The lattice generated by hesitant fuzzy filters in pseudo-BCI algebras

Abstract

The notion of hesitant fuzzy set is introduced by V. Torra, which is a very useful tool to express peoples’ hesitancy in daily life. The notion of pseudo-BCI algebra is introduced by W. A. Dudek and Y. B. Jun, which is a kind of nonclassical logic algebra and close connection with various non-commutative fuzzy logic algebras. In this paper, hesitant fuzzy theory is applied to pseudo-BCI algebras. The new concepts of hesitant fuzzy filter and anti-grouped hesitant fuzzy filter in pseudo-BCI algebras are proposed, and their characterizations are presented. Also, the relationships between fuzzy filters and hesitant fuzzy filters are discussed. Moreover, by introducing the notion of tip-extended pair of hesitant fuzzy filters, a new union operation (generated by the union of two hesitant fuzzy filters) is defined and it is proved that the set of all hesitant fuzzy filters in pseudo-BCI algebras forms a bounded distributive lattice about intersection and the new union.

Keywords

Hesitant fuzzy set pseudo-BCI algebra hesitant fuzzy filter anti-grouped hesitant fuzzy filter lattice

1 Introduction

In order to express and deal with uncertainty, many theories were established, such as probability theory, fuzzy set theory [28 , 38], intuitionistic fuzzy set theory [1], soft set theory [12 , 29], rough set theory [2 , 40], et al. The notions of Atanassov’s intuitionistic fuzzy sets [1], type 2 fuzzy sets and fuzzy multisets etc. are a generalization of fuzzy sets. As another generalization of fuzzy sets, V. Torra [20] introduced the notion of hesitant fuzzy sets which is a very useful tool to express peoples’ hesitancy in daily life. The hesitant fuzzy set is very useful to deal with uncertainty, which can be accurately and perfectly described in terms of the opinions of decision makers. Also, hesitant fuzzy set theory is used in decision making problem etc. [17 , 27], and is applied to MTL-algebras [10].

The notion of pseudo-BCI algebra is introduced by W. A. Dudek and Y. B. Jun in [3]. Pseudo-BCI algebra is a generalization of BCI-algebra [7 , 19], and it is also a generalization of pseudo-BCK algebra which is close connection with various non-commutative fuzzy logic formal systems [4 , 41]. In 2006, the notion of pseudo-BCI filter (ideal) of pseudo-BCI algebras is introduced in [8], then some researchers focus on this topic and published some articles related filter theory in pseudo-BCI algebras [11 , 35].

In recent years, various uncertainty mathematical theories are applied to many algebraic systems [13 , 37]. In this paper, the applications of hesitant fuzzy sets to pseudo-BCI algebras are studied. The new concepts of hesitant fuzzy filter and anti-grouped hesitant fuzzy filter in pseudo-BCI algebras are introduced and their properties are studied. Also, the relationships between fuzzy filters and hesitant fuzzy filters are discussed. Moreover, by introducing the notion of tip-extended pair of hesitant fuzzy filters, a new union operation is defined and it is proved that the set of all hesitant fuzzy filters in pseudo-BCI algebras forms a bounded distributive lattice about intersection and the new union.

2 Preliminary

In this section, we review several basic concepts of hesitant fuzzy sets and pseudo-BCI algebras.

Definition 2.1. [20] A hesitant fuzzy set (HFS) A on the universe X is an object of the form $A = {< x, h_{A} (x) > | h_{A} (x) \in ρ ([0, 1]), x \in X},$ where ρ([0, 1]) is the power set of [0, 1].

So, we can define a set of hesitant fuzzy sets by union of their membership functions.

Definition 2. 2. [20] Let A = {μ₁, μ₂, …, μ_n} be a set of n membership functions. The HFS that is associated with A, h_A, is defined as $\begin{matrix} h_{A} : X \to ρ ([0, 1]) \\ h_{A} (x) = ⋃_{μ \in A} {μ (x)} . \end{matrix}$

It is remarkable that this definition is quite suitable to decision making, when experts should access a set of alternatives. In such a case, A represents the assessments of the experts for each alternative and h_A the assessments of the set of experts. However, note that it only allows to recover those HFSs whose memberships are given by sets of cardinalities less than or equal to n.

For convenience, Xia and Xu [25] named the set h = h_A(x) as a hesitant fuzzy element HFE. The family of all hesitant elements defined on X was denoted by HFE(X).

Proposition 2. 1. [21] Let h, h₁, h₂ ∈ HFE (X) and λ∈ [0, 1]. Then the operations complement, union, intersection are defined as follows:

h^c = {1 - γ : γ ∈ h};

h₁ ∪ h₂ = {max (γ₁, γ₂) : γ₁ ∈ h₁, γ₂ ∈ h₂};

h₁ ∩ h₂ = {min (γ₁, γ₂) : γ₁ ∈ h₁, γ₂ ∈ h₂};

h₁ ⊕ h₂ = {γ₁ + γ₂ - γ₁γ₂ : γ₁ ∈ h₁, γ₂ ∈ h₂};

h₁ ⊗ h₂ = {γ₁γ₂ : γ₁ ∈ h₁, γ₂ ∈ h₂};

λh = {1 - (1 - γ) ^λ : γ ∈ h} .

Suppose that the number of values in the hesitant elements is limited, let n(h_A(x)) be the number of values of h_A(x), and arrange them in descending order, and let $h_{A}^{i} (x)$ be the ith smallest value of h_A(x). In many cases, for two hesitant fuzzy sets A and B, n (h_A (x)) ≠ n (h_B (x)). To operate correctly, it is requested that two HFEs have the same length when they are compared. Thus, we should extend the shorter one such that their length is the same. For this, Xu and Xia [26] give the following regulation:

If n (h_A (x)) > n (h_B (x)), then h_B (x) should be extended by adding the minimum value in it until it has the same length with h_A (x); If n (h_A (x)) < n (h_B (x)), then h_A (x) should be extended by adding the minimum value in it until it has the same length with h_B (x). For instance, let h_A (x) = {0.5, 0.3} , h_B (x) = {0.7, 0.3, 0.1}. Clearly, n (h_A (x)) < n (h_B (x)), so we should extend h_A (x) to h_A (x) = {0.5, 0.3, 0.3}.

In fact, we can extend the shorter HFE by adding any value in it until it has the same length with the longer one according to the decision makers’ preferences and actual situations. In this paper, we assume that the decision makers all adopt the above regulation.

Definition 2.3. [3] A pseudo-BCI algebra is a structure (X ≤ , → , ⇝ , 1), where “≤” is a binary relation on X, “→” and “⇝” are binary operations on X and “1” is an element of X, verifying the axioms: for all x, y, z ∈ X,

y → z ≤ (z → x) ⇝ (y → x) , y ⇝ z ≤ (z ⇝ x) → (y ⇝ x);

x ≤ (x → y) ⇝ y, x ≤ (x ⇝ y) → y;

x ≤ x;

x ≤ y, y ≤ x ⇒ x = y;

x ≤ y ⇔ x → y = 1 ⇔ x ⇝ y = 1 .

Proposition 2.2. [3 , 32] Let (X ≤ , → , ⇝ , 1) be a pseudo-BCI algebra, then X satisfy the following properties (∀ x, y, z ∈ X):

1 ≤ x ⇒ x = 1;

x ≤ y ⇒ y → z ≤ x → z, y ⇝ z ≤ x ⇝ z;

x ≤ y, y ≤ z ⇒ x ≤ z;

x ⇝ (y → z) = y → (x ⇝ z);

x ≤ y → z ⇔ y ≤ x ⇝ z;

x → y ≤ (z → x) → (z → y) , x ⇝ y ≤ (z ⇝ x) ⇝ (z ⇝ y);

x ≤ y ⇒ z → x ≤ z → y, z ⇝ x ≤ z ⇝ y;

1 → x = x, 1 ⇝ x = x;

((y → x) ⇝ x) → x = y → x, ((y ⇝ x) → x) ⇝ x = y ⇝ x;

x → y ≤ (y → x) ⇝1, x ⇝ y ≤ (y ⇝ x) →1;

(x → y) →1 = (x → 1) ⇝ (y ⇝ 1) ,

(x⇝ y) ⇝1 = (x ⇝ 1) → (y → 1)

x → 1 = x ⇝ 1.

Definition 2. 4. [8] A nonempty subset F of pseudo-BCI algebra X is called a pseudo-BCI filter (briefly, filter) of X if it satisfies:

1 ∈ F,

x ∈ F, x → y ∈ F ⇒ y ∈ F,

x ∈ F, x ⇝ y ∈ F ⇒ y ∈ F.

Definition 2. 5. [35] A pseudo-BCI algebra X is said to be anti-grouped pseudo-BCI algebra if it satisfies the following identity:

∀ x, y, z ∈ X, (x → y) → (x → z) = y → z,

∀ x, y, z ∈ X, (x ⇝ y) ⇝ (x ⇝ z) = y ⇝ z.

Proposition 2.3. [35] A pseudo-BCI algebra X is an anti-grouped pseudo-BCI algebra if and only if it satisfies: $\forall x \in X, (x \to 1) \to 1 = x or (x ⇝ 1) ⇝ 1 = x .$

Definition 2.6. [8, 32] A fuzzy set A on pseudo-BCI algebra X is called fuzzy filter of X if it satisfies:

∀ x ∈ X, μ_A (x) ≤ μ_A (1);

∀ x, y ∈ X, min {μ_A (x) , μ_A (x → y)} ≤ μ_A (y);

∀ x, y ∈ X, min {μ_A (x) , μ_A (x ⇝ y)} ≤ μ_A (y).

Definition 2.7. [32] A fuzzy set A on pseudo-BCI algebra X is called fuzzy anti-grouped filter if it satisfies:

∀ x ∈ X, μ_A (x) ≤ μ_A (1);

∀ x, y, z ∈ X, min {μ_A (y) , μ_A ((x → y) → (x → z))} ≤ μ_A (z);

∀ x, y, z ∈ X, min {μ_A (y) , μ_A ((x ⇝ y) ⇝ (x ⇝ z))} ≤ μ_A (z).

Proposition 2.4. [32] Let A be a fuzzy filter of pseudo-BCI algebra X. Then A is a fuzzy anti-grouped filter of X if and only if it satisfies: ∀ x ∈ X, $\begin{matrix} μ_{A} (x) \geq μ_{A} ((x \to 1) \to 1), μ_{A} (x) \\ \geq μ_{A} ((x ⇝ 1) ⇝ 1) . \end{matrix}$

3 Hesitant fuzzy filters in pseudo-BCI algebras

In this section, we propose the concept of hesitant fuzzy filter in pseudo-BCI algebra and study their characterizations.

Definition 3.1. A hesitant fuzzy set A in pseudo-BCI algebra X is called a hesitant fuzzy filter of X if it satisfies:

∀x ∈ X, h_A (x) ⊆ h_A (1);

∀ x, y ∈ X, h_A (x) ∩ h_A (x → y) ⊆ h_A (y);

∀ x, y ∈ X, h_A (x) ∩ h_A (x ⇝ y) ⊆ h_A (y).

Proposition 3.1. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X, then

∀ x, y ∈ X, x ≤ y ⇒ h_A (x) ⊆ h_A (y).

Proof. Suppose x ≤ y, then x → y = 1 (by Definition 2.3 (5)). It follows that h_A (x → y) = h_A (1). From this and using Definition 3.1 (HFF1) and (HFF2) we get $\begin{matrix} h_{A} (x) = h_{A} (x) \cap h_{A} (1) = h_{A} (x) \\ \cap h_{A} (x \to y) \subseteq h_{A} (y) . \end{matrix}$

That is, x ≤ y ⇒ h_A (x) ⊆ h_A (y) .

By Definition 2.2 and Definition 3.1, we can easy to verify that the following proposition is true.

Example 3.1. Let X = {a, b, c, d, e, 1} with two binary operations given in Tables 1 and 2. Then (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, where x ≤ y if and only if x → y = 1.

Table 1

→ a b c d e 1

a 1 a b e c b

b b 1 e c d a

c d e 1 a b c

d e c b 1 a d

e c d a b 1 e

1 a b c d e 1

→	a	b	c	d	e	1
a	1	a	b	e	c	b
b	b	1	e	c	d	a
c	d	e	1	a	b	c
d	e	c	b	1	a	d
e	c	d	a	b	1	e
1	a	b	c	d	e	1

Table 2

⇝	a	b	c	d	e	1
a	1	a	e	c	d	b
b	b	1	d	e	c	a
c	e	d	1	b	a	c
d	c	e	a	1	b	d
e	d	c	b	a	1	e
1	a	b	c	d	e	1

Define a hesitant fuzzy set A in X as following: $\begin{matrix} h_{A} (a) = h_{A} (b) = h_{A} (d) = h_{A} (e) = {0.2, 0.3}, \\ h_{A} (c) = {0.4, 0.6}, h_{A} (1) = {0.7, 0.9} . \end{matrix}$ Then A is a hesitant fuzzy filter in X.

Proposition 3.2. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X, denote that A (h) = {x ∈ X|h_A (x) = h_A (1)}. Then A(h) is a filter of X.

Proof. Obviously, 1 ∈ A (h). Assume that x ∈ A (h) and x → y ∈ A (h), then h_A (x) = h_A (1) , h_A (x → y) = h_A (1). From this, using Definition 3.1, we have $h_{A} (1) = h_{A} (x) \cap h_{A} (x \to y) \subseteq h_{A} (y) \subseteq h_{A} (1) .$

It follows that h_A (y) = h_A (1), that is, y ∈ A (h). In the same way, we can get x ∈ A (h) , x ⇝ y ∈ A (h) ⇒ y ∈ A (h). By Definition 2.4 we know that A(h) is a filter of X.

Proposition 3.3. Let A be a hesitant fuzzy set in pseudo-BCI algebra X, then A is a hesitant fuzzy filter in pseudo-BCI algebra X if and only if ∀ x, y, z ∈ X, $\begin{matrix} z \leq x \to y \Rightarrow h_{A} (x) \cap h_{A} (z) \subseteq h_{A} (y); \\ z \leq x ⇝ y \Rightarrow h_{A} (x) \cap h_{A} (z) \subseteq h_{A} (y) . \end{matrix}$

Proof. Assume that A is a hesitant fuzzy filter in X. Let x, y, z ∈ X be such that z ≤ x → y, z ≤ x ⇝ y. By proposition 3.1 (HFF4) and definition 3.1 (HFF2), we have $h_{A} (x) \cap h_{A} (z) \subseteq h_{A} (x) \cap h_{A} (x \to y) \subseteq h_{A} (y) .$

Also, by proposition 3.1 (HFF4) and definition 3.1 (HFF3), we have $h_{A} (x) \cap h_{A} (z) \subseteq h_{A} (x) \cap h_{A} (x ⇝ y) \subseteq h_{A} (y) .$

Conversely, suppose that A is a hesitant fuzzy set in pseudo-BCI algebra X, and ∀ x, y, z∈ X, z ≤ x → y ⇒ h_A (x) ∩ h_A (z) ⊆ h_A (y) z ≤ x ⇝ y ⇒ h_A (x) ∩h_A (z) ⊆ h_A (y). By definition 2.3 (5), we have x ≤ x → 1 =1, x ≤ x ⇝ 1 =1, let z = x, y = 1, then h_A (x) ⊆ h_A (1). Also, by definition 2.3 (2), we have $x \leq (x \to y) ⇝ y, x \leq (x ⇝ y) \to y,$ so $\begin{matrix} h_{A} (x) \cap h_{A} (x \to y) \subseteq h_{A} (y), h_{A} (x) \\ \cap h_{A} (x ⇝ y) \subseteq h_{A} (y) . \end{matrix}$

Therefore, A is a hesitant fuzzy filter.

Proposition 3.4. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X, then $\begin{matrix} \prod_{i = 1}^{n} a_{i} \to x = 1 \Rightarrow ⋂_{i = 1}^{n} h_{A} (ai) \subseteq h_{A} (x); \\ ⨿_{i = 1}^{n} a_{i} ⇝ x = 1 \Rightarrow ⋂_{i = 1}^{n} h_{A} (ai) \subseteq h_{A} (x), \end{matrix}$ for all x, a₁, …, a_n ∈ X, $\prod_{i = 1}^{n}$ a_i → x = a_n → (a_n-1 → ⋯ → (a₁ → x) ⋯)), $⨿_{i = 1}^{n}$ a_i ⇝ x = a_n ⇝ (a_n-1 ⇝ ⋯ ⇝ (a₁ ⇝ x) ⋯)).

Proof. The proof is by induction on n. Let A be a hesitant fuzzy filter in X, by Proposition 3.1 (HFF4) and Proposition 3.3, we know that the conclusion is true for n = 1, 2. Assume that the conclusion is true for n = k, i.e., $\prod_{i = 1}^{k} a_{i} \to x = 1 \Rightarrow ⋂_{i = 1}^{k} h_{A} (a_{i}) \subseteq h_{A} (x); ⨿_{i = 1}^{k} a_{i} ⇝ x = 1 \Rightarrow ⋂_{i = 1}^{k} h_{A} (a_{i}) \subseteq h_{A} (x)$ , for all x, a₁, …, a_k ∈ X. Suppose that $\prod_{i = 1}^{k + 1} a_{i} \to x = 1; ⨿_{i = 1}^{k + 1} a_{i} ⇝ x = 1$ , for all x, a₁, …, a_k, a_k+1 ∈ X, then $\begin{matrix} h_{A} (a_{1} \to x) \supseteq ⋂_{i = 2}^{k + 1} h_{A} (a_{i}), \\ h_{A} (a_{1} ⇝ x) \supseteq ⋂_{i = 2}^{k + 1} h_{A} (a_{i}) . \end{matrix}$

Since A is a hesitant fuzzy filter in X, by Definition 3.1 (HFF2) and (HFF3), we have $\begin{matrix} h_{A} (x) \supseteq h_{A} (a_{1} \to x) \cap h_{A} (a_{1}) \supseteq ⋂_{i = 2}^{k + 1} h_{A} (a_{i}) \cap \\ h_{A} (a_{1}) = ⋂_{i = 1}^{k + 1} h_{A} (a_{i}); \\ h_{A} (x) \supseteq h_{A} (a_{1} ⇝ x) \cap h_{A} (a_{1}) \supseteq ⋂_{i = 2}^{k + 1} h_{A} (a_{i}) \cap \\ h_{A} (a_{1}) = ⋂_{i = 1}^{k + 1} h_{A} (a_{i}) . \end{matrix}$

This completes the proof.

Proposition 3.5. Let A be a hesitant fuzzy set in pseudo-BCI algebra X, if for all $x, a_{1}, \dots, a_{n} \in X, \prod_{i = 1}^{n} a_{i} \to x = 1 \Rightarrow ⋂_{i = 1}^{n} h_{A} (a_{i}) \subseteq h_{A} (x); ⨿_{i = 1}^{n} a_{i} ⇝ x = 1 \Rightarrow ⋂_{i = 1}^{n} h_{A} (a_{i}) \subseteq h_{A} (x)$ , where $\prod_{i = 1}^{n} a_{i} \to x = a_{n} \to (a_{n - 1} \to L \dots \to (a_{1} \to x) \dots)), ⨿_{i = 1}^{n} a_{i} ⇝ x = a_{n} ⇝ (a_{n - 1} ⇝ \dots \to (a_{1} \to x) \dots))$ , then A is a hesitant fuzzy filter in pseudo-BCI algebra X.

Proof. Let x, y, z ∈ X, and z ≤ x → y, z ≤ x ⇝ y, then z → (x → y) =1, z ⇝ (x ⇝ y) =1, so h_A (x) ∩ h_A (z) ⊆ h_A (y), therefore according proposition 3.3, A is a hesitant fuzzy filter in pseudo-BCI algebra X.

4 Anti-grouped hesitant fuzzy filters in pseudo-BCI algebras

In this section, we propose the concept of anti-grouped hesitant fuzzy filter in pseudo-BCI algebra and study their characterizations.

Definition 4.1. A hesitant fuzzy set A in pseudo-BCI algebra X is called anti-grouped hesitant fuzzy filter if it satisfies:

∀ x ∈ X, h_A (x) ⊆ h_A (1);

∀ x, y, z ∈ X, h_A (y) ∩ h_A ((x → y) → (x → z)) ⊆ h_A (z);

∀ x, y, z ∈ X, h_A (y) ∩ h_A ((x ⇝ y) ⇝ (x ⇝ z)) ⊆ h_A (z).

When x = y in Definition 4.1 (2) and (3), we can get (HFF2) and (HFF3) in Definition 3.1, this means that the following proposition is true.

Proposition 4.1. Let A be an anti-grouped hesitant fuzzy filter in pseudo-BCI algebra X. Then A is a hesitant fuzzy filter of X.

Example 4.1. Let X = {a, b, c, d, 1} with two binary operations given in Tables 3 and 4. Then (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, where x ≤ y if and only if x → y = 1.

Table 3

→ a b c d 1

a 1 1 1 d 1

b b 1 1 d 1

c b b 1 d 1

d d d d 1 d

1 a b c d 1

→	a	b	c	d	1
a	1	1	1	d	1
b	b	1	1	d	1
c	b	b	1	d	1
d	d	d	d	1	d
1	a	b	c	d	1

Table 4

⇝	a	b	c	d	1
a	1	1	1	d	1
b	c	1	1	d	1
c	a	b	1	d	1
d	d	d	d	1	d
1	a	b	c	d	1

Define hesitant fuzzy set A in X as following: $\begin{matrix} h_{A} (a) = h_{A} (b) = h_{A} (c) = {0.5, 0.6}, \\ h_{A} (d) = {0.2, 0.4}, h_{A} (1) = {0.7, 0.9} . \end{matrix}$

Then A is an anti-grouped hesitant fuzzy filter in X.

Proposition 4.2. Let A be an anti-grouped hesitant fuzzy filter in pseudo-BCI algebra X. Then A satisfies the following condition, ∀ x ∈ X,

h_A (x) ⊇ h_A ((x → 1) →1);

h_A (x) ⊇ h_A ((x ⇝ 1) ⇝1).

Proof. Putting z = x and y = 1 in Definition 4.1 (2) and (3), we can get the results.

Proposition 4.3. Let A be an anti-grouped hesitant fuzzy set in pseudo-BCI algebra X, denote that A (h) = {x ∈ X|h_A (x) = h_A (1)}. Then A(h) is an anti-grouped filter of X.

It follows that h_A (y) = h_A (1), that is, y ∈ A (h). In the same way, we can get x ∈ A (h) , x ⇝ y ∈ A (h) ⇒ y ∈ A (h). According the Definition 2.4, we know that A(h) is an anti-grouped filter in X.

Lemma 4.1. [32] Let X be a pseudo-BCI algebra X. Then X satisfies the following properties:

∀ x, y, z ∈ X, ((x → y) → (x → z)) →1 = (y → z) →1;

∀ x, y, z ∈ X, ((x ⇝ y) ⇝ (x ⇝ z)) ⇝1 = (y ⇝ z) ⇝1;

∀ x, y ∈ X, (x → y) → (x → 1) = y → 1;

∀ x, y ∈ X, (x ⇝ y) ⇝ (x ⇝ 1) = y ⇝ 1.

Theorem 4.1. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X. Then A is an anti-grouped hesitant fuzzy filter in X if and only if it satisfies: ∀ x ∈ X,

h_A (x) ⊇ h_A ((x → 1) →1);

h_A (x) ⊇ h_A ((x ⇝ 1) ⇝1);

Proof. If A is an anti-grouped hesitant fuzzy filter in X, then by Proposition 4.2 we know that the condition (1) and (2) hold.

Conversely, suppose that A satisfies the condition (1) and (2). For any x, y, z ∈ X, by Definition 2.3 (2) and Lemma 3.1 (1) we have $\begin{matrix} (x \to y) \to (x \to z) \leq (((x \to y) \to (x \to z)) \\ \to 1) \to 1 = (y \to z) \to 1 \to 1 . \end{matrix}$

From this, using Proposition 3.1 (HFF4) and condition (1), we have $\begin{matrix} h_{A} ((x \to y) \to (x \to z)) \subseteq h_{A} (((y \to z) \to 1) \\ \to 1) \subseteq h_{A} (y \to z) . \end{matrix}$

From this, using Definition 3.1 (HFF2) we get $\begin{matrix} h_{A} (y) \cap h_{A} ((x \to y) \to (x \to z)) \subseteq h_{A} (y) \\ \cap h_{A} (y \to z) \subseteq h_{A} (z) . \end{matrix}$

This means that Definition 4.1 (2) holds. By the same way, we can prove that Definition 4.1 (3) holds. Therefore, A is an anti-grouped hesitant fuzzy filter in X.

Proposition 4.4. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X. Then there exist an anti-grouped hesitant fuzzy filter B of X such that h_A (t) ⊆ h_B (t) for any t ∈ [0, 1].

Proof. ∀ x ∈ X, define B : X → [0, 1] by h_B (x) = h_A ((x → 1) →1) .

Firstly, we prove that B is a hesitant fuzzy filter of X. Obviously, ∀ x ∈ X, h_B (x) ⊆ h_B (1), that is, (HFF1) holds for hesitant fuzzy set B. Moreover, ∀ x, y ∈ X, by Proposition 2.2 and Lemma 4.1 we have $\begin{matrix} h_{A} (((x \to y) \to 1) \to 1) = h_{A} (((x \to 1) \to 1) \\ \to ((y \to 1) \to 1)) . \end{matrix}$

By Definition 3.1 (HFF2) we have $\begin{matrix} h_{A} ((y \to 1) \to 1) \supseteq h_{A} (((x \to 1) \to 1) \\ \to ((y \to 1) \to 1)) \cap h_{A} ((x \to 1) \to 1) . \end{matrix}$

Thus, $\begin{matrix} h_{A} ((y \to 1) \to 1) \supseteq h_{A} (((x \to y) \to 1) \to 1) \\ \cap h_{A} ((x \to 1) \to 1) . \end{matrix}$

By the definition of B we have $\begin{matrix} h_{B} (y) = h_{A} ((y \to 1) \to 1) \\ \supseteq h_{A} (((x \to y) \to 1) \to 1) \cap h_{A} ((x \to 1) \to 1) \\ = h_{B} (x \to y) \cap h_{B} (x) . \end{matrix}$

This means that (HFF2) holds for hesitant fuzzy set B. Similarly, we can get that (HFF3) holds for hesitant fuzzy set B. Therefore, B is a hesitant fuzzy filter of X.

Finally, we prove that B is an anti-grouped hesitant fuzzy filter of X. ∀ x ∈ X, by Proposition 2.2 we have $(x \to 1) \to 1 = (((x \to 1) \to 1) \to 1) \to 1 .$

Hence, $\begin{matrix} h_{B} (x) \\ = h_{A} ((x \to 1) \to 1) \\ = h_{A} ((((x \to 1) \to 1) \to 1) \to 1) \\ = h_{B} ((x \to 1) \to 1) . \end{matrix}$

It follows that Proposition 4.2 (1) for hesitant fuzzy filter B. Similarly, we can get Proposition 4.2 (2) holds for hesitant fuzzy filter B. By Theorem 4.1 we know that B is an anti-grouped hesitant fuzzy filter of X.

5 The lattice generated by hesitant Fuzzy filters

In this section, we propose the concept of generated hesitant fuzzy filter in pseudo-BCI algebra and prove that the set of all hesitant fuzzy filters forms a bounded distributive lattice.

Proposition 5.1. If A and B are two hesitant fuzzy filters in pseudo-BCI algebra X, then A ∩ B is also a hesitant fuzzy filter in X.

Proof. Suppose that A and B are two hesitant fuzzy filters in pseudo-BCI algebra X. By Definition 3.1 (HFF1), we have ∀ x ∈ X, h_A (x) ⊆ h_A (1) and h_B (x) ⊆ h_B (1). It follows that ∀ x ∈ X, h_A (x) ∩ h_B (x) ⊆ h_A (1) ∩ h_B (1). From this, using Definition 2.2, we get that ∀ x ∈ X, h_A∩B (x) ⊆ h_A∩B (1). That is, A ∩ B satisfies (HFF1).

Moreover, by Definition 3.1 (HFF2), we have $\begin{matrix} \forall x, y \in X, h_{A} (x) \cap h_{A} (x \to y) \subseteq h_{A} (y); h_{B} (x) \\ \cap h_{B} (x \to y) \subseteq h_{B} (y) . \end{matrix}$

Then, ∀ x, y ∈ X, $\begin{matrix} h_{A \cap B} (x) \cap h_{A \cap B} (x \to y) = (h_{A} (x) \cap h_{B} (x)) \\ \cap (h_{A} (x \to y) \cap h_{B} (x \to y)) \subseteq h_{A} (x) \\ \cap h_{A} (x \to y) \subseteq h_{A} (y); \end{matrix}$ $\begin{matrix} h_{A \cap B} (x) \cap h_{A \cap B} (x \to y) = (h_{A} (x) \cap h_{B} (x)) \\ \cap (h_{A} (x \to y) \cap h_{B} (x \to y)) \subseteq h_{B} (x) \\ \cap h_{B} (x \to y) \subseteq h_{B} (y); \end{matrix}$

It follows that $\begin{matrix} h_{A \cap B} (x) \cap h_{A \cap B} (x \to y) \subseteq h_{A} (y) \\ \cap h_{B} (y) = h_{A \cap B} (y) . \end{matrix}$

That is, A ∩ B satisfies (HFF2).

Similarly, we can prove that A ∩ B satisfies (HFF3). Therefore, A ∩ B is a hesitant fuzzy filter in X.

The following example shows that the union of two hesitant fuzzy filters may be not a hesitant fuzzy filter.

Example 5.1. Let X = {a, b, c, d, e, 1} with two binary operations given in Tables 1 and 2 (see Example 3.1). Then (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, where x ≤ y if and only if x → y = 1. Define hesitant fuzzy sets A and B in X as following: $\begin{matrix} h_{A} (a) = h_{A} (b) = h_{A} (d) = h_{A} (e) = {0.2, 0.3}, \\ h_{A} (c) = {0.4, 0.6}, h_{A} (1) = {0.7, 0.9} . \\ h_{B} (a) = h_{B} (b) = h_{B} (c) = h_{B} (d) = {0.3, 0.4}, \\ h_{B} (e) = {0.6, 0.7}, h_{B} (1) = {0.8, 0.9} . \end{matrix}$

Then A, B are hesitant fuzzy filters in X. Let C = A ∩ B and D = A ∪ B, then $\begin{matrix} h_{C} (a) = h_{A \cap B} (a) = h_{C} (b) = h_{A \cap B} (b) = h_{C} (d) = \\ h_{A \cap B} (d) = h_{C} (e) = h_{A \cap B} (e) = {0.2, 0.3}, \\ h_{C} (c) = h_{A \cap B} (c) = {0.3, 0.4}, \\ h_{C} (1) = h_{A \cap B} (1) = {0.7, 0.8}; \\ h_{D} (a) = h_{A \cup B} (a) = h_{D} (b) = h_{A \cup B} & (b) = h_{D} (d) = h_{A \cup B} & (d) = {0.3, 0.4}, h_{D} (e) = h_{A \cup B} & (e) = {0.6, 0.7}, \\ h_{D} (c) = h_{A \cup B} (c) = {0.4, 0.6}, \\ h_{D} (1) = h_{A \cup B} (1) = {0.8, 0.9} . \end{matrix}$

We can verify that C = A ∩ B is a hesitant fuzzy filter in X. But D = A ∪ B is not a hesitant fuzzy filter in X, since $\begin{matrix} h_{A \cup B} (c) \cap h_{A \cup B} (c \to b) = h_{A \cup B} (c) \cap h_{A \cup B} \\ (e) = {0.6, 0.7} ⊄ {0.3, 0.4} = h_{A \cap B} (b) . \end{matrix}$

Lemma 5.1. Let (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, for any x ∈ X, if exist a_i ∈ X can be used $\prod_{i = 1}^{n} a_{i} \to x = 1$ , then exist b_j ∈ X can be used $⨿_{j = 1}^{n} b_{j} ⇝ x = 1$ . Conversely, for any x ∈ X, if exist b_j ∈ X can be used $⨿_{j = 1}^{n} b_{j} ⇝ x = 1$ , then exist a_i ∈ X can be used $⨿_{i = 1}^{n} a_{i} \to x = 1$ , where $\prod_{i = 1}^{n} a_{i} \to x = a_{n} \to (a_{n - 1} \to \dots \to (a_{1} \to x) \dots))$ , $⨿_{j = 1}^{n} b_{j} ⇝ x = b_{n} ⇝ (b_{n - 1} ⇝ \dots ⇝ (b_{1} ⇝ x) \dots))$ .

Proof. The proof is by induction on n. We first prove the part one, that is if exist a_i ∈ X can be used $\prod_{i = 1}^{n} a_{i} \to x = 1$ , then exist b_j ∈ X can be used $⨿_{j = 1}^{n} b_{j} ⇝ x = 1$ . Obviously, the conclusion is true for n = 1. When n = 2, by the Proposition 2.2, if a₂ → (a₁ → x) =1, then a₂ → (a₁ → x) =1 ⇔ a₂ ⇝ (a₁ → x) =1 ⇔ a₁ → (a₂ ⇝ x) =1 ⇔ a₂ ⇝ (a₁ ⇝ x) =1, so let b₁ = a₂, b₂ = a₁, we can get the conclusion. Assume that the conclusion is true for n = k, when n = k + 1, suppose that for any x ∈ X, if exist a_i ∈ X can be used $\prod_{i = 1}^{k + 1} a_{i} \to x = 1$ , then $\begin{matrix} \prod_{i = 1}^{k + 1} a_{i} \to x = a_{k + 1} \to (a_{k} \to \dots \\ \to (a_{1} \to x) \dots)) = 1 \\ \Leftrightarrow a_{k + 1} ⇝ (a_{k} \to \dots \to (a_{1} \to x) \dots)) = 1 \\ \Leftrightarrow a_{k} \to (a_{k + 1} ⇝ \dots \to (a_{1} \to x) \dots)) = 1 \\ \Leftrightarrow a_{k} ⇝ (a_{k + 1} ⇝ \dots \to (a_{1} \to x) \dots)) = 1 \\ \Leftrightarrow a_{k - 1} \to (a_{k} ⇝ (a_{k + 1} ⇝ \dots \to (a_{1} \to x) \dots))) \\ \Leftrightarrow a_{k - 1} ⇝ (a_{k} ⇝ (a_{k + 1} ⇝ \dots \to (a_{1} \to x) \dots))) \\ \Leftrightarrow a_{1} ⇝ (a_{2} ⇝ \dots \to (a_{k + 1} ⇝ x) \dots))) . \end{matrix}$

Thus, let b₁ = a_k+1, b₂ = a_k, …, b_k = a₂, b_k+1 = a₁, then $⨿_{j = 1}^{k + 1}$ b_j ⇝ x = 1. So, we get the proof.

Also, the proof of part two is similar with the proof of part one. That is, for any x ∈ X, if exist b_j ∈ X can be used $⨿_{j = 1}^{n} b_{j} ⇝ x = 1$ , then exist a_i ∈ X can be used $\prod_{i = 1}^{n} a_{i} \to x = 1$ .

This completes the proof.

Similar to Lemma 5.1, we can get the following results (the proofs are omitted).

Lemma 5.2. Let (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, for any x ∈ X, if exist a_i ∈ X can be used $\prod_{i = 1}^{n} a_{i} \to x = 1$ , then exist b_j ∈ X can be used $\prod_{j = 1}^{n} b_{j} * x = 1$ (* represent → or ⇝). Conversely, for any x ∈ X, if exist b_j ∈ X can be used $\prod_{j = 1}^{n} b_{j} * x = 1$ , then exist a_i ∈ X can be used $\prod_{i = 1}^{n} a_{i} \to x = 1$ , where $\prod_{i = 1}^{n} a_{i} \to x = a_{n} \to (a_{n - 1} \to \dots \to (a_{1} \to x) \dots))$ , $\prod_{j = 1}^{n} b_{j} * x = b_{n} * (b_{n - 1} * \dots * (b_{1} * x) \dots))$ .

Lemma 5.3. Let (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, for any x ∈ X, if exist a_i ∈ X can be used $\prod_{i = 1}^{n} a_{i} ⇝ x = 1$ , then exist b_j ∈ X can be used $\prod_{j = 1}^{n} b_{j} * x = 1$ (* represent → or ⇝). Conversely, for any x ∈ X, if exist b_j ∈ X can be used $\prod_{j = 1}^{n} b_{j} * x = 1$ , then exist a_i ∈ X can be used $⨿_{i = 1}^{n} a_{i} ⇝ x = 1$ , where $⨿_{i = 1}^{n} a_{i} ⇝ x = a_{n} ⇝ (a_{n - 1} ⇝ \dots ⇝ (a_{1} ⇝ x) \dots))$ , $\prod_{j = 1}^{n} b_{j} * x = b_{n} * (b_{n - 1} * \dots * (b_{1} * x) \dots))$ .

Lemma 5.4. Let (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, for any x, y ∈ X, if exist a_i, b_j ∈ X can be used $\prod_{i = 1}^{n} a_{i} \to x = 1, \prod_{j = 1}^{m} b_{i} \to (x \to y) = 1 (m, n \in N)$ , then b_m → (b_m-1 → ⋯ → (b₁ → ⋯ → (a_n → (a_n-1 → ⋯ → (a₁ → y) ⋯)))) ⋯)) =1, where $\prod_{i = 1}^{n} a_{i} \to x = a_{n} \to (a_{n - 1} \to \dots \to (a_{1} \to x)$ $\dots)), \prod_{j - 1}^{m} b_{j} \to (x \to y) = b_{m} \to (b_{m - 1} \to \dots \to (b_{1} \to (x \to y))$ ⋯)) .

Proof. The proof is by induction on n.

Case 1. m = n. When n = 1, let a₁ → x = 1, b₁ → x → y) =1, then we can get a₁ ≤ x, b₁ ≤ x → y. By Proposition 2.2 (2), we have x → y ≤ a₁ → y. So b₁ ≤ a₁ → y, that is b₁ → a₁ → y) =1. Hence the conclusion is true for n = 1. When n = 2, let a₂ → (a₁ → x) =1, b₂ → (b₁ → x → y)) =1, then we can get a₂ ≤ a₁ → x, b₂ ≤ b₁ → x → y). By Proposition 2.2 (2) and (6), we have x → y ≤ (a₁ → x) → (a₁ → y) ≤ a₂ → (a₁ → y).

By Proposition 2.2 (7), we can get b₁ → x → y) ≤ b₁ → (a₂ → (a₁ → y)). So b₂ ≤ b₁ → x → y) ≤ b₁ → (a₂ → (a₁ → y)), that is b₂ → (b₁ → (a₂ → (a₁ → y))) =1. Hence the conclusion is true for n = 2. Assume that the conclusion is true for n = k, when n = k + 1, suppose that $\begin{matrix} a_{k + 1} \to (a_{k} \to \dots \to (a_{1} \to x) \dots)) = 1, \\ b_{k + 1} \to (b_{k} \to \dots \to (b_{1} \to x \to y)) \dots)) = 1 . \end{matrix}$

Then we have a_k+1 ≤ a_k → ⋯ → (a₁ → x) ⋯), b_k+1 ≤ b_k → ⋯ → (b₁ → x → y)) ⋯). By Proposition 2.2 (2) and (6), we have

$\begin{matrix} x \to y \leq (a_{1} \to x) \to (a_{1} \to y) \\ \leq (a_{2} \to (a_{1} \to x)) \to (a_{2} \to (a_{1} \to y)) \\ \leq (a_{3} \to (a_{2} \to (a_{1} \to x))) \to (a_{3} (a_{2} \to (a_{1} \to y))) \\ \leq \dots \leq (a_{k} \to (a_{k - 1} \to \dots \to (a_{1} \to x) \dots))) \\ \to (a_{k} \to (a_{k - 1} \to \dots \to (a_{1} \to y) \dots))) \\ \leq a_{k + 1} \to (a_{k} \to \dots \to (a_{1} \to y) \dots)) . \end{matrix}$

By Proposition 2.2 (7), we can get $\begin{matrix} b_{1} \to (x \to y) \leq b_{1} \to (a_{k + 1} \to (a_{k} \to (\dots \to \\ (a_{1} \to y) \dots))), \end{matrix}$ $\begin{matrix} b_{2} \to (b_{1} \to (x \to y)) \leq b_{2} \to (b_{1} \to (a_{k + 1} \\ \to (a_{k} \to (\dots \to (a_{1} \to y) \dots)))), \dots, \end{matrix}$ $b_{k} \to (\dots \to (b_{1} \to (x \to y)) \dots)$ $\begin{matrix} \leq b_{k} \to (b_{k - 1} \to (\dots \to (b_{1} \to (a_{k + 1} \\ \to (a_{k} \to (\dots \to (a_{1} \to y) \dots)))) \dots)) . \end{matrix}$

So $b_{k + 1} \leq b_{k} \to (\dots \to (b_{1} \to (x \to y)) \dots)$ $\begin{matrix} \leq b_{k} \to (b_{k - 1} \to (\dots \to (b_{1} \to (a_{k + 1} \\ \to (a_{k} \to (\dots \to (a_{1} \to y) \dots)))) \dots)), \end{matrix}$ that is $\begin{matrix} b_{k + 1} \to (b_{k} \to (b_{k - 1} \to (\dots \to (b_{1} \to (a_{k + 1} \\ \to (a_{k} \to (\dots \to (a_{1} \to y) \dots)))) \dots))) = 1 . \end{matrix}$

Hence the conclusion is true for n = k + 1.

Therefore, the conclusion is true for Case 1.

Case 2. m ≠ n. Let n < m and a_n → (a_n-1 → (⋯ → (a₁ → x) ⋯)) =1, b_m → (b_m-1 → (⋯ → (b₁ → (x → y)) ⋯)). So, 1 → (1 → ⋯ → (1 → (a_n → (a_n-1 → (⋯ → (a₁ → x) ⋯)))) ⋯) =1 (the number of (1 →) is m−n), then the conclusion is true and the proof is similar with Case 1. Also, if m < n, then the conclusion is true.

This completes the proof.

Lemma 5.5. Let (X ≤ , → , ⇝ , 1) is a pseudo-BCI algebra, for any x, y ∈ X, if exist a_i, b_j ∈ X can be used $\prod_{i = 1}^{n} a_{i} \to x = 1, \prod_{j = 1}^{m} b_{i} \to (x ⇝ y) = 1 (m, n \in N)$ , then b₁ ⇝ (b₂ (⋯ ⇝ (b_m ⇝ (a₁ ⇝ (a₂ ⇝ (⋯ ⇝ (a_n ⇝ y) ⋯)))) ⋯)) =1, where $\prod_{i = 1}^{n} a_{i} \to x = a_{n} \to (a_{n - 1} \to (\dots \to (a_{1} \to x) \dots$ $)), \prod_{j = 1}^{m} b_{j} \to (x ⇝ y) = b_{m} \to (b_{m - 1} \to (\dots \to (b_{1} \to (x ⇝ y)) \dots)) .$

Proof. According Lemma 5.1 and Lemma 5.4, we can get the proof.

Definition 5.1. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X, the intersection of all hesitant fuzzy filters containing A is called the generated hesitant fuzzy filter by A, denoted as <A>.

Proposition 5.2. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X, B is a hesitant fuzzy set of X where

$\begin{matrix} h_{B} (x) \\ = \underset{a_{i}, x \in X; i = 1, 2, \dots, n; n \in N}{⋃_{a_{n} \to (\dots \to (a_{1} \to x) \dots) = 1}} {h_{A} (a_{1}) \cap \dots \cap h_{A} (a_{n})} \end{matrix}$ for all x ∈ X. Then B = <A>.

Proof. We first verify that B is a hesitant fuzzy filter. Obviously, h_B (y) ⊆ h_B (1). For all x, y ∈ X, wehave

$\begin{matrix} h_{B} (x) \cap h_{B} (x \to y) \\ = (\underset{a_{i}, x \in X; i = 1, 2, \dots, n; n \in N}{⋃_{a_{n} \to (\dots \to (a_{1} \to x) \dots) = 1}} {h_{A} (a_{1}) \cap \dots \cap h_{A} (a_{n})}) \\ \cap (\underset{b_{j}, y \in X; j = 1, 2, \dots, m; m \in N}{⋃_{b_{m} \to (\dots \to (b_{1} \to (x \to y)) \dots) = 1}} {h_{A} (b_{1}) \cap \dots \cap h_{A} (b_{m})}) \\ = \underset{a_{i}, x \in X; i = 1, 2, \dots, n; n \in N}{⋃_{a_{n} \to (\dots \to (a_{1} \to x) \dots) = 1}} \underset{b_{j}, y \in X; j = 1, 2, \dots, m; m \in N}{⋃_{b_{m} \to (\dots \to (b_{1} \to (x \to y)) \dots) = 1}} \\ {h_{A} (a_{1}) \cap \dots \cap h_{A} (a_{n}) \cap h_{A} (b_{1}) \cap \dots \cap h_{A} (b_{m})} . \end{matrix}$

By Lemma 5.4, the above can be written as $\begin{matrix} \subseteq \underset{a_{i}, b_{j}, x, y \in X; i = 1, 2, \dots n; j = 1, 2, \dots, m; m, n \in N}{⋃_{a_{n} \to (\dots \to (a_{1} \to (b_{m} \to (\dots \to (b_{1} \to y) \dots))) \dots) = 1}} \\ {h_{A} (a_{1}) \cap \dots \cap h_{A} (a_{n}) \cap h_{A} (b_{1}) \cap \dots \cap h_{A} (b_{m})} \\ = h_{B} (y) . \end{matrix}$

Similarly, we can get $h_{B} (x) \cap h_{B} (x \in y) \subseteq h_{B} (y) .$

Thus, B is a hesitant fuzzy filter in X.

Then we need prove that B is the minimal hesitant fuzzy filter containing A of X. Let C be a hesitant fuzzy filter of X, and A ⊆ C, then $\begin{matrix} h_{B} (x) = \underset{a_{i} \in X, n \in N}{⋃_{a_{n} \to (\dots \to (a_{1} \to x) \dots) = 1}} {h_{A} (a_{1}) \cap \dots \cap h_{A} (a_{n})} \\ \subseteq \underset{a_{i} \in X, n \in N}{⋃_{a_{n} \to (\dots \to (a_{1} \to x) \dots) = 1}} {h_{C} (a_{1}) \cap \dots \cap h_{C} (a_{n})} \subseteq h_{C} (x) \end{matrix}$

So, B ⊆ C.

Therefore, B = <A>.

Definition 5.2. [5, 6] Let f and g be fuzzy sets on L. Then tip-extended pair of f and g can be defined by $f^{g} (x) = {\begin{matrix} f (x), & x \neq 1, \\ f (1) \lor g (1), & x = 1, \end{matrix}$ $g^{f} (x) = {\begin{matrix} g (x), & x \neq 1, \\ g (1) \lor f (1), & x = 1 . \end{matrix}$

Theorem 5.1. Let A be a hesitant fuzzy filter in pseudo-BCI algebra X and t ∈ [0, 1]. Then A^t is a hesitant fuzzy filter in X, where $h_{A}^{t} (x) = {\begin{matrix} h_{A} (x), x \neq 1, \\ h_{A} (1) \cup t, x = 1 . \end{matrix}$

Proof. Obviously, A^t is a hesitant fuzzy set, now we prove A^t is a hesitant fuzzy filter. Let x, y ∈ X, it is obvious that $h_{A}^{t} (x) \subseteq h_{A}^{t} (1)$ , then we consider the following two cases:

Case i: x → y = 1, x ⇝ y = 1.

Case 1. if x = 1, y = 1, then $h_{A}^{t} (x) \cap h_{A}^{t} (x \to y) = h_{A}^{t} (1) = h_{A}^{t} (y) \subseteq h_{A}^{t} (y), h_{A}^{t} (x) \cap h_{A}^{t} (x ⇝ y) = h_{A}^{t} (1) = h_{A}^{t} (y) \subseteq h_{A}^{t} (y);$

Case 2. if x = 1, y ≠ 1, it is a contradiction;

Case 3. if x ≠ 1, y = 1, then $h_{A}^{t} (x) \cap h_{A}^{t} (x \to y) = h_{A}^{t} (x) \cap h_{A}^{t} (1) \subseteq h_{A}^{t} (1) = h_{A}^{t} (y), h_{A}^{t} (x) \cap h_{A}^{t} (x ⇝ y) = h_{A}^{t} (x) \cap h_{A}^{t} (1) \subseteq h_{A}^{t} (1) = h_{A}^{t} (y);$

Case 4. if x ≠ 1, y ≠ 1, then $h_{A}^{t} (x) \cap h_{A}^{t} (x \to y) = h_{A}^{t} (x) \cap h_{A}^{t} (1) \subseteq h_{A}^{t} (x) = h_{A} (x) \subseteq h_{A} (y) = h_{A}^{t} (y), h_{A}^{t} (x) \cap h_{A}^{t} (x ⇝ y) = h_{A}^{t} (x) \cap h_{A}^{t} (1) \subseteq h_{A}^{t} (x) = h_{A} (x) \subseteq h_{A} (y) = h_{A}^{t} (y) .$

Case ii: x → y ≠ 1, x ⇝ y ≠ 1.

Case 1. if x = 1, y = 1, it is a contradiction;

Case 2. if x = 1, y ≠ 1, $h_{A}^{t} (x) \cap h_{A}^{t} (x \to y) = h_{A}^{t} (1) \cap h_{A}^{t} (y) \subseteq h_{A}^{t} (y), h_{A}^{t} (x) \cap h_{A}^{t} (x ⇝ y) = h_{A}^{t} (1) \cap h_{A}^{t} (y) \subseteq h_{A}^{t} (y);$

Case 3. if x ≠ 1, y = 1, $h_{A}^{t} (x) \cap h_{A}^{t} (x \to y) = h_{A} (x) \cap h_{A} (x \to y) \subseteq h_{A} (y) = h_{A} (1) = h_{A}^{t} (y), h_{A}^{t} (x) \cap h_{A}^{t} (x ⇝ y) = h_{A} (x) \cap h_{A} (x ⇝ y) \subseteq h_{A} (y) = h_{A} (1) = h_{A}^{t} (y);$

Case 4. if x ≠ 1, y ≠ 1, then $h_{A}^{t} (x) \cap h_{A}^{t} (x \to y) = h_{A} (x) \cap h_{A} (x \to y) \subseteq h_{A} (y) = h_{A}^{t} (y), h_{A}^{t} (x) \cap h_{A}^{t} (x ⇝ y) = h_{A} (x) \cap h_{A} (x ⇝ y) \subseteq h_{A} (y) = h_{A}^{t} (y) .$

All in all, it holds that $h_{A}^{t} (x) \cap h_{A}^{t} (x \to y) \subseteq h_{A}^{t} (y), h_{A}^{t} (x) \cap h_{A}^{t} (x ⇝ y) \subseteq h_{A}^{t} (y)$ .

Thus, A^t is a hesitant fuzzy filter.

Definition 5.3. For hesitant fuzzy sets A and B, the operation * is defined by $A * B = {〈 x, (h_{A} ⊙ h_{B}) (x) 〉 | x \in X} .$ where $h_{A} ⊙ h_{B} (x) = \underset{y, z, x \in X}{\underset{y \to (z \to x) = 1}{\cup}} {h_{A} (y) \cap h_{B} (z)} .$

Furthermore, the tip-extended pair for hesitant fuzzy sets A and B are defined by $\begin{matrix} A^{B} = {〈 x, h_{A}^{h_{B}} (x) 〉 | x \in X}, \\ B^{A} = {〈 x, h_{B}^{h_{A}} (x) 〉 | x \in X} . \end{matrix}$ where $\begin{matrix} h_{A}^{h_{B}} (x) = {\begin{matrix} h_{A} (x), & x \neq 1 \\ h_{A} (1) \cup h_{B} (1), & x = 1 \end{matrix}, \\ h_{B}^{h_{A}} (x) = {\begin{matrix} h_{B} (x), & x \neq 1 \\ h_{B} (1) \cup h_{A} (1), & x = 1 \end{matrix} . \end{matrix}$

Definition 5.4. Let A and B be hesitant fuzzy filters in pseudo-BCI algebra X, then $A^{B} * B^{A} = 〈 A \cup B 〉 .$

Proof. Let x, y ∈ X, then $\begin{matrix} h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (x) \cap h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (x \to y) \\ = (\underset{a, b, x \in X}{⋃_{a \to (b \to x) = 1}} {h_{A}^{h_{B}} (a) \cap h_{A}^{h_{B}} (b)}) \cap \end{matrix}$

$\begin{matrix} (\underset{c, d, x, y \in X}{⋃_{c \to (d \to (x \to y)) = 1}} {h_{A}^{h_{B}} (c) \cap h_{A}^{h_{B}} (d)}) \\ = \underset{a, b, x \in X}{⋃_{a \to (b \to x) = 1}} \underset{c, d, x, y \in X}{⋃_{c \to (d \to (x \to y)) = 1}} \\ {h_{A}^{h_{B}} (a) \cap h_{A}^{h_{B}} (b) \cap h_{A}^{h_{B}} (c) \cap h_{A}^{h_{B}} (d)} \\ \subseteq \underset{a, b, c, d, x, y \in X}{⋃_{c \to (d \to (a \to (b \to y))) = 1}} \\ {h_{A}^{h_{B}} (a) \cap h_{A}^{h_{B}} (b) \cap h_{A}^{h_{B}} (c) \cap h_{A}^{h_{B}} (d)} \\ = h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (y) . \end{matrix}$

Also, we can get $h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (x) \cap h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (x \in y) \subseteq h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (y)$

Thus, A^B *B^A is a hesitant fuzzy filter in X.

It is easy to prove that A, B ⊆ A^B * B^A, hence A ∪ B ⊆ A^B * B^A. Thus 〈A ∪ B 〉 ⊆ A^B * B^A.

Let C be a hesitant fuzzy filter in X and A ∪ B ⊆ C, then $\begin{matrix} h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (x) \\ = \underset{a, b, x \in X}{⋃_{a \to (b \to x) = 1}} {h_{A}^{h_{B}} (a) \cap h_{A}^{h_{B}} (b)} \\ = \underset{a, b, x \in X; a \neq 1, b \neq 1}{⋃_{a \to (b \to x) = 1}} {h_{A}^{h_{B}} (a) \cap h_{A}^{h_{B}} (b)} \cup \\ ⋃_{a \to x = 1} {h_{A} (a)} \cup ⋃_{b \to x = 1} {h_{B} (b)} \\ = \underset{a, b, x \in X; a \neq 1, b \neq 1}{⋃_{a \to (b \to x) = 1}} {h_{A} (a) \cap h_{B} (b)} \cup \end{matrix}$

$\begin{matrix} ⋃_{a \to x = 1} {h_{A} (a)} \cup ⋃_{b \to x = 1} {h_{B} (b)} \\ \subseteq \underset{a, b, x \in X; a \neq 1, b \neq 1}{⋃_{a \to (b \to x) = 1}} {h_{C} (a) \cap h_{C} (b)} \cup \\ ⋃_{a \to x = 1} {h_{C} (a)} \cup ⋃_{b \to x = 1} {h_{C} (b)} \\ = \underset{a, b, x \in X}{⋃_{a \to (b \to x) = 1}} {h_{C} (a) \cap h_{C} (b)} \subseteq h_{C} (x) . \end{matrix}$

So, A^B * B^A ⊆ C.

Thus A^B * B^A=<A∪B>B.

For A, Bare two hesitant fuzzy filters in pseudo-BCI algebra X, the operations ⊓ and ⊔ on hesitant fuzzy filters in X are defined by $A ⊓ B = A \cap B, A ⊔ B = A^{B} * B^{A} .$

Theorem 5.2. (HFF (X) , ⊔ , ⊓ , ∅ , X) is a bounded distributive lattice.

Proof. We only verify the distributivity. Let A, B, C be hesitant fuzzy filters in X. Obviously, C ⊓ (A ⊔ B) ⊇ (C ⊓ A) ⊔ (C ⊓ B). So, we only prove that C ⊓ (A ⊔ B) ⊆ (C ⊓ A) ⊔ (C ⊓ B). Assume that x ∈ X, we consider the following two cases:

Case 1. x = 1. We have

$h_{C} \cap h_{A ⊔ B} (x) = h_{C} \cap h_{A ⊔ B} (1) = h_{C} (1) \cap h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (1) = h_{C} (1) \cap (h_{A} (1) \cup h_{B} (1)) = (h_{C} (1) \cap h_{A} (1)) \cup (h_{C} (1) \cap h_{B} (1)) = (h_{C} \cap h_{A})^{hC \cap hB} ⊙ (h_{C} \cap h_{B})^{hC \cap hA} (1);$

Case 2: x ≠ 1. We have

$\begin{matrix} h_{C} \cap h_{A ⊔ B} (x) \\ = h_{C} (x) \cap h_{A}^{h_{B}} ⊙ h_{B}^{h_{A}} (x) \\ = h_{C} (x) \cap (⋃_{y \to (z \to x) = 1} {h_{A}^{h_{B}} (y) \cap h_{B}^{h_{A}} (z)}) \\ = ⋃_{y \to (z \to x) = 1} {h_{C} (x) \cap h_{A}^{h_{B}} (y) \cap h_{B}^{h_{A}} (z)} \\ = (\underset{y \neq 1, z \neq 1}{⋃_{y \to (z \to x) = 1}} {h_{C} (x) \cap h_{A} (y) \cap h_{B} (z)}) \cup {h_{C} (x) \cap \end{matrix}$

$\begin{matrix} h_{A}^{h_{B}} (1) \cap h_{B} (x)} \cup {h_{C} (x) \cap h_{A} (x) \cap h_{B}^{h_{A}} (1)} \\ = (\underset{y \neq 1, z \neq 1}{⋃_{y \to (z \to x) = 1}} {h_{C} (x) \cap h_{A} (y) \cap h_{C} (x) \cap h_{B} (z)}) \\ \cup {h_{C} (x) \cap h_{C} (1) \cap h_{A}^{h_{B}} (1) \cap h_{B} (x)} \\ \cup {h_{C} (x) \cap h_{C} (1) \cap h_{A} (x) \cap h_{B}^{h_{A}} (1)} \\ = (\underset{y \neq 1, z \neq 1}{⋃_{y \to (z \to x) = 1}} {h_{C} (x) \cap h_{A} (y) \cap h_{C} (x) \cap h_{B} (z)}) \\ \cup {h_{C} (1) \cap (h_{A} (1) \cup h_{B} (1)) \cap \\ [(h_{C} (x) \cap h_{B} (x)) \cup (h_{C} (x) \cap h_{A} (x))]} \\ = (\underset{y \neq 1, z \neq 1}{⋃_{y \to (z \to x) = 1}} {h_{C} (x) \cap h_{A} (y) \cap h_{C} (x) \cap h_{B} (z)}) \\ \cup {[(h_{C} (1) \cap h_{A} (1)) \cup (h_{C} (1) \cap h_{B} (1))] \cap \\ [(h_{C} (x) \cap h_{B} (x)) \cup (h_{C} (x) \cap h_{A} (x))]} \\ \subseteq \underset{y \neq 1, z \neq 1}{⋃_{y \to (z \to x) = 1}} {{(h_{C} \cap h_{A})}^{h_{C} \cap h_{B}} (y \lor x) \cap \\ {(h_{C} \cap h_{B})}^{h_{C} \cap h_{A}} (z \lor x)} \\ \cup [{(h_{C} \cap h_{A})}^{h_{C} \cap h_{B}} (1) \cap (h_{C} \cap h_{A}) (y \lor x)] \\ \cup [{(h_{C} \cap h_{B})}^{h_{C} \cap h_{A}} (1) \cap (h_{C} \cap h_{B}) (z \lor x)] \\ = ⋃_{y \to (z \to x) = 1} {{(h_{C} \cap h_{A})}^{h_{C} \cap h_{B}} (y \lor x) \cap \\ {(h_{C} \cap h_{B})}^{h_{C} \cap h_{A}} (z \lor x)} . \end{matrix}$

Let y ∨ x = y′ and z ∨ x = z′ It is easy to verify that y′ → z′ → x)=1, and then the above can be written as $\begin{matrix} = ⋃_{y^{'} \to (z^{'} \to x) = 1} {{(h_{C} \cap h_{A})}^{h_{C} \cap h_{B}} (y^{'}) \\ \cap {(h_{C} \cap h_{B})}^{h_{C} \cap h_{A}} (z^{'})} \\ = {(h_{C} \cap h_{A})}^{h_{C} \cap h_{B}} ⊙ {(h_{C} \cap h_{B})}^{h_{C} \cap h_{A}} (x) \end{matrix}$

Thus, C ⊓ (A ⊔ B) ⊆ (C ⊓ A) ⊔ (C ⊓ B).

Hence, C ⊓ (A ⊔ B) = (C ⊓ A) ⊔ (C ⊓ B), that is, the distributivity holds.

6 Conclusion

In this paper, we introduce the new concepts of hesitant fuzzy filter and anti-grouped hesitant fuzzy filter in pseudo-BCI algebras and study their characterizations. And by introducing the notion of tip-extended pair of hesitant fuzzy filters, a new union operation is defined and it is proved that the set of all hesitant fuzzy filters in pseudo-BCI algebras forms a bounded distributive lattice about intersection and the new union.

In terms of the future direction, we will study the fusions of algebra structure and neutrosophic sets based on paper [34, 39]. Also we will address these issues in our forthcoming research.

Footnotes

Acknowledgments

This work was supported by National Natural Science Foundation of China (Grant No. 61573240, 61473239) and Graduate Student Innovation Project of Shanghai Maritime University 2017ycx082.

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The lattice generated by hesitant fuzzy filters in pseudo-BCI algebras

Abstract

Keywords

1 Introduction

2 Preliminary

3 Hesitant fuzzy filters in pseudo-BCI algebras

Table 1 → a b c d e 1 a 1 a b e c b b b 1 e c d a c d e 1 a b c d e c b 1 a d e c d a b 1 e 1 a b c d e 1

Table 3 → a b c d 1 a 1 1 1 d 1 b b 1 1 d 1 c b b 1 d 1 d d d d 1 d 1 a b c d 1

6 Conclusion

Footnotes

Acknowledgments

References

Table 1

→ a b c d e 1

a 1 a b e c b

b b 1 e c d a

c d e 1 a b c

d e c b 1 a d

e c d a b 1 e

1 a b c d e 1

Table 3

→ a b c d 1

a 1 1 1 d 1

b b 1 1 d 1

c b b 1 d 1

d d d d 1 d

1 a b c d 1