Co-annihilators and fuzzy co-annihilators in EQ-algebras

1 Introduction

Logic algebras are the corresponding algebraic semantics with all sorts of propositional calculus, which are the algebraic foundations of reasoning mechanism of many fields such as computer sciences, information sciences, cybernetics, artificial intelligence and so on. EQ-algebra is a new class of logic algebra which was proposed by Nov a ´ k in [10] which generalizes the residuated lattice. One of the motivations is to introduce a special algebra as the correspondence of truth values for high-order fuzzy type theory (FTT). Another motivation is from the equational style of proof in logic. It has three connectives: meet ∧, product ⊗ and fuzzy equality ∼. The implication operation → is the derived of the fuzzy equality ∼ and it together with ⊗ no longer strictly form the adjoint pair. About EQ-algebras, one can see [3, 14].

The filter theory of logic algebras plays an important role in studying logic algebras [5, 18]. From a logic point of view, various filters have natural interpretation as various sets of provable formulas, which has a very close relationship with decision-making [15–17]. Turunen [13] and Leustean [6] introduced and studied the co-annihilator of a non-void set and the co-annihilator of an element a relative to a filter in a BL-algebra, respectively. Then Meng [8] introduced generalized co-annihilators in BL-algebras and give characterizations of prime filters and minimal prime filters. Recently, as the generalization of the co-annihilator in a BL-algebra, Saeid [12] introduced the co-annihilator of a set relative to another set in a residuated lattice where they give some relations between filters and co-annihilators. It is helpful for the co-annihilators to study structures and properties in algebraic systems.

In this paper, we mainly introduce two types of co-annihilators in an EQ-algebra ɛ. Furthermore, as an application, we define and study Δ-co-annihilators in EQ_Δ-algebra ɛ_Δ and two types of fuzzy co-annihilators in an EQ-algebra ɛ. This paper is organized as follows: In section 2, we review some basic definitions and results about EQ-algebras. In section 3, we introduce two types of the co-annihilators in an EQ-algebra and Δ-co-annihilators in an EQ_Δ-algebra ɛ_Δ. We mainly study the structures of PF ( ɛ ) , COA ( ɛ ) and PFP Δ ( ɛ Δ ) . In section 4, we introduce and investigate fuzzy co-annihilators in EQ-algebras.

2 Preliminaries

In this section, we recollect some definitions and results which will be used in the following.

Definition 2.1. [10] An EQ-algebra is an algebra ɛ = (E, ∧ , ⊗ , ∼ , 1) of type (2, 2, 2, 0) such that for all x, y, z, t ∈ E:

(E, ∧ , 1) is a commutative idempotent monoid (i.e. ∧-semilattice with top element 1);

In every EQ-algebra ɛ, define →, ¬ by x → y : = (x ∧ y) ∼ x, ¬ x : = x ∼ 0.

Example 2.2. [10] Let E = [0, 1]. Define operators ⊗, ∼ by x ⊗ y = 0 ∨ (x + y - 1) as the Łukasiewicz conjunction and x ∼ y = 1 - |x - y|. Then ɛ_Ł = ([0, 1] , ∧ , ⊗ , ∼ , 1) is an EQ-algebra, where the implication operation → is the Łukasiewicz implication.

Definition 2.3. [3, 10] Let ɛ be an EQ-algebra. For all x, y, z, m ∈ E, we say that ɛ is

bounded if it has a a bottom element 0;

According to [3, 10], every residuated EQ-algebra is good and every good EQ-algebra is separated; Every prelinear and good EQ-algebra is an ℓEQ-algebra; Every finite EQ-algebra is a lattice-ordered EQ-algebra; In any separated EQ-algebra, x ≤ y iff x → y = 1.

Proposition 2.4. [3, 10] Let ɛ be an EQ-algebra. For any x, y, z, u ∈ E, we have:

x ⊗ y ≤ x, y, x ⊗ y ≤ x ∧ y ≤ x → y;

Definition 2.5. [4] Given an EQ-algebra ɛ, for x, y, z ∈ E, F ⊆ E is called a prefilter of ɛ if it satisfies:

1 ∈ F;

A prefilter F is proper if F ≠ E. If ɛ is bounded, F is proper iff 0 ∉ F. {1} is a prefilter in a semi-separated EQ-algebra and is a filter in a separated EQ-algebra. Denote by PF ( ɛ ) the set of all prefilters of ɛ. If F is a nonempty subset of E, for all x, y ∈ E, denote: (F4) x ∈ F, x ≤ y imply y ∈ F; (F5) x, y ∈ F imply x ⊗ y ∈ F and (F6) x, y ∈ F implyx ∧ y ∈ F.

Lemma 2.6. Let ɛ be an EQ-algebra and ∅ ≠ F ⊆ E. We have the following:

If F is a prefilter of ɛ, then (F4) and (F6) follow;

Proof. (1) (F4) is clear and (F6) follows from (F4) and y ≤ x → y = x → x ∧ y.

By 1 → y ∈ F and x → (x ⊗ y) = (x ⊗ 1) → (x ⊗ y) ∈ F for any x, y ∈ F.

Definition 2.7. [7] A prefilter F of an EQ-algebra ɛ is called a positive implicative prefilter if x → y ∈ F, x → (y → z) ∈ F imply x → z ∈ F for all x, y, z ∈ E.

Let F be a prefilter of an EQ-algebra ɛ and ∅ ≠ A ⊆ E, a ∈ E. Denote by <A> (<F∪ {a} >) the generated prefilter by A (F, a). We abbreviate < {a}> (<F∪ {a} >) by <a> (F _a ), respectively, where <a> is called the principal prefilter. We also denote by PFP ( ɛ ) the set of all principle prefilters of ɛ. Write x → ⁰y = y,x →  ⁿ y = x → (x → ^n-1y). By [9], <A > = {x ∈ E : a₁ → (a₂ → (⋯ → (a _n  → x) ⋯)) =1 for some a _i  ∈ A, n ≥ 1}, F _a  = < F ∪ {a} > = {x ∈ E : a →  ⁿ x ∈ F for some n ≥ 1}. In particular, if F is positive implicative, then F _a  = {x ∈ E : a → x ∈ F} , < a > = < {a} > = {x ∈ E : a →  ⁿ x = 1 for some n ≥ 1}.

Lemma 2.8. [9] Given an EQ-algebra ɛ and a, b ∈ E, we have:

<a∧ b > = < a > ∨ < b >, where <a> ∨ <b> = << a > ∪ <b>>;

Definition 2.9. [14] Given a fuzzy set μ of ɛ, for all x, y, z ∈ E, μ is called a fuzzy prefilter if it satisfies:

μ (1) ≥ μ (x);

Proposition 2.10. [14] (1) A fuzzy prefilter μ of an EQ-algebra ɛ is isotone.

(2) If μ is a fuzzy filter of an EQ-algebra ɛ, then μ (x ⊗ y) = μ (x) ∧ μ (y) , μ (x ∧ y) ≥ μ (x) ∧ μ (y).

Definition 2.11. [4] An EQ_Δ-algebra is a structure ɛ_Δ = (E, ∧ , ⊗ , ∼ , Δ, 0, 1) that is a good EQ-algebra with a bottom element 0 expanded by a unary operation Δ : E → E fulfilling the following axioms: for all a, b ∈ E,

Δ1 = 1;

Let ɛ_Δ be an EQ_Δ-algebra. It is well known that Δ is isotone and Δ (a → b) ≤ Δa → Δb. A prefilter F of ɛ_Δ is said to be a Δ-prefilter if Δa ∈ F whenever a ∈ F. Clearly, {1} is a Δ-filter of ɛ_Δ. For every ∅ ≠ A ⊆ E, denote by <A > _Δ the generated Δ-prefilter of ɛ_Δ, then <A > _Δ = {a ∈ E : Δb₁ → (Δb₂ → (⋯ → (Δb _n  → a) ⋯)) =1 for some b₁, …, b _n  ∈ A, n ∈ N}; <a > _Δ = {x ∈ E : Δa → x = 1}; <F ∪ G > _Δ = {x ∈ E : f → (g → x) =1, f ∈ F, g ∈ G}, for any Δ-prefilters F, G of ɛ_Δ (see [4]).

3 Co-annihilators in EQ-algebras

In the following sequels, we assume that ɛ = (E, ∧ , ⊗ , ∼ , 1) is a lattice-ordered EQ-algebra unless otherwise stated.

Definition 3.1. Let F be a prefilter (filter) of ɛ and let a ∈ E. The co-annihilators of a relative to F is the set (F : a) = {x ∈ E : a ∨ x ∈ F}.

Example 3.2. Let E = {0, a, b, c, d, 1} with 0 < a < b < d < 1, a < c < d. Define operations ⊗, ∼ on E as follows:

Then ɛ = (E, ∧ , ⊗ , ∼ , 1) is an EQ-algebra [10]. One can check that F₁ = {d, 1} is a prefilter and F₂ = {1} is a filter of ɛ, and hence (F₁ : 0) = (F₁ : a) = F₁, (F₁ : b) = {c, d, 1} , (F₁ : c) = {b, d, 1} , (F₁ : d) = (F₁ : 1) = E, (F₂ : 0) = (F₂ : a) = (F₂ : b) = (F₂ : c) = (F₂ : d) = F₂, (F₂ : 1) = E.

Definition 3.3. A lattice-ordered EQ-algebra ɛ is said to be distributive if (E, ∧ , ∨) is a distributive lattice.

According to [3], every prelinear and separated ℓEQ-algebra is a distribute EQ-algebra.

Definition 3.4. A proper prefilter F of ɛ is called prime if x ∨ y ∈ F implies x ∈ F or y ∈ F for any x, y ∈ E.

Example 3.5. Consider Example 3.2 and one can check that the set {b, d, 1} is a prime prefilter of ɛ.

Proposition 3.6. Let F, G be prefilters of ɛ and let a, b ∈ E. We have:

If x ∈ (F : a) and x ≤ y, then y ∈ (F : a);

Proof. We shall prove (7) and (10) and the other proofs are easy.

(7) By (6), (F : a ∧ b) ⊆ (F : a) ∩ (F : b). Conversely, if x ∈ (F : a) ∩ (F : b), then x ∨ a, x ∨ b ∈ F. It follows from Lemma 2.6 that x ∨ (a ∧ b) = (x ∨ a) ∧ (x ∨ b) ∈ F.

(10) We have that x ∈ ((F : a) : b) iff x ∨ b ∈ (F : a) iff (x ∨ b) ∨ a ∈ F iff x ∨ (a ∨ b) ∈ F iff x ∈ (F : a ∨ b). In a similar, x ∈ ((F : b) : a) iff x ∈ (F : a ∨ b).

Proposition 3.7. For any a, b ∈ E, we have:

(< a > : a) = E;

Proof. (1) By (5) of Proposition 3.6.

(2) By (1), Lemma 2.8 and (9) of Proposition 3.6, (< a ∨ b > : a) = (< a > ∩ < b > : a) = (< a > : a) ∩ (< b > : a) = E ∩ (< b > : a) = (< b > : a).

Proposition 3.8. Let F be a positive implicative prefilter of an ℓEQ-algebra ɛ and let a ∈ E \ F. Then F = F _a  ∩ (F : a).

Proof. If x ∈ F _a  ∩ (F : a), then x ∨ a, a → x ∈ F since F is positive implicative. Thus a → x = (a ∨ x) → x ∈ F and so x ∈ F.

Theorem 3.9. Let F be a prefilter of an ℓEQ-algebra ɛ and let a ∈ E. We have:

(F : a) is a prefilter of ɛ;

Proof. (1) Clearly, 1 ∈ (F : a). If x, x → y ∈ (F : a), then x ∨ a, (x → y) ∨ a ∈ F. Denote y ∨ a = t, then y, a ≤ t and so x → y ≤ x → t, a ≤ t ≤ x → t. Thus (x → y) ∨ a ≤ (x → t) ∨ a = x → t. By (x → y) ∨ a ∈ F, x → t ∈ F. Since ɛ is an ℓEQ-algebra, then x → t ≤ (x ∨ a) → (t ∨ a). Combining x ∨ a, x → t ∈ F, we obtain y ∨ a = (y ∨ a) ∨ a = t ∨ a ∈ F, which shows y ∈ (F : a).

(2) We have a ∨ 0 = a ∉ F, that is, 0 ∉ (F : a).

(3) By (p9) a ∨ ((x ⊗ z) → (y ⊗ z)) ≥ a ∨ (x → y).

Definition 3.10. Let F be a prefilter (filter) of ɛ. The co-annihilators of a nonempty set A of E is the set ^⊥A = {x ∈ E: a ∨ x = 1 for any a ∈ A}. If A = {a}, denote ^⊥ {a} = {x ∈ E: a ∨ x = 1} by ^⊥a.

In a semi-separated EQ-algebra, ({1} ^:a)=⊥a.

Example 3.11. Let E = {0, a, b, c, d, e, f, 1} with 0 < a < c < d < e < 1, 0 < b < c < d < f < 1. Define operations ⊗, ∼ on E as follows:

Then ɛ = (E, ∧ , ⊗ , ∼ , 1) is an EQ-algebra [10]. It is easy to see that ^⊥A = ^⊥ {e, 1} = {f, 1}.

Proposition 3.12. For any x, y ∈ E, A, B, A _i  ⊆ E,

if x ∈ ^⊥A and x ≤ y, then;

Proof. It can be easily proven.

Proposition 3.13. For any a, b ∈ E, we have:

^⊥⊥a = {x ∈ E : x ∨ y = 1 for any y ∈ E such that y ∨ a = 1};

Proof. (5) If a ≤ b and ɛ is semi-separated, then {1} is a prefilter of ɛ, and hence by Proposition 3.6, ^⊥a = ({1} : a) ⊆ ({1} : b) = ^⊥b. Further from Proposition 3.12, ^⊥a ∩ ^⊥b = ^⊥ (a ∧ b).

(6) By x ∈ ^⊥ (a ∧ b) iff x ∨ (a ∧ b) = (x ∨ a) ∧ (x ∨ b) =1 iff x ∨ a = 1 and x ∨ b = 1 iff ^⊥a ∩ ^⊥b.

The other proofs are easy and here omit them.

Theorem 3.14. Let A ⊆ E. We have:

^⊥ A is a prefilter of ɛ if one of the following conditions holds:

ɛ is a separated EQ-algebra;

Proof. (1) (a) Clearly, 1 ∈ ^⊥ A. If x, x → ∈ ^⊥ A, then for any a ∈ A, x ∨ a = 1, (x → y) ∨ a = 1. Denote y ∨ a = z, we have y, a ≤ z, x → y ≤ x → z, a ≤ z ≤ x → z. Hence 1 = (x → y) ∨ a ≤ x → z and so x → z = 1. Since ɛ is separated, then x ≤ z. Combining 1 = a ∨ x ≤ z, we obtain z = 1.

(b) As {1} is a prefilter of ɛ, it follows from the proof of Theorem 3.9.

(2) By A ≠ {1}, there is t ∈ A such that t ≠ 1. It implies 0 ∨ t = t ≠ 1 and so 0 ∉ ^⊥A.

(3) By 1 = a ∨ (x → y) ≤ a ∨ (x ⊗ z → y ⊗ z) whenever a ∈ A, x → y ∈ ^⊥A.

Proposition 3.15. If F is a prefilter of ɛ, then F ∩ ^⊥F = {1}.

Proof. Trivial.

We shall introduce and investigate several pre-filters which have a help for the research of co-annihilators.

Definition 3.16. A proper prefilter F of an EQ-algebra ɛ is called

irreducible if F = F₁ ∩ F₂ implies F = F₁ or F = F₂ for any proper prefilters F₁, F₂ of ɛ;

Example 3.17. In Example 3.2, {c, d, 1} is an irreducible prefilter and {b, c, d, 1} is a maximal prefilter of ɛ.

Theorem 3.18. In any ℓEQ-algebra ɛ, each irreducible prefilter is a prime prefilter.

Proof. Let F be an irreducible prefilter of ɛ and let a ∨ b ∈ F. By Lemma 2.8 F = F _a  ∩ F _b , which implies F = F _a or F = F _b . Therefore a ∈ F or b ∈ F.

Lemma 3.19. Given an EQ-algebra ɛ and F ∈ PF ( ɛ ) , F is irreducible if and only if for every a, b ∉ F, there exist c ∉ F, n ≥ 0 such that a →  ⁿ c, b →  ⁿ c ∈ F.

Proof. Let F be an irreducible prefilter and a, b ∉ F. Since F ≠ F _a , F ≠ F _b , then F ⊂ F _a  ∩ F _b . Hence there exists c ∈ E such that c ∈ F _a  ∩ F _b , but c ∉ F. It follows that a →  ^m c ∈ F and b →  ⁿ c ∈ F for some n ≥ m. Applying a →  ^m c ∈ F ≤ b →  ⁿ c ∈ F, we have a →  ⁿ c, b →  ⁿ c ∈ F. Conversely, let F , F 1 , F 2 ∈ PF ( ɛ ) such that F = F₁ ∩ F₂, but F₁ ≠ F and F₂ ≠ F. Then there are a ∈ F₁ - F and b ∈ F₂ - F. Hence by the assumption there exist c ∉ F and n ≥ 0 such that a →  ⁿ c, b →  ⁿ c ∈ F. This implies a →  ⁿ c, b →  ⁿ c ∈ F₁ and a →  ⁿ c, b →  ⁿ c ∈ F₂. Considering a ∈ F₁, b ∈ F₂, it follows that c ∈ F₁ ∩ F₂ = F, which is a contradiction.

Definition 3.20. Let ɛ be an EQ-algebra. A subset I ⊆ E is called an ideal of ɛ if for any x, y, z ∈ E:

x ≤ y and y ∈ I imply x ∈ I;

Example 3.21. In Example 3.2, it is easy to see that the set {0, a, c} is an ideal of ɛ.

We denote the set of all ideals of ɛ by I ( ɛ ) .

Theorem 3.22. Let ɛ be an EQ-algebra and let F ∈ PF ( ɛ ) , I ∈ I ( ɛ ) . If F∩ I = ∅, then there exists an irreducible prefilter J such that F ⊆ J and J∩ I = ∅.

Proof. Set H = { K ∈ PF ( ɛ ) : F ⊆ K , K ∩ I = ∅ } . First, F≠ ∅ from F ∈ H . Since each chain of H has an upper bound in H , by Zorn’lemma, there exists a maximal element M of H . Now, we prove M is an irreducible prefilter of ɛ. Let a, b ∉ M. Then M ⊂ M _a  ∩ M _b and hence M a , M b ∉ H . This implies M _a ∩ I ≠ ∅ , M _b  ∩ I ≠ ∅. Thus there exist x, y ∈ I such that a →  ⁿ x ∈ M, b →  ⁿ y ∈ M for some n ≥ m ≥ 0. Considering b →  ^m y ≤ b →  ⁿ y ∈ M and I is an ideal, we have that there is c ∈ I such that x ≤ c and y ≤ c. Therefore a →  ⁿ x ≤ a →  ⁿ c ∈ M and b →  ⁿ y ≤ n →  ⁿ c ∈ M. It follows from Lemma 3.19 that M is an irreducible prefilter of ɛ.

Corollary 3.23. Let ɛ be an EQ-algebra and let F ∈ PF ( ɛ ) . Then

for each a ∉ F, there is an irreducible prefilter R such that a ∉ R and F ⊆ R.

We have the below theorem by the above results.

Theorem 3.24. Given a separated ℓEQ-algebra ɛ and A ⊆ E, we have ^⊥<A> = ^⊥A.

Proof. Clearly, ⊥ 〈 A 〉 = ⊥ A . Then we shall prove that x ∈ ^⊥ A implies x ∨ a = 1 for any a∈ < A >. If there is a∈ < A > such that a ∨ x ≠ 1, then there exist a₁, ⋯ , a _n  ∈ A such that a₁ → (a₂ → (⋯ → (a _n  → x) ⋯)) =1. As x ∈ ^⊥ <A>, x ∨ a_i = 1 for every x _i  ∈ A, i = 1, 2, ⋯ , n. Considering a ∨ x ≠ 1 and {1} is a prefilter, it follows from Corollary 3.23 and Theorem 3.18 that there exists an irreducible prefilter R of ɛ such that a ∨ x ∉ R. This results in x, a ∉ R. On the other hand, since x ∨ a _i  = 1 and R is a prime prefilter, we have a _i  ∈ F, i = 1, ⋯ , n. Applying a₁ → (a₂ → (⋯ → (a _n  → x) ⋯)) =1 ∈ R, a ∈ R, a contradiction. Therefore a ∨ x = 1 for any a∈ < A > and consequently x ∈ ^⊥ <A>.

Theorem 3.25. Let ɛ be a separated and distributive EQ-algebra. If F is a proper and linear prefilter of ɛ, then ^⊥ F is a prime prefilter of ɛ.

Proof. By Theorem 3.14, ^⊥F is a prefilter of ɛ. Let a, b ∉ ^⊥F such that a ∨ b ∈ ^⊥ F. Then there exist x, y ∈ F such that a ∨ x ≠ 1, b ∨ y ≠ 1. Hence z = x ∧ y ∈ F from Lemma 2.6. Since ɛ is distributive, we have a ∨ z = a ∨ (x ∧ y) = (a ∨ x) ∧ (a ∨ y) ≠1, b ∨ z = b ∨ (x ∧ y) = (b ∨ x) ∧ (b ∨ y) ≠1. On the other hand, z ≤ a ∨ z, b ∨ z follow from z ≤ a ∨ z, b ∨ z ∈ F. As F is linear, without loss of generality, we suppose b ∨ z ≤ a ∨ z. Thus, 1 = (a ∨ b) ∨ z = a ∨ (b ∨ z) ≤ a ∨ (a ∨ z) = a ∨ z and so a ∨ z = 1, which is a contradiction. Therefore ^⊥F is a prime prefilter of ɛ.

It is easy to see that ( PF ( ɛ ) , ∧ , ∨ , { 1 } , E } is a bounded lattice whence F∧ G = F ∩ G, F ∨ G = < F ∪ G > for any F , G ∈ PF ( ɛ ) .

Theorem 3.26. Let ɛ be a separated EQ-algebra. Then ( PF ( ɛ ) , ∧ , ∨ , { 1 } , E ) is a pseudo-complementedlattice, where ^⊥F is the pseudo-complemented of F ∈ PF ( ɛ ) .

Proof. By Proposition 3.15, F ∩ ⊥ F = { 1 } . Assume that J ∈ PF ( ɛ ) such that F ∩ J = {1}. If a ∈ J, then for any x ∈ F, x, a ≤ a ∨ x, which conclude x ∨ a ∈ F ∩ J = {1}. Thus x ∈ ^⊥F, i.e.,J ⊆ ⊥ F .

The set of all co-annihilators of ɛ is denoted by COA ( ɛ ) . That is, COA ( ɛ ) = { ⊥ A : A ⊆ E } . Clearly, { 1 } , E ∈ COA ( ɛ ) . If ɛ is a separated ℓEQ-algebra, it follows from Theorem 3.24 that COA ( ɛ ) = { ⊥ F : F ∈ PF ( ɛ ) } .

Proposition 3.27. Let F be a prefilter of a separated ℓEQ-algebra ɛ. Then F ∈ COA ( ɛ ) iff ^⊥⊥F = F.

Proof. If F ∈ COA ( ɛ ) , then there exists G ∈ PF ( ɛ ) such that F = ^⊥G. Hence ^⊥⊥F = ^⊥⊥⊥G = ^⊥G = F. The converse is true since ^⊥F is a prefilter of ɛ.

Theorem 3.28. Given a separated ℓEQ-algebra ɛ, we denote F ∧  _C J = F ∩ J, F ∨  _C J = ^⊥ (^⊥F ∩ ^⊥J) for all F , J ∈ COA ( ɛ ) . Then ( COA ( ɛ ) , ∧ C , ∨ C , ⊥ , { 1 } , E ) is a Boolean algebra.

Proof. Firstly, ( COA ( ɛ ) , ∧ c , ∨ c , { 1 } , E ) is a bounded lattice. Indeed, if F , J ∈ COA ( ɛ ) , then there exist A, B ⊆ E such that F = ^⊥A, J = ^⊥B. Hence F ∩ J = ⊥ A ∩ ⊥ B = ⊥ ( A ∪ B ) ∈ COA ( ɛ ) , and so COA ( ɛ ) is a ∧-semilattice. Since ^⊥F ∩ ^⊥J ⊆ ^⊥F, ^⊥J, we have F = ^⊥⊥F ⊆ ^⊥ (^⊥F ∩ ^⊥J) , J = ^⊥⊥J ⊆ ^⊥ (^⊥F ∩ ^⊥J). So ^⊥ (^⊥F ∩ ^⊥J) is an upper bound of {F, J}. Suppose now that H ∈ COA ( ɛ ) is any upper bound of {F, J}, i.e., F ∪ J ⊆ H. It follows that ^⊥ (^⊥F ∩ ^⊥J) = ^⊥⊥ (F ∪ J) ⊆ ^⊥⊥H = H. Therefore F ∨  _C J = ^⊥ (^⊥F ∩ ^⊥J).

Next, ^⊥F is the complement of F ∈ COA ( ɛ ) in COA ( ɛ ) , as F ∧ C ⊥ F = F ∩ ⊥ F = { 1 } , F ∨ C ⊥ F = ⊥ ( ⊥ F ∩ ⊥ ⊥ F ) = ⊥ { 1 } = E .

Finally, we shall show that ( COA ( ɛ ) , ∧ C , ∨ C ) is a distributive lattice. Let H , F , J ∈ COA ( ɛ ) such that L = (H ∧  _C F) ∨  _C  (H ∧  _C J). Then L ∈ COA ( ɛ ) , H ∧ C F ⊆ ⊥ ⊥ L . Further H ∧ C F ∧ C ⊥ L = { 1 } and hence H ∧ C ⊥ L ⊆ ⊥ F . Similarly, H ∧ C ⊥ L ⊆ ⊥ J . This implies H ∧ C ⊥ L ⊆ ⊥ F ∧ C ⊥ J = ⊥ ( ⊥ F ∩ ⊥ J ) and so H ∧ C ⊥ L ∧ C ( ⊥ F ∧ C ⊥ J ) = { 1 } . Therefore we see that H ∧ C ( F ∨ C J ) = H ∧ C ⊥ ( ⊥ F ∩ ⊥ J ) ⊆ ⊥ ⊥ L = L . It follows that H ∧  _C  (F ∨  _C F) ⊆ (H ∧  _C F) ∨  _C  (H ∧  _C J).

Given an EQ_Δ-algebra ɛ_Δ and a Δ-prefilter F of ɛ_Δ, we define (A : F) _Δ = {a ∈ E : Δa ∨ x ∈ F, for all x ∈ A}, which is called a Δ-co-annihilator of F with respect to A. In particular, denote (x : F) _Δ : = ({x} : F) _Δ = {a ∈ E : Δa ∨ x ∈ F}; ^⊥_ΔA : = (A : {1}) _Δ = {a ∈ E : Δa ∨ x = 1, for all x ∈ A}; ^⊥_Δa : = ({x} : {1}) _Δ = {b ∈ E : Δb ∨ a = 1}.

Example 3.29. Let E = {0, a, b, c, d, e, f, 1} with 0 < a, b < d, a < c, d < f, c < e, f < 1. Define operations ⊗, ∼ on E as follows:

Then ɛ = (E, ∧ , ⊗ , ∼ , 1) is a good EQ-algebra. Define an operation Δ on E by Δ0 = Δa = Δc = 0, Δb = Δd = Δf = b, Δe = e, Δ1 = 1. Then ɛ_Δ is an EQ_Δ-algebra and F = {d, f, 1} is a Δ-prefilter of ɛ_Δ. Take A = {a, c, d, e}, it is routine to verify that (A : F) _Δ = {b, d, f, 1} , ^⊥_ΔA = {1} , ^⊥_Δb = (b : F) _Δ = {e, 1}.

Theorem 3.30. Let F be a Δ-prefilter of an ℓEQ_Δ-algebra ɛ_Δ and let A ⊆ E. Then (A : F) _Δ is a Δ-prefilter of ɛ_Δ.

Proof. Firstly, 1 ∈ (A : F) _Δ as Δ1 ∨ x = 1 ∨ x = 1 ∈ F for all x ∈ A. Next, we put a, a → b ∈ (A : F) _Δ. Then Δa ∨ x ∈ F and Δ (a → b) ∨ x ∈ F. Hence it follows from Δ (a → b) ∨ x ≤ (Δa → Δb) ∨ x that (Δa → Δb) ∨ x ∈ F. Assume Δb ∨ x = t. Then Δb, x ≤ t and thus Δa → Δb ≤ Δa → t, x ≤ t ≤ Δa → t. This implies (Δa → Δb) ∨ x ≤ Δa → t and so Δa → t ∈ F. Since ɛ is an ℓEQ-algebra, we have Δa → t ≤ (Δa ∨ x) → (t ∨ x). Combing that F is a prefilter of ɛ_Δ and Δa → t, Δa ∨ x ∈ F, we obtain t = t ∨ x ∈ F. Therefore (A : F) _Δ is a prefilter of ɛ_Δ. Finally, let a ∈ (A : F) _Δ. By item (Δ3), Δa ∨ x ≤ ΔΔa ∨ x for all x ∈ A, and hence, ΔΔa ∨ x ∈ F. That is, Δa ∈ (A : F) _Δ. Consequently, (A : F) _Δ is a Δ-prefilter of ɛ_Δ.

Corollary 3.31. Let F be a Δ-prefilter of an ℓEQ_Δ-algebra ɛ_Δ and let a ∈ E. Then ^⊥_ΔF and ^⊥_Δa are prefilters of ɛ_Δ.

Corollary 3.32. Let F be a Δ-prefilter of an ℓEQ_Δ-algebra ɛ_Δ and let A ⊆ E. Then (A : F) is a prefilter of ɛ, where (A : F) = {a ∈ E : a ∨ x ∈ F, for all x ∈ A} is called the co-annihilator of A w.r.t. F.

We denote by PF Δ ( ɛ Δ ) the set of all Δ-prefilters of an EQ_Δ-algebra ɛ_Δ. For any F , G ∈ PF Δ ( ɛ Δ ) , define F ∧ _ΔG = F ∩ G, F ∨ _ΔG = < F ∪ G > _Δ. Then ( PF Δ ( ɛ ) , ∧ Δ , ∨ Δ , { 1 } , E ) is a bounded lattice.

Theorem 3.33. Let ɛ_Δ be an ℓEQ_Δ-algebra. Then ( PF Δ ( ɛ Δ ) , ∧ Δ , ∨ Δ , { 1 } , E ) is a relative pseudo-complemented lattice, where (F : G) _Δ is the relative pseudo-complement of F w.r.t. G in PF Δ ( ɛ Δ ) .

Proof. By Theorem 3.30, (F : G) _Δ is a Δ-prefilter of ɛ_Δ. Let F , G ∈ PF Δ ( ɛ Δ ) and a ∈ F ∩ (F : G) _Δ. Then a ∈ F and Δa ∨ x ∈ G for all x ∈ F. Hence taking x = a, we get a = Δa ∨ a ∈ G, which implies F ∩ (F : G) _Δ ⊆ G. Suppose that b ∈ M such that F ∩ M ⊆ G, we shall prove M ⊆ (F : G) _Δ. In fact, set b ∈ M. Then Δb ∈ M as M is a Δ-prefilter of ɛ_Δ. Considering Δb, x ≤ Δb ∨ x for all x ∈ F, we can obtain Δb ∨ x ∈ F ∩ M and further Δb ∨ x ∈ G. That is, M ⊆ (F : G) _Δ and this proof is complete.

Corollary 3.34. Let ɛ be an ℓEQ_Δ-algebra. Then ( PF Δ ( ɛ Δ ) , ∧ Δ , ∨ Δ , { 1 } , E ) is a pseudo-complemented lattice, where ^⊥_ΔF is the pseudo-complement of F in PF Δ ( ɛ Δ ) .

Corollary 3.35. Let ɛ be an ℓEQ-algebra. Then ( PF ( ɛ ) , ∧ , ∨ , { 1 } , E ) is a relative pseudo-complemented lattice, where (F : G) is the relative pseudo-complement of F w.r.t. G in PF ( ɛ ) .

Lemma 3.36. Let ɛ_Δ be an EQ_Δ-algebra. We have:

If a ≤ b, then <b > _Δ ⊆ < a > _Δ;

Proof. (1) and (2) are evident.

(3) By (1) <a > _Δ ∨ < b > _Δ ⊆ < a ∧ b > _Δ. Put now x ∈ < a ∧ b > _Δ. Then Δ (a ∧ b) → x = 1 and hence (Δa ∧ Δb) → x = 1. By (p14), a ≤ 1 → a = 1 ⊗ (1 → a) = [(Δa ∧ Δb) → x] ⊗ (1 → a) ≤ (1 ∧ Δb) → x = Δb → x. It follows from Δa ≤ a that Δa → (Δb → x) =1, that is, x ∈ < {a, b} > _Δ. From (2) x ∈ < a > _Δ ∨ < b > _Δ.

(4) Let x ∈ ^⊥_Δ < a > _Δ. Then for all t ∈ < a > _Δ, Δx ∨ t = 1. Hence Δx ∨ a = 1. That is, x ∈ < a > _Δ. Conversely, if x∈ ^⊥_Δa, y ∈ < a >, then Δx ∨ a = 1, Δa → y = 1. Thus Δx ∨ Δa = Δ (Δx ∨ a) = Δ1 = 1, Δa ≤ y. It follows that Δx ∨ y ≥ Δx ∨ Δa = 1, which results in Δx ∨ y = 1 for all y ∈ < a > _Δ, and consequently, x ∈ ^⊥_Δ < a > _Δ.

(5) By Theorem 4 in [4], ɛ_Δ is a prelinear ℓEQ_Δ-algebra. Firstly, we shows ^⊥_Δ < a > _Δ ⊆ < ¬ Δa > _Δ. Set x ∈ ^⊥_Δ < a > _Δ. Then Δx ∨ a = 1 and hence Δx ∨ Δa = Δ (Δx ∨ a) = Δ1 = 1. Applying (p8) and (p15), we have that 1 = (Δa ∨ Δx) ∨ x = Δa ∨ (Δx ∨ x) ≤ (Δa → 0) → (0 ∨ (x ∨ Δx)) ≤ Δ (¬ Δa) → x. Thus, Δ (¬ Δa) → x = 1, that is, x ∈ < ¬ Δa > _Δ. On the other hand, we put x ∈ < ¬ Δa > _Δ. Then Δ (¬ Δa) → x = 1. Since Δa ≤ a, then ¬Δa ≥ ¬ a and so Δ (¬ Δa) ≥ Δ (¬ a). Therefore 1 = Δ (¬ Δa) → x ≤ Δ (¬ a) → x, which implies Δ (¬ a) → x = 1. Considering 1 = Δ1 = Δ (Δ (¬ a) → x) ≤ Δ (¬ a) → Δx, we see that Δ (¬ a) → Δx = 1. As ɛ_Δ is a prelinear ℓEQ_Δ-algebra with ¬a ∨ a = 1, it follows from (p8), (p16) that 1 = (Δ¬ a → Δx) ∨ (Δ (¬ a) → a) = Δ (¬ a) → (Δ x ∨ a) ≤ (Δ (¬ a) ∨ Δ a) → ((Δ x ∨ a) ∨ Δ a) = Δ (a ∨ ¬ a) → (Δ x ∨ a) = Δ 1 → (Δ x ∨ a) = 1 → (Δ x ∨ a) = Δ x ∨ a. Therefore Δx ∨ t = 1 and further x ∈ ^⊥_Δ < a > _Δ.

The following example indicates that there exist prelinear EQ_Δ-algebras with ¬a ∨ a = 1 for any a ∈ E.

Example 3.37. Let E = {0, a, b, c, d, 1} with 0 < a, b < d, a < c, d < 1. Define operations ⊗, ∼ on E as follows:

Then ɛ = (E, ∧ , ⊗ , ∼ , 1) is an EQ-algebra and further it is not residuated as d ⊗ b ≤ 0 can’t imply d ≤ b → 0. Now define an operation Δ on E by Δ0 = Δa = 0, Δb = Δd = b, Δc = c, Δ1 = 1. Routine calculation shows that ɛ_Δ is a prelinear EQ_Δ-algebra with ¬a ∨ a = 1 for all a ∈ E.

Given an EQ_Δ-algebra ɛ_Δ, denote by <a > _Δ the principle Δ-prefilters of ɛ_Δ, and PFP Δ ( ɛ Δ ) the set of all principle Δ-prefilters of ɛ_Δ.

Theorem 3.38. Let ɛ_Δ be a prelinear EQ_Δ-algebra with ¬a ∨ a = 1 for all a ∈ E. Then ( PFP Δ ( ɛ Δ ) , ∧ Δ , ∨ Δ , ⊥ Δ , < 1 > Δ , E ) is a Boolean algebra.

Proof. By Theorem 4 in [4] and Theorem 3.24 in [9], we first know that ɛ_Δ is a prelinear ℓEQ_Δ-algebra and <1 > _Δ = {1} is a principle Δ-prefilter of ɛ_Δ, respectively. Then it follows from Lemma 3.36 that operations ∧_Δ, ∨ _Δ, ^⊥_Δ are well-defined in PFP Δ ( ɛ Δ ) , where <a > _Δ ∧ _Δ < b > _Δ = < a > _Δ ∩ < b > _Δ = <a ∨ b > _Δ, <a > _Δ ∨_Δ <b > _Δ = <a > _Δ ∨ <b > _Δ = <a ∧ b > _Δ, ^⊥_Δ < a > _Δ = < ¬ Δa > _Δ. Thus, according to Proposition 12 of [4], we see that ( PFP Δ ( ɛ Δ ) , ∧ Δ , ∨ Δ , < 1 > Δ , E ) is a bounded distributive lattice. Also, by Corollary 3.34, we get that ( PFP Δ ( ɛ Δ ) , ∧ Δ , ⊥ Δ , < 1 > Δ , E ) is a pseudo-complemented lattice, where ^⊥_Δ < a > _Δ is the pseudo-complement of <a > _Δ. In order to prove ( PFP Δ ( ɛ Δ ) , ∧ Δ , ∨ Δ , ⊥ Δ , < 1 > Δ , E ) is a Boolean algebra, it suffice to show that for any a ∈ E, ^⊥_Δ < a > _Δ = <1 > _Δ implies <a > _Δ = E (see [1]). Indeed, 1 = Δ1 = Δ (a ∨ ¬ a) = Δa ∨ Δ (¬ a) ≤ a ∨ Δ (¬ a) from a ∨ ¬ a = 1, and hence, Δ (¬ a) ∨ a = 1. This implies ¬a ∈ ^⊥_Δ < a > _Δ = <1 > _Δ = {1}, which concludes ¬a = 1. Then applying Δa → 0 ≥ a → 0 =1, we obtain Δa → 0 =1 and therefore 0 ∈ < a > _Δ. Consequently <a > _Δ = E.

Corollary 3.39. Let ɛ be a prelinear and good EQ-algebra with ¬a ∨ a = 1 for all a ∈ E. Then ( PFP ( ɛ ) , ∧ , ∨ , ⊥ , < 1 > , E ) is a Boolean algebra.

4 Fuzzy co-annihilators in EQ-algebras

In this section, we introduce two types of fuzzy co-annihilators in EQ-algebras.

Definition 4.1. Let ɛ be an EQ-algebra. A fuzzy relation R on E is called a fuzzy congruence relation if R is a fuzzy equivalence relation such that R (xΘu, yΘv) ≥ R (x, y) ∧ R (u, v) for any x, y, u, v ∈ E, whereby Θ ∈ {⊗ , ∧ , ∼}.

Clearly, R (x → u, y → v) ≥ R (x, y) ∧ R (u, v).

Definition 4.2. Let μ be a fuzzy set of E and R be a fuzzy congruence relation on ɛ. Define the fuzzy co-annihilator of μ w.r.t. R by for x ∈ E, Ann _R  (μ) (x) = ∧ _z∈E {μ (z) → R (1, z ∨ x)}, where → is the residuated implication w.r.t. a left-continuoust-norm.

Note that if A is a classic subset of E and R is the identity relation on A, we obtain the co-annihilator of A: ^⊥A = {x ∈ E: a ∨ x = 1 for any a ∈ A}.

Lemma 4.3. Let μ be a fuzzy filter of ɛ with μ (1) =1. Define a fuzzy relation R on E by R (x, y) = μ (x ∼ y) for x, y ∈ E. Then R is a fuzzy congruence relation on ɛ, which is called the generated fuzzy congruence relation by μ.

Proof. Firstly, R is a fuzzy equivalence relation on ɛ. Indeed, the Reflexivity and Symmetry of R hold by R (x, x) = μ (x ∼ x) = μ (1) =1, R (x, y) = R (y, x). Since μ is a fuzzy filter of ɛ, by (p11) and Proposition 2.10, the transitivity of R is as follows: R∘ R (x, z) = ∨ _y∈E {R (x, y) ∧ R (y, z)} = ∨ _y∈E {μ (x ∼ y) ∧ μ (y ∼ z)} ≤ ∨_y∈E {μ ((x ∼ y) ⊗ (y ∼ z))} ≤ ∨_y∈E μ (x ∼ z) = μ (x ∼ z) = R (x, z). Secondly, we shall show that R is a fuzzy congruence relation on ɛ. Since μ is a fuzzy filter of ɛ, it follows from (E5),(p12), Proposition 2.10 that R (xΘu, yΘv) = μ ((xΘu) ∼ (yΘv)) ≥ μ ((x ∼ y) ⊗ (u ∼ v)) ≥ μ (x ∼ y) ∧ μ (u ∼ v) = R (x, y) ∧ R (u, v), where Θ ∈ {∼ , ∧}. In what follows, by using (p11), (p13) and Proposition 2.10, we shall show μ ((x ⊗ u ∼ y ⊗ v) ≥ μ (x ∼ y) ∧ μ (u ∼ v). In fact, since x ∼ y ≤ x ↔ y ≤ x → y, y → x, then μ (x ∼ y) ≤ μ (x → y), μ (y → x). Also, μ (x → y) ≤ μ (x ⊗ u → y ⊗ u) and μ (y → x) ≤ μ (y ⊗ u → x ⊗ u). Considering μ (x ⊗ u ∼ y ⊗ u) ≥ μ (x ⊗ u ⇌ y ⊗ u) ≥ μ ((x ⊗ u → y ⊗ u) ∧ (y ⊗ u → x ⊗ u)), we can obtain μ (x ⊗ u ∼ y ⊗ u) ≥ μ (x ∼ y). Similarly, μ (y ⊗ u ∼ y ⊗ v) ≥ μ (u ∼ v). Therefore μ ((x ⊗ u ∼ y ⊗ v) ≥ μ ((x ⊗ u ∼ y ⊗ u) ⊗ (y ⊗ u ∼ y ⊗ v)) ≥ μ (x ⊗ u ∼ y ⊗ u) ∧ μ (y ⊗ u ∼ y ⊗ v) ≥ μ (x ∼ y) ∧ μ (u ∼ v). Combining the above argument, R is a fuzzy congruence relationon ɛ.

Example 4.4. Let E = {0, a, b, 1} with 0 < a < b < 1. Define operations ⊗, ∼ as follows:

Then ɛ = (E, ∧ , ⊗ , ∼ , 1) is an EQ-algebra [7]. Take a fuzzy set μ by μ (0) = μ (a) =0.2, μ (b) =0.7, μ (1) =1, then μ is a fuzzy filter of ɛ and the fuzzy congruence relation R is generated by μ. Hence it can be calculated that Ann _R  (μ) = E, where the residuated implication → is taken as Łukasiewicz implication.

Proposition 4.5. Let R be a fuzzy congruence relation on ɛ and μ, ν be two fuzzy sets of E. We have:

If μ ⊆ ν, then Ann _R  (ν) ⊆ Ann _R  (μ);

Proof. (2) Ann _R  (μ ∪ ν) (x) = ∧ _z∈E {μ (z) ∨ ν (z) → R (1, z ∨ x)} = ∧ _z∈E {(μ (z) → R (1, z ∨ x)) ∧ (ν (z) → R (1, z ∨ x))} = Ann _R (μ) (x) ∧ Ann _R (ν) (y).

(3) Ann _R  (μ ∩ ν) (x) = ∧ _z∈E {μ (z) ∧ ν (z) → R (1, z ∨ x)} = ∧_z∈E {(μ (z) → R (1, z ∨ x)) ∨ (ν (z) → R (1, z ∨ x))} = Ann _R (μ) (x) ∨ Ann _R (ν) (y).

The other proofs are easy.

Lemma 4.6. Let ɛ be a good EQ-algebra and μ be a fuzzy set of E. If μ satisfies (FF1), (FF4) x ≤ y implies μ (x) ≤ μ (y) and (FF5) μ (x ⊗ y) ≥ μ (x) ∧ μ (y), for any x, y ∈ E, then μ is a fuzzy prefilter of ɛ.

Proof. (FF2) follows from μ (y) ≥ μ (x ⊗ (x → y)) ≥ μ (x) ∧ μ (x → y).

Lemma 4.7. Let R be a fuzzy congruence relation on a good EQ-algebra ɛ. Define a fuzzy set μ (x) = R (1, x) for x ∈ E. Then μ is a fuzzy prefilter of ɛ.

Proof. (FF1) is clear and (FF2) is true since μ (x) ∧ μ (x → y) = R (1, x) ∧ R (1, x → y) ≤ R (1, x ⊗ (x → y)) ≤ R (y, y) ∧ R (x ⊗ (x → y) , 1) ≤ R (y ∨ (x ⊗ (x → y) , y ∨ 1) = R (1, y) = μ (y).

Theorem 4.8. Given a good EQ-algebra ɛ with (z ∨ x) ⊗ (z ∨ y) ≤ z ∨ (x ⊗ y) for all x, y ∈ E. We have:

Ann _R  (μ) is a fuzzy prefilter;

Proof. First, from Ann _R  (μ) (1) = ∧ _z∈E {μ (z) → R (1, z ∨ 1)} = ∧ _z∈E {μ (z) → R (1, 1)} =1 ≥ Ann _R  (μ) (x), (FF1) holds.

(1) If x ≤ y, then R (1, z ∨ x) ≤ R (1, z ∨ y), and hence, Ann _R  (μ) (x) = ∧ _z∈E [(μ (z) → R (1, z ∨ x)] ≤ ∧ _z∈E [(μ (z) → R (1, z ∨ x)] = Ann _R  (μ) (y), which shows (FF4). In the following, we shall prove (FF5). In fact, since (z ∨ x) ⊗ (z ∨ y) ≤ z ∨ (x ⊗ y), by Lemma 4.7 we get that Ann _R  (μ) (x ⊗ y) = ∧ _z∈E [(μ (z) → R (1, z ∨ (x ⊗ y))] ≥ ∧ _z∈E [(μ (z) → R (1, (z ∨ x) ⊗ (z ∨ y))] ≥ ∧_z∈E [(μ (z) → R (1, z ∨ x) ∧ R (1, z ∨ y)] = ∧_z∈E [(μ (z) → R (1, z ∨ x)) ∧ (μ (z) → R (1, z ∨ y))] = ∧_z∈E {μ (z) → R (1, z ∨ x)} ∧ ∧_z∈E {μ (z) → R (1, z ∨ y))} = Ann _R  (μ) (x) ∧ Ann _R  (μ) (y). It implies by Lemma 4,6 that Ann _R  (μ) is a fuzzy prefilter of ɛ.

(2) Since ɛ is residuated, then Ann _R  (μ) (x ⊗ u → y ⊗ u) = ∧ _z∈E [(μ (z) → R (1, x ⊗ u → y ⊗ u)] ≤ ∧ _z∈E [(μ (z) → R (1, z ∨ (x → y)) = Ann _R  (μ) (x → y), which implies (FF3).

Example 4.9. Consider ɛ = {0, a, b, c, 1} with 0 < a < b < c < 1 in which operations ⊗, ∼ are defined as follows:

Then ɛ is a good EQ-algebra [4]. Furthermore we can check that (z ∨ x) ⊗ (z ∨ y) ≤ z ∨ (x ⊗ y) for all x, y ∈ E.

Definition 4.10. Let μ, ν be two fuzzy sets of E. Define the fuzzy co-annihilator of μ w.r.t. ν as follows: for x ∈ E, Ann (μ, ν) (x) = ∧ _z∈E {μ (z) → ν (z ∨ x)}, where → is the residuated implication w.r.t. a left-continuous t-norm.

Example 4.11. In Example 3.2, take two fuzzy sets μ, ν by μ (0) = μ (a) = 0.4, μ (b) = 0.5, μ (c) = 0.8, μ (d) = μ (1) = 1; ν (0) = ν (a) = ν (b) = 0.2, ν (c) = ν (d) = ν (1) = 0.9. Then Ann (μ, ν) (0) = Ann (μ, ν) (a) = Ann (μ, ν) (b) = 0.7, Ann (μ, ν) (c) = Ann (μ, ν) (d) = Ann (μ, ν) (1) = 0.9, where the residuated implication → is Łukasiewicz implication.

Proposition 4.12. Let μ, ν be two fuzzy sets of E. We have:

Ann (χ_{0}, ν) = ν;

Proof. It is not hard from the definition of Ann (μ, ν).

Definition 4.13. For all x, y ∈ E, a fuzzy prefilter μ of ɛ is called a

fuzzy prime prefilter if it is nonconstant and μ (x ∨ y) = μ (x) ∨ μ (y);

Example 4.14. In Example 3.2. Give two fuzzy sets μ, ν as follows: μ (0) = μ (a) =0.1, μ (b) = μ (d) =0.8, μ (c) =0.5, μ (1) =1;ν (0) = ν (a) =0.2, ν (b) = ν (c) = ν (d) = ν (1) =0.8. Then μ is a fuzzy prime prefilter and ν is a fuzzy Boolean prefilter of ɛ.

Theorem 4.15. Assume that ɛ is a good EQ-algebra with (z ∨ x) ⊗ (z ∨ y) ≤ z ∨ (x ⊗ y) for all x, y ∈ E, and μ, ν is two fuzzy sets of E. We have:

If ν is a fuzzy filter, then Ann (μ, ν) is a fuzzy prefilter;

Proof. (1) According to Ann (μ, ν) (1) = ∧_z∈E {μ (z) → ν (z ∨ 1)} = ∧_z∈E {μ (z) → ν (1)} ≥ ∧_z∈E {μ (z) → ν (z ∨ x)} = Ann (μ, ν) (x), (FF1) follows. If x ≤ y, (FF4) follows from Ann (μ, ν) (x) = ∧_z∈E {μ (z) → ν (z ∨ x)} ≤ ∧_z∈E {μ (z) → ν (z ∨ y)} = Ann (μ, ν) (y). Also, Ann (μ, ν) (x ⊗ y) = ∧_z∈E {μ (z) → ν (z ∨ (x ⊗ y))} ≥ ∧_z∈E {μ (z) → ν ((z ∨ x) ⊗ (z ∨ y))} ≥ ∧_z∈E {μ (z) → ν (z ∨ x) ∧ ν (z ∨ y)} = ∧_z∈E {(μ (z) → ν (z ∨ x)) ∧ (μ (z) → ν (z ∨ y))} = ∧_z∈E {μ (z) → ν (z ∨ x)} ∧ ∧_z∈E {μ (z) → ν (z ∨ y)} = Ann (μ, ν) (x) ∧ Ann (μ, ν) (y), which shows (FF5). By Lemma 4.6 Ann (μ, ν) is a fuzzy prefilter.

(2) If ɛ is residuated, then Ann (μ, ν) (x ⊗ u → y ⊗ u) = ∧ _z∈E {μ (z) → ν (z ∨ (x ⊗ u → y ⊗ u))} ≥ ∧ _z∈E {μ (z) → ν (z ∨ (x → y)}. This shows (FF3) and so Ann (μ, ν) is a fuzzy filter.

(3) Since ν is a fuzzy prime filter, then Ann (μ, ν) (x ∨ y) = ∧_z∈E {μ (z) → ν (z ∨ (x ∨ y))} = ∧_z∈E {μ (z) → ν ((z ∨ x) ∨ (z ∨ y))} = ∧_z∈E {μ (z) → ν (z ∨ x) ∨ ν (z ∨ y)} = ∧_z∈E {(μ (z) → ν (z ∨ x)) ∨ (μ (z) → ν (z ∨ y))} = ∧_z∈E {μ (z) → ν (z ∨ x)} ∨ ∧_z∈E {μ (z) → ν (z ∨ y)} = Ann (μ, ν) (x) ∨ Ann (μ, ν) (y).

(4) Since ν is a fuzzy Boolean filter, it follows that Ann (μ, ν) (x ∨ y) = ∧ _z∈E {μ (z) → ν (z ∨ (x ∨ ¬ x))} ≥ ∧ _z∈E {μ (z) → ν (x ∨ ¬ x)} = ∧ _z∈E {μ (z) → ν (1)} = ∧ _z∈E {μ (z) → ν (1 ∨ z)} = Ann (μ, ν) (1).

Corollary 4.16. Let ɛ be a residuated ℓEQ-algebra and μ be a fuzzy set of E. If ν be a fuzzy filter of ɛ, then Ann (μ, ν) is a fuzzy filter of ɛ.

Proof. Since ɛ is residuated, we have that ɛ is good. By Theorem 4.15 it suffice to show (z ∨ x) ⊗ (z ∨ y) ≤ z ∨ (x ⊗ y). In fact, it follows from (p5), (p8) and (p9) that x ≤ y → (x ⊗ y) ≤ (z ∨ y) → (z ∨ (x ⊗ y)) and z ≤ z ∨ (x ⊗ y) ≤ (z ∨ y) → (z ∨ (x ⊗ y)). Hence, z ∨ x ≤ (z ∨ y) → (z ∨ (x ⊗ y)) and therefore (z ∨ x) ⊗ (z ∨ y) ≤ z ∨ (x ⊗ y).

The set of all fuzzy prefilters of an EQ-algebra ɛ is denoted by FPF ( ɛ ) ( FF ( ɛ ) ) . If μ is a fuzzy set of E, then the fuzzy prefilter generated by μ is defined as < μ > = ⋂ ν ∈ FPF ( ɛ ) , μ ⊆ ν ν and <μ> can be further represented as <μ> (x) = ∨ {μ (a₁) ∧ ⋯ ∧ μ (a _n ) | a₁, ⋯, a _n ∈ E, a₁ → (a₂ → (⋯ → (a _n → x) ⋯)) = 1}. In FPF ( ɛ ) , μ ⊆ ν iff μ (x) ≤ ν (x) and denote μ∧ ν = μ ∩ ν, μ ∨ ν = < μ ∪ ν >. Then ( FPF ( ɛ ) , ∧ , ∨ , ∅ , E ) is a bounded lattice (see [2]).

Lemma 4.17. Let ɛ be a residuated EQ-algebra. Then ( FF ( ɛ ) , ∧ , ∨ , ∅ , E ) is a bounded lattice.

Proof. To prove that μ ∨ ν is a fuzzy filter, from Theorem 4.15 it suffice to show (μ ∨ ν) (x ⊗ z → y ⊗ z) ≥ (μ ∨ ν) (x → y). Indeed, (μ ∨ ν) (x ⊗ z → y ⊗ z) = < μ ∪ ν > (x ⊗ z → y ⊗ z) = ∨ {(μ (a₁) ∨ ν (a₁)) ∧ ⋯ ∧ (μ (a _n ) ∨ ν (a _n )) | a₁, ⋯, a _n ∈ E, a₁ → (a₂ → (⋯ → (a _n → (x ⊗ z → y ⊗ z) ⋯)) = 1} ≥ ∨ {(μ (a₁) ∨ ν (a₁)) ∧ ⋯ ∧ (μ (a _n ) ∨ ν (a _n )) | a₁, ⋯, a _n ∈ E, a₁ → (a₂ → (⋯ → (a _n → (x → y) ⋯)) = 1} = (μ ∨ ν) (x → y).

Consider the residuated implication → as the Gödel residuated implication in the definition of St ^r  (μ, ν). We have the following theorem.

Theorem 4.18. Given a residuated ℓEQ-algebra ɛ, we have that ( FF ( ɛ ) , ∧ , ∨ , ∅ , E ) is a relative pseudo-complemented lattice, where Ann (μ, ν) is the relative pseudo-complement of μ w.r.t. ν in FF ( ɛ ) .

Proof. By Corollary 4.16 and Lemma 4.17, it suffice to prove Ann (μ, ν) ∩ μ ⊆ ν. Indeed, put μ , ν ∈ FF ( ɛ ) . Then for x ∈ E, (Ann (μ, ν) ∩ μ) (x) = Ann (μ, ν) (x) ∧ μ (x) = ∧_z∈E{μ (z) → ν (z ∨ x)} ∧ μ (x) ≤ μ (x) ∧ (μ (x) → ν (x)) ≤ ν (x). Now let λ ∈ FF ( ɛ ) such that λ ∩ μ ⊆ ν. Since λ is a fuzzy filter of ɛ, by Proposition 4.12, Ann (μ, ν) = ∧ _z∈E {μ (z) → ν (x ∨ z)} ≥ ∧_z∈E {μ (z) → λ (x ∨ z) ∧ μ (x ∨ z)} = ∧_z∈E {μ (z) → λ (x ∨ z)} ∧ ∧_z∈E {μ (z) → μ ((x ∨ z)} = Ann (μ, λ) (x) ∧ Ann (μ, μ) (x) = Ann (μ, λ) (x) ≥ λ (x). This shows λ ⊆ Ann (μ, ν). Therefore Ann (μ, ν) is the relative pseudo-complement of μ with respect to ν in FF ( ɛ ) .

5 Conclusions

In this paper, motivated by the previous research of co-annihilators in logic algebras, we introduce two types of (fuzzy) co-annihilators in an EQ-algebra and Δ-co-annihilators in an EQ_Δ-algebra. Several important results have been obtained. For example, we prove that the set of all prefilters in a separated EQ-algebra forms a pseudo-complemented lattice and the set of all fuzzy filters in a residuated ℓEQ-algebra forms a relative pseudo-complemented lattice, etc.