I (T _i , Y)⊆ (  ⋃ _i∈IT _i , Y).

(vii) ((T, Y) , Z) = ((T, Z) , Y) = ⋂ _z∈Z,y∈Y (T, z∨y).

(viii) Y ⊆ (T, (T, Y)).

(ix) Y ∩ (T, Y) ⊆ T.

(x)  (T, (T, (T, Y))  = (T, Y).

(xi) (T, Y) = ⋂ _y∈Y (T, y).

Proof. (i) Let x ∈ (T₁, Y). Then x ∨ y ∈ T₁ ⊆ T₂, for all y ∈ Y. Hence x ∈ (T₂, Y).

(ii) Let x ∈ (T, Y₂). Then x ∨ z ∈ T, for all z ∈ Y₂. We have Y₁ ⊆ Y₂, so x ∈ (T, Y₁).

(iii) Let x ∈ (T₁, Y) ∩ (T₂, Z). Then x ∨ y ∈ T₁ and x ∨ z ∈ T₂, for all y ∈ Y and z ∈ Z. Since Y ∩ Z ⊆ Y, Z, we have x ∨ u ∈ T₁ ∩ T₂, for all u ∈ Y ∩ Z, hence x ∈ (T₁ ∩ T₂, Y ∩ Z).

(iv) We have Y ∩ Z ⊆ Y, Z, by (ii), (T, Y) , (T, Z) ⊆ (T, Y ∩ Z) and so (T, (T, Y ∩ Z)) ⊆ (T, (T, Y)) , (T, (T, Z)). Hence (T, (T, Y ∩ Z)) is a lower bound for (T, (T, Y)) , (T, (T, Z)), therefore (T, (T, Y ∩ Z)) ⊆ (T, (T, Y)) ∩ (T, (T, Z)).

(v) By (ii), we have (T, ⋃ _i∈IY _i ) ⊆ (T, Y _i ), for every i ∈ I. So (T, ⋃ _i∈IY _i ) is an lower bound for (T, Y _i ), for every i ∈ I, therefore (T, ⋃ _i∈IY _i ) ⊆ ⋂ _i∈I (T, Y _i ).

Conversely, let x ∈ ⋂ _i∈I (T, Y _i ). Then x ∨ y _i  ∈ T, for all y _i  ∈ Y _i . Now let z be an arbitrary element in ⋃_i∈IY _i . So there is i ∈ I such that z ∈ Y _i . Hence the result holds due to the assumption.

Once again by (ii), we have (T, Y _i ) ⊆ (T, ⋂ _i∈IY _i ), for every i ∈ I. So (T, ⋂ _i∈IY _i ) is an upper bound for (T, Y _i ) for every i ∈ I, hence ⋃_i∈I (T, Y _i ) ⊆ (T, ⋂ _i∈IY _i ).

(vi) We have ⋂_i∈IT _i  ⊆ T _i , for every i ∈ I. By (i), (  ⋂ _i∈IT _i , Y) ⊆ (T _i , Y), for every i ∈ I. So (  ⋂ _i∈IT _i , Y) ⊆ ⋂ _i∈I (T _i , Y).

Conversely, let x ∈ ⋂ _i∈I (T _i , Y). Then x ∨ y ∈ T _i , for every i ∈ I, y ∈ Y, hence x ∨ y ∈ ⋂ _i∈IT _i , for all y∈Y. Thus x ∈ (⋂ _i∈IT _i , Y). The other part is similar.

(vii) We have that x ∈ ((T, Y) , Z) iff x ∨ z ∈ (T, Y), for all z ∈ Z iff (x ∨ z) ∨ y ∈ T, for all z ∈ Z, y ∈ Y iff x ∨ (z ∨ y) ∈ T, for all z ∈ Z, y ∈ Y iff x ∈ ⋂ _z∈Z,y∈Y (T, z ∨ y).

(viii) Let x be an arbitrary element in Y. We must show that x ∨ z ∈ T, for all z ∈ (T, Y) that is, y ∨ z ∈ T, for all y ∈ Y, so x ∨ z ∈ T.

(ix) Let x ∈ Y ∩ (T, Y). Then x ∈ Y and x ∨ y ∈ T, for all y ∈ Y, hence x = x ∨ x ∈ T. Thus Y ∩ (T, Y) ⊆ T.

(x) By (viii), we have (T, Y) ⊆  (T, (T, (T, Y)) ). Let x ∈  (T, (T, (T, Y)) ). Then z ∨ x ∈ T, for all z ∈ (T, (T, Y)). Since Y ⊆ (T, (T, Y)), we conclude that x ∨ z ∈ T, for all z ∈ Y. Hence x ∈ (T, Y).

(xi) x ∈ (T, Y) ⇔ (∀ y ∈ Y) :x ∨ y ∈ T ⇔ (∀ y∈Y) :x ∈ (T, y) ⇔ x ∈ ⋂ _y∈Y (T, y).□

By an upset U of L, we mean a subset U of L with the property that, if x ∈ U, y ∈ L and x ≤ y, then y ∈ U.

Proposition 3.4. Let T be an upset and Y be a non-empty subset of L. Then

(i) T ⊆ (T, Y).

(ii) (T, Y) = L iff Y ⊆ T, (So (L, Y) = L).

(iii) (T, L) = T and (T, T) = L.

(iv) (T, (T, T)) = T and ((T, T) , T) = L.

(v) T ⊆ Y implies (T, Y) ∩ Y = T.

(vi) Y ∩ (T, Y) = Y ∩ T.

(vii) (T, (T, Y)) ∩ (T, Y) = T.

Proof. (i) Let a ∈ T. Then a ∨ y ≥ a ∈ T, for all y ∈ Y, hence a ∨ y ∈ T, for all y ∈ Y. That is, a ∈ (T, Y).

(ii) If (T, Y) = L, then 0 ∈ (T, Y), hence y = 0 ∨ y ∈ T, for all y ∈ Y. Let y be an arbitrary element in Y. Then for any x ∈ L, y ≤ x ∨ y, hence x ∨ y ∈ T. That is, for any x ∈ L, we have that x ∈ (T, Y).

(iii) By (i) and Proposition 3.2 (viii), (T, L) = T. Now, Let x be an arbitrary element in L. We have t ≤ x ∨ t, for all t ∈ T, so by the definition of an upset x ∨ t ∈ T, for all t ∈ T. Hence x ∈ (T, T).

(iv) It is a straight conclusion from (ii) and (iii).

(v) An immediate consequence of (i) and Proposition 3.3 (ix).

(vi) By (i), we have T ⊆ (T, Y). So Y ∩ T ⊆ Y ∩ (T, Y). We have Y ∩ (T, Y) ⊆ Y, also by Proposition 3 (ix), Y ∩ (T, Y) ⊆ T which means Y ∩ (T, Y) is a lower bound for Y and T. Hence Y ∩ (T, Y) ⊆ Y ∩ T.

(vii) It follows when we put (T, Y) instead of Y in (vi).

From Proposition 3.4 (ii) and (iii), we can conclude that:

Remark 3.5. Let T be an upset and Y be a non-empty subset of L such that Y ⊆ T. Then (T, (T, Y)) = T.

Let T₁ and T₂ be filters of L. We recall from [3] that T₁ ⟶ T₂ = {x ∈ L ∣ T₁ ∩ [x) ⊆ T₂}, where [x) = {a ∈ L ∣ a ≥ x ⁿ  for some n ≥ 1}.

Theorem 3.6. Let T, T₁ and T₂ be filters and Y _i be a non-empty subsets of L, for all i ∈ I. Then

(i) (T, Y) is a filter of L.

(ii) (T, ⋁ _i∈IY _i ) = ⋂ _i∈I (T, Y _i ).

(iii) (T₁, Y) ∩ (T₂, Y) ⊆ (T₁ ⟶ T₂, Y).

By ⋁_i∈IY _i , we mean the filter generated by ⋃_i∈IY _i and denoted by [⋃ _i∈IY _i ). If A is a non-empty subset of L, then [A) = {x ∈ L ∣ x ≥ a₁ ∗ . . . ∗ a _n ,  for some n ≥ 1 and a₁, . . . , a _n  ∈ A}.

Proof. (i) We have y ∨ 1 =1 ∈ T for all y ∈ Y, hence 1 ∈ (T, Y). If a ≤ b and a ∈ (T, Y), then a ∨ y ∈ T and a ∨ y ≤ b ∨ y, for all y ∈ Y. Hence b ∨ y ∈ T, for all y ∈ Y, i.e. b ∈ (T, Y). Assume that a, b ∈ (T, Y), i.e. a ∨ y, b ∨ y ∈ T, for all y ∈ Y. By Theorem 2.2 (2), y ∨ (a ∗ b) ≥ (y ∨ a) ∗ (y ∨ b) ∈ T, it follows that y ∨ (a ∗ b) ∈ T, for all y ∈ Y, so a ∗ b ∈ (T, Y).

(ii) We know Y _i  ⊆ ⋁ _i∈IY _i  = [⋃ _i∈IY _i ), for all i ∈ I, so by Proposition 3.3 (ii), (T, ⋁ _i∈IY _i ) ⊆ (T, Y _i ), for all i ∈ I. So (T, ⋁ _i∈IY _i ) ⊆ ⋂ _i∈I (T, Y _i ). Let x be an arbitrary element in ⋂_i∈I (T, Y _i ) which means x ∈ (T, Y _i ), for all i ∈ I. We must show that x ∨ z ∈ T, for all z ∈ ⋁ _i∈IY _i  = [⋃ _i∈IY _i ). Let z be an arbitrary element in ⋁_i∈IY _i . Hence there exist i₁, i₂, . . . , i _m in I, y _{i j}  ∈ Y _j (1 ≤ j ≤ m) such that z ≥ y _{i 1}  ∗ y _{i 2}  ∗ . . . ∗ y _{i m} . By Theorem 2.2 (2), we obtain x ∨ z ≥ x ∨ (y _{i 1}  ∗ y _{i 2}  ∗ . . . ∗ y _{i m} ) ≥ (x ∨ y _{i 1} ) ∗ (x ∨ y _{i 2} ) ∗ . . . ∗ (x ∨ y _{i m} ). Since x ∨ y _{i j}  ∈ T, for every 1 ≤ j ≤ m, by Definition 2.3, we get x ∨ z ∈ T.

(iii) Let x ∈ (T₁, Y) ∩ (T₂, Y). Then x ∨ y ∈ T₁, T₂, for all y ∈ Y. It is sufficient to show that T₁ ∩ [x ∨ y) ⊆ T₂, for all y ∈ Y. We have x ∨ y ∈ T₂ for all y ∈ Y, so [x ∨ y) ⊆ T₂, for all y ∈ Y and therefore T₁ ∩ [x ∨ y) ⊆ T₂, for all y ∈ Y, which means x ∈ (T₁ ⟶ T₂, Y).

In the following we give examples to show that some of the statements in Proposition 3.3, 3.4 and Theorem 3.6 are true only under necessary conditions such as upset, filter, etc, and just one side of some of them holds.

Example 3.7. [9] Let L = {0, a, b, c, 1, 2, 3}, with 0 < a < c < 1 <2 < 3 and 0 < b < c < 1 <2 < 3, and the elements a, b are incomparable. Then L = (L, ∨ , ∧ , ∗ , ⟶ , 0, 1) becomes a distributive residuated lattice relative to the operation defined in Table 2, also the lattice structure of L is shown in Fig. 2:

Put T₁ = {b, c, 1}, T₂ = {a, c, 1} which are not filters, Y = {c, a, b} and Z = {c, 1}, then (T₁, Y) = {b, c, 1} and (T₂, Z) = {0, a, b, c, 1}, hence (T₁ ∩ T₂, Y ∩ Z) = ({c, 1} , {c}) = {0, a, b, c, 1} ⊈ (T₁, Y) ∩ (T₂, Z) = {b, c, 1}. This shows that the other side of Proposition 3.3 (iii) fails.

By setting Y₁ = {b, 2}, Y₂ = {a, 2} and T = {0, c, 2}, then (T, Y₁) = {a, c, 2}, (T, Y₂) = {b, c, 2} and (T, Y₁ ∩ Y₂) = {0, a, b, c, 1, 2}, so (T, Y₁ ∩ Y₂) ⊈ (T, Y₁) ∪ (T, Y₂). This shows that converse of Proposition 3.3 (v) fails.

If we choose T = {c, 1}, Y = {c, 1, 2} and Z = {c, b}, then (T, Y) =∅ and (T, Z) = {a, c, 1}, so (T, (T, Y)) ∩ (T, (T, Z)) = L ∩ {b, c, 1} ⊈ (T, (T, Y∩Z)) = {c, 1}. This shows that the other side of Proposition 3.3 (iv) fails.

If we choose T₁ = {b, c, 1}, T₂ = {a, c, 1} and Y = {c, a, b}, then (T₁, Y) = {b, c, 1} and (T₂, Y)= {a, c, 1}, hence (T₁ ∪ T₂, Y) = {0, a, b, c, 1} ⊈ (T₁, Y) ∪ (T₂, Y). This shows that the other side of Proposition 3.3 (vi) fails.

By choosing T = {b, c, 1} and Y = {c, a, b}, we have (T, Y) = {b, c, 1} and (T, (T, Y)) = {0, a, b, c, 1}. Hence (T, (T, Y)) ⊈ Y and T ⊈ Y ∩ (T, Y) = {b, c}. This shows that converse of Proposition 3.3 (viii) , (ix) fails.

Also if we choose T₁ = {c, 1, 2, 3}, T₂ ={b, c, 1, 2, 3} and Y = {0, a}, then (T₁, Y) = {c, 1, 2, 3}, (T₂, Y) = {b, c, 1, 2, 3} and T₁ ⟶ T₂ = L, hence (T₁ ⟶ T₂, Y) = L ⊈ (T₁, Y) ∩ (T₂, Y). This shows that the other side of Theorem 3.6 (iii) fails.

Example 3.8. [11] Let L = {0, a, b, c, d, e, 1}, where 0 < a < b < e < 1 and 0 < c < d < e < 1.

Then L = (L, ∨ , ∧ , ∗ , ⟶ , 0, 1) is a residuated lattice to the operation defined in Table 3, also the lattice structure of L is shown in Fig. 3. Put T = {1, b, c, d, e} and Y = L. We see that Y is a filter and T is not a filter. We have (T, Y) = {1, b, c, d, e}, that is not a filter. This shows that T being filter is a necessary condition for Theorem 3.6 (i). Once again, put T = {0, a, e, 1} and Y = {0, c, d, e, 1}. We see that T is not an upset. We have (T, Y) = {a, 1} and T ⊈ (T, Y). This shows that T being upset is a necessary condition for Proposition 3.4 (i).

Also if we choose T₁ = {1, b, c, d, e}, T₂ = {0, a, e, 1} and Y = {0, c, d, e, 1}, then (T₁, Y) = {1, b, c, d, e} and (T₂, Y) = {a, 1}, hence (T₁, Y) ⊈ (T₂, Y). This shows that Proposition 3.3 (i) fails without the condition T₁ ⊆ T₂.

By choosing Y₁ = {1, b, c, d, e}, Y₂ = {0, a, e, 1} and T = {0, c, e, 1}, then (T, Y₁) = {1, e} and (T, Y₂) = {c, e, 1}, so (T, Y₂) ⊈ (T, Y₁). This shows that Proposition 3.3 (ii) fails without the condition Y₁ ⊆ Y₂.

In Theorem 3.6 (i), we showed that if T is a filter, then (T, Y) is a filter but it does not important on the set Y, to be a filter or not. In the other words, it doesn’t need filter restriction on the set Y. It is interesting to know that if Y is replaced by [Y), we get the same co-annihilator.

Theorem 3.9. Let T be a filter and Y be a non-empty subset of L. Then (T, Y) = (T, [Y)), where [Y) = {x ∈ L|:x ≥ y₁ ∗ y₂ ∗ . . . ∗ y _n , :for:some:n ≥ 1:and::y₁, y₂, . . . , y _n  ∈ Y}.

Proof. Since Y ⊆ [Y), by Proposition 3.3 (ii), (T, [Y)) ⊆ (T, Y). Let x ∈ (T, Y) and z ∈ [Y). Then x ∨ y ∈ T, for all y ∈ Y and z ≥ y₁ ∗ y₂ ∗ . . . ∗ y _n for some y _i  ∈ Y, i = 1, . . . , n and some n ≥ 1. Using Theorem 2.2 (2), we have x ∨ z ≥ x ∨ (y₁ ∗ y₂ ∗ . . . ∗ y _n ) ≥ (x ∨ y₁) ∗ (x ∨ y₂) ∗ . . . ∗ (x ∨ y _n ). Since each x ∨ y _i  ∈ T, i = 1, . . . , n, and T is a filter, x ∨ z∈T.□

Recall from [12] that an element a ∈ L is called a node, if and only if for every x ∈ L either x ≤ a or a ≤ x.

Proposition 3.10. Let T be a filter of L and d be a node in L. Then (T, d) = T or (T, d) = L.

Proof. Let (T, d) ≠ L. Then by Proposition 3.4 (ii), d ∉ T. Since T ⊆ (T, d), Proposition 3.4 (i), it is sufficient to show that (T, d) ⊆ T. We assume that x ∈ (T, d), so x ∨ d ∈ T. Using hypothesize x ∨ d = x, hence the result holds. □

Now we can extend previous Proposition to all elements of L, specifically we assume that L is a linear residuated lattice.

Proposition 3.11. Let L be a linear residuated lattice, T be a filter and Y be a non-empty subset of L. Then (T, Y) = T or (T, Y) = L.

Proof. Let (T, Y) ≠ L. By Proposition 3.4 (ii), we have Y ⊈ T, so there exist b ∈ Y \ T. We assume that a ∈ (T, Y), so a ∨ y ∈ T, for all y ∈ Y. So we have a ∨ b ∈ T. Now using hypothesize a ∨ b = a, therefore the result holds. □

We recall that in an MTL-algebra L (a residuated lattice with the condition (x → y) ∨ (y → x) =1) we have B (L) = Rg (L) ∩ Id (L) [2]. So it is worth to note that (B (L) , Y) = (Rg (L) , Y) ∩ (Id (L) , Y), for any Y ⊆ L, where Rg (L) = {x ∈ L|: ¬¬ x = x} and B (L) = {x ∈ L|:x ∨ ¬ x = 1}.

4 Relationship between relative co-annihilator and filters

Now, we investigate the relationship between the notions of various types of filters and their relative co-annihilators in a residuated lattice L.

Theorem 4.1. Let T and Y be two non-empty subsets of L. Then the following assertions hold:

(i) If T is a Boolean filter of the second kind (BF₂), then (T, Y) is so. (ii) If T is a pseudocomplementation filter (PSF), of L, then (T, Y) is so. (iii) If T is an implicative filter (IF), of L, then (T, Y) is so.

Proof. (i) Let a be an arbitrary element of L. We show that a ∈ (T, Y) or ¬a ∈ (T, Y) which means that y ∨ ¬ a ∈ T or y ∨ a ∈ T, for all y ∈ Y. Since a ≤ a ∨ y and ¬a ≤ ¬ a ∨ y, for all y ∈ Y. So by assumption, the result holds. (ii) Let a be an arbitrary element of L. We show that ¬ (a ∧ ¬ a) ∈ (T, Y) which means y ∨ ¬ (a ∧ ¬ a) ∈ T, for all y ∈ Y. Since ¬ (a ∧ ¬ a) ∈ T and ¬ (a ∧ ¬ a) ≤ y ∨ ¬ (a ∧ ¬ a), for all y ∈ Y. Hence y ∨ ¬ (a ∧ ¬ a) ∈ T, for all y ∈ Y. (iii) Let a be an arbitrary element of L. By Proposition 3.5, it is sufficient to show that a → a²∈ (T, Y). We have a → a²∈ T and a → a²≤ y ∨ (a → a²), for all y ∈ Y and therefore by assumption, the result holds.□

Corollary 4.2. Let T and Y be two non-empty subsets of L. If T is a Boolean filter of the second kind of L, then (T, Y) is an obstinate filter (OF) of L and so (T, Y) is (MF), (BF), (PF), (PF₂), (FF), (VF) and (PIF).

Proof. This is a straight conclusion from Theorem 4 (i).□

Theorem 4.3. Let T be a prime filter of the second kind and Y ⊈ T. Then the following assertions hold:

(i) (T, Y) is a prime filter of the second kind of L.

(ii) (T, Y) = T.

(iii) (T, (T, Y)) = L.

Proof. (i) Suppose that a ∨ b ∈ (T, Y) , a, b ∉ (T, Y) for arbitrary element a, b in L, hence y ∨ (a ∨ b) ∈ T, for all y ∈ Y. So (y ∨ a) ∨ b ∈ T, for all y ∈ Y. By using assumption twice, we have y ∈ T, for all y ∈ Y, so Y ⊆ T, which is a contradiction.

(ii) By Proposition 3.4 (i), we have T ⊆ (T, Y). It is sufficient to show that (T, Y) ⊆ T. By contrary, we assume that a ∈ (T, Y) and a ∉ T, so a ∨ y ∈ T, for all y ∈ Y. By assumption, we have y ∈ T, for all y ∈ Y, hence Y ⊆ T, which is a contradiction.

(iii) An immediate consequence of (ii) and Proposition 3.4 (iii).

Proposition 4.4. Let T be a maximal filter of L and Y ⊈ T. Then (T, Y) = T.

Proof. By Proposition 3.4 (i) and (ii), we have T ⊆ (T, Y) ⊊ L, hence (T, Y) = T. □

Finally, we give conditions on subsets T and Y of residuated lattices L containing elements such as atoms and molecules, for which (T, Y) is a special subset of L.

Recall that an element of L is an atom, if there aren’t any element between that element and bottom element and is a molecule, if there aren’t any element between that element and top element in a residuated lattice.

Theorem 4.5. Let T and Y be non-empty subsets of L. Then

(i) If L contains only one molecule c, 1 ∈ T, c ∉ T and c ∈ Y, then (T, Y) = {1}.

(ii) If L contains at least one molecule c, c ∈ T, 1 ∈ T and Y only contains the molecule c, then (T, Y) = L.

(iii) If L contains at least one atom, T = AT (L) ∪ {1} and Y = AT (L), where AT (L) is the set of all atoms of L, then (T, Y) is a subset of AT (L) ∪ {0, 1}.

In particular, if L consists of at least two atoms and one molecule, then (T, Y) = {0, 1}, and if L consists of one atom b, then (T, Y) = {0, b, 1}.

(iv) If L contains only one atom b and at least two molecules, T = {1, b}, Y consists of at least one molecule, m, and 1 ∈ Y, then M = {x ∈ L| x is a molecule of L and x ∉ Y} ⊆ (T, Y) ⊆ ({1} , m). In particular, if Y = {1, m}, then (T, Y) = ({1} , m).

Proof. (i) Let c be the molecule of L. If 1 ≠ a ∈ L, then either a ∨ c = 1 or a ∨ c = c. The first case dose not happen, since otherwise L has at least one molecule other than c. In the latter case c ∉ T implies that a ∉ (T, Y). So (T, Y) = {1}.

(ii) Let c be the molecule of L. For all a ∈ L, a ∨ c = c ∈ T or a ∨ c = 1 ∈ T, hence (T, Y) = L.

(iii) It is obvious that 0, 1 ∈ (T, Y). Let AT (L) = {a _i } _i∈I and x be an element of L other than 0, 1 and {a _i }, for all i ∈ I. Then either x ∧ a _j  = 0 or x ∧ a _j  = a _j , for some j ∈ I. The first case does not happen, since otherwise x ∈ AT (L) that is not true. In the latter case x is comparable with some a _j  (j ∈ I), thus x ∨ a _j  = x which is not in T. Therefore x ∉ (T, Y).

(iv) Let x be an element of M. Then x ∨ y = 1 ∈ T, for all y ∈ Y, so x ∈ (T, Y). Now let x ∈ (T, Y). Then either x ∨ m = m or x ∨ m = 1. The first case does not happen, since otherwise x ∉ (T, Y). Hence x ∈ ({1} , m). □

5 Conclusion and future research

We introduced the extension of co-annihilator in residuated lattices and investigated its properties. It is extended to two subsets of L such as T, Y and denoted by (T, Y). In this extension, T has a more important role than Y, as expected. It is proved if T is a filter, then (T, Y) is a filter but it is not important that Y be a filter or not. In the other words, it doesn’t need filter restriction on the set Y for being filter. So we showed that (T, Y) = (T, [Y)). Examples are given for better understanding the results. In addition, we showed that if T is a Boolean filter of the second kind, then (T, Y) is so. In other words, (T, Y) is an obstinate filter and therefore it is (MF), (BF), (PF), (PF₂), (FF), (VF) and (PIF). We proved that if T is (PF₂) and Y ⊈ T, then (T, Y) is so and (T, Y) = T. We also showed that if T is an upset of L and Y ⊆ T, then (T, (T, Y)) = T. and other results obtained. Finally we have given one of our main theorems consisting of statements concerning some constraints on subsets T and Y as well as on L (e.g. containing special elements such as atoms or molecules) and found co-annihilators as special subsets.

In our future work, we will continue our study of algebraic properties of this extension, especially singleton subsets, Y, of L. We will use these filters to define congruence relations on L. It seems that the residuated lattice can be partitioned in a very “nice" way.