Filter theory of pseudo equality algebras

1 Introduction

As we all known, various fuzzy logic algebras have been widely introduced and studied as non-classical logic semantic systems. The well-known algebraic structure is residuated lattice. BL-algebras, MTL-algebras, MV-algebras, etc., are the best known classes of residuated lattice [11, 12]. With the further development of fuzzy logic, EQ-algebras [17] were proposed by V. Novák with the intent to develop an algebraic structure of truth values for fuzzy type theory. An EQ-algebra has three binary operations, meet, multiplication, fuzzy equality, and a unit element. And it generalized the residuated lattice. As S. Jenei mentioned in [14], if the product operation in EQ-algebras is replaced by another binary operation smaller or equal than the original product we still obtain an EQ-algebra, and this fact might make it difficult to obtain certain algebraic results. For this reason, S. Jenei introduced a new structure in [14] called equality algebras which is similar to EQ-algebras, but without a product. These algebras were assumed for a possible algebraic semantics of fuzzy type theory. So since the equality algebra was proposed, it has attracted extensive attention of many scholars [3, 19] and it is meaningful to study equality algebras. As a generalization of equality algebras, S. Jenei and L. Kóródi introduced the notion of pseudo equality algebras and proved that the pseudo equality algebras were term equivalent to pseudo BCK-meet-semilattices in [16]. Then L.C. Ciungu found a gap in the proof of this result and presented a counterexample and a correct version of the theorem. Also, the correct version of the corresponding result for equality algebras was given in [7]. Moreover, Dvurečenskij and Zahiri showed in [10] that every pseudo equality algebra in the sense of [16] was an equality algebra and they defined and investigated a new concept of pseudo equality algebras (called JK-algebras) and established a connection between pseudo equality algebras and a special class of pseudo BCK-meet-semilattices. Apart from their logical interest, equality algebras as well as pseudo equality algebras seem to have important algebraic properties and it is worth studying them from an algebraic point of view.

Filter is an important concept in the field of non-classical logical algebra. They play an important role in the study of the completeness of various logical systems. From a logical point of view, various filters correspond to various sets of provable formulas. In recent years, filters on equality algebras have been widely studied and many important conclusions have been obtained. In 2012, S. Jenei introduced the concept of filters in equality algebras and obtained some important properties in [14, 15]. Subsequently, in [18], M. Zarean and R.A. Borzooei introduced the notions of prime filters, maximal filters and principal filters in equality algebras, and discussed their properties and relations among them. For the first time, R.A. Borzooei systematically studied the concepts and properties of several special types of filters (implicative filters, positive implicative filters, fantastic filters and Boolean filters) in equality algebras, and discussed the relations among these filters in [4].

Now, in this paper, we give the generating formula of filters in pseudo equality algebras, then we introduce the notions of prelinear pseudo equality algebras and divisible pseudo equality algebras, and we investigate some characterizations of them. Moreover, we give the notions of (positive) implicative, fantastic filters in pseudo equality algebras inspired by [1, 9]. Then we investigate some properties of these filters and study their equivalent definitions. Also, the relations among them are discussed. This paper is organized as follows: In Section 2, we review some necessary notions and properties in pseudo equality algebras, which will be used in the next sections. In Section 3, we prove some properties that will be used in subsequent sections, and we give the generating formula of filters in pseudo equality algebras. Then we introduce prelinear pseudo equality algebras and divisible pseudo equality algebras and give some characterizations of them. Moreover, we investigate algebraic structures of the set F (X) of all filters in pseudo equality algebras X and obtain that F (X) can form a Heyting algebra. In Section 4, we define (positive) implicative filters, and investigate some properties and relations of them. Moreover, we give the notion of strong normal filters and discuss some properties. By using the notion of strong normal filters, we get the new relations between implicative filters and positive implicative filters. In Section 5, we introduce the notion of fantastic filters, then investigate some properties and the relations between fantastic filters and (positive) implicative filters.

2 Preliminaries

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

Definition 2.1. [10] An algebra (X, ∧ , ∼, ∽ , 1) of type (2, 2, 2, 0) is called a pseudo equality algebra (or a JK-algebra) if it satisfies the following conditions, for all x, y, z ∈ X:

(X, ∧ , 1) is a meet-semilattice with top element 1,

The operation ∧ is called meet (infimum) and ∼, ∽ are called pseudo equality operations. Define the relation x ≤ y if and only if x ∧ y = x, for all x, y ∈ X. Also, two other operations are defined, called implications: x → y = ( x ∧ y ) ∼ x x ⇝ y = x ∽ ( x ∧ y )

A pseudo equality algebra X is bounded if there exists an element 0 ∈ X, such that 0 ≤ x, for all x ∈ X. And we define two unary operations, x^- = x → 0, x^∼ = x ⇝ 0, for all x ∈ X.

Remark 2.2. [10] If (X, ∧ , ∼, 1) is an equality algebra, then (X ; ∧ , ∼, ∼, 1) is a pseudo equality algebra.

Proposition 2.3. [10] Let X be a pseudo equality algebra. Then the following properties hold for all x, y, z ∈ X:

x ∽ y ≤ x ⇝ y and y∼x ≤ x → y,

Proposition 2.4. [10] Let X be a pseudo equality algebra. Then the following properties hold for all x, y, z ∈ X:

y ≤ z implies that x → y ≤ x → z and x ⇝ y ≤ x ⇝ z,

Definition 2.5. [5] Let (X, ∧ , ∼, ∽ , 1) be a pseudo equality algebra. A subset F ⊆ X is called a filter (or deductive system) of X if for all x, y ∈ X:

1 ∈ F,

( XF 3 ′ ) If x, y∼x ∈ F, then y ∈ F.

Proposition 2.6. [5] Let (X, ∧ , ∼, ∽ , 1) be a pseudo equality algebra. Then the following hold:

F is a filter of X if and only if for all x, y ∈ X, it satisfies condition (XF₁) and the condition:

Proposition 2.7. [10] If F is a filter of a pseudo equality algebra (X, ∧ , ∼, ∽ , 1) and a, b ∈ F, then a → b, a ⇝ b, a ∧ b ∈ F.

A filter F of a pseudo equality algebra X is proper if F ≠ X. A proper filter is called maximal if it is not strictly contained in any other proper filter of X. We will denote by F (X) the set of filters of X.

Definition 2.8. [2] A Heyting algebra is an algebra (H, ∨ , ∧ , → , 0, 1) of type (2, 2, 2, 0, 0) for which (H, ∨ , ∧ , 0, 1) is a bounded lattice and for all a, b ∈ H, a → b is the relative pseudocomplement of a with respect to b, i.e., a ∧ c ≤ b if and only if c ≤ a → b.

3 New types of pseudo equality algebras

In this section, we firstly prove some properties that can be used in the following sections, and we give the generating formulas of filters in pseudo equality algebras. Then we introduce prelinear pseudo equality algebras and divisible pseudo equality algebras, also we give some characterizations of them. Moreover, we investigate algebraic structures of the set F (X) of all filters in pseudo equality algebras.

Proposition 3.1. Let X be a pseudo equality algebra. Then the following properties hold for all x, y, z ∈ X:

x → y ≤ (z → x) → (z → y),

Proof. We prove (1), the proofs of (2) is similar. (1) By using (X5)(X7) and Proposition 2.3(1), 2.4(1), we have: x∼y ≤ (x ∧ z) ∼ (y ∧ z)

≤ ((x ∧ z) ∼z) ∼ ((y ∧ z) ∼z)

= (z → x) ∼ (z → y)

≤ (z → y) → (z → x) Hence, x → y = (x ∧ y) ∼x

≤ (z → x) → (z → (x ∧ y))

≤ (z → x) → (z → y))

Definition 3.2. A lattice pseudo equality algebra is a pseudo equality algebra in which (X, ≤) is a lattice, as well.

Proposition 3.3. Let X be a lattice pseudo equality algebra. Then the following properties hold, for all x, y, z ∈ X:

For all indexed families {x _i } _i∈I in X, we have (∨ _i∈Ix _i ) → y = ∧ _i∈I (x _i  → y) and (∨ _i∈Ix _i )  ⇝ y = ∧ _i∈I (x _i  ⇝ y), provided that the infimum and suprimum of {x _i } _i∈I exist in X,

{ Proof. (1) Suppose ∨_i∈Ix _i exist in X. Since x _i  ≤ ∨ _i∈Ix _i , by Proposition 2.4(2), (∨ _i∈Ix _i ) → y ≤ x _i  → y, for all x _i  ∈ X. Hence, (∨ _i∈Ix _i ) → y is a lower bound in X for the indexed family {x _i  → y} _i∈I. Let δ be an another lower bound for this family. Then δ ≤ x _i  → y, for all i ∈ I. By Proposition 2.3(10), x _i  ≤ δ ⇝ y. Then ∨_i∈Ix _i  ≤ δ ⇝ y. Again, by Proposition 2.3(10), δ ≤ (∨ _i∈Ix _i ) → y. Thus (∨ _i∈Ix _i ) → y is the infimum in X for the indexed family {x _i  → y} _i∈I. The proof of the second equation is similar.

The proof is straightforward consequence of (1).

Definition 3.4. A pseudo equality algebra X is called prelinear if 1 is the unique upper bound of the set {x → y, y → x} and {x ⇝ y, y ⇝ x}, i.e., (x → y) ∨ (y → x) =1 = (x ⇝ y) ∨ (y ⇝ x), for all x, y ∈ X.

Theorem 3.5. Any prelinear pseudo equality algebra is a distributive lattice.

Proof. Let (X, ∧ , ∼, ∽ , 1) be a prelinear pseudo equality algebra. By (X1), (X, ∧ , 1) is a meet-semilattice with top element 1, we prove that, for any x, y ∈ X, x ∨ y = ((x → y)  ⇝ y) ∧ ((y → x)  ⇝ x) = ((x ⇝ y) → y) ∧ ((y ⇝ x) → x). Now we prove the first equation, and the second equation is similar. For this, suppose ((x → y)  ⇝ y) ∧ ((y → x)  ⇝ x) = α, then by Proposition 2.3(7) and (8), α is an upper bound of {x, y}. Let β be another upper bound of x and y, then by Proposition 2.3(8) and Proposition 2.4(1), we have x → y ≤ ((x → y)  ⇝ y) → y

≤ ((x → y)  ⇝ y) → β

≤ (((x → y)  ⇝ y) ∧ ((y → x)  ⇝ x)) → β

= α → β y → x ≤ ((y → x)  ⇝ x) → x

≤ ((y → x)  ⇝ x) → β

≤ (((x → y)  ⇝ y) ∧ ((y → x)  ⇝ x)) → β

= α → β Since X is prelinear, we have (x → y) ∨ (y → x) =1, then α → β = 1. So α ≤ β. Thus, α is the least upper bound of {x, y}. Hence, L (X) = (X, ∧ , ∨ , 1) is a lattice.

Now, it is enough to prove that x ∧ (y ∨ z) ≤ (x ∧ y) ∨ (x ∧ z). By Proposition 5.8(3) and Proposition 2.3(12), y → z = (y ∨ z) → z ≤ (x ∧ (y ∨ z)) → (x ∧ z). By the similar way, z → y = (y ∨ z) → y ≤ (x ∧ (y ∨ z)) → (x ∧ y). Since X is prelinear, 1 = (y → z) ∨ (z → y) ≤ ((x ∧ (y ∨ z)) → (x ∧ z)) ∨ ((x ∧ (y ∨ z)) → (x ∧ y)) ≤ ((x ∧ (y ∨ z)) → ((x ∧ z) ∨ (x ∧ y)). Hence x ∧ (y ∨ z) ≤ (x ∧ z) ∨ (x ∧ y). The proof of other side is clear. Thus x ∧ (y ∨ z) = (x ∧ z) ∨ (x ∧ y).

Proposition 3.6. Let X be a prelinear pseudo equality algebra. Then the following statements are hold, for all x, y, z ∈ X:

(x ∧ y) → z = (x → z) ∨ (y → z) and (x ∧ y)  ⇝ z = (x ⇝ z) ∨ (y ⇝ z),

Proof. (1) We first proof (x ∧ y) → z = (x → z) ∨ (y → z). By Proposition 2.3(13)(9) and Proposition 2.4(1), we have x → y = x → (x ∧ y) ≤ ((x ∧ y) → z)  ⇝  (x → z) ≤ ((x ∧ y) → z)⇝ ((x → z) ∨ (y → z). By the similar way, y → x ≤ ((x ∧ y) → z)  ⇝  ((x → z) ∨ (y → z)). Since X is prelinear, (x → y) ∨ (y → x) =1, and so ((x ∧ y) → z)  ⇝  ((x → z) ∨ (y → z)) =1. Then (x ∧ y) → z ≤ (x → z) ∨ (y → z). Conversely, since x ∧ y ≤ x, by Proposition 2.4(2), x → z ≤ (x ∧ y) → z. By the similar way, y → z ≤ (x ∧ y) → z. Thus, (x → z) ∨ (y → z) ≤ (x ∧ y) → z. The proof of second equation is similar.

(2) We first proof x → (y ∧ z) = (x → y) ∧ (x → z). By Proposition 2.3(13), Proposition 3.1(1) and Proposition 2.4(2), we have y → z = y → (y ∧ z) ≤ (x → y) → (x → (y ∧ z)) ≤ ((x → y) ∧ (x → z))→ (x → (y ∧ z)). By the similar way, z → y ≤ ((x → y) ∧ (x → z)) → (x → (y ∧ z)). Since X is prelinear, (z → y) ∨ (y → z) =1, and so ((x → y) ∧ (x → z)) → (x → (y ∧ z)) =1. Then (x → y) ∧ (x → z) ≤ x → (y ∧ z). Conversely, by Proposition 2.4(1), we have x → (y ∧ z) ≤ (x → y) ∧ (x → z). The proof of second equation is similar.

Definition 3.7. A pseudo equality algebra X is called divisible if it satisfies x ∧ y = (x → y) ∧ x = x ∧ (x ⇝ y) = (y → x) ∧ y = y ∧ (y ⇝ x), for any x, y ∈ X.

Example 3.8. [8] Suppose X = {0, a, b, 1}, with 0 < a, b < 1. We define ∼ and ∽ as follows:

Routine calculations show that (X, ∧ , ∼, ∽ , 1) is a divisible pseudo equality algebra.

Example 3.9. [10] Suppose X = {0, a, b, 1}, with 0 < a < b < 1. We define ∼ and ∽ as follows:

Routine calculations show that X is a pseudo equality algebra which is not divisible, since (b → 0) ∧ b = a ≠ 0 = b ∧ (b ⇝ 0).

Definition 3.10. Let X be a pseudo equality algebra. For every subset D ⊆ X, the smallest filter of X containing D (i.e., the intersection of all filters F ⊆ F (X) such that D ⊆ F) is called the filter generated by D and will be denoted by 〈D〉.

Theorem 3.11. Let D be a subset of a pseudo equality algebra X, then 〈 D 〉 = { a ∈ X | x 1 → ( ⋯ ( x n → a ) ⋯ ) = 1 , for some n ∈ N and x 1 , x 2 , ⋯ , x n ∈ D } = { a ∈ X | x 1 ⇝ ( ⋯ ( x n ⇝ a ) ⋯ ) = 1 , for some n ∈ N and x 1 , x 2 , ⋯ , x n ∈ D }

Proof. Let S = {a ∈ X|x₁ → (⋯ (x _n  → a) ⋯) =1, for some n ∈ N and x₁, x₂, ⋯ , x _n  ∈ D}. First, we show that D ⊆ S. For any d ∈ D, there exist x₁ = d ∈ D, n = 1, such that x₁ → d = d → d = 1, so d ∈ S. Hence D ⊆ S. Now we prove that S is a filter of X. Since for all x ∈ D, x → 1 =1, by the definition of S, 1 ∈ S. Now, let a ∈ S and a ≤ b. Then there exist n ∈ N and x₁, x₂, ⋯ , x _n  ∈ D such that x₁ → (⋯ (x _n  → a) ⋯) =1. By Proposition 2.4(1), we have x₁ → (⋯ (x _n  → a) ⋯) ≤ x₁ → (⋯ (x _n  → b) ⋯), hence x₁ → (⋯ (x _n  → b) ⋯) =1, so b ∈ S. Then let a, a ∽ b ∈ S. So there are m, n ∈ N and x₁, x₂, ⋯ , x _m  ∈ D and y₁, y₂, ⋯ , y _n  ∈ D such that x₁ → (⋯ (x _m  → a) ⋯) =1 and y₁ → (⋯ (y _n  → (a ∽ b)) ⋯) =1. By Proposition 2.3(1), a ∽ b ≤ a ⇝ b, so y₁ → (⋯ (y _n  → (a ⇝ b)) ⋯) =1. Then by Proposition 2.3(6)(8) and Proposition 3.1(1), we have

y₁ → (⋯ (y _n  → b) ⋯) = 1 → (y₁ → (⋯ (y _n  → b) ⋯)) = (y₁ → (⋯ (y _n  → (a ⇝ b)) ⋯)) → (y₁ → (⋯ (y _n  → b) ⋯)) ≥ (a ⇝ b) → b ≥a Hence x₁ → (⋯ (x _m  → a) ⋯) ≤ x₁ → (⋯ (x _m → (y₁ → (⋯ (y _n  → b) ⋯)) ⋯)) and so x₁ → (⋯ (x _m  → (y₁ → (⋯ (y _n  → b) ⋯)) ⋯)) =1. By the definition of S, b ∈ S. Then by Definition 2.5, S is a filter of X. Finally, we show that S is the smallest filter of X containing D. Suppose that F is a filter of X containing D and a ∈ S, then there exist x₁, x₂, ⋯ , x _n  ∈ D ⊆ F, such that x₁ → (⋯ (x _n  → a) ⋯) =1 ∈ F. Since x₁, x₂, ⋯ , x _n  ∈ F, then by repeated application of (XF₄), we have a ∈ F. Hence S ⊆ F. The proof of the second equation is similar.

Remark 3.12. As can be seen above, the generation of filters is formally the same as that of the equality algebra, but its proof is different.

For each d belonging to a pseudo equality algebra X, the filter generated by {d} is called principal filters. Clearly, 〈d〉 = {a ∈ X|d →  _n a = 1, for some n ∈ N}   = {a ∈ X|d ⇝  _n a = 1, for some n ∈ N},

where d → ₀a = a, d →  _n a = d → (d → _n-1a), d ⇝ ₀a = a, d ⇝  _n a = d ⇝  (d ⇝ _n-1a).

Theorem 3.13. Let X be a divisible pseudo equality algebra, in which (X, ≤) is a distributive lattice. Then (F (X) , ∧ , ∨ , → , {1} , X) is a heyting algebra, and the operations are given by F₁ ∧ F₂ = F₁ ∩ F₂, F₁ ∨ F₂ = 〈F₁ ∪ F₂〉, F₁ → F₂ = {x ∈ X|f ∨ x ∈ F₂, for any f ∈ F₁}, for any F₁, F₂ ∈ F (X).

Proof. It is clear that if F₁ and F₂ are filters of X then so is F₁ ∩ F₂, which is the largest filter of X and is contained in both F₁ and F₂. Hence F₁ ∧ F₂ exists in F (X) and is F₁ ∩ F₂. And we can see F₁ ∨ F₂ = 〈F₁ ∪ F₂〉, thus F₁ ∨ F₂ is the supremum of F₁ and F₂. Therefore, (F (X) , ∧ , ∨ , → , {1} , X) is a bounded lattice.

Moreover, we show that F₁ → F₂ is a filter of X. Clearly, 1 ∈ F₁ → F₂. Let x ∈ F₁ → F₂ and x ≤ y, then for any f ∈ F₁, f ∨ x ∈ F₂. Since f ∨ x ≤ f ∨ y, by (XF₂), f ∨ y ∈ F₂. So y ∈ F₁ → F₂. Now, assume x, x ∽ y ∈ F₁ → F₂, then for any f ∈ F₁, f ∨ x ∈ F₂, f ∨ (x ∽ y) ∈ F₂. By Proposition 2.3(1) and (XF₂), we can get f ∨ (x ⇝ y) ∈ F₂. Since X is a distributive lattice, then by Proposition 2.7, f₁ ∨ (x ∧ (x ⇝ y)) = (f₁ ∨ x) ∧ (f₁ ∨ (x ⇝ y)) ∈ F₂. Since X is divisible, then x ∧ y = x ∧ (x ⇝ y). So f ∨ (x ∧ (x ⇝ y)) = f ∨ (x ∧ y) ≤ f ∨ y. Then by (XF₂), f ∨ y ∈ F₂, hence y ∈ F₁ → F₂. And so F₁ → F₂ is a filter of X.

Finally, in order to prove that (F (X) , ∧ , ∨ , → , {1} , X) is a heyting algebra, we only to prove that F₁ ∧ F₂ ≤ F₃ if and only if F₁ ≤ F₂ → F₃. Assume that F₁ ∧ F₂ ≤ F₃. Let x ∈ F₁, then for any f ∈ F₂, x ∨ f ∈ F₁ ∧ F₂, and so x ∨ f ∈ F₃. By the definition of F₂ → F₃, x ∈ F₂ → F₃, then F₁ ≤ F₂ → F₃. Conversely, assume that F₁ ≤ F₂ → F₃. Let x ∈ F₁ ∧ F₂, then x ∈ F₁ ⊆ F₂ → F₃, so for any f₂ ∈ F₂, f₂ ∨ x ∈ F₃. Taking f₂ = x ∈ F₂, we have f₂ ∨ x = x ∈ F₃. Thus F₁ ∧ F₂ ≤ F₃.

By Definition 2.8, (F (X) , ∧ , ∨ , → , {1} , X) is a heyting algebra.

From now on, X denotes a pseudo equality algebra, unless otherwise stated.

4 (Positive) Implicative filters

In this section, we define (positive) implicative filters, and investigate some properties and relations of them. Moreover, we give the notion of strong normal filters and discuss some properties. And by using the notion of strong normal filters, we get the new relations between implicative filters and positive implicative filters.

Definition 4.1. Let F be a subset of X. Then F is called an implicative filter of X if it satisfies the following conditions:

1 ∈ F,

Example 4.2. In Example 3.8, routine calculations show that F₁ = {1}, F₂ = {b, 1}, F₃ = {a, 1}, F₄ = {0, a, b, 1} are the filters of X. Also they are implicative filters of X.

Example 4.3. In Example 3.9, routine calculations show that F₁ = {1}, F₂ = {b, 1}, F₃ = {0, a, b, 1} are the filters of X. We can easily check that F₁ = {1} and F₃ = {0, a, b, 1} are implicative filters of X, but F₂ = {b, 1} is not an implicative filter of X, because, 1 → ((a → 0)  ⇝ a) =1 ∈ F₁ and 1 ∈ F₁, but a ∉ F₁.

Proposition 4.4. Every implicative filter of X is a filter.

Proof. Let F be an implicative filter and x, x → y ∈ F, for any x, y ∈ X. By Proposition 2.3(6), x → ((y → 1)  ⇝ y) = x → y ∈ F. Since x ∈ F and F is an implicative filter, by (IF2), y ∈ F. Therefore, F is a filter.

Remark 4.5. The converse of Proposition 4.4 may not be true. In Example 4.3, we can see F₂ = {b, 1} is a filter, but is not an implicative filter of X.

Now we proof a new proposition in pseudo equality algebras, which will be used by the following sections.

Proposition 4.6. Let F be a filter of pseudo equality algebra X.Then for any x, y, z ∈ X, the following properties hold:

x∼y ∈ F and y∼z ∈ F imply x∼z ∈ F,

Proof. We prove (1) and (2), the proofs of (1)′ and (2)′ are similar.

If x∼y ∈ F and y∼z ∈ F, then by (X7), x∼y ≤ (x∼z) ∼ (y∼z). since F is a filter, by (XF₂), (x∼z) ∼ (y∼z) ∈ F, by (XF₃), x∼z ∈ F.

Theorem 4.7. Let X be bounded and F be a filter of X. Then the following conditions are equivalent:

F is an implicative filter of X;

Proof. (1) ⇒ (2) Let F be an implicative filter and (x → y)  ⇝ x ∈ F. By Proposition 2.3(6), 1 → ((x → y)  ⇝ x) = (x → y)  ⇝ x ∈ F, since 1 ∈ F and F is an implicative filter, by(IF2), x ∈ F. The proof of other case is similar.

(2) ⇒ (1) Suppose that (2) holds. Let x → ((y → z)  ⇝ y) ∈ F and x ∈ F. Since F is a filter, by (XF₄), (y → z)  ⇝ y ∈ F. Now, by (2), y ∈ F. By a similar argument, we can show that x ⇝  ((y ⇝ z) → y) ∈ F and x ∈ F imply y ∈ F. Therefore, F is an implicative filter.

(2) ⇒ (3) If x^- ⇝ x ∈ F, then x^- ⇝ x = (x → 0)  ⇝ x ∈ F and by (2), x ∈ F. The proof of other case is similar.

(3) ⇒ (2) Let (x → y)  ⇝ x ∈ F, since 0 ≤ y, by Proposition 2.4(1)(2), (x → y)  ⇝ x ≤ (x → 0)  ⇝ x = x^- ⇝ x. Since F is a filter of X, by (XF₂), x^- ⇝ x ∈ F. So by (3), x ∈ F. The proof of other case is similar.

(3) ⇒ (4) Let x^- ⇝  (y ⇝ z) ∈ F and z ⇝ x ∈ F. By Proposition 2.3(9), x^- ⇝  (y ⇝ z) ≤ ((y ⇝ z)  ⇝  (y ⇝ x)) → (x^- ⇝  (y ⇝ x)). Since F is a filter, by (XF₂), ((y ⇝ z)  ⇝  (y ⇝ x)) → (x^- ⇝  (y ⇝ x)) ∈ F, by Proposition 3.1(2), z ⇝ x ≤ (y ⇝ z)  ⇝  (y ⇝ x) and since z ⇝ x ∈ F, by (XF₂) and (XF₄), x^- ⇝  (y ⇝ x) ∈ F. Thus, by Proposition 2.3(7), x ≤ y ⇝ x. So by Proposition 2.4(2), x^- ⇝  (y ⇝ x) ≤ (y ⇝ x) ^- ⇝  (y ⇝ x). Then (y ⇝ x) ^- ⇝  (y ⇝ x) ∈ F. By (3), y ⇝ x ∈ F. The proof of other case is similar.

(4) ⇒ (5) If x^- ⇝  (y ⇝ x) ∈ F, let z = x in (4), then since x ⇝ x = 1 ∈ F, so y ⇝ x ∈ F. The proof of other case is similar.

(5) ⇒ (3) If x^- ⇝ x ∈ F, then by Proposition 2.3(6), x^- ⇝ x = x^- ⇝ (1 ⇝ x) ∈ F, then by (5), x = 1 ⇝ x ∈ F.

(3) ⇒ (6) Let α = (x^∼ → x)  ⇝ x, then by Proposition 2.3(11) and 2.3(9), we have α^∼ → α = (α ⇝ 0) → α = (((x^∼ → x)  ⇝ x)  ⇝ 0) → ((x^∼ → x)  ⇝ x) = (x^∼ → x)  ⇝  ((((x^∼ → x)  ⇝ x)  ⇝ 0) → x) ≥ (((x^∼ → x)  ⇝ x)  ⇝ 0) → (x ⇝ 0) ≥x ⇝  ((x^∼ → x)  ⇝ x). By Proposition 2.3(7), x ≤ (x^∼ → x)  ⇝ x, hence x ⇝  ((x^∼ → x)  ⇝ x) =1 ∈ F. Hence by (XF₂), α^∼ → α ∈ F. Then by(3), α ∈ F. Hence, (x^∼ → x)  ⇝ x ∈ F. By the similar way, we can get (x^- ⇝ x) → x ∈ F.

(6) ⇒ (3) Suppose x^- ⇝ x ∈ F. By assumption, (x^- ⇝ x) → x ∈ F, then by (XF₄), x ∈ F. By a similar argument, we show that x^∼ → x ∈ F imply x ∈ F.

Corollary 4.8. Let F and G be two filters of X such that F ⊆ G. If F is an implicative filter, then G is an implicative filter, too.

Proof. By the equivalence of (1) and (6) in Theorem 4.7, the proof is clear.

Corollary 4.9. Every filter of X is an implicative filter if and only if {1} is an implicative filter.

Definition 4.10. Let F be a subset of X. Then F is called a positive implicative filter of X if it satisfies the following conditions:

1 ∈ F,

Example 4.11. In Example 3.8, F₁ = {1}, F₂ = {b, 1}, F₃ = {a, 1}, F₄ = {0, a, b, 1} are positive implicative filters of X. In Example 3.9, F₁ = {1} and F₃ = {0, a, b, 1} are positive implicative filters of X, but F₂ = {b, 1} is not a positive implicative filter of X. Since a → (0 → 0) =1 ∈ F and a ⇝ 0 = b ∈ F, but a → 0 = a ∉ F.

Proposition 4.12. Every positive implicative filter of X is a filter.

Proof. Suppose that F is a positive implicative filter and x, x ⇝ y ∈ F, for any x, y ∈ X. By Proposition 2.3(6), 1 ⇝  (x ⇝ y) = x ⇝ y ∈ F and 1 → x = x ∈ F. Since F is a positive implicative filter, we have 1 ⇝ y = y ∈ F. Therefore, F is a filter.

Remark 4.13. The converse of Proposition 4.12 may not be true. In Example 4.11, we can see F₂ = {b, 1} is a filter, but is not a positive implicative filter of X.

Proposition 4.14. Let F be a positive implicative filter of X. Then for all x, y, z ∈ X, the following statements are hold: (1) if x → (y → z) ∈ F, then (x ⇝ y) → (x ⇝ z) ∈ F, (2) if x ⇝  (y ⇝ z) ∈ F, then (x → y)  ⇝  (x → z) ∈ F.

Proof. (1) Let F be a positive implicative filter and x → (y → z) ∈ F. By Proposition 4.12, F is a filter. Then by Proposition 2.3(9), (x ⇝ y) → y) ≤ (y → z)  ⇝  ((x ⇝ y) → z). By Proposition 2.3(5) and (XF₁), ((x ⇝ y) → y)  ⇝  ((y → z)  ⇝  ((x ⇝ y) → z)) =1 ∈ F. By Proposition 2.3(8), x ≤ (x ⇝ y) → y, so we have x ⇝  ((x ⇝ y) → y) =1 ∈ F. So by Proposition 4.6(2) ′, we have x ⇝  ((y → z)  ⇝  ((x ⇝ y) → z)) ∈ F. Since x → (y → z) ∈ F and F is a positive implicative filter, we have x ⇝  ((x ⇝ y) → z)) ∈ F. Then by Proposition 2.3(11), (x ⇝ y) → (x ⇝ z) ∈ F.

(2) We can see the prove of (2) is similar.

Proposition 4.15. Let F be a positive implicative filter of X. Then for all x, y ∈ X, the following statements are hold:

If x ⇝  (x ⇝ y) ∈ F, then x ⇝ y ∈ F,

Proof. (1) Since F is a positive implicative filter, by Proposition 4.12, F is a filter. Now, let x ⇝  (x ⇝ y) ∈ F, for all x, y ∈ X. Then, by Proposition 2.3(6), x → x = 1 ∈ F, thus, by (PIF3), x ⇝ y ∈ F.

(2) We can see the prove of (2) is similar.

In the following, we investigate the relations between positive implicative filters and implicative filters.

Example 4.16. [4] Suppose X = {0, a, b, 1}, with 0 < a < b < 1. We define ∼ =∽ as follows:

By routine calculations, we can see that (X, ∧ , ∼, ∽, 1) is a pseudo equality algebra. Moreover, F = {b, 1} is a positive implicative filter which is not an implicative filter of X, because, 1 → ((a → 0)  ⇝ a) =1 ∈ F and 1 ∈ F, but a ∉ F.

From Example 4.16, we can see that not all positive implicative filters are implicative filters. In the following, we give the conditions under which the positive implicative filters become the implicative filters.

Theorem 4.17. Let X be bounded and F be a positive implicative filter of X. Then the following conditions are equivalent:

F is an implicative filter of X;

Proof. (1) ⇒ (2) Suppose that F is an implicative filter. By Proposition 4.4, F is a filter. If x^-∼ ∈ F, by Proposition 2.4(1), x^-∼ = x^- ⇝ 0 ≤ x^- ⇝ x. Then by (XF₂), x^- ⇝ x ∈ F. In the similar way, we can show that x^∼ → x ∈ F implies x ∈ F. Hence by Theorem 4.7, x ∈ F.

(2) ⇒ (1) Suppose that (2) holds and F be a positive implicative filter. By Proposition 4.12, F is a filter. Suppose that x^- ⇝ x ∈ F, By Proposition 2.3(8), x ≤ x^-∼. Then by Proposition 2.4(1), we have x^- ⇝ x ≤ x^- ⇝ x^-∼ = x^- ⇝  (x^- ⇝ 0). Since F is a filter and by (XF₂), x^- ⇝  (x^- ⇝ 0) ∈ F. Since F is a positive implicative filter, then by Proposition 4.15, x^- ⇝ 0 = x^-∼ ∈ F, then by (2), x ∈ F. By a similar argument, we can show that x^∼ → x ∈ F implies x ∈ F. Then by Theorem 4.7, F is an implicative filter.

(1) ⇒ (3) Let F be an implicative filter of X and (x → y)  ⇝ y ∈ F. By Proposition 2.3(7), x ≤ (y → x)  ⇝ x, by Proposition 2.4, ((y → x)  ⇝ x) ^- ≤ x^- = x → 0 ≤ x → y, hence by by Proposition 2.4, (x → y)  ⇝ y ≤ ((y → x)  ⇝ x) ^- ⇝ y. Then by Proposition 2.3(8), y ≤ (y → x)  ⇝ x, then by Proposition 2.4(1), ((y → x)  ⇝ x) ^- ⇝ y ≤ ((y → x)  ⇝ x) ^- ⇝  ((y → x)  ⇝ x). Thus, (x → y)  ⇝ y ≤ ((y → x)  ⇝ x) ^- ⇝  ((y → x)  ⇝ x). By Proposition 4.4, F is a filter. Then by (XF₂), ((y → x)  ⇝ x) ^- ⇝  ((y → x)  ⇝ x) ∈ F. Hence, by Theorem 4.7, (y → x)  ⇝ x ∈ F. The proof of other case is similar.

(3) ⇒ (1) Let F be a positive implicative filter and x^- ⇝ x ∈ F. By Proposition 2.3(8) and Proposition 2.4(1), x^- ⇝ x ≤ x^- ⇝  (x^- ⇝ 0). Then by Proposition 4.12, F is a filter. By (XF₂), x^- ⇝  (x^- ⇝ 0) ∈ F. Then by Proposition 4.15, x^- ⇝ 0 ∈ F. By assumption, we have (0 → x)  ⇝ x = x ∈ F. By a similar argument, we can show that x^∼ → x ∈ F implies x ∈ F. Then by Theorem 4.7, F is an implicative filter.

Next, we give the notion of strong normal filters and discuss some properties. By using the notion of strong normal filters, we give the equivalent definition of (positive) implicative filters and get the new relations between implicative filters and positive implicative filters.

Definition 4.18. A filter F of a pseudo equality algebra X is called a strong normal filter if for all x, y ∈ X, we have x∼y ∈ F if and only if y ∽ x ∈ F.

Example 4.19. In Example 3.8, F₁ = {1}, F₂ = {b, 1}, F₃ = {a, 1}, F₄ = {0, a, b, 1} are strong normal filters of X. In Example 3.9, F₂ = {b, 1} is not a strong normal filter. Because a ∽ 0 = b ∈ F, but 0∼a = a ∉ F.

Proposition 4.20. Let F be a strong normal filter of X. Then, for all x, y ∈ X, x → y ∈ F if and only if x ⇝ y ∈ F.

Proof. Let x → y ∈ F. Then x → y = (x ∧ y) ∼x ∈ F. By assumption, x ∽ (x ∧ y) ∈ F. That is, x ⇝ y ∈ F. By a similar argument, we can show that x ⇝ y ∈ F implies x → y ∈ F.

Theorem 4.21. Let X be bounded and F is a strong normal filter of X. Then the following conditions are equivalent: (1) F is an implicative filter of X; (2) If x → (y^∼ → y) ∈ F, then x → y ∈ F and if x ⇝  (y^- ⇝ y) ∈ F, then x ⇝ y ∈ F, for all x, y ∈ X; (3) If x → (z^∼ → y) ∈ F and y → z ∈ F, then x → z ∈ F and if x ⇝  (z^- ⇝ y) ∈ F and y ⇝ z ∈ F, then x ⇝ z ∈ F, for all x, y, z ∈ X.

Proof. (1) ⇒ (2) Let F be an implicative filter and x → (y^∼ → y) ∈ F. By Proposition 2.3(7), y ≤ x ⇝ y, then by Proposition 2.4(2), y^∼ → y ≤ (x ⇝ y) ^∼ → y. Now, by Proposition 2.4(1), x → (y^∼ → y) ≤ x → ((x ⇝ y) ^∼ → y). Hence by (XF₂), x → ((x ⇝ y) ^∼ → y) ∈ F. Since F is a strong normal filter, then by Proposition 4.20, we have x ⇝  ((x ⇝ y) ^∼ → y) ∈ F. By Proposition 2.3(11), we get (x ⇝ y) ^∼ → (x ⇝ y) ∈ F, thus by Proposition 2.3(6), 1 ⇝  ((x ⇝ y) ^∼ → (x ⇝ y)) ∈ F. Since F is an implicative filter and 1 ∈ F, by (IF2), x ⇝ y ∈ F. By a similar argument, we can show that x ⇝  (y^- ⇝ y) ∈ F implies x ⇝ y ∈ F.

(2) ⇒ (3) Suppose that x → (z^∼ → y) ∈ F and y → z ∈ F. By Proposition 3.1(1), y → z ≤ (z^∼ → y) → (z^∼ → z). Since F is a filter, by (XF₂), (z^∼ → y) → (z^∼ → z) ∈ F. By Proposition 4.6(2), x → (z^∼ → z) ∈ F. By (6), x → z ∈ F. The proof of other case is similar.

(3) ⇒ (1) Suppose that (3) holds. If x^∼ → x ∈ F, then 1 → (x^∼ → x) ∈ F, x → x = 1 ∈ F, by (3), 1 → x ∈ F. By Proposition 2.3(6), x ∈ F. By a similar argument, we can show that x^- ⇝ x ∈ F implies x ∈ F. Then by the equivalence of (1) and (3) in Theorem 4.7, F is an implicative filter of X.

Corollary 4.22. Let X be bounded and F is a strong normal implicative filter of X. Then for all x ∈ X, x^∼- → x ∈ F and x^-∼ ⇝ x ∈ F.

Proof. Suppose F is an implicative filter of X. Since X is a bounded pseudo equality algebras, we have 0 ≤ x, for all x ∈ X. Then by Proposition 2.4(1), x^- ⇝ 0 ≤ x^- ⇝ x, and by Proposition 2.3(5), x^-∼ ⇝  (x^- ⇝ x) =1. Thus, by (IF1), x^-∼ ⇝  (x^- ⇝ x) =1 ∈ F. Hence, by the equivalence of (1) and (3) in Theorem 4.7, x^∼- ⇝ x ∈ F. By the similar way, we get x^-∼ → x ∈ F.

The inverse of Proposition 4.15 is not hold unless F be a strong normal filter of X, and the next proposition will proof it.

Proposition 4.23. Let F be a strong normal filter of X. Then for any x, y ∈ X, the following conditions are equivalent:

F is a positive implicative filter of X,

Proof. (1) ⇒ (2) By Proposition 4.15 is trivial.

(2) ⇒ (1) Suppose that (2) holds. Let x ⇝  (y ⇝ z) ∈ F and x → y ∈ F, for any x, y ∈ X. Since F is a strong normal filter and x ⇝  (y ⇝ z) ∈ F, by Proposition 4.20, we have x → (y ⇝ z) ∈ F. By Proposition 2.3(11), y ⇝  (x → z) ∈ F. Since F is a strong normal filter, by Proposition 4.20, we have y → (x → z) ∈ F and x → y ∈ F. By Proposition 4.6(2), x → (x → z) ∈ F. Then, by (2), x → z ∈ F. Since F is a strong normal filter, by Proposition 4.20, x ⇝ z ∈ F. Then we have (PIF3). The proof of (PIF2) is similar. Therefore, F is a positive implicative filter.

Proposition 4.24. Let F be a strong normal filter of X. Then F is a positive implicative filter if and only if for all x, y ∈ X, (x ∧ (x → y)) → y ∈ F and (x ∧ (x ⇝ y))  ⇝ y ∈ F.

Proof. (⇒) Since x ∧ (x → y) ≤ x → y and x ∧ (x → y) ≤ x, we have (x ∧ (x → y)) → (x → y) =1 ∈ F and (x ∧ (x → y))  ⇝ x = 1 ∈ F. Since F is a positive implicative filter, (x ∧ (x → y)) → y ∈ F. We can see the prove of the second equation is similar.

(⇐) Let x → (y → z) ∈ F and x ⇝ y ∈ F. By Proposition 2.3(12), x → (y → z) ≤ (x ∧ y) → (y ∧ (y → z)). Since F is a filter, (x ∧ y) → (y ∧ (y → z)) ∈ F. By Proposition 2.3(13), x ⇝  (x ∧ y) = x ⇝ y ∈ F. Since F is a strong normal filter, by Proposition 4.20, x → (x ∧ y) ∈ F. So by Proposition 4.6(2), x → (y ∧ (y → z)) ∈ F. By assumption, (y ∧ (y → z)) → z ∈ F, thus, x → z ∈ F. By the similar way, we can see that x ⇝  (y ⇝ z) ∈ F and x → y ∈ F imply x ⇝ z ∈ F. So we can see that F is a positive implicative filter.

Corollary 4.25. Let F and G be two strong normal filters of X such that F ⊆ G. If F is a positive implicative filter of X, then G is a positive implicative filter of X, too.

Proof. By Proposition 4.24, the proof is clear.

Corollary 4.26. Every strong normal filter of X is a positive implicative filter if and only if {1} is a strong normal positive implicative filter.

Proposition 4.27. Let F be a strong normal filter of X. If F is an implicative filter, then F is a positive implicative filter.

Proof. By Proposition 4.24, it is enough to prove that (x ∧ (x → y)) → y ∈ F and (x ∧ (x ⇝ y))  ⇝ y ∈ F, for all x, y ∈ X. For this, x ∧ (x → y) ≤ x, x ∧ (x → y) ≤ x → y and by Proposition 2.4(2), x ∧ (x → y) ≤ x → y ≤ (x ∧ (x → y)) → y. Thus ((x ∧ (x → y)) → y) → y ≤ (x ∧ (x → y)) → y, and so (((x ∧ (x → y)) → y) → y)  ⇝  ((x ∧ (x → y)) → y) =1 ∈ F. Since F is an implicative filter, by Theorem 4.7(2), (x ∧ (x → y)) → y ∈ F. By a similar argument, we have (x ∧ (x ⇝ y))  ⇝ y ∈ F.

Next, we give the conditions in which the positive implicative filters become the implicative filters under the notion of strong normal filters.

Theorem 4.28. Let F be a strong normal positive implicative filter of X. Then the following conditions are equivalent:

F is an implicative filter of X,

Proof. (1) ⇒ (2) By Proposition 4.27 is trivial.

(2) ⇒ (1) Let (x → y)  ⇝ x ∈ F, for any x, y ∈ X. By Proposition 2.3(8), x ≤ (x → y)  ⇝ y. Then by Proposition 2.4(1), (x → y)  ⇝ x ≤ (x → y)  ⇝  ((x → y)  ⇝ y). Since (x → y)  ⇝ x ∈ F, then by (XF₂), (x → y)  ⇝  ((x → y)  ⇝ y) ∈ F. Since F is a positive implicative filter, by Proposition 4.15, (x → y)  ⇝ y ∈ F, and so by assumption, (y → x)  ⇝ x ∈ F. Also, by Proposition 2.3(7), y ≤ x → y. Then, by Proposition 2.4(2), (x → y)  ⇝ x ≤ y ⇝ x, thus by (XF₂), y ⇝ x ∈ F. Since F ia a strong normal filter, by Proposition 4.20, y → x ∈ F. Then, by ( XF 4 ′ ) , x ∈ F. By the similar way, if (x ⇝ y) → x ∈ F, we get x ∈ F. By Theorem 4.7(2), F is an implicative filter.

5 Fantastic filters

In this section, we first introduce the notion of fantastic filters, and then investigate some properties and the relations with (positive) implicative filters.

Definition 5.1. Let F be a subset of X. Then F is called a fantastic filter of X if it satisfies the following conditions:

1 ∈ F,

Example 5.2. In Example 3.8, F₁ = {1}, F₂ = {b, 1}, F₃ = {a, 1}, F₄ = {0, a, b, 1} are fantastic filters of X. In Example 3.9, F₁ = {1} and F₃ = {0, a, b, 1} are fantastic filters of X, but F₂ = {b, 1} is not a fantastic filter of X. Since b → (0 → a) =1 ∈ F and b ∈ F, but ((a → 0)  ⇝ 0) → a = a ∉ F.

Proposition 5.3. Every fantastic filter of X is a filter.

Proof. Let F be a fantastic filter and x, x → y ∈ F. By Proposition 2.3(6), x → y = x → (1 → y) ∈ F. Since x ∈ F, and F is a fantastic filter, by (FF2), ((y → 1)  ⇝ 1) → y ∈ F. Then by Proposition 2.3(6), y ∈ F. Therefore, F is a filter.

Remark 5.4. The converse of Proposition 5.3 may not be true. In Example 5.2, we can see F₂ = {b, 1} is a filter, but is not a fantastic filter of X.

Proposition 5.5. Let F be a filter of X. Then the following conditions are equivalent:

F is a fantastic filter of X;

Proof. (1) ⇒ (2) Let F be a fantastic filter and y → x ∈ F, for any x, y ∈ X. By Proposition 2.3(6), y → x = 1 → (y → x) ∈ F. Since F is a filter, 1 ∈ F, and by (FF2), ((x → y)  ⇝ y) → x ∈ F. By a similar argument, we can show that y ⇝ x ∈ F implies ((x ⇝ y) → y)  ⇝ x ∈ F.

(2) ⇒ (1) Since F is a filter, (FF1) holds. Now, let z → (x → y) ∈ F and z ∈ F, for any x, y ∈ X. Since F is a filter, by (XF₄), x → y ∈ F. Then by assumption, ((y → x)  ⇝ x) → y ∈ F. By a similar argument, we can show that z ⇝  (x ⇝ y) ∈ F and z ∈ F imply ((y ⇝ x) → x)  ⇝ y ∈ F. Thus F is a fantastic filter.

Proposition 5.6. Let F and G be two filters of X such that F ⊆ G. If F is a fantastic filter, then G is a fantastic filter, too.

Proof. Let F and G be two filters of X such that F ⊆ G. Suppose that y → x ∈ G, for any x, y ∈ X. By Proposition 2.3(6), (y → x)  ⇝  (y → x) =1. Then by Proposition 2.3(11), y → ((y → x)  ⇝ x) =1 ∈ F. Since F is a fantastic filter and by Proposition 5.5 and 2.3(11), ((((y → x)  ⇝  x) → y)  ⇝  y) → ((y → x)   ⇝  x) = (y → x)   ⇝  (((((y → x)  ⇝  x) → y)  ⇝ y) → x) ∈ F. Since F ⊆ G, (y → x)  ⇝  (((((y → x)  ⇝ x) → y)  ⇝ y) → x) ∈ G. Since y → x ∈ G, by ( XF 4 ′ ) , (((((y → x)  ⇝ x) → y)  ⇝ y) → x) ∈ G. Let α = ((((y → x)  ⇝ x) → y)  ⇝ y) → x ∈ G. Now, by Proposition 2.3(11), 2.3(9), 2.3(5) and 2.3(6), we get α ⇝  (((x → y)  ⇝ y) → x) = ((x → y)  ⇝ y) → (α ⇝  x) ≥ ((x → y)  ⇝ y) → ((((y → x)  ⇝ x) → y)  ⇝ y) = (((y →x)  ⇝ x) → y)  ⇝  (((x → y)  ⇝ y) → y) ≥ (((y → x)  ⇝ x) → y)  ⇝  (x → y) = x → ((((y → x)  ⇝ x) → y)  ⇝ y) ≥x → ((y → x)  ⇝ x) = (y → x)  ⇝  (x → x) = 1. Then by assumption, G is a filter, then 1 ∈ F and by (XF₂), α ⇝  (((x → y)  ⇝ y) → x) ∈ G. Since α ∈ G, by ( XF 4 ′ ) , ((x → y)  ⇝ y) → x ∈ G. By a similar argument, we can show that y ⇝ x ∈ G implies ((x ⇝ y) → y)  ⇝ x ∈ G. Thus G is a fantastic filter.

Lemma 5.7. In any pseudo equality algebra X, for any x, y ∈ X, the following are equivalent: (1) ((x → y)  ⇝ y) → x = y → x and ((x ⇝ y) → y)  ⇝ x = y ⇝ x, (2) (x → y)  ⇝ y = (y → x)  ⇝ x and (x ⇝ y) → y = (y ⇝ x) → x.

Proof. (1) ⇒ (2) Let ((x → y)  ⇝ y) → x = y → x for all x, y ∈ X, by Proposition 2.3(11), we have ((x → y)  ⇝ y) → ((y → x)  ⇝ x) = (y → x)  ⇝  (((x → y)  ⇝ y) → x) = (y → x)  ⇝  (y → x) = 1, that is (x → y)  ⇝ y ≤ (y → x)  ⇝ x, similarly (x → y)  ⇝ y ≥ (y → x)  ⇝ x. Therefore, (x → y)  ⇝ y = (y → x)  ⇝ x. By the similar way, we can prove that (x ⇝ y) → y = (y ⇝ x) → x.

(2) ⇒ (1) Assume that (x → y)  ⇝ y = (y → x)  ⇝ x for all x, y ∈ X, then by Proposition 2.3(11) we have (y→ x) → (((x → y)  ⇝ y) → x) = ((x → y)  ⇝ y) → ((y → x) → x) = 1, therefore (y → x) ≤ ((x → y)  ⇝ y) → x. Now by Proposition 2.3(8) and Proposition 2.4(1), (((x→ y)  ⇝ y) → x)  ⇝  (y → x) ≥ y → ((x → y)  ⇝ y) = (x → y)  ⇝  (y → y) = (x → y)  ⇝ 1 =1, hence (((x → y)  ⇝ y) → x)  ⇝  (y → x) =1, i.e., (((x → y)  ⇝ y) → x) ≤ y → x. Then (x → y)  ⇝ y = (y → x)  ⇝ x. The proof of the second equation is similar.

Proposition 5.8. In any pseudo equality algebra X, the following conditions are equivalent: (1) {1} is a fantastic filter, (2) Every filter of X is a fantastic filter, (3) ((x → y)  ⇝ y) → x = y → x and ((x ⇝ y) → y)  ⇝ x = y ⇝ x, for any x, y ∈ X.

Proof. (1) ⇔ (2): By Proposition 5.6 is trivial.

(1) ⇒ (3): Assume that {1} is a fantastic filter and a = (y → x)  ⇝ x for every x, y ∈ X. Then by Proposition 2.3(11), y → a = y → ((y → x)  ⇝ x) = (y → x)  ⇝  (y → x) =1 ∈ {1}, by Proposition 5.5, ((a → y)  ⇝  y) → a = 1, i.e., (a → y)  ⇝ y ≤ a. Since x ≤ a, by Proposition 2.4(2), a → y ≤ x → y and so (x → y)  ⇝ y ≤ (a → y)  ⇝ y. Then we have 1 = ((a → y)  ⇝  y) → a ≤ ((x → y)  ⇝  y) → a then ((x→ y)  ⇝  y) → ((y → x)  ⇝ x) = ((x → y)  ⇝ y) → a = 1, i.e., ((x → y)  ⇝ y) ≤ ((y → x)  ⇝ x) and similarly ((y → x)  ⇝ x) ≤ ((x → y)  ⇝ y), therefore ((x → y)  ⇝ y) = ((y → x)  ⇝ x). By Lemma 5.7, ((x → y)  ⇝ y) → x = y → x. And the second equation is similar.

(3) ⇒ (1) By Proposition 5.5 is trivial.

Theorem 5.9. Let X be a lattice pseudo equality algebra, and F be a filter of X. Then F is a fantastic filter if and only if ((x → y)  ⇝ y) → ((y → x)  ⇝ x) ∈ F and ((x ⇝ y) → y)  ⇝  ((y ⇝ x) → x) ∈ F, for all x, y ∈ X.

Proof. (⇒) Let X be a lattice pseudo equality algebra and F be a fantastic filter of X. By Proposition 5.8(3), (x ∨ y) → y = x → y and y → (x ∨ y) =1 ∈ F, for any x, y ∈ X. Then by Proposition 5.5, we have (((x ∨ y) → y)  ⇝ y) → (x ∨ y) ∈ F, and so ((x → y)  ⇝ y) → (x ∨ y) ∈ F. Moreover, by Proposition 2.3(7) and (8), x ≤ (y → x)  ⇝ x, y ≤ (y → x)  ⇝ x. Thus, x ∨ y ≤ (y → x)  ⇝ x, and so (x ∨ y) → ((y → x)  ⇝ x) =1 ∈ F. Then by Proposition 4.6(2), ((x → y)  ⇝ y) → ((y → x)  ⇝ x) ∈ F. By the similar way, we can show that ((x ⇝ y) → y)  ⇝  ((y ⇝ x) → x) ∈ F.

(⇐) Let x, y ∈ X and y → x ∈ F. By Proposition 2.3(11), (y → x)  ⇝  (((x → y)  ⇝ y) → x) = ((x → y)  ⇝ y) → ((y → x)  ⇝ x) ∈ F and F is a filter, by (XF₅), we have ((x → y)  ⇝ y) → x ∈ F. By a similar argument, we can show that y ⇝ x ∈ F implies ((x ⇝ y) → y)  ⇝ x ∈ F. Hence by Proposition 5.5, F is a fantastic filter.

In the following, we investigate the relations between fantastic filters and (positive) implicative filters.

Theorem 5.10. Let X be a pseudo equality algebras and F be a strong normal filter. If F is an implicative filter, then F is a fantastic filter.

Proof. Let y → x ∈ F, for all x, y ∈ X. By Proposition 2.3(7), x ≤ ((x → y)  ⇝ y) → x. Then by Proposition 2.4(2), (((x → y)  ⇝ y) → x) → y ≤ x → y. By Proposition 2.4(2), 2.3(11), ((((x → y)  ⇝ y) → x) → y)  ⇝  (((x → y)  ⇝ y) → x) ≥ (x → y)  ⇝  (((x → y)  ⇝ y) → x) = ((x → y)  ⇝ y) → ((x → y)  ⇝ x) Since F be a strong normal filter and y → x ∈ F, then by Proposition 4.20, y ⇝ x ∈ F. By Proposition 3.1, y ⇝ x ≤ ((x → y)  ⇝ y)  ⇝  ((x → y)  ⇝ x) ∈ F. Then by Proposition 4.20 again, ((x → y)  ⇝ y) → ((x → y)  ⇝ x) ∈ F. So by (XF₂), ((((x → y)  ⇝ y) → x) → y)  ⇝  (((x → y)  ⇝ y) → x) ∈ F. Since F is an implicative filter, by Theorem 4.7(2), ((x → y)  ⇝ y) → x ∈ F. By a similar argument, we have y ⇝ x ∈ F implies ((x ⇝ y) → y)  ⇝ x ∈ F. Then by Proposition 5.5, F is a fantastic filter.

The following example shows that not every fantastic filter of X is an implicative filter.

Example 5.11. [4] Suppose X = {0, a, b, 1}, with 0 < a < b < 1. We define ∼ =∽ as follows:

By routine calculations, we can see that (X, ∧ , ∼, ∽ , 1) is a pseudo equality algebra. Moreover, F = {b, 1} is a fantastic filter, which is not an implicative filter of X, because, 1 → ((a → 0)  ⇝ a) =1 ∈ F and 1 ∈ F, but a ∉ F.

Example 5.12. [4] Suppose X = {0, a, b, c, d, 1} be a lattice with the following diagram. We define ∼ =∽ on X by

By routine calculations, we can see that (X, ∧ , ∼, ∽ , 1) is a pseudo equality algebra, and we can prove that G = {c, 1} is a fantastic filter, but it is not a positive implicative filter. Because a → (a → b) =1 ∈ G and a ⇝ a = 1 ∈ G, but a → b = a ∉ G. Moreover, in Example 4.16, F = {b, 1} is a positive implicative filter, but it is not a fantastic filter. Because 0 → a = 1 ∈ F, but ((a → 0)  ⇝ 0) → a = a ∉ F. This shows that positive implicative and fantastic filters do not coincide, in general.

Theorem 5.13. Let F be a strong normal filter of X, then F is an implicative filter if and only if F is a fantastic and positive implicative filter.

Proof. (⇒) By Proposition 4.27 and 5.10, the proof is clear.

(⇐) Let x, y ∈ X and (x → y)  ⇝ x ∈ F. By Proposition 2.3(9), (x → y)  ⇝ x ≤ (x → y)  ⇝  ((x → y) → y). Since F is a strong normal filter, then by (XF₂), (x → y)  ⇝  ((x → y) → y) ∈ F. By Proposition 4.20, (x → y) → ((x → y) → y) ∈ F. Since F is a positive implicative filter, by Proposition 4.15(2), (x → y) → y ∈ F, and so (x → y)  ⇝ y ∈ F. Moreover, by Proposition 2.3(7) and Proposition 2.4(2), (x → y)  ⇝ x ≤ y ⇝ x. By (XF₂), y ⇝ x ∈ F. Thus, y → x ∈ F. Since F is a fantastic filter, ((x → y)  ⇝ y) → x ∈ F. Since (x → y)  ⇝ y ∈ F and F is a filter, then x ∈ F. By the similar way, we can show that (x ⇝ y) → x ∈ F implies x ∈ F. By Theorem 4.7(2), F is an implicative filter.

6 Conclusion

It is well-known that using filters with special properties plays an important role in investigating the structure of a logical system. From a logical point of view, the sets of provable formulas can be described by filters of those algebraic semantics. Moreover, the properties of filters have a strong influence on the structure properties of algebras. The aim of this paper is to introduce the notions of (positive) implicative and fantastic filters in pseudo equality algebras and to investigate their properties. Several characterizations of these filters are derived and in this paper we try to investigate the relations among them. We want to set up quotient algebra in pseudo equality algebras associated with these filters in the future.