δ-relation on dual hyper K

Abstract

In this paper, we are going to introduce a fundamental relation “δ” on (hyper BE-algebra) dual hyper K-algebra and investigate some properties. We show that quotient of any dual hyper K-algebra via a regular relation is a hyper BE-algebra and this quotient, via any strongly regular relation is a BE-algebra. Furthermore, it shows that “δ” under some condition is transitive and quotient of any weak commutative dual hyper K-algebra via “δ^*” is a dual BCK-algebra.

Keywords

fundamental relation

1 Introduction

H. S. Kim and Y. H. Kim introduced the notion of a BE-algebra as a generalization of a dual BCK–algebra [8]. A. Rezaei et al. in [11, 12] have studied commutative ideals in BE-algebras and have given some properties. They showed that a commutative implicative BE-algebra is equivalent to the commutative self distributive BE-algebra. Also, they proved that every Hilbert algebra is a self distributive BE-algebra and commutative self distributive BE-algebra is a Hilbert algebra.

The hyper algebraic structure theory was introduced in 1934 [9], by F. Marty at the 8th congress of Scandinavian Mathematicians. Hyperstructures have many applications to several sectors of both pure and applied sciences as geometry, graphs and hypergraphs, fuzzy sets and rough sets, automata, cryptography, codes, relation algebras, C-algebras, artificial intelligence, probabilities, chemistry, physics, especially in atomic physics and in harmonic analysis [4 –6]. R.A. Borzooei et al. defined the notion of a hyper K-algebra, bounded hyper K-algebra and considered the zero condition in hyper K-algebras. They showed that every hyper K-algebra with the zero condition can be extended to a bounded hyper K-algebra [1, 15].

Recently, in [10], A. Radfar et al. introduced the notion of hyper BE-algebra and investigated some properties. Furthermore, they showed that under some condition hyper BE-algebras are equivalent to dual hyper K-algebras. Fundamental relations are one of the main tools in algebraic hyperstructures theory. In [2, 3], Hamidi et al. introduced the concept of fundamental relation on hyper BCK–algebras and obtained some result on connection between hyper BCK-algebras, Q-algebras and BCK-algebras.

In this paper, we try to find a connection between algebraic structures and hyper algebraic structures and for construction a (BE-algebra) dual BCK-algebra from a (hyper BE-algebra) dual hyper K-algebra, we need to introduce such relation. We show that quotient of any dual hyper K-algebra on a regular relation is a hyper BE-algebra and this quotient on any good strongly regular relation is a dual BCK-algebra. We introduce “δ” as a relation on (hyper BE-algebra) dual hyper K-algebras and “δ ^*” as a transitive closure of “δ” in such a way that is the smallest equivalence relation that contains “δ”. We show “δ ^*" is strongly regular and under some conditions “δ” is a transitiverelation.

2 Preliminaries

Definition 2.1. [7] Let X be a set with a binary operation “ * " and a constant “0 ". Then, (X, * , 0) is called a BCK-algebra if for all x, y, z ∈ X it satisfies the following conditions:

((x * y) * (x * z)) * (z * y) =0,

(x * (x * y)) * y = 0,

x * x = 0,

x * y = 0 and y * x = 0 imply x = y,

0 * x = 0.

We define a binary relation “≤” on X by x ≤ y if and only if x * y = 0. Then, (X, * , 0) is a BCK-algebra if and only if it satisfies the following conditions:

((x * y) * (x * z)) ≤ (z * y),

(x * (x * y)) ≤ y,

x ≤ x,

x ≤ y and y ≤ x imply x = y,

0 ≤ x.

Definition 2.2. [8] An algebra (X ; ∗ , 1) of type (2, 0) is called a BE-algebra if for all x, y, z ∈ X, the following axioms hold:

x ∗ x = 1,

x ∗ 1 =1,

1 ∗ x = x,

x ∗ (y ∗ z) = y ∗ (x ∗ z) .

The BE-algebra (X ; ∗ , 1) is said to be commutative, if for all x, y ∈ X, (x * y) * y = (y * x) * x. We introduce a relation “≤” on X by x ≤ y if and only if x ∗ y = 1 .

Proposition 2.3. a[8] Let X be a BE-algebra. Then for all x, y ∈ X

x ∗ (y ∗ x) =1,

y ∗ ((y ∗ x) ∗ x) =1.

Definition 2.4. [14] An algebra (X ; ∗ , 1) of type (2, 0) is called a dual BCK-algebra if for all x, y, z ∈ X

x ∗ x = 1,

x ∗ 1 =1,

x ∗ y = y ∗ x = 1 ⇒ x = y,

(x ∗ y) ∗ ((y ∗ z) ∗ (x ∗ z)) =1,

x ∗ ((x ∗ y) ∗ y) =1.

The dual BCK-algebra (X ; ∗ , 1) is said to be commutative, if for all x, y ∈ X,

(x * y) * y = (y * x) * x .

Lemma 2.5. [14] Let (X ; ∗ , 1) be a dual BCK-algebra. Then for all x, y, z ∈ X

x ∗ (y ∗ z) = y ∗ (x ∗ z),

1 ∗ x = x.

Proposition 2.6. [14] Any dual BCK-algebra is a BE-algebra and any commutative BE-algebra is a dual BCK-algebra.

Example 2.7. [13] Let X = {1, 2, …} and the operation ∗ be defined as follows: $x * y = {\begin{matrix} 1 & if y \leq x \\ y & otherwise . \end{matrix}$ Then (X ; ∗ , 1) is a BE-algebra, but it is not a dual BCK-algebra.

Definition 2.8. [1] Let H be a nonempty set and cc : H × H → P ^* (H) be a hyperoperation. Then (H ; cc, 0) is called a hyper K-algebra, if for all x, y, z ∈ H it satisfies in the following axioms:

(xccz) cc (yccz) < xccy,

(xccy) ccz = (xccz) ccy,

x < x,

x < y and y < x imply that x = y,

0 < x.

Where x < y is defined by 0 ∈ xccy. For every A, B ⊆ H, A < B if and only if there exist a ∈ A and b ∈ B in such a way that a < b. Note that if A, B ⊆ H then by AccB we mean the subset dis ⋃ _a∈A,b∈B accb of H.

Theorem 2.9. [1] Let H be a hyper K-algebra. Then for all x ∈ H

x ∈ xcc0,

0 ∈ 0ccx.

Definition 2.10. [10] Let H be a nonempty set and cc : H × H → P ^* (H) be a hyperoperation. Then (H ; cc, 1) is called a hyper BE–algebra, if for all x, y, z ∈ H it satisfies in the following axioms:

x < 1 and x < x,

xcc (yccz) = ycc (xccz),

x ∈ 1ccx,

1 < x implies x = 1.

(H ; cc, 1) is called a dual hyper K-algebra if it satisfies (HBE ₁), (HBE ₂) and the following axioms:

xccy < (yccz) cc (xccz),

x < y and y < x imply that x = y.

Where the relation “<” is defined by x < y ⇔ 1 ∈ xccy.

Theorem 2.11. [10] Let H be a hyper BE-algebra. Then for all x, y, z ∈ H and A, B, C ⊆ H

Acc (BccC) = Bcc (AccC),

A < A,

1 < A implies 1 ∈ A,

x < yccx,

x < yccz implies y < xccz,

x < (xccy) ccy,

z ∈ xccy implies x < zccy,

y ∈ 1ccx implies y < x.

Proposition 2.12. [10] Every dual hyper K–algebra is a hyper BE–algebra.

Definition 2.13. [10] Let F be a nonempty subset of hyper BE-algebra X, x, y ∈ X and 1 ∈ F. Then F is called

a weak hyper filter of X if xoy ⊆ F and x ∈ F imply y ∈ F,

a hyper filter of X if xoy ≈ F and x ∈ F imply y ∈ F, where xoyapF means that xoy∩ F ≠ ∅.

3 Fundamental relation on (hyper BE-algebra) dual hyper K-algebras

In this section, we show that strongly regular relations have played important role in connection between hyper algebraic structures and algebraic structures.

In the following definition we are trying to generalize relations between elements to relation between sets.

Definition 3.1. Let (X;∘) be a dual hyper K-algebra and R be an equivalence relation on X. If A and B are nonempty subsets of X, then

$A \bar{R} B$ means that for all a ∈ A, there exists b ∈ B in such a way that aRb and for all b′ ∈ B, there exists a′ ∈ A in such a way that b′Ra′,

$A \bar{\bar{R}} B$ means that for all a ∈ A, and b ∈ B, we have aRb,

R is called right regular (left regular) if for all x of X, from aRb, it follows that $(a \circ x) \bar{R} (b \circ x)$ $((x \circ a) \bar{R} (x \circ b))$ ,

R is called strongly right regular (strongly left regular) if for all x of X, from aRb, it follows that $(a \circ x) \bar{\bar{R}} (b \circ x)$ $((x \circ a) \bar{\bar{R}} (x \circ b))$ ,

R is called (strongly) regular if it is (strongly) right regular and (strongly) left regular,

R is called good, if a ∘ b R 1 and b ∘ a R 1 imply aRb, for all a, b ∈ X.

Example 3.1. Let X = {1, a, b}. Define the hyperoperation “cc” as follows: $\begin{matrix} \begin{matrix} o & 1 & a & b \\ 1 & {1} & {a, b} & {b} \\ a & {1} & {1, a, b} & {b} \\ b & {1, b} & {1, a, b} & {1, a, b} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a dual hyper K-algebra. It is easy to see that R = {(1, 1) , (a, a) , (b, b) , (a, b) , (b, a) , (1, b) , (b, 1) , (a, 1) , (1, a)} is a good strongly regular relation on X and for any A, B ∈ P ^* (X), $A \bar{\bar{R}} B$ .

Definition 3.2. Let (X ; ∘ , 1) be a (hyper BE-algebra) dual hyper K-algebra and R be a regular relation on X. We denote the set of all equivalence classes of R by X/R. Hence $X / R = {\bar{x} | x \in X}$ . For any $\bar{x}, \bar{y} \in X / R$ , define a hyperoperation “*” on X/R by $\bar{x} * \bar{y} = {\bar{z} | z \in x \circ y}$ and a binary relation “<” on X/R by $\bar{x} < \bar{y} \Leftrightarrow \bar{1} \in \bar{x} * \bar{y} .$

Lemma 3.2. Let (X ; ∘ , 1) be a hyper BE-algebra and R be a regular relation on X. Then (X/R ; *) is a hypergroupoid.

Proof. Let $\bar{x_{1}} = \bar{x_{2}}$ and $\bar{y_{1}} = \bar{y_{2}}$ , where $\bar{x_{1}}, \bar{x_{2}}, \bar{y_{1}}, \bar{y_{2}} \in X / R$ . Then x ₁ Rx ₂ and y ₁ Ry ₂. Since R is a regular relation, we have $(x_{1} \circ y_{1}) \bar{R} (x_{2} \circ y_{2})$ . Let $\bar{r} \in \bar{x_{1}} * \bar{y_{1}}$ , then there exists s ∈ x ₁ ∘ y ₁ in such a way that $\bar{r} = \bar{s}$ . Now, s ∈ x ₁ ∘ y ₁ and $(x_{1} \circ y_{1}) \bar{R} (x_{2} \circ y_{2})$ , then there exists t ∈ (x ₂ ∘ y ₂) in such a way that $\bar{s} = \bar{t}$ and so $\bar{r} = \bar{t}$ . Thus $\bar{x_{1}} * \bar{y_{1}} \subseteq \bar{x_{2}} * \bar{y_{2}}$ and in a similar way we get $\bar{x_{2}} * \bar{y_{2}} \subseteq \bar{x_{1}} * \bar{y_{1}}$ . Therefore * is well-defined and (X/R ; *) is a hypergroupoid.□

Theorem 3.3. Let (X ; ∘ , 1) be a hyper BE-algebra and R be an equivalence relation on X. Then, R is a regular relation on X if and only if $(X / R; *, \bar{1})$ is a hyper BE-algebra.

Proof. Let R be a regular relation on X.

$(\underline{HBE 1}) :$ Let x ∈ X. Then $\bar{x} \circ \bar{1} = {\bar{t} | t \in x \circ 1}$ . Since X is a hyper BE–algebra, we can see that 1 ∈ x ∘ 1 and so $\bar{1} \in \bar{x} * \bar{1}$ . Therefore, $\bar{x} < \bar{1}$ . Since $\bar{x} \circ \bar{x} = {\bar{t} | t \in x \circ x}$ and 1 ∈ x ∘ x, then we get $\bar{x} < \bar{x}$ .

$(\underline{HBE 2}) :$ Let x, y, z ∈ X and $\bar{r} \in \bar{x} * (\bar{y} * \bar{z})$ . Then there exists s ∈ y ∘ z, in such a way that $\bar{r} \in \bar{x} * \bar{s}$ . So there exists t ∈ x ∘ s, in such a way that $\bar{r} = \bar{t}$ . Since X is a hyper BE-algebra, we have x ∘ (y ∘ z) = y ∘ (x ∘ z) and so t ∈ x ∘ s ⊆ x ∘ (y ∘ z) = y ∘ (x ∘ z). Now, by definition there exists m ∈ x ∘ z in such a way that t ∈ y ∘ m and then, $\bar{r} = \bar{t} \in \bar{y} * \bar{m} \subseteq \bar{y} * (\bar{x} * \bar{z}) .$ Hence $\bar{x} * (\bar{y} * \bar{z}) \subseteq \bar{y} * (\bar{x} * \bar{z})$ . In a similar way we have $\bar{y} * (\bar{x} * \bar{z}) \subseteq \bar{x} * (\bar{y} * \bar{z})$ .

$(\underline{HBE 3}) :$ Let x ∈ X. Then $\bar{1} * \bar{x} = {\bar{t} | t \in 1 \circ x}$ . Since X is a hyper BE-algebra, we have x ∈ 1 ∘ x and so $\bar{x} \in \bar{1} * \bar{x}$ .

$(\underline{HBE 4}) :$ Let x ∈ X and $\bar{1} < \bar{x}$ . Then $\bar{1} \in \bar{1} * \bar{x}$ . Hence 1 ∈ 1 ∘ x and then 1 < x. Since X is a hyper BE-algebra, we have x = 1 and so $\bar{x} = \bar{1}$ .

Conversely, let x, y ∈ X and xRy. Then $\bar{x} = \bar{y}$ and for any a ∈ X and r ∈ x ∘ a by Definition 3.2, $\bar{r} \in \bar{x} * \bar{a}$ . Since (X/R ; *) is a hypergroupoid by Lemma 3.2, $\bar{x} = \bar{y}$ implies that $\bar{x} * \bar{a} = \bar{y} * \bar{a}$ . Hence there exists s ∈ y ∘ a in such a way that $\bar{r} = \bar{s}$ and then $(x \circ a) \bar{R} (y \circ a)$ . Therefore R is right regular and in a similar way R is a left regular relation.□

Corollary 3.4. Let (X ; ∘ , 1) be a dual hyper K-algebra and R be an equivalence relation on X. Then, R is a regular relation on X if and only if $(X / R; *, \bar{1})$ is a hyper BE-algebra.

Theorem 3.5. Let (X ; ∘ , 1) be a (hyper BE-algebra) dual hyper K-algebra and R be an equivalence relation on X. If R is a strongly regular relation on X, then $(X / R; *, \bar{1})$ is a BE-algebra.

Proof. Since R is strongly regular then for any $\bar{x}$ , $\bar{y} \in X / R, | \bar{x} * \bar{y} | = 1$ and so by Theorem 3.3, $(X / R; *, \bar{1})$ is a BE-algebra.□

Example 3.6. Let X = {1, a, b, c, d, e}. Define the hyperoperation “cc” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c & d & e \\ 1 & {1, c} & {a} & {b} & {c} & {d} & {e} \\ a & {1, c} & {1, c} & {a} & {1, c} & {c} & {d} \\ b & {1, c} & {1, c} & {1, c} & {1, c} & {c} & {c} \\ c & {1, c} & {a} & {b} & {1, c} & {a} & {b} \\ d & {1, c} & {1, c} & {a} & {1, c} & {1, c} & {a} \\ e & {1, c} & {1, c} & {1, c} & {1, c} & {1, c} & {1, c} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a hyper BE-algebra. It is easy to see that R = {(1, 1) , (a, a) , (b, b) , (c, c) , (d, d) , (e, e) , (1, c) , (c, 1) , (e, b) , (b, e) , (a, d) , (d, a)} is a good strongly regular relation on X and $X / R = {{1, c}, {a, d}, {e, b}} = {R (1), R (a), R (b)} .$ Now, we have: $\begin{matrix} \begin{matrix} * & R (1) & R (a) & R (b) \\ R (1) & R (1) & R (a) & R (b) \\ R (a) & R (1) & R (1) & R (a) \\ R (b) & R (1) & R (1) & R (1) \end{matrix} \end{matrix}$ Clearly, (X/R ; * , R (1)) is a BE-algebra.

Lemma 3.7. Let (X ; ∘ , 1) be a dual hyper K-algebra and R be a regular relation on X. Then for any $\bar{x}, \bar{y}, \bar{z} \in X / R$ , $(\bar{x} * \bar{y}) < (\bar{y} * \bar{z}) * (\bar{x} * \bar{z})$ .

Proof. Let $\bar{r} \in \bar{x} * \bar{y}$ . Then there exists r′ ∈ x ∘ y in such a way that $\bar{r} = \bar{r^{'}}$ . Since (X ; ∘ , 1) is a dual hyper K–algebra then there exists t ₁ ∈ y ∘ z and t ₂ ∈ x ∘ z in such a way that 1 ∈ r′ ∘ (t ₁ ∘ t ₂). For t ₁ ∈ y ∘ z and t ₂ ∈ x ∘ z there exists $t_{1}^{'} \in y \circ z$ and $t_{2}^{'} \in x \circ z$ in such a way that $\bar{t_{1}} = \bar{t_{1}^{'}}$ and $\bar{t_{2}} = \bar{t_{2}^{'}}$ . Hence, $\bar{1} \in \bar{r^{'}} * \bar{m}$ that m ∈ t ₁ ∘ t ₂ and $\bar{m} \in \bar{t_{1}^{'}} * \bar{t_{2}^{'}} \subseteq (\bar{y} * \bar{z}) * (\bar{x} * \bar{z})$ . Therefore there exists $\bar{m} \in (\bar{y} * \bar{z}) * (\bar{x} * \bar{z})$ in such a way that $\bar{r} = \bar{r^{'}} < \bar{m}$ .□

Theorem 3.8. Let (X ; ∘ , 1) be a dual hyper K-algebra and R be a regular relation on X. If R is a good relation, then $(X / R; *, \bar{1})$ is a dual hyper K-algebra.

Proof. Let $\bar{x}, \bar{y} \in X / R$ , $\bar{x} < \bar{y}$ and $\bar{y} < \bar{x}$ . Then $\bar{1} \in \bar{x} * \bar{y}$ and $\bar{1} \in \bar{y} * \bar{x}$ . Hence there exists t ∈ x ∘ y and s ∈ y ∘ x in such a way that tR1Rs so x ∘ y R 1 and y ∘ x R 1. Now, since R is a good relation, we can see that $\bar{x} = \bar{y}$ . Therefore by Corollary 3.4 and Lemma 3.7, $(X / R; *, \bar{1})$ is a dual hyper K-algebra.□

Theorem 3.9. Let (X ; ∘ , 1) be a dual hyper K-algebra and R be a strongly regular relation on X. If R is a good relation, then $(X / R; *, \bar{1})$ is a dual BCK-algebra.

Proof. Let x, y ∈ X. If $\bar{x} * \bar{y} = \bar{1}$ and $\bar{y} * \bar{x} = \bar{1}$ , then respectively we have x ∘ y R 1 and y ∘ x R 1. Now, since R is good, we have $\bar{x} = \bar{y}$ . Therefore, by Theorem 3.5, and Lemma 3.7, $(X / R; *, \bar{1})$ is a dual BCK-algebra.□

Example 3.10. Let X = {1, a, b, c, d, e}. Define the hyperoperation “∘” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c & d & e \\ 1 & {1, e} & {a} & {b} & {c} & {d} & {e} \\ a & {1, e} & {1, e} & {b} & {c} & {d} & {e} \\ b & {1, e} & {a} & {1, e} & {c} & {d} & {e} \\ c & {1, e} & {a} & {b} & {1, e} & {d} & {e} \\ d & {1, e} & {a} & {b} & {c} & {1, e} & {e} \\ e & {1, e} & {a} & {b} & {c} & {d} & {1, e} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a dual hyper K-algebra. It is easy to see that $R = {(1, 1), (a, a), (b, b), (c, c), (d, d), (e, e), (1, e), (e, 1)}$ is a good strongly regular relation on X and $\begin{matrix} X / R & = & {{1, e}, {a}, {b}, {c}, {d}} \\ = & {R (1), R (a), R (b), R (c), R (d)} . \end{matrix}$ Now, we have $\begin{matrix} \begin{matrix} * & R (1) & R (a) & R (b) & R (c) & R (d) \\ R (1) & R (1) & R (a) & R (b) & R (c) & R (d) \\ R (a) & R (1) & R (1) & R (b) & R (c) & R (d) \\ R (b) & R (1) & R (a) & R (1) & R (c) & R (d) \\ R (c) & R (1) & R (a) & R (b) & R (1) & R (d) \\ R (d) & R (1) & R (a) & R (b) & R (c) & R (1) \end{matrix} \end{matrix}$ Clearly, (X/R ; * , R (1)) is a dual BCK-algebra.

4 Relation δ on (hyper BE-algebra) dual hyper K-algebra

Let (X ; ∘) be a (hyper BE-algebra) dual hyper K-algebra and A be a subset of X. $L (A)$ will denote the set of all finite combinations of elements A with ∘ and $⨀_{i = 1}^{n} a_{i} = a_{1} \circ a_{2} \circ \dots a_{n}$ .

Definition 4.1. Let (X ; ∘ , 1) be a (hyper BE-algebra) dual hyper K-algebra. Then we set: $δ_{1} = {(x, x) | x \in X}$ and for every integer n ≥ 1, δ _n is defined as follows: $x δ_{n} y \Leftrightarrow \exists (a_{1}, a_{2}, \dots, a_{n}) \in X^{n},$ $\exists u \in L (a_{1}, a_{2}, \dots, a_{n})$ in such a way that {x, y} ⊆ u .

Obviously for every n ≥ 1 the relations δ _n are symmetric, the relation δ = di ⋃ _n≥1 δ _n is a reflexive and symmetric relation. Let δ ^* be the transitive closure of δ (the smallest transitive relation in such a way that contains δ). Then in the following theorem we show that δ ^* is a strongly regular relation.

Theorem 4.1. Let (X ; ∘ , 1) be a hyper BE-algebra. Then δ ^* is a strongly regular relation on X.

Proof. Let x, y ∈ X and x δ ^* y . Then there exist a ₀, a ₁, …, a _n ∈ X in such a way that a ₀ = x, a _n = y and there exist $δ_{q_{1}}, δ_{q_{2}}, \dots, δ_{q_{n}} \in ℕ$ , in such a way that $x = a_{0} δ_{q_{1}} a_{1} δ_{q_{2}} a_{2} \dots a_{n - 2} δ_{q_{n - 1}} a_{n - 1} δ_{q_{n}} a_{n} = y,$ where $n \in ℕ$ . Since for any 1 ≤ i ≤ n, a _i-1 δ _{q
_i} a _i, then there exist $z_{t}^{j} \in X$ in such a way that ${a_{i}, a_{i + 1}} \subseteq z_{1}^{i + 1} \circ z_{2}^{i + 1} \circ \dots \circ z_{q_{i + 1}}^{i + 1},$ where for 1 ≤ m ≤ n - 1, we have 1 ≤ t ≤ q _m and 1 ≤ j ≤ n - 1. Now, let s ∈ X. Then for all 0 ≤ i ≤ n - 1, $a_{i} \circ s \subseteq z_{1}^{i + 1} \circ z_{2}^{i + 1} \circ \dots \circ z_{q_{i + 1}}^{i + 1} \circ s .$ In a similar way, we get that $a_{i + 1} \circ s \subseteq z_{1}^{i + 1} \circ z_{2}^{i + 1} \circ \dots \circ z_{q_{i + 1}}^{i + 1} \circ s .$ Then for all 0 ≤ i ≤ n and for all u ∈ a _i ∘ s, v ∈ a _i+1 ∘ s, we have u δ _{q
_i+1} v, and so for all z ∈ a ₀ ∘ s = x ∘ s, w ∈ a _n ∘ s = y ∘ s, we have z δ ^* w. Then δ ^* is a strongly right regular and similarly is a strongly left regular relation. Therefore δ ^* is a strongly regular relation.□

Corollary 4.2. Let (X ; ∘ , 1) be a dual hyper K-algebra. Then δ ^* is a strongly regular relation on X.

Theorem 4.3. Let (X ; ∘ , 1) be a hyper BE-algebra. Then $(X / δ^{*}; *, \bar{1})$ is a BE-algebra.

Proof. By Theorems 3.5 and 4.1, the proof isobvious.□

Example 4.4. Let X = {1, a, b, c, d}. Define the hyperoperation “cc” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c & d \\ 1 & {1, c} & {a} & {b} & {c} & {d} \\ a & {1, c} & {1, c} & {1, c} & {1, c} & {1, c} \\ b & {1, c} & {1, c} & {1, c} & {1, c} & {1, c} \\ c & {1, c} & {1, c} & {1, c} & {1, c} & {1, c} \\ d & {1, c} & {1, c} & {1, c} & {1, c} & {1, c} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a hyper BE-algebra. We have (a ∘ b) ∘ a = {1, a, c} , (a ∘ b) ∘ b = {1, b, c} and (a ∘ b) ∘ d = {1, d, c} . Then for any x ∈ X, 1 δ ^* x and so δ ^* (1) = X = δ ^* (x). Hence X/δ ^* = {δ ^* (1)} and clearly (X/δ ^* ; * , δ ^* (1)) is a trivial BE-algebra.

Example 4.5. Let X = {1, a, b, c, d}. Define the hyperoperation “cc” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c & d \\ 1 & {1} & {a} & {b} & {c} & {d} \\ a & {1} & {1} & {1} & {1} & {1} \\ b & {1} & {a} & {1} & {c} & {1} \\ c & {1} & {a} & {1, b} & {1} & {1} \\ d & {1} & {1} & {1} & {1} & {1} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a hyper BE-algebra. Now, we have: $\begin{matrix} \begin{matrix} * & δ^{*} (1) & δ^{*} (a) & δ^{*} (c) & δ^{*} (d) \\ δ^{*} (1) & δ^{*} (1) & δ^{*} (a) & δ^{*} (c) & δ^{*} (d) \\ δ^{*} (a) & δ^{*} (1) & δ^{*} (1) & δ^{*} (1) & δ^{*} (1) \\ δ^{*} (c) & δ^{*} (1) & δ^{*} (a) & δ^{*} (1) & δ^{*} (1) \\ δ^{*} (d) & δ^{*} (1) & δ^{*} (1) & δ^{*} (1) & δ^{*} (1) \end{matrix} \end{matrix}$ Clearly, (X/δ ^* ; * , δ ^* (1)) = {{1, b} , {a} , {c} , {d}} is a BE-algebra but is not a dual BCK-algebra. (Since δ ^* (a) * δ ^* (d) = δ ^* (1) and δ ^* (d) * δ ^* (a) = δ ^* (1) but δ ^* (a) ≠ δ ^* (d)).

Remark 4.2. Let (X ; ∘ , 1) be a dual hyper K-algebra. Then for the strongly regular relation δ ^*, the structure X/δ ^* is a dual BCK-algebra, if it satisfies in the condition (DHK ₁). For this, we need if δ ^* (x) * δ ^* (y) = δ ^* (1) and δ ^* (y) * δ ^* (x) = δ ^* (1), then 1 δ ^* (x ∘ y) and 1 δ ^* (y ∘ x), and so δ ^* (x) = δ ^* (y). Hence, we introduced some conditions for the structure X/δ ^* to be a dual BCK-algebra.

Definition 4.3. Let (X ; ∘ , 1) be a dual hyper K-algebra. Then, (X ; ∘ , 1) is called a weak commutative dual hyper K-algebra, if for any x, y ∈ X, $((x \circ y) \circ y) ⋂ ((y \circ x) \circ x) \neq \emptyset .$

Theorem 4.6. Let (X ; ∘ , 1) be a weak commutative dual hyper K-algebra and R be a strongly regular relation on X. If R is a good relation, then $(X / R; *, \bar{1})$ is a commutative dual BCK-algebra.

Proof. By Theorem 3.9, $(X / R; *, \bar{1})$ is a dual BCK-algebra. Let x, y ∈ X. Since (X ; ∘ , 1) is a weak commutative dual hyper K-algebra, there exists t ∈ (x ∘ y) ∘ y ⋂ (y ∘ x) ∘ x and so there exist m ∈ x ∘ y, n ∈ y ∘ x in such a way that R (x) * R (y) = R (m) and R (y) * R (x) = R (n). Hence R (m) * R (y) = R (t) = R (n) * R (x) and so (R (x) * R (y)) * R (y) = (R (y) * R (x)) * R (x). Therefore $(X / R; *, \bar{1})$ is a commutative dual BCK-algebra.□

Corollary 4.7. Let (X ; ∘ , 1) be a weak commutative hyper BE-algebra and R be an equivalence relation on X. If R is a strongly regular relation on X, then $(X / R; *, \bar{1})$ is a commutative BE-algebra.

Example 4.8. Let X = {1, a, b, c, d, e}. Define the hyperoperation “cc” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c & d \\ 1 & {1, d} & {a} & {b} & {c} & {d} \\ a & {1, d} & {1, d} & {b} & {c} & {d} \\ b & {1, d} & {a} & {1, d} & {c} & {d} \\ c & {1, d} & {a} & {b} & {1, d} & {d} \\ d & {1, d} & {a} & {b} & {c} & {1, d} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a weak commutative dual hyper K-algebra.

Theorem 4.9. Let (X ; ∘ , 1) be a weak commutative hyper BE-algebra. Then, δ ^* is a good strongly regular relation on X.

Proof. Since, (X ; ∘ , 1) is a hyper BE-algebra, then by Corollary 4.2, δ ^* is a strongly regular relation on X. Now, we show that δ ^* is good. Let 1 δ ^* (a ∘ b) and 1 δ ^* (b ∘ a). Since, δ ^* is a strongly regular relation, we have $(1 \circ b) \bar{\bar{δ^{*}}} ((a \circ b) \circ b)$ and in a similar way we have $(1 \circ a) \bar{\bar{δ^{*}}} ((b \circ a) \circ a)$ . Since (X ; ∘) is a weak commutative hyper BE-algebra, there exists t ∈ ((a ∘ b) ∘ b) ⋂ ((b ∘ a) ∘ a).

By definition, we have, $a \in (1 \circ a) \bar{\bar{δ^{*}}} ((b \circ a) \circ a), b \in (1 \circ b) \bar{\bar{δ^{*}}} ((a \circ b) \circ b)$ and there exists, t ∈ ((a ∘ b) ∘ b) ⋂ ((b ∘ a) ∘ a). Since, δ ^* is a strongly regular and using Definition 3.1, we have a δ ^* t δ ^* b. Now, since δ ^* is a transitive relation, we get a δ ^* b. Therefore, δ ^* is a good strongly regular relation on X.□

Corollary 4.10. Let (X ; ∘ , 1) be a weak commutative dual hyper K-algebra. Then, δ ^* is a good strongly regular relation on X.

Corollary 4.11. Let (X ; ∘ , 1) be a dual hyper K-algebra. If (X ; ∘ , 1) is weak commutative, then $(X / δ^{*}; *, \bar{1})$ is a commutative dual BCK-algebra.

Example 4.12. Let X = {1, a, b, c}. Define the hyperoperation “∘” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c \\ 1 & {1, c} & {a} & {b} & {c} \\ a & {1, c} & {1, c} & {b} & {c} \\ b & {1, c} & {a} & {1, c} & {c} \\ c & {1, c} & {a} & {b} & {1, c} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a weak commutative dual hyper K-algebra, $X / R = {{1, c}, {a}, {b}} = {δ^{*} (1), δ^{*} (a), δ^{*} (b)} .$ Now, we have $\begin{matrix} \begin{matrix} * & δ^{*} (1) & δ^{*} (a) & δ^{*} (b) \\ δ^{*} (1) & δ^{*} (1) & δ^{*} (a) & δ^{*} (b) \\ δ^{*} (a) & δ^{*} (1) & δ^{*} (1) & δ^{*} (b) \\ δ^{*} (b) & δ^{*} (1) & δ^{*} (a) & δ^{*} (1) \end{matrix} \end{matrix}$ Clearly, (X/R ; * , R (1)) is a dual BCK-algebra.

Theorem 4.13. Let (X ; ∘ , 1) be a (hyper BE-algebra) weak commutative dual hyper K-algebra. Then δ ^* is the smallest strongly regular equivalence relation on X, in such a way that X/δ ^* is a (commutative BE-algebra) commutative dual BCK-algebra.

Proof. By Corollaries 4.2 and 4.11, δ ^* is a strongly regular equivalence relation on X and so X/δ ^* is a dual BCK-algebra. Now, we show that it is the smallest. Let R be a strongly regular equivalence relation on X, in such a way that X/R is a dual BCK-algebra. By induction, we prove that for all $n \in ℕ, δ_{n} \subseteq R$ . Since R is an equivalence relation, we have δ ₁ ⊆ R. Let for all k < n, δ _k ⊆ R, then we show that δ _n ⊆ R. For x, y ∈ X, if x δ _n y, then there exist a ₁, a ₂, …, a _n ∈ X, in such a way that ${x, y} \subseteq ⨀_{i = 1}^{n} a_{i} = ⨀_{i = 1}^{k} a_{i} \circ ⨀_{i = k + 1}^{n} a_{i} .$ Then there exist ${u, u^{'}} \subseteq ⨀_{i = 1}^{k} a_{i}$ and ${v, v^{'}} \subseteq ⨀_{i = k + 1}^{n} a_{i}$ , in such a way that x ∈ u ∘ v and y ∈ u′ ∘ v′. Moreover, u δ _k u′ and v δ _n-k v′. Now, since k < n and n - k < n, we have u R u′ and v R v′ by induction hypothesis. But R is a strongly regular relation then $u \circ v \bar{\bar{R}} u^{'} \circ v^{'}$ , and then x R y. Hence, for all n ≥ 1, δ _n ⊆ R and so δ = di ⋃ _n≥1 δ _n ⊆ R.

By Theorem 4.1, $δ^{*} = \bar{δ}$ and since R is a transitive relation, we can see that $δ^{*} = \bar{δ} \subseteq \bar{R} = R .$ Therefore, δ ^* is the smallest strongly regular equivalence relation on X, in such a way that X/δ ^* is a dual BCK-algebra.□

Remark 4.4. We consider the following dual hyper K-algebra (X ; ∘): $\begin{matrix} \begin{matrix} \circ & 1 & b & c & d \\ 1 & {1} & {b} & {c} & {d} \\ b & {1} & {1} & {1} & {1} \\ c & {1} & {b} & {1} & {d} \\ d & {1} & {b} & {1, c} & {1} \end{matrix} \end{matrix}$ X/δ ^* = {{1, c} , {b} , {d}} = {δ ^* (1) , δ ^* (b) , δ ^* (d)}. Now, we have: $\begin{matrix} \begin{matrix} * & δ^{*} (1) & δ^{*} (b) & δ^{*} (d) \\ δ^{*} (1) & δ^{*} (1) & δ^{*} (b) & δ^{*} (d) \\ δ^{*} (b) & δ^{*} (1) & δ^{*} (1) & δ^{*} (1) \\ δ^{*} (d) & δ^{*} (1) & δ^{*} (b) & δ^{*} (1) \end{matrix} \end{matrix}$ Clearly, (X/δ ^* ; * , δ ^* (1)) is a dual BCK-algebra, while in (X ; ∘ , 1) , $(b \circ c) \circ c \cap (c \circ b) \circ b = \emptyset .$ Therefore, the converse of Theorem 4.9, is not necessarily valid.

5 Transitivity and some conditions on δ

In this section, we determine how the relation δ is a transitive relation.

Definition 5.1. Let X be a dual hyper K-algebra and M a non-empty subset of X. M is called δ-part if for any $n \in ℕ,$ a _i ∈ X, if $L (a_{1}, a_{2}, \dots, a) \cap M \neq \emptyset$ then $L (a_{1}, a_{2}, \dots, a) \subseteq M$ .

Example 5.2. Let X = {1, a, b, c}. Define the hyperoperation “cc” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c \\ 1 & {1, a} & {1, a} & {b} & {c} \\ a & {1, a} & {1, a} & {b} & {c} \\ b & {1, a} & {1, a} & {1, a} & {c} \\ c & {1, a} & {1, a} & {1, a} & {1, a} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a dual hyper K-algebra. It is easy to verify that for any M ⊆ X that M ≠ {1} and M ≠ {a}, M is a δ-part.

Lemma 5.3. Let M be a non-empty subset of a dual hyper K-algebra X. Then the following conditions are equivalent:

M is a δ-part of X,

x ∈ M, x δ y imply y ∈ M,

x ∈ M, x δ ^* y imply y ∈ M.

Proof. (i) ⇒ (ii). Let x ∈ M, y ∈ X and xδy. Then there exist $n \in ℕ,$ a _i ∈ X in such a way that ${x, y} \subseteq (⨀_{i = 1}^{n} a_{i}) \in L (a_{1}, a_{2}, \dots, a_{n}) .$ Since M is a δ-part and $(⨀_{i = 1}^{n} a_{i}) \cap M \neq \emptyset$ , we havey ∈ M.

(ii) ⇒ (iii). Let x ∈ M, y ∈ X and x δ ^* y. Then there exist $n \in ℕ, a_{i} \in X$ in such a way that x δ a ₁ δ a ₂ δ … δ a _n δ y. Now x ∈ M and x δ a ₁, then by (ii) , a ₁ ∈ M. Since for any 1 ≤ i ≤ n, a _i δ a _i+1, a ₁ ∈ M and by using (ii), a _n ∈ M and so y ∈ M.

(iii) ⇒ (ii). Let $n \in ℕ, a_{i} \in X$ and $(⨀_{i = 1}^{n} a_{i}) \cap M \neq \emptyset$ . Then there exists $y \in (⨀_{i = 1}^{n} a_{i}) \cap M$ and for any $t \in ⨀_{i = 1}^{n} a_{i}$ , since ${y, t} \subseteq ⨀_{i = 1}^{n} a_{i}$ we have t δ y. Therefore t δ ^* y, y ∈ M implies that t ∈ M.□

Theorem 5.4. Let X be a dual hyper K-algebra. Then the following conditions are equivalent:

δ is a transitive relation,

for any x ∈ X, δ ^* (x) is a δ-part.

Proof. (i) ⇒ (ii). Let $n \in ℕ,$ a _i ∈ X and $(⨀_{i = 1}^{n} a_{i}) \cap δ^{*} (x) \neq \emptyset$ . For any $y \in (⨀_{i = 1}^{n} a_{i})$ since there exists $t \in (⨀_{i = 1}^{n} a_{i}) \cap δ^{*} (x)$ , we can see that ${t, y} \subseteq (⨀_{i = 1}^{n} a_{i})$ and so y δ t. Now, x δ t δ y and by (i), δ is a transitive relation, then y ∈ δ ^* (x).

(ii) ⇒ (i). Let x δ y and y δ z. Then there exist $m, n \in ℕ$ , a _i, b _j ∈ X in such a way that ${x, y} \subseteq (⨀_{i = 1}^{n} a_{i})$ and ${z, y} \subseteq (⨀_{j = 1}^{m} b_{j})$ . Now, δ ^* (x) is a δ-part, $x \in δ^{*} (x) \cap (⨀_{i = 1}^{n} a_{i})$ and $y \in (⨀_{i = 1}^{n} a_{i}) \cap (⨀_{j = 1}^{m} b_{j})$ , then $\begin{matrix} (⨀_{i = 1}^{n} a_{i}) \subseteq δ^{*} (x) & \Rightarrow & y \in (⨀_{j = 1}^{m} b_{j}) \cap δ^{*} (x) \\ \Rightarrow & (⨀_{j = 1}^{m} b_{j}) \subseteq δ^{*} (x) \\ \Rightarrow & z \in δ^{*} (x) . \end{matrix}$ Now, z ∈ δ (z) and z δ ^* x, then by Lemma 5.3, xδz.□

Theorem 5.5. Let (X ; ∘ , 1) be a (hyper BE-algebra) dual hyper K-algebra and δ ₀ = {(1, x) , (x, 1) | x ∈ X}. Then

If δ′ = di ⋃ _n≥0 δ _n, then δ ^′* ⊇ δ ^* and is a strongly regular on X.

For all x, y ∈ X, δ ^′* (x) = δ ^′* (y).

$(X / δ^{' *}, *, \bar{1})$ is a trivial dual BCK-algebra.

Proof. (i) Clearly δ ^′* is an equivalence relation and by Theorem 4.1, is a strongly regular relation.

(ii) Let x, y ∈ X. Since δ ₀ ⊆ δ′, we have x δ ^′* 1, 1 δ ^′* y and so x δ ^′* y.

(iii) By using (ii), since for any x, y ∈ X, δ ^′* (x) = δ ^′* (y), we have $X / δ^{' *} = {δ^{' *} (1)} . □$

Corollary 5.6. Let (X ; ∘ , 1) be a (hyper BE-algebra) dual hyper K-algebra. Then δ ^′* = δ′.

Definition 5.1. Let (X ; ∘) be a hypergroupoid. Then for every integer n ≥ 1, $δ_{n}^{″}$ , define as follows: $x δ_{n}^{″} y \Leftrightarrow \exists n \in ℕ, σ \in S_{n}, (a_{1}, a_{2}, \dots, a_{n}) \in X^{n},$ in such a way that $x \in ⨀_{i = 1}^{n} a_{i} and y \in ⨀_{i = 1}^{n} a_{σ (i)} .$

Obviously, for every n ≥ 1, the relations $δ_{n}^{″}$ are symmetric, the relation $δ^{″} = ⋃_{n \geq 1} δ_{n}^{″}$ is a reflexive and symmetric relation. Let δ ^″* be the transitive closure of δ″. Then in the following theorem we show that δ ^″* is a strongly regular relation.

Theorem 5.7. Let (X ; ∘) be a hypergroupoid. Then δ ^″* is a strongly regular on X and for any x, y ∈ X, δ ^″* (x) * δ ^″* (y) = δ ^″* (y) * δ ^″* (x).

Proof. Let x, y ∈ X and x δ″y . Then there exist $n \in ℕ, σ \in S_{n}, (a_{1}, a_{2}, \dots, a_{n}) \in X^{n}$ in such a way that $x \in ⨀_{i = 1}^{n} a_{i} and y \in ⨀_{i = 1}^{n} a_{σ (i)}$ . Let t ∈ X. Then for any v ∈ x ∘ t and for any w ∈ y ∘ t, we have $v \in ⨀_{i = 1}^{n} a_{i} \circ t$ and $w \in ⨀_{i = 1}^{n} a_{σ (i)} \circ t$ . Hence $v δ_{n + 1}^{″} w$ , v δ″ w and so $x \circ t \bar{\bar{δ^{″}}} y \circ t$ . Let x δ ^″* y, then there exist $m \in ℕ$ and w ₀, w ₁, …, w _m in such a way that $x = w_{0} δ^{″} w_{1} δ^{″} w_{2} \dots δ^{″} w_{m} = y$ Then, $x \circ t = w_{0} \circ t \bar{\bar{δ^{″}}} w_{1} \circ t \bar{\bar{δ^{″}}} \dots \bar{\bar{δ^{″}}} w_{m} \circ t = y \circ t$ Now, for all v ∈ x ∘ t = w ₀ ∘ t, w ∈ y ∘ a = w _m ∘ t, z ₁ ∈ w ₁ ∘ t, …, z _m-1 ∈ w _m-1 ∘ t, we have v δ″ z ₁ δ″ z ₂ δ″ … δ″ z _m-1 δ″ w then v δ″ w and so x ∘ t δ ^″* y ∘ t. Then δ ^″* is a strongly right regular and similarly is a strongly left regular relation. Therefore, δ ^″* is a strongly regular relation. Let x ₁, x ₂ ∈ X and z ∈ x ₁ ∘ x ₂. We define σ = (1 2), then for any w ∈ x _σ(1) * x _σ(2), we have z δ″ w and so δ ^″* (x ₁) * δ ^″* (x ₂) = δ ^″* (x ₂) * δ ^″* (x ₁).□

Corollary 5.8. Let (X ; ∘ , 1) be a (hyper BE-algebra) dual hyper K-algebra. Then δ ^″* is a strongly regular on X.

Theorem 5.9. Let (X ; ∘ , 1) be a hyper BE-algebra. Then δ ^′* = δ ^″*.

Proof. For any x ∈ X by Theorem 5.5, x δ ^′* 1. Since for any x ∈ X, x ∈ 1 ∘ x, we have n = 2 and get σ = (1 x), then 1 ∈ 1_σ(1) ∘ x _σ(x) = x ∘ 1 and so 1δ ^″* x. Therefore, δ ^′* = δ . ^″*□

Corollary 5.10. Let (X ; ∘ , 1) be a hyper BE-algebra. Then $(X / δ^{″ *}; *, \bar{1})$ is a trivial dual BCK-algebra.

Proof. Based on Theorems 3.5, 4.1 and 5.9, $(X / δ^{″ *}; *, \bar{1})$ is a trivial dual BCK-algebra.□

Theorem 5.11. Let (X ; ∘ , 1) be a hyper BE-algebra. Then δ ^″* is the smallest strongly regular equivalence relation on X, in such a way that X/δ ^″* is a trivial dual BCK-algebra.

Proof. Let R be a strongly regular equivalence relation on X, in such a way that X/R is trivial dual BCK-algebra. Let $x δ_{n}^{″} y$ , then there exist $n \in ℕ, σ \in S_{n}, (a_{1}, a_{2}, \dots, a_{n}) \in X^{n},$ $in such a way that x \in ⨀_{i = 1}^{n} a_{i} and y \in ⨀_{i = 1}^{n} a_{σ (i)}$ . Now, we have $\begin{matrix} δ^{″ *} (x) = δ^{″ *} (⨀_{i = 1}^{n} a_{i}) & = & ⨀_{i = 1}^{n} δ^{″ *} (a_{i}) \\ = & ⨀_{i = 1}^{n} δ^{″ *} (a_{σ (i)}) \\ = & δ^{″ *} (⨀_{i = 1}^{n} a_{σ (i)}) \\ = & δ^{″ *} (y) . \end{matrix}$ Hence, x R y implies any $n \in ℕ$ that $δ_{n}^{″} \subseteq R$ and so δ″ ⊆ R. Since R and δ″ are transitive, we have $\bar{δ^{″}} \subseteq \bar{R}$ . Therefore, δ ^″* is the smallest strongly regular equivalence relation on X, in such a way that X/δ ^″* is a dual BCK-algebra.□

Corollary 5.12. Let X be a dual hyper K-algebra. Then, δ ^″* = δ″.

6 Hyper filters and filters

In this section, we apply the concept of hyper filter and weak hyper filter in hyper BE-algebras that is defined in [10] and we show that in hyper BE-algebras the quotient of any hyper filter via any strongly regular equivalence relation is a filter in corresponding BE-algebra. Moreover, we introduce the concept of homomorphism in hyper BE-algebras and present some hyper filters and weak hyper filters via this concept and investigate the relations between them.

Theorem 6.1. Let (X ; ∘ , 1) be a hyper BE-algebra and F be a nonempty subset of X. If R is a strongly regular relation on X and F is a hyper filter of X, then F/R is a filter of X/R

Proof. Let F be a hyper filter of X. We have 1 ɛ F and so R(1) ɛ F/R. Now, let R(x) . R(y) ɛ F/R and R(x) ɛ F/R. Then for any t ɛ x . y,R(t) ɛ F/R and so there exists s ɛ F in such a way that R(t) = R(s). Then R(t) . R(s) = R(1) and then x .y < F. Since F is a hyper filter, we have y ɛ F and so R(y) ɛ F/R. Therefore, F/R is a filter of X/R.

Corollary 6.2. Let (X ; ∘ , 1) be a hyper BE-algebra and F be a nonempty subset of X. If F is a hyper filter of X then F/δ ^* is a filter of X/δ ^*.

Example 6.3. 5.2 Let X = {1, a, b, c}. Define the hyperoperation “cc” as follows: $\begin{matrix} \begin{matrix} \circ & 1 & a & b & c \\ 1 & {1} & {a} & {b} & {c} \\ a & {1} & {1} & {1} & {1} \\ b & {1} & {a} & {1, b} & {c} \\ c & {1} & {a} & {1, b} & {1, b} \end{matrix} \end{matrix}$ Then (X ; cc, 1) is a hyper BE-algebra. Now, we have $\begin{matrix} \begin{matrix} * & δ^{*} (1) & δ^{*} (a) & δ^{*} (c) \\ δ^{*} (1) & δ^{*} (1) & δ^{*} (a) & δ^{*} (c) \\ δ^{*} (a) & δ^{*} (1) & δ^{*} (1) & δ^{*} (1) \\ δ^{*} (c) & δ^{*} (1) & δ^{*} (a) & δ^{*} (1) \end{matrix} \end{matrix}$ Then (X/δ ^* ; * , δ ^* (1)) is a BE-algebra, F = {1, b} is a weak hyper filter of X and F/δ ^* = {{1, b}} is a filter of X/δ ^* = {{1, b} , {a} , {c}}.

Remark 6.1. (i) In Example 6.1, we consider F = {1, c}. It is easy to see that F is not a hyper filter of X, while F/δ ^* = {δ ^* (1) , δ ^* (c)} is a filter. Therefore, the converse of Corollary 6.2, is not necessarily valid.

(ii) In Example 5.2, we consider F = {1, c}. It is easy to see that F is a weak hyper filter of X. But in the corresponding BE-algebra $\begin{matrix} \begin{matrix} * & δ^{*} (1) & δ^{*} (b) & δ^{*} (c) \\ δ^{*} (1) & δ^{*} (1) & δ^{*} (b) & δ^{*} (c) \\ δ^{*} (b) & δ^{*} (1) & δ^{*} (1) & δ^{*} (c) \\ δ^{*} (c) & δ^{*} (1) & δ^{*} (1) & δ^{*} (1) \end{matrix} \end{matrix}$ F/δ ^* = {δ ^* (1) , δ ^* (c)} is not a filter in X/δ ^*.

Corollary 6.4. Let (X ; ∘ , 1) be a hyper BE-algebra and F be a nonempty subset of X. F is a hyper filter of X if and only if F/δ ^″* is a filter of X/δ ^″*.

Theorem 6.5. Let X be a hyper BE-algebra, F be a hyper filter of X and y ∈ X. If x < y and x ∈ F, then δ ^* (y) ∈ F/δ ^*.

Proof. Since x < y, we have 1 ∈ x ∘ y and δ ^* (1) = δ ^* (x) * δ ^* (y). Now, F is a hyper filter and x ∈ F, then δ ^* (x) * δ ^* (y) ∈ F/δ ^* and δ ^* (x) ∈ F/δ ^*. By Theorem 6.1, F/δ ^* is a filter and so δ ^* (y) ∈ F/δ ^*.□

Definition 6.2. Let (X ; ∘ , 1) and (X′ ; ∘ ′, 1′) be two hyper BE-algebras. A mapping f : X → X′ is called a homomorphism from X into X′, if for any x, y ∈ X, f (x ∘ y) = f (x) ∘ ′f (y) and f (1) =1′. The homomorphism f, is called an isomorphism, if it is onto and one to one.

Theorem 6.6. Let f : (X ; ∘ , 1) → (X′ ; ∘ ′, 1′) be a homomorphism and x, y ∈ X. Then

if x < y then φ (x) < φ (y),

mapping $\bar{f} : X / δ^{*} ⟶ X^{'} / δ^{*}$ by $\bar{f} (δ^{*} (x)) = δ^{*} (f (x))$ is a homomorphism.

Ker (f) = {x ∈ X | f (x) =1′} is a weak hyper filter of X and is called Kernel of φ.

Proof. (i) Since x < y, we have 1 ∈ x ∘ y and so 1′ = φ (1) ∈ φ (x) ∘ ′φ (y). Hence, φ (x) < φ (y).

(ii) Since f is a homomorphism, we have $\begin{matrix} \bar{f} (δ^{*} (x) * δ^{*} (x)) & = & \bar{f} (δ^{*} (x \circ y)) \\ = & δ^{*} (f (x \circ y)) \\ = & δ^{*} (f (x) \circ f (y)) \\ = & δ^{*} (f (x)) *^{'} δ^{*} (f (y)) \\ = & \bar{f} (δ^{*} (x)) *^{'} \bar{f} (δ^{*} (y)) \end{matrix}$

(iii) Clearly 1 ∈ Ker (f). Let x ∘ y ⊆ Ker (f) and x ∈ Ker (f). Then f (x) ∘ ′f (y) = f (x ∘ y) =1′ and since f (x) =1′, we get f (y) =1′ and then y ∈ Ker (f). Therefore, Ker (f) is a weak hyper filter of X.□

Corollary 6.7. Let f : (X ; ∘ , 1) → (X′ ; ∘ ′, 1′) be a homomorphism and x, y ∈ X. Then

$Ker (f) / δ^{*} \subseteq Ker (\bar{f})$ .

$Ker (f) / δ^{″ *} = Ker (\bar{f})$ .

Example 6.8. Let (X ; ∘ ₁, 1) and (Y ; ∘ ₂, 1) be hyper BE-algebras by the following tables: $\begin{matrix} \begin{matrix} \circ_{1} & 1 & a & b & c & d \\ 1 & {1, d} & {a} & {b} & {c} & {d} \\ a & {1, d} & {1, d} & {b} & {c} & {d} \\ b & {1, d} & {a} & {1, d} & {c} & {d} \\ c & {1, d} & {a} & {b} & {1, d} & {d} \\ d & {1, d} & {a} & {b} & {c} & {1, d} \end{matrix} \\ and \\ \begin{matrix} \circ_{2} & 1 & a & b & c \\ 1 & {1} & {a} & {b} & {c} \\ a & {1} & {1} & {b} & {c} \\ b & {1} & {a} & {1} & {c} \\ c & {1} & {a} & {b} & {1} \end{matrix} \end{matrix}$

Now, we define a map f : (X ; ∘ ₁, 1) ⟶ (Y ; ∘ ₂, 1) by f (c) = b, f (b) = c and f (1) = f (a) = f (d) =1. Clearly f is a homomorphism, Ker (f) = {1, a, d} is a weak hyper filter, $\begin{matrix} \begin{matrix} *_{1} & δ^{*} (1) & δ^{*} (a) & δ^{*} (b) & δ^{*} (c) \\ δ^{*} (1) & δ^{*} (1) & δ^{*} (a) & δ^{*} (b) & δ^{*} (c) \\ δ^{*} (a) & δ^{*} (1) & δ^{*} (1) & δ^{*} (b) & δ^{*} (c) \\ δ^{*} (b) & δ^{*} (1) & δ^{*} (a) & δ^{*} (1) & δ^{*} (c) \\ δ^{*} (c) & δ^{*} (1) & δ^{*} (a) & δ^{*} (b) & δ^{*} (c) \end{matrix} \\ and \\ \begin{matrix} *_{2} & δ^{*} (1) & δ^{*} (a) & δ^{*} (b) \\ δ^{*} (1) & δ^{*} (1) & δ^{*} (a) & δ^{*} (b) \\ δ^{*} (a) & δ^{*} (1) & δ^{*} (1) & δ^{*} (b) \\ δ^{*} (b) & δ^{*} (1) & δ^{*} (a) & δ^{*} (1) \end{matrix} \end{matrix}$

are isomorphic corresponding BE-algebras and $\begin{matrix} Ker (f) / δ^{*} & = & {δ^{*} (1), δ^{*} (a), δ^{*} (d)} \\ \subseteq & {δ^{*} (1), δ^{*} (a), δ^{*} (d)} \\ = & Ker (\bar{f}) . \end{matrix}$

Theorem 6.9. Let (X, ∘ , 1) be a hyper BE-algebra. Then

φ : X ⟶ X/δ ^* by φ (x) = δ ^* (x) is a homomorphism and is called Canonical homomorphism.

w = {x ∈ X | φ (x) = δ ^* (1)} is a hyper filter of X and is called Heart of φ.

w/δ ^* = {δ ^* (1)}.

w/δ ^″* = {X}.

Ker (φ)/δ ^* is a filter in X/δ ^*.

Proof. (i) Let x, y ∈ X. Then $φ (x \circ y) = δ^{*} (x \circ y) = δ^{*} (x) * δ^{*} (y) = φ (x) * φ (y)$ and so φ is a homomorphism.

(ii) Clearly 1 ∈ w. Let (x∘ y) ∩ w ≠ ∅ and x ∈ w. Then δ ^* (x) * δ ^* (y) = δ ^* (1) and δ ^* (x) = δ ^* (1). Hence, δ ^* (y) = δ ^* (1) and so y ∈ w. Therefore, w is a hyper filter of X.

(iii) w/δ ^* = {δ ^* (x) | x ∈ w} = {δ ^* (x) | δ ^* (x) = δ ^* (1)} = {δ ^* (1)}.

(iv) By using (ii), and Theorem 5.9, w/δ ^″* = {X}.

(v) Since 1 ∈ Ker (φ), we get δ ^* (1) ∈ Ker (φ)/δ ^*. Let δ ^* (x) * δ ^* (y) and δ ^* (x) ∈ Ker (φ)/δ ^* . Then for any t ∈ x ∘ y there exists s ∈ Ker (φ) in such a way that δ ^* (t) = δ ^* (s). Then δ ^* (s) = δ ^* (1) and so t ∈ Ker (φ). Hence we have (x∘ y) ∩ Ker (φ) ≠ ∅ and by Theorem 6.6, Ker (φ) is a weak hyper filter, so y ∈ Ker (φ). Therefore δ ^* (y) ∈ Ker (φ) and Ker (φ)/δ ^* is a filter.□

Corollary 6.10. Let (X ; ∘ , 1) be a hyper BE-algebra. Then w = Ker (φ) and so Ker (φ)/δ ^* = w/δ ^*.

7 Conclusion

In the present paper, we introduced the notion of fundamental relation δ on dual hyper K-algebra and investigated some of their useful properties. We introduced δ ^″* as an extension of δ and showed that

The quotient of any dual hyper K-algebra on a regular relation is a hyper BE-algebra.

Quotient on any good strongly regular relation is a dual BCK-algebra.

The relation δ is transitive.

Quotient of any weak commutative dual hyper K-algebra on δ ^* is a commutative dual BCK-algebra and this quotient on δ ^″* is trivial dual BCK-algebra.

The quotient of any hyper filter on δ ^* is a filter.

By defining the concept of homomorphism between dual hyper K-algebras we showed that kernel and heart of the canonical homomorphism is weak hyper filter.

We think such results are very useful for studying in hyper structures. In our future work, we should obtain more results in hyper BE-algebras and its application.

Acknowledgments

The authors would like to express their gratitude to anonymous referees for their comments and suggestions which improved the paper.

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δ-relation on dual hyper K –algebras

Abstract

Keywords

1 Introduction

2 Preliminaries

3 Fundamental relation on (hyper BE-algebra) dual hyper K-algebras

4 Relation δ on (hyper BE-algebra) dual hyper K-algebra

5 Transitivity and some conditions on δ

6 Hyper filters and filters

7 Conclusion

Acknowledgments

References