Characterizations of semisimple ordered semihypergroups in terms of fuzzy hyperideals

Abstract

In this paper, the concept of ordered fuzzy points of ordered semihypergroups is introduced. By using this new concept, we define and study the fuzzy left, right and two-sided hyperideals of an ordered semihypergroup. In particular, we investigate the properties of fuzzy hyperideals generated by ordered fuzzy points of an ordered semihypergroup. Furthermore, we introduce the concepts of prime, semiprime, weakly prime and weakly semiprime fuzzy hyperideals of ordered semihypergroups and establish the relationship between the four classes of fuzzy hyperideals. Finally, we give some characterizations of semisimple ordered semihypergroups in terms of fuzzy hyperideals. Especially, we prove that an ordered semihypergroup S is semisimple if and only if every fuzzy hyperideal of S can be expressed as the intersection of all weakly prime fuzzy hyperideals of S containing it.

Keywords

Ordered semihypergroup ordered fuzzy point fuzzy hyperideal prime (semiprime) fuzzy hyperideal weakly prime (weakly semiprime) fuzzy hyperideal semisimple ordered semihypergroup

1 Introduction

It is well known that fuzzy set theory and its applications in several branches of science are growing day by day. These applications can be found in various fields such as artificial intelligence, computer science, expert systems, control engineering and many others. After the introduction of fuzzy sets by Zadeh [31], reconsideration of the concept of classical mathematics began. For instance, because of the importance of group theory in mathematics, as well as its many areas of application, the notion of fuzzy subgroups was defined by Rosenfeld [24] and its structure was investigated. Later on, many researchers used the idea of fuzzy sets and gave several results in different branches of algebra. In [18], Kehayopulu and Tsingelis applied the concept of fuzzy sets to the theory of ordered semigroups. Then they defined “fuzzy” analogous of several notations, which appeared to be useful in the theory of ordered semigroups. The theory of fuzzy sets on ordered semigroups has been recently developed. For more details, the reader is referred to [17 , 30].

The algebraic hyperstructure is a natural generalization of the classical algebraic structures which was first proposed by Marty [22] in 1934. After the pioneering work of Marty, algebraic hyperstructures have been intensively studied, both from the theoretical point of view and especially for their applications in other fields such as Euclidean and non-Euclidean geometries, graphs and hypergraphs, fuzzy sets, automata, cryptography, artificial intelligence, codes, probabilities, lattices and so on (see [3]). Algebraic hyperstructures have been recently developed by many researchers. A lot of papers and several books have been written on algebraic hyperstructure theory, see [5 , 16]. There are some books on the general theory of algebraic hyperstructures: one by Corsini [2] on the basic theory of hypergroups, another by Vougiouklis [27], mostly on representations of hypergroups and on H_v-structures, which are hyperstructures satisfying conditions weaker than the classic ones.

We noticed that the relationships between the fuzzy sets and algebraic hyperstructures have been already considered by Corsini, Davvaz, Leoreanu, Dudek, Zhan, Hila and others, for instance, the reader can refer to [4 , 32–34]. Recently, Heidari and Davvaz applied the theory of hyperstructures to ordered semigroups and introduced the concept of ordered semihypergroups (see [14]), which is a generalization of ordered semigroups. Also see [1, 9]. It is now natural to investigate similar type of the existing fuzzy subsystems of ordered semihypergroups. In [23], Pibaljommee and Davvaz discussed the properties of fuzzy bi-hyperideals in ordered semihypergroups.

Motivated by the study of fuzzy hyperideals in hyperrings and semihypergroups, and also motivated by Davvaz’s works in ordered hyperstructures, we attempt in the present paper to study fuzzy hyperideals of ordered semihypergroups in detail. The rest of this paper is organized as follows. InSection 2, we recall some basic definitions and results of ordered semihypergroups which will be used throughout this paper. In Section 3, we introduce the concept of ordered fuzzy points of an ordered semihypergroup, and discuss the properties of fuzzy hyperideals generated by ordered fuzzy points of an ordered semihypergroup. In Sections 4 and 5, we introduce the concepts of prime, semiprime, weakly prime and weakly semiprime fuzzy hyperideals in ordered semihypergroups, and give some characterizations of them. In particular, we establish the relationship between the four classes of fuzzy hyperideals. In Section 6, some characterizations of semisimple ordered semigroups in terms of fuzzy hyperideals are given. It is also proven that an ordered semihypergroup S is semisimple if and only if every fuzzy hyperideal of S can be expressed as the intersection of all weakly prime fuzzy hyperideals of S containing it.

2 Preliminaries and some notations

In this section, we present some definitions and results which will be used throughout this paper.

As it is well known a hypergroupoid (S, ∘) is a nonempty set S together with a hyperoperation, that is a map ∘ : S × S → P^∗ (S) , where P^∗ (S) denotes the set of all nonempty subsets of S . The image of the pair (x, y) is denoted by x ∘ y . If x ∈ S and A, B are nonempty subsets of S, then A ∘ B is defined by $A \circ B = ⋃_{a \in A, b \in B} a \circ b .$ Also A ∘ x is used for A ∘ {x} and x ∘ A for {x} ∘ A . A hypergroupoid (S, ∘) is called a semihypergroup if (x ∘ y) ∘ z = x ∘ (y ∘ z) for all x, y, z ∈ S (see [2]).

Recall that an ordered semigroup (S, · , ≤) is a semigroup (S, ·) with an order relation “ ≤ ” such that a ≤ b implies xa ≤ xb and ax ≤ bx for any x ∈ S . In the following, we shall extend the concept of ordered semigroups to the hyper version, and introduce the concept of ordered semihypergroups.

Definition 2.1. An algebraic hyperstructure (S, ∘ , ≤) is called an ordered semihypergroup (also calledpo-semihypergroup in [14]) if (S, ∘) is a semihypergroup and (S, ≤) is a partially ordered set such that the monotone condition holds as follows:

x ≤ y ⇒ a ∘ x ⪯ a ∘ y and x ∘ a ⪯ y ∘ a for all x, y, a ∈ S, where, if A, B ∈ P^∗ (S) , then we say that A ⪯ B if for every a ∈ A there exists b ∈ B such that a ≤ b . If A = {a} , then we write a ⪯ B instead of {a} ⪯ B .

Throughout this paper, unless stated otherwise S stands for an ordered semihypergroup.

Definition 2.2. Let (S, ∘ , ≤) be an ordered semihypergroup. Then S is called commutative if a ∘ b = b ∘ a for any a, b ∈ S .

Clearly, every ordered semigroup is an ordered semihypergroup. In the following we give two examples of ordered semihypergroups.

Example 2.3. Let (S, ≤) be a partially ordered set. If for every x, y ∈ S, we define x ∘ y = {x, y} , then (S, ∘ , ≤) is a commutative ordered semihypergroup.

Example 2.4 ([14]). Let (S, · , ≤) be an ordered semigroup. If for each x, y ∈ S, we define x ∘ y = < x, y > , where <x, y> is the ideal of S generated by {x, y} , then (S, ∘ , ≤) is an ordered semihypergroup.

For ∅ ≠ H ⊆ S, we define

(H] : = {t ∈ S | t ≤ h forsome h ∈ H} .

For H = {a} , we write (a] instead of ({a}] .

By a subsemihypergroup of an ordered semihypergroup S we mean a nonempty subset A of S such that A ∘ A ⊆ A . A nonempty subset A of S is called a left (resp. right) hyperideal of S if (1) S ∘ A ⊆ A (resp . A ∘ S ⊆ A) and (2) If a ∈ A and S ∋ b ≤ a, then b ∈ A . If A is both a left and a right hyperideal of S, then it is called a (two-sided) hyperideal of S (see [14]). We denote by L (a) (resp. R (a) , I (a)) the left (resp. right, two-sided) hyperideal of S generated by a (a ∈ S) . One can easily prove that L (a) = (a ∪ S ∘ a] , R (a) = (a ∪ a ∘ S] and I (a) = (a ∪ S ∘ a ∪ a ∘ S ∪ S ∘ a ∘ S] .

Lemma 2.5. Let S be an ordered semihypergroup. Then the following statements hold:

A ⊆ (A] , ∀ A ⊆ S .

If A ⊆ B ⊆ S, then (A] ⊆ (B] .

(A] ∘ (B] ⊆ (A ∘ B] and

((A] ∘ (B]] = (A ∘ B] , ∀ A, B ⊆ S .

((A]] = (A] , ∀ A ⊆ S .

For every left (resp. right, two-sided) hyperideal T of S, we have (T] = T .

If A, B are left (resp. right, two-sided) hyperideals of S, then (A ∘ B] and A ∩ B are left (resp. right, two-sided) hyperideals of S.

For every a ∈ S, (S ∘ a] , (a ∘ S] and (S ∘ a ∘ S] are a left hyperideal, a right hyperideal and a hyperideal of S, respectively.

If T is a left (resp. right, two-sided) hyperideal of S and A, B are two nonempty subsets of S such that A ⪯ B ⊆ T, then A ⊆ T .

For any two nonempty subsets A, B of S such that A ⪯ B, we have C ∘ A ⪯ C ∘ B and A ∘ C ⪯ B ∘ C for any nonempty subset C of S .

Proof. Straightforward.

Let I be a hyperideal of an ordered semihypergroup S . I is called prime if for any two elements a, b of S such that a ∘ b ⊆ I, then a ∈ I or b∈ I ; I is called weakly prime if for any two hyperideals A, B of S such that A ∘ B ⊆ I, then A ⊆ I or B⊆ I ; I is called semiprime if for any nonempty subset A of S such that A ∘ A ⊆ I, we have A ⊆ I (Equivalent Definition: a ∘ a ⊆ I ⇒ a ∈ I, ∀ a ∈ S); I is called weakly semiprime if for any hyperideal A of S such that A ∘ A ⊆ I, then A ⊆ I .

We next state some fuzzy logic concepts.

Let S be an ordered semihypergroup. By a fuzzy subset of S, we mean a function from S into the real closed interval [0,1], that is, f : S → [0, 1] . The fuzzy subset 1 of S is defined as follows: $1 : S \to [0, 1], x \mapsto 1 (x) : = 1, \forall x \in S .$

Let f and g be two fuzzy subsets of S. Then the inclusion relation f ⊆ g is defined by f (x) ≤ g (x) for all x ∈ S, and f ∩ g, f ∪ g are defined by $\begin{matrix} (f \cap g) (x) & = & min {f (x), g (x)} = f (x) \land g (x), \\ (f \cup g) (x) & = & max {f (x), g (x)} = f (x) \lor g (x) \end{matrix}$ for all x ∈ S, respectively. We denote by F (S) the set of all fuzzy subsets of S . One can easily show that (F (S) , ⊆ , ∩ , ∪) forms a complete lattice with the maximum element 1 and the minimum element 0, which is a mapping from S into [0, 1] defined by $0 : S \to [0, 1], x \mapsto 0 (x) : = 0, \forall x \in S .$

Let (S, ∘ , ≤) be an ordered semihypergroup. For x ∈ S, we define H_x : = {(y, z) ∈ S × S| x ⪯ y ∘ z} . For any f, g ∈ F (S) , the product f ∗ g of f and g is defined by $(f * g) (x) = {\begin{matrix} \overset{(y, z) \in H_{x}}{⋁} [f (y) \land g (z)], & if H_{x} \neq \emptyset, \\ 0, & if H_{x} = \emptyset, \end{matrix}$ for all x ∈ S . One can easily prove that the multiplication “∗” on F (S) is well defined. Furthermore, we have the following lemma:

Lemma 2.6. (F (S) , ∗ , ⊆) forms an ordered semigroup.

Proof. We first show that the multiplication “∗” on F (S) is associative. Indeed, for any f, g, h ∈ F (S) , we shall prove that $((f * g) * h) (x) = (f * (g * h)) (x)$ for any x ∈ S . We consider the following two cases:

Case 1. If H_x = ∅ , then we have $((f * g) * h) (x) = 0 = (f * (g * h)) (x) .$

Case 2. Let H_x ≠ ∅ . Then there exist y, z ∈ S such that x ⪯ y ∘ z, and we have $\begin{matrix} ((f * g) * h) (x) & = & \underset{x ⪯ y \circ z}{⋁} [(f * g) (y) \land h (z)], \\ (f * (g * h)) (x) & = & \underset{x ⪯ y \circ z}{⋁} [f (y) \land (g * h) (z)] . \end{matrix}$

We put $\begin{matrix} s & = & \underset{x ⪯ y \circ z}{⋁} [(f * g) (y) \land h (z)], \\ t & = & \underset{x ⪯ y \circ z}{⋁} [f (y) \land (g * h) (z)] . \end{matrix}$

Now we prove that s ≥ f (y) ∧ (g ∗ h) (z) for every y, z ∈ S such that x ⪯ y ∘ z . Then we have $s \geq \underset{x ⪯ y \circ z}{⋁} [f (y) \land (g * h) (z)] = t .$

Similarly we can show that t ≥ s, and thus s = t .

Let u, v ∈ S such that x ⪯ u ∘ v . Then s ≥ f (u) ∧ (g ∗ h) (v) . In fact,

(A) Let s ≥ f (u) . Then, since f (u)≥ f (u) ∧(g ∗ h) (v) , we have s ≥ f (u) ∧ (g ∗ h) (v) .

(B) Let s < f (u) . Then

(α) If H_v = ∅ , then (g ∗ h) (v) =0 . Since f is a fuzzy subset of S, we have f (u) ≥0 . Then s ≥ 0 = f (u) ∧ (g ∗ h) (v) .

(β) Let H_v ≠ ∅ . Then there exist p, q ∈ S such that v ⪯ p ∘ q . We prove that s ≥ g (p) ∧ h (q) for every p, q ∈ S such that v ⪯ p ∘ q . Then we have $s \geq \underset{v ⪯ p \circ q}{⋁} [g (p) \land h (q)] = (g * h) (v),$ and we have $s \geq (g * h) (v) \geq f (u) \land (g * h) (v) .$

Let p, q ∈ S such that v ⪯ p ∘ q . Then s ≥ g (p) ∧ h (q) . Indeed,

(i) If s ≥ g (p) , then we have s ≥ g (p) ≥ g (p) ∧ h (q) .

(ii) Let s < g (p) . Since x ⪯ u ∘ v, v ⪯ p ∘ q, we have $x ⪯ u \circ v \subseteq u \circ (p \circ q) = (u \circ p) \circ q,$ and there exists w ∈ u ∘ p such that x ⪯ w ∘ q . Thus we have s ≥ (f ∗ g) (w) ∧ h (q) . On the other hand, since s < f (u) and s < g (p) , we have f (u) ∧ g (p) > s . Also, since w ≤ w ∈ u ∘ p, we have w ⪯ u ∘ p, and thus we have $(f * g) (w) \geq f (u) \land g (p) > s,$ from which we can deduce that s ≥ h (q) . Otherwise, if h (q) > s, then (f ∗ g) (w) ∧ h (q) > s . Impossible. Consequently, s ≥ h (q) ≥ g (p) ∧ h (q) .

Thus we have shown that the multiplication “∗” on F (S) is associative. Furthermore, it is obvious that the order relation “⊆” on F (S) is compatible with the multiplication “∗”. Therefore, (F (S) , ∗ , ⊆) is an ordered semigroup.

Lemma 2.7. Let S be an ordered semihypergroup and f, g, h ∈ F (S) . Then the following statements are true:

f ∗ (g ∪ h) = (f ∗ g) ∪ (f ∗ h) .

(g ∪ h) ∗ f = (g ∗ f) ∪ (h ∗ f) .

f ∗ (g ∩ h) ⊆ (f ∗ g) ∩ (f ∗ h) .

(g ∩ h) ∗ f ⊆ (g ∗ f) ∩ (h ∗ f) .

Proof. Straightforward.

Definition 2.8. Let S be an ordered semihypergroup and f ∈ F (S) . The set $f_{t} : = {x \in S | f (x) \geq t}, where t \in (0, 1]$ is called a level subset of f .

Let A be a nonempty subset of S . We define a fuzzy subset λf_A (λ ∈ (0, 1]) of S as follows: $(\forall x \in S) λ f_{A} (x) = {\begin{matrix} λ & if x \in A, \\ 0 & if x \notin A . \end{matrix}$

Clearly, λf_A is a generalization of the characteristic mapping f_A of A (see [29]).

Lemma 2.9. Let A, B be any nonempty subsets of an ordered semihypergroup S. Then the following statements are true:

A ⊆ B if and only if λf_A ⊆ λf_B .

λf_A ∗ λf_B = λf_(A∘B].

Proof. The proofs of (1) and (2) are straightforward verification, and hence we omit the details.

The reader is referred to [3, 28] for notation and terminology not defined in this paper.

3 Fuzzy hyperideals of ordered semihypergroups

In this section, we introduce and study fuzzy hyperideals of ordered semihypergroups. In particular, we discuss the properties of fuzzy hyperideals generated by ordered fuzzy points of an ordered semihypergroup.

Definition 3.1. Let S be an ordered semihypergroup. A fuzzy subset f of S is called a fuzzy left (resp. right) hyperideal of S if

$\underset{z \in x \circ y}{⋀} f (z) \geq f (y)$ (resp. $\underset{z \in x \circ y}{⋀} f (z) \geq f (x)$ ) for all x, y ∈ S .

x ≤ y implies f (x) ≥ f (y) .

A fuzzy (two-sided) hyperideal of S is a fuzzy subset of S which is both a fuzzy left and a fuzzy right hyperideal of S .

Proposition 3.2. Let S be an ordered semihypergroup. A nonempty subset A of S is a left (resp. right, two-sided) hyperideal of S if and only if λf_A is a fuzzy left (resp. right, two-sided) hyperideal of S .

Proof. The proof is straightforward verification, and we omit it.

By Proposition 3.2, we immediately obtain the following corollary:

Corollary 3.3. Let S be an ordered semihypergroup. A nonempty subset A of S is a hyperideal of S if and only if the characteristic function f_A of A is a fuzzy hyperideal of S .

As a generalization of Corollary 3.3, we have the following results.

Proposition 3.4. Let S be an ordered semihypergroup and f a fuzzy subset of S. Then f is a fuzzy hyperideal of S if and only if the level subset f_t (t ∈ (0, 1]) of f is a hyperideal of S for f_t ≠ ∅ .

Proof. (⇒) Let x ∈ f_t, S ∋ y ≤ x . Then y ∈ f_t . Indeed, since x ∈ f_t, f (x) ≥ t and f is a fuzzy hyperideal of S, we have f (y) ≥ f (x) ≥ t, so y ∈ f_t .

(⟸) Let x, y ∈ S and x ≤ y . Then f (x) ≥ f (y) . In fact, let t = f (y) . Then y ∈ f_t . Since f_t is a hyperideal of S, we have x ∈ f_t . Then f (x) ≥ t = f (y) .

The rest of the proof is a consequence of Proposition 2.12 in [4].

Example 3.5. We consider a set S : = {a, b, c, d, e} with the following hyperoperation “ ∘ ″ and the order “ ≤ ”:

o	a	b	c	d	e
a	{a, b}	{a, b}	{a, b}	{a, b}	{a, b}
b	{a, b}	{a, b}	{a, b}	{a, b}	{a, b}
c	{a, b}	{a, b}	{c}	{c}	{e}
d	{a, b}	{a, b}	{c}	{d}	{e}
d	{a, b}	{a, b}	{c}	{c}	{e}

≤ : = {(a, a) , (a, c) , (a, d) , (a, e) , (b, b) , (b, c) , (b, d) , (b, e) , (c, c) , (c, d) , (c, e) , (d, d) , (e, e)} .

Then (S, ∘ , ≤) is an ordered semihypergroup, and all

hyperideals of S are the sets {a, b} , {a, b, c, e} and S . Now let f be a fuzzy subset of S such that f (a) = f (b) =0.8, f (c) = f (d) = f (e) =0.6 . Then $f_{t} = {\begin{matrix} S, & if t \in (0, 0.6], \\ {a, b}, & if t \in (0.6, 0.8], \\ \emptyset, & if t \in (0.8, 1] . \end{matrix}$

Thus all nonempty level subsets f_t (t ∈ (0, 1]) of f are hyperideals of S and by Proposition 3.4, f is a fuzzy hyperideal of S .

Lemma 3.6. LetS be an ordered semihypergroup and f a fuzzy subset of S . Then f is a fuzzy left hyperideal of S if and only if f satisfies the following conditions:

(1) x ≤ y ⇒ f (x) ≥ f (y) , for all x, y ∈ S .

(2) 1 ∗ f ⊆ f .

Proof. Suppose that f is a fuzzy left hyperideal of S . By Definition 3.1, the condition (1) holds. To prove the condition (2) holds, it is enough to prove that (1 ∗ f) (x) ≤ f (x) for any x ∈ S . Indeed, let x ∈ S . If H_x = ∅ , then (1 ∗ f) (x) =0 . Since f (x) ≥0 for all x ∈ S, we have (1 ∗ f) (x) ≤ f (x) . Let H_x ≠ ∅ . Then there exist y, z ∈ S such that x ⪯ y ∘ z, and there exists w ∈ y ∘ z such that x ≤ w . Since f is a fuzzy left hyperideal of S, we have $\begin{matrix} f (z) & \leq & \underset{w \in y \circ z}{⋀} f (w) \leq \underset{\binom{w \in y \circ z}{x \leq w}}{⋀} f (w) \\ \leq & \underset{\binom{w \in y \circ z}{x \leq w}}{⋀} f (x) = f (x) . \end{matrix}$

Hence we have f (z) ≤ f (x) for any x ⪯ y ∘ z . Thus $\begin{matrix} (1 * f) (x) & = & \underset{x ⪯ y \circ z}{⋁} [1 (y) \land f (z)] = \underset{x ⪯ y \circ z}{⋁} [1 \land f (z)] \\ = & \underset{x ⪯ y \circ z}{⋁} f (z) \leq f (x) . \end{matrix}$

Thus (1 ∗ f) (x) ≤ f (x) for any x ∈ S .

Conversely, assume that the conditions (1) and (2) hold. Let y, z ∈ S . Then we can prove that f (x) ≥ f (z) for any x ∈ y ∘ z . In fact, since x ∈ y ∘ z, x ≤ x, we have x ⪯ y ∘ z . Thus, by hypothesis, we have $\begin{matrix} f (x) & \geq & (1 * f) (x) = \underset{x ⪯ u \circ v}{⋁} [1 (u) \land f (v)] \\ \geq & 1 (y) \land f (z) = 1 \land f (z) = f (z) \end{matrix}$ for every y, z ∈ S such that x ∈ y ∘ z, from which we can conclude that $\underset{x \in y \circ z}{⋀} f (x) \geq f (z) .$ By Definition 3.1, f is a fuzzy left hyperideal of S .

Similar to Lemma 3.6, we have the following lemma.

Lemma 3.7. Let S be an ordered semihypergroup and f a fuzzy subset of S . Then f is a fuzzy right hyperideal of S if and only if f satisfies the following conditions:

x ≤ y ⇒ f (x) ≥ f (y) , for all x, y ∈ S .

f ∗ 1 ⊆ f .

Lemma 3.8. Let S be an ordered semihypergroup and f a fuzzy subset of S . Then f is a fuzzy hyperideal of S if and only if f satisfies the following conditions:

x ≤ y ⇒ f (x) ≥ f (y) , for all x, y ∈ S .

1 ∗ f ⊆ f, f ∗ 1 ⊆ f .

Proof. It is obvious by Lemmas 3.6 and 3.7.

Proposition 3.9. If f is a fuzzy left hyperideal and g a fuzzy right hyperideal of an ordered semihypergroup S, then f ∗ g is a fuzzy hyperideal of S .

Proof. The proof comes from Lemmas 2.6, 3.6, 3.7 and 3.8, and we omit the details.

Remark 3.10. The proposition above shows that the set of fuzzy hyperideals of an ordered semihypergroup forms a semigroup.

Theorem 3.11. Let S be an ordered semihypergroup. Then the intersection and union of fuzzy hyperideals of S are both fuzzy hyperideals of S .

Proof. The proof is straightforward verification, and hence we omit the details.

We now introduce the concept of ordered fuzzy points in an ordered semihypergroup.

Definition 3.12. Let S be an ordered semihypergroup, a ∈ S and λ ∈ [0, 1]. An ordered fuzzy point a_λ of S is defined by the rule that $(\forall x \in S) a_{λ} (x) = {\begin{matrix} λ & if x \in (a], \\ 0 & if x \notin (a] . \end{matrix}$

It is accepted that a_λ is a mapping from S into [0, 1] , that is, every ordered fuzzy point of S is a fuzzy subset of S . For any fuzzy subset f of S, we also denote a_λ ⊆ f by a_λ ∈ f in the sequel.

Definition 3.13. Let f be a fuzzy subset of an ordered semihypergroup S . We define (f] by the rule that $(f] (x) = \underset{y \geq x}{⋁} f (y)$ for all x ∈ S. A fuzzy subset f of S is called strongly convex if f = (f] .

Lemma 3.14. Let f be a fuzzy subset of an ordered semihypergroup S . Then f is a strongly convex fuzzy subset of S if and only if x ≤ y implies f (x) ≥ f (y) , for all x, y ∈ S .

Proof. The proof is similar to that of Theorem 3.3 in [29], and hence we omit the details.

Clearly, every fuzzy hyperideal of an ordered semihypergroup S is strongly convex. Moreover, each ordered fuzzy point of S is also strongly convex.

Lemma 3.15. If f is a strongly convex fuzzy subset of an ordered semihypergroup S, then $f = ⋃_{a_{λ} \in f} a_{λ} .$

Proof. The proof is analogous to that of Proposition 3.4 in [29], we omit it.

Proposition 3.16. Let a_λ be an ordered fuzzy point of an ordered semihypergroup S . Then

(1) The fuzzy left hyperideal generated by a_λ, denoted by L (a_λ), is $(\forall x \in S) L (a_{λ}) (x) = {\begin{matrix} λ, & if x \in L (a), \\ 0, & if x \notin L (a), \end{matrix}$ where L (a) is a left hyperideal of S generated by a .

(2) The fuzzy hyperideal generated by a_λ, denoted by I (a_λ) , is $(\forall x \in S) I (a_{λ}) (x) = {\begin{matrix} λ, & if x \in I (a), \\ 0, & if x \notin I (a), \end{matrix}$ where I (a) is a hyperideal of S generated by a .

Proof. Let f be a fuzzy subset of S defined by $f : S \to [0, 1], x \mapsto f (x) : = {\begin{matrix} λ, & if x \in L (a), \\ 0, & if x \notin L (a) . \end{matrix}$

To prove that (1) holds, we now consider the following three steps:

(A) f is a fuzzy left hyperideal of S . In fact, let x, y ∈ S . If f (y) =0, it is obvious that f (z) ≥ f (y) for any z ∈ x ∘ y, then $\underset{z \in x \circ y}{⋀} f (z) \geq f (y) .$ Let f (y) ≠0 . Then f (y) = λ and y ∈ L (a) = (a ∪ S ∘ a] , and we have y ≤ a or y ⪯ b ∘ a for some b ∈ S . Hence x ∘ y ⪯ x ∘ a or x ∘ y ⪯ x ∘ (b ∘ a) = (x ∘ b) ∘ a . It implies that x ∘ y ⊆ L (a) . Thus, for any z ∈ x ∘ y, we have z ∈ L (a) , and $\underset{z \in x \circ y}{⋀} f (z) = λ \geq f (y) .$ Furthermore, if x, y ∈ S such that x ≤ y, then f (x) ≥ f (y) . Indeed, if y ∉ L (a), then f (x) ≥ f (y) =0. If y ∈ L (a), since S ∋ x ≤ y, we have x ∈ L (a) , and then f (x) = λ ≥ f (y). Summarizing all cases above, we have shown that f is a fuzzy left hyperideal of S.

(B) For each x ∈ S and λ ∈ [0, 1] , a_λ ∈ f . Indeed, if x ∈ (a] , then x ∈ L (a) and a_λ (x) = f (x) = λ . If x ∉ (a] , then a_λ (x) =0 ≤ f (x) .

(C) Let g be a fuzzy left hyperideal of S containing the ordered fuzzy point a_λ . We claim that g ⊇ f . To prove our claim, it is enough to show that g (y) ≥ f (y) for any y ∈ S . In fact, if y ∉ L (a) , then f (y) =0 ≤ g (y) . Let y ∈ L (a) . Then we have y ≤ a or y ⪯ b ∘ a for some b ∈ S .

(α) If y ≤ a, then, by hypothesis, we have $g (y) \geq g (a) \geq a_{λ} (a) = λ = f (y) .$

(β) Let y ⪯ b ∘ a for some b ∈ S . Then there exists c ∈ b ∘ a such that y ≤ c . Since g is a fuzzy left hyperideal of S, we have $\begin{matrix} g (y) & \geq & \underset{\binom{c \in b \circ a}{y \leq c}}{⋀} g (c) \geq \underset{c \in b \circ a}{⋀} g (c) \geq g (a) \\ \geq & a_{λ} (a) = λ = f (y) . \end{matrix}$

Thus it is shown, from (A),(B) and (C), that f = L (a_λ) is the fuzzy left hyperideal generated by a_λ .

(2) The proof is similar to that of (1), we omit it.

For the sake of simplicity, throughout this paper, we denote fⁿ = f ∗ f ∗ ⋯ ∗ f (n-copies).

Proposition 3.17. Let a_λ, b_μ be ordered fuzzy points of an ordered semihypergroup (S, ∘ , ≤) , λ > 0, μ > 0 . Then the following statements hold:

$(1 * a_{λ} * 1) (x) = {\begin{matrix} λ & if x \in (S \circ a \circ S], \\ 0 & if x \notin (S \circ a \circ S], \end{matrix}$ for any x ∈ S, and 1 ∗ a_λ ∗ 1 is a fuzzy hyperideal of S .

$(\forall x \in S) (1 * a_{λ}) (x) = {\begin{matrix} λ & if x \in (S \circ a], \\ 0 & if x \notin (S \circ a], \end{matrix}$ and 1 ∗ a_λ is a fuzzy left hyperideal of S .

$(\forall x \in S) (a_{λ} * 1) (x) = {\begin{matrix} λ & if x \in (a \circ S], \\ 0 & if x \notin (a \circ S], \end{matrix}$ and a_λ ∗ 1 is a fuzzy right hyperideal of S .

$a_{λ} * b_{μ} = ⋃_{c \in (a \circ b]} c_{λ \land μ} .$

I (a_λ) = a_λ ∪ 1 ∗ a_λ ∪ a_λ ∗ 1 ∪1 ∗ a_λ ∗ 1, L (a_λ) = a_λ ∪ 1 ∗ a_λ .

(I (a_λ)) ³ ⊆ 1 ∗ a_λ ∗ 1 .

b_μ ∈ 1 ∗ a_λ ∗ 1 if and only if b ∈ (S ∘ a ∘ S] , μ ≤ λ .

a_λ ∗ b_μ = b_μ ∗ a_λ if and only if (a ∘ b] = (b ∘ a] .

If S is commutative, then 1 ∗ a_λ = a_λ ∗ 1 for every ordered fuzzy point a_λ of S .

Proof. (1) If x ∈ (S ∘ a ∘ S], then there exist y, z ∈ S such that x ⪯ y ∘ a ∘ z = (y ∘ a) ∘ z, and there exists w ∈ y ∘ a such that x ⪯ w ∘ z . Thus we have $\begin{matrix} (1 * a_{λ} * 1) (x) & = & \underset{x ⪯ u \circ v}{⋁} [(1 * a_{λ}) (u) \land 1 (v)] \\ \geq & (1 * a_{λ}) (w) \land 1 (z) = (1 * a_{λ}) (w) \\ = & \underset{w ⪯ p \circ q}{⋁} [1 (p) \land a_{λ} (q)] \\ \geq & 1 (y) \land a_{λ} (a) = 1 \land λ = λ . \end{matrix}$

On the other hand, since 1 (b) ∧ a_λ (c) ∧1 (d) = a_λ (c) ≤ λ for all b, c, d ∈ S, we have $(1 * a_{λ} * 1) (x) \leq \underset{x ⪯ b \circ c \circ d}{⋁} [1 (b) \land a_{λ} (c) \land 1 (d)] \leq λ$ for all x ∈ S . It follows that (1 ∗ a_λ ∗ 1) (x) = λ for all x ∈ (S ∘ a ∘ S] .

If x ∉ (S ∘ a ∘ S] , we consider the following cases:

(A) If H_x = ∅ , then, by the definition of the product of two fuzzy subsets of S, (1 ∗ a_λ ∗ 1) (x) =0 .

(B) Let H_x ≠ ∅ . If (1 ∗ a_λ ∗ 1) (x) ≠0, then we have $\begin{matrix} (1 * a_{λ} * 1) (x) & = & \underset{x ⪯ y \circ z}{⋁} [(1 * a_{λ}) (y) \land 1 (z)] \\ = & \underset{x ⪯ y \circ z}{⋁} (1 * a_{λ}) (y) \neq 0, \end{matrix}$ which implies that there exist b, c ∈ S such that x ⪯ b ∘ c and (1 ∗ a_λ) (b) ≠0 . Then H_b ≠ ∅ , and $(1 * a_{λ}) (b) = \underset{b ⪯ u \circ v}{⋁} [1 (u) \land a_{λ} (v)] = \underset{b ⪯ u \circ v}{⋁} a_{λ} (v) \neq 0,$ which implies that there exist p, q ∈ S such that b ⪯ p ∘ q and a_λ (q) ≠0 . Then q ∈ (a] , and $x ⪯ b \circ c ⪯ p \circ q \circ c ⪯ p \circ a \circ c,$ from which we can deduce that x ∈ (S ∘ a ∘ S] . Impossible. Consequently, (1 ∗ a_λ ∗ 1) (x) =0 .

Furthermore, we can prove that 1 ∗ a_λ ∗ 1 is a fuzzy hyperideal of S . Indeed, by Lemma 2.6, we have $1 * (1 * a_{λ} * 1) = (1 * 1) * a_{λ} * 1 \subseteq 1 * a_{λ} * 1$ and $(1 * a_{λ} * 1) * 1 = 1 * a_{λ} * (1 * 1) \subseteq 1 * a_{λ} * 1 .$

Moreover, let x, y ∈ S, x ≤ y . Then (1 ∗ a_λ ∗ 1) (x) ≥ (1 ∗ a_λ ∗ 1) (y) . In fact, if y ∉ (S ∘ a ∘ S] , then (1 ∗ a_λ ∗ 1) (y) =0 . Since (1 ∗ a_λ ∗ 1) (x) ≥0, we have (1 ∗ a_λ ∗ 1) (x) ≥ (1 ∗ a_λ ∗ 1) (y) . Let y ∈ (S ∘ a ∘ S] . Then, since x ≤ y, we have x ∈ (S ∘ a ∘ S] . Thus, in this case, (1 ∗ a_λ ∗ 1) (x) = λ = (1 ∗ a_λ ∗ 1) (y) . Therefore, 1 ∗ a_λ ∗ 1 is a fuzzy hyperideal of S by Lemma 3.8.

Since the proofs of (2) and (3) are similar to that of (1), we omit them.

(4) Let x ∈ S . If x ∈ (a ∘ b] , then x ⪯ a ∘ b, and we have $\begin{matrix} (a_{λ} * b_{μ}) (x) & = & \underset{x ⪯ y \circ z}{⋁} [a_{λ} (y) \land b_{μ} (z)] \\ \geq & a_{λ} (a) \land b_{μ} (b) = λ \land μ . \end{matrix}$

Since a_λ (y) ∧ b_μ (z) ≤ λ ∧ μ for any y, z ∈ S, we have (a_λ ∗ b_μ) (x) ≤ λ ∧ μ, and so $(a_{λ} * b_{μ}) (x) = λ \land μ = (⋃_{c \in (a \circ b]} c_{λ \land μ}) (x) .$

If x ∉ (a ∘ b] , then x ∉ (c] for any c ∈ (a ∘ b] . Thus we have $(⋃_{c \in (a \circ b]} c_{λ \land μ}) (x) = \underset{c \in (a \circ b]}{⋁} (c_{λ \land μ} (x)) = 0 .$

On the other hand, (a_λ ∗ b_μ) (x) =0 . In fact, if (a_λ ∗ b_μ) (x) ≠0, then $(a_{λ} * b_{μ}) (x) = \underset{x ⪯ y \circ z}{⋁} [a_{λ} (y) \land b_{μ} (z)] \neq 0 .$

Thus there exist u, v ∈ S such that x ⪯ u ∘ v and a_λ (u) ∧ b_μ (v) ≠0 . Hence u ∈ (a] , v ∈ (b] . It follows that x ∈ (u ∘ v] ⊆ ((a] ∘ (b]] = (a ∘ b] , which is impossible. Thus, in this case, $(a_{λ} * b_{μ}) (x) = (⋃_{c \in (a \circ b]} c_{λ \land μ}) (x) .$

(5) Let x ∈ S . If x ∉ I (a) , then x ∉ (a] ∪ (S ∘ a] ∪ (a ∘ S] ∪ (S ∘ a ∘ S] . Thus, by (1), (2) and (3), we have (a_λ ∪ 1 ∗ a_λ ∪ a_λ ∗ 1 ∪1 ∗ a_λ ∗ 1) (x) = a_λ (x) ∨ (1 ∗ a_λ) (x) ∨ (a_λ ∗ 1) (x) ∨ (1 ∗ a_λ ∗ 1) (x) =0 . If x∈I (a) , then x ∈ (a] ∪ (S ∘ a] ∪ (a ∘ S] ∪ (S ∘ a ∘ S] , and x ∈ (a] or x ∈ (S ∘ a] or x ∈ (a ∘ S] or x ∈ (S ∘ a ∘ S] . Hence by (1), (2) and (3), we have (a_λ ∪ 1 ∗ a_λ ∪ a_λ ∗ 1 ∪1 ∗ a_λ ∗ 1) (x) = a_λ (x) ∨ (1 ∗ a_λ) (x) ∨ (a_λ ∗ 1) (x) ∨ (1 ∗ a_λ ∗ 1) (x) = λ . By Proposition 3.16, I (a_λ) = a_λ ∪ 1 ∗ a_λ ∪ a_λ ∗ 1 ∪1 ∗ a_λ ∗ 1 .

Similarly, it can be shown that L (a_λ) = a_λ ∪ 1 ∗ a_λ .

(6) By (5), we have $\begin{matrix} (I (a_{λ}))^{2} \\ = I (a_{λ}) * (a_{λ} \cup 1 * a_{λ} \cup a_{λ} * 1 \cup 1 * a_{λ} * 1) \\ \subseteq 1 * (a_{λ} \cup 1 * a_{λ} \cup a_{λ} * 1 \cup 1 * a_{λ} * 1) \\ \subseteq 1 * a_{λ} \cup 1 * a_{λ} * 1 (By Lemmas 2.6 and 2.7) . \end{matrix}$

Thus we have $\begin{matrix} (I (a_{λ}))^{3} \\ = (I (a_{λ}))^{2} * (a_{λ} \cup 1 * a_{λ} \cup a_{λ} * 1 \cup 1 * a_{λ} * 1) \\ \subseteq (1 * a_{λ} \cup 1 * a_{λ} * 1) * 1 (By Lemma 2.6) \\ \subseteq 1 * a_{λ} * 1 (By Lemma 2.7) . \end{matrix}$

(7) If b_μ ∈ 1 ∗ a_λ ∗ 1, then $(1 * a_{λ} * 1) (b) \geq b_{μ} (b) = μ > 0 .$

Thus, by (1), b ∈ (S ∘ a ∘ S] and (1 ∗ a_λ ∗ 1) (b) = λ ≥ μ .

Conversely, let x ∈ S . If x ∈ (b] , then, by b ∈ (S ∘ a ∘ S] , we have $x \in (b] \subseteq ((S \circ a \circ S]] = (S \circ a \circ S] .$

Thus, by (1), we have $(1 * a_{λ} * 1) (x) = λ \geq μ = b_{μ} (x) .$

If x ∉ (b] , then b_μ (x) =0 ≤ (1 ∗ a_λ ∗ 1) (x) . Therefore, b_μ ∈ 1 ∗ a_λ ∗ 1 .

(8) Let x ∈ S . By (4), we have $\begin{matrix} (a_{λ} * b_{μ}) (x) & = & (⋃_{c \in (a \circ b]} c_{λ \land μ}) (x) = \underset{c \in (a \circ b]}{⋁} (c_{λ \land μ} (x)), \\ (b_{μ} * a_{λ}) (x) & = & (⋃_{d \in (b \circ a]} d_{μ \land λ}) (x) = \underset{d \in (b \circ a]}{⋁} (d_{μ \land λ} (x)) . \end{matrix}$

Assume a_λ ∗ b_μ = b_μ ∗ a_λ holds. Then $\underset{c \in (a \circ b]}{⋁} (c_{λ \land μ} (x)) = \underset{d \in (b \circ a]}{⋁} (d_{μ \land λ} (x))$ for all x ∈ S . Let x ∈ (a ∘ b] . Then there exists c ∈ (a ∘ b] such that x = c, and x ∈ (c] . Thus $\underset{d \in (b \circ a]}{⋁} (d_{μ \land λ} (x)) = \underset{c \in (a \circ b]}{⋁} (c_{λ \land μ} (x)) = λ \land μ = μ \land λ,$ which implies that there exists d ∈ (b ∘ a] such that d_μ∧λ (x) = μ ∧ λ > 0 . Thus we have $x \in (d] \subseteq ((b \circ a]] = (b \circ a] .$

It follows that (a ∘ b] ⊆ (b ∘ a] . In a similar way, we can prove that (b ∘ a] ⊆ (a ∘ b] . Consequently, (a ∘ b] = (b ∘ a] .

Conversely, if (a ∘ b] = (b ∘ a] , then, by (4), it is easy to show that a_λ ∗ b_μ = b_μ ∗ a_λ .

(9) By (8) and Lemma 3.15, it is obvious.

4 Prime and semiprime fuzzy hyperideals of ordered semihypergroups

In the current section we define and study the prime and semiprime fuzzy hyperideals of ordered semihypergroups, and discuss their related properties.

Definition 4.1. A fuzzy hyperideal f of an ordered semihypergroup S is called prime if for any two ordered fuzzy points a_λ, b_μ (λ, μ ∈ (0, 1]) of S, a_λ ∗ b_μ ∈ f implies a_λ ∈ f or b_μ∈ f ; f is called semiprime if for any ordered fuzzy point a_λ (λ ∈ (0, 1]) of S, a_λ ∗ a_λ ∈ f implies a_λ ∈ f .

Theorem 4.2. Let S be an ordered semihypergroup. Then a fuzzy hyperideal f of S is prime if and only if for any strongly convex fuzzy subset g and h of S, g ∗ h ⊆ f implies g ⊆ f or h ⊆ f .

Proof. Let g, h be strongly convex fuzzy subsets of S such that g ∗ h ⊆ f . If g⊈f, then, by Lemma 3.15, there exists an ordered fuzzy point a_λ ∈ g such that a_λ ∉ f . For any b_μ ∈ h, we have a_λ ∗ b_μ ∈ g ∗ h ⊆ f . By hypothesis, b_μ ∈ f . By Lemma 3.15, $h = ⋃_{b_{μ} \in h} b_{μ},$ and thus h ⊆ f .

The reverse implication is obvious.

Theorem 4.3. Let S be an ordered semihypergroup. Then a fuzzy hyperideal f of S is semiprime if and only if for any strongly convex fuzzy subset g of S, g ∗ g ⊆ f implies g ⊆ f .

Proof. The proof is similar to that of Theorem 4.2 with a slight modification, and hence we omit the details.

Theorem 4.4. Let f be a fuzzy hyperideal of an ordered semihypergroup S . Then the following statements are equivalent:

f is semiprime.

$f (a) \geq \underset{x \in a \circ a}{⋀} f (x)$ for any a ∈ S .

$f (a) = \underset{x \in a \circ a}{⋀} f (x)$ for any a ∈ S .

Proof. (1) ⇒ (2) . Assume that f is a semiprime fuzzy hyperideal of S . Let a ∈ S . Then $f (a) \geq \underset{x \in a \circ a}{⋀} f (x) .$ Indeed, if $f (a) < \underset{x \in a \circ a}{⋀} f (x),$ then there exists λ ∈ (0, 1] such that $f (a) < λ \leq \underset{x \in a \circ a}{⋀} f (x) .$ Then for any x ∈ a ∘ a, f (x) ≥ λ . Since f is strongly convex, by Lemma 3.14 we can easily deduce that f (y) ≥ λ for any y ∈ (a ∘ a] , and we have y_λ ∈ f . Thus by Proposition 3.17(4), we have $a_{λ} * a_{λ} = ⋃_{y \in (a \circ a]} y_{λ} \in f .$ Since f is a semiprime fuzzy hyperideal of S, we have a_λ ∈ f, and f (a) ≥ λ, which is a contradiction.

(2) ⇒ (3) . Let a ∈ S . Then, since f is a fuzzy hyperideal of S, by hypothesis we have $f (a) \geq \underset{x \in a \circ a}{⋀} f (x) \geq f (a),$ which implies that $f (a) = \underset{x \in a \circ a}{⋀} f (x) .$

(3) ⇒ (1) . Let f be a fuzzy hyperideal of S and a ∈ S . By (3), $f (a) = \underset{x \in a \circ a}{⋀} f (x) .$ If a_λ ∗ a_λ ∈ f, λ ∈ (0, 1] , then $⋃_{x \in (a \circ a]} x_{λ} \in f .$ It follows that x_λ ∈ f for any x ∈ (a ∘ a] . Thus we have $f (a) = \underset{x \in a \circ a}{⋀} f (x) \geq \underset{x \in (a \circ a]}{⋀} f (x) \geq λ .$

Then a_λ ∈ f because f is strongly convex. In other words, f is indeed a semiprime fuzzy hyperideal of S .

Theorem 4.5. Let A be a hyperideal of an ordered semihypergroup S . Then A is semiprime if and only if the characteristic function f_A of A is a semiprime fuzzy hyperideal of S .

Proof. Let A be a semiprime hyperideal of S . By Corollary 3.3, f_A is a fuzzy hyperideal of S . To prove that f_A is semiprime, it is enough to show that $f_{A} (a) \geq \underset{x \in a \circ a}{⋀} f_{A} (x)$ for any a ∈ S . In fact, if a ∘ a ⊆ A, then, since A is semiprime, we have a ∈ A . Thus $f_{A} (a) = 1 = \underset{x \in a \circ a}{⋀} f_{A} (x) .$ If there exists y ∈ a ∘ a such that y ∉ A, then we have $f_{A} (a) \geq 0 = \underset{x \in a \circ a}{⋀} f_{A} (x) .$ Consequently, $f_{A} (a) \geq \underset{x \in a \circ a}{⋀} f_{A} (x)$ for any a ∈ S . By Theorem 4.4, f_A is semiprime.

Conversely, let a ∈ S such that a ∘ a ⊆ A . Then x ∈ A for any x ∈ a ∘ a . Since f_A is a semiprime fuzzy hyperideal of S, by Theorem 4.4 we have $f_{A} (a) \geq \underset{x \in a \circ a}{⋀} f_{A} (x) = 1 .$ On the other hand, since f_A is a fuzzy subset of S, we have f_A (a) ≤1 for all a ∈ S . Hence f_A (a) =1, which implies that a ∈ A . It thus follows that A is semiprime.

As a consequence of the so-called Transfer Principle for Fuzzy Sets in [20], we have the following theorem.

Theorem 4.6. Let S be an ordered semihypergroup and f a fuzzy subset of S. Then f is a semiprime fuzzy hyperideal of S if and only if the level subset f_t (t ∈ (0, 1]) of f is a semiprime hyperideal of S for f_t ≠ ∅ .

Proposition 4.7. If f is a semiprime fuzzy hyperideal of an ordered semihypergroup S, then $\underset{x \in a \circ b}{⋀} f (x) = \underset{y \in b \circ a}{⋀} f (y)$ for all a, b ∈ S .

Proof. Suppose that f is a semiprime fuzzy hyperideal of S . Let x be any element of a ∘ b . Then, by hypothesis and Theorem 4.4, we have $\begin{matrix} f (x) & = & \underset{u \in x \circ x}{⋀} f (u) \geq \underset{u \in a \circ b \circ a \circ b}{⋀} f (u) = \underset{\binom{u \in a \circ v}{v \in b \circ a \circ b}}{⋀} f (u) \\ \geq & \underset{u \in a \circ v}{⋀} f (u) \geq f (v) \geq \underset{v \in b \circ a \circ b}{⋀} f (v) \\ = & \underset{\binom{v \in y \circ b}{y \in b \circ a}}{⋀} f (v) \geq \underset{v \in y \circ b}{⋀} f (v) \geq f (y) \\ \geq & \underset{y \in b \circ a}{⋀} f (y) . \end{matrix}$

It follows that $\underset{x \in a \circ b}{⋀} f (x) \geq \underset{y \in b \circ a}{⋀} f (y) .$ By symmetry, it can be shown that $\underset{y \in b \circ a}{⋀} f (y) \geq \underset{x \in a \circ b}{⋀} f (x) .$ This completes the proof.

In what follows, for the sake of simplicity, we denote Aⁿ = A ∘ A ∘ ⋯ ∘ A (n-copies).

Definition 4.8. Let (S, ∘ , ≤) be an ordered semihypergroup. S is called intra-regular if, for each element a of S, there exist x, y ∈ S such that a ⪯ x ∘ a² ∘ y . Equivalently, a ∈ (S ∘ a² ∘ S] , ∀ a ∈ S .

Example 4.9. We consider a set S : = {a, b, c, d} with the following hyperoperation “ ∘ ″ and theorder “ ≤ ”:

o	a	b	c	d
a	{a}	{a, b}	{a, c}	{S}
b	{b}	{b}	{b, d}	{b, d}
c	{c}	{c, d}	{c}	{c, d}
d	{d}	{d}	{d}	{d}

\leq : = {(a, a), (a, b), (b, b), (c, c), (c, d), (d, d)} .

Then we can easily verify that (S, ∘ , ≤) is an intra-regular ordered semihypergroup.

Theorem 4.10. Let S be an ordered semihypergroup. Then the following statements are equivalent:

S is intra-regular.

Every fuzzy hyperideal of S is semiprime.

Every hyperideal of S is semiprime.

Proof. (1) ⇒ (2) . Let f be a fuzzy hyperideal of S and a ∈ S . Then, by (1), there exist x, y ∈ S such that a ⪯ x ∘ a ∘ a ∘ y . Thus there exists z ∈ x ∘ a ∘ a ∘ y such that a ≤ z . Since f is a fuzzy hyperideal of S, we have $\begin{matrix} f (a) & \geq & \underset{\binom{z \in x \circ a \circ a \circ y}{a \leq z}}{⋀} f (z) \geq \underset{z \in x \circ a \circ a \circ y}{⋀} f (z) \\ = & \underset{\binom{z \in x \circ u}{u \in a \circ a \circ y}}{⋀} f (z) \geq \underset{u \in a \circ a \circ y}{⋀} f (u) \\ = & \underset{\binom{u \in v \circ y}{v \in a \circ a}}{⋀} f (u) \geq \underset{v \in a \circ a}{⋀} f (v) \geq f (a), \end{matrix}$ which implies that $f (a) = \underset{v \in a \circ a}{⋀} f (v) .$ By Theorem 4.4, f is a semiprime fuzzy hyperideal of S .

(2) ⇒ (3) . Let A be a hyperideal of S . Then, by Corollary 3.3, the characterization f_A of A is a fuzzy hyperideal of S . By (2), f_A is a semiprime fuzzy hyperideal of S . Hence A is a semiprime hyperideal of S by Theorem 4.5.

(3) ⇒ (1) . Assume that every hyperideal of S is semiprime. Let a ∈ S . Since a² ∘ a² = a ∘ a² ∘ a ∈ (S ∘ a² ∘ S] and (S ∘ a² ∘ S] is a hyperideal of S, by hypothesis we have a² ⊆ (S ∘ a² ∘ S] , and a ∈ (S ∘ a² ∘ S] . Therefore, S is intra-regular.

Definition 4.11. Let (S, ∘ , ≤) , (T, ∘ ₁, ≤ ₁) be two ordered semihyergroups. A mapping φ : S → T is called isotone if x, y ∈ S, x ≤ y implies φ (x) ≤ ₁φ (y) in T . φ is called a strong homomorphism if it is isotone and satisfies that $⋃_{z \in x \circ y} φ (z) = φ (x) \circ_{1} φ (y)$ for all x, y ∈ S .

Theorem 4.12. Let (S, ∘ , ≤) , (T, ∘ ₁, ≤ ₁) be two ordered semihyergroups and φ : S → T is a strong homomorphism. If f is a semiprime fuzzy hyperideal of T, then φ^-1 (f) is a semiprime fuzzy hyperideal of S, where φ^-1 (f) (x) = f (φ (x)) for all x ∈ S .

Proof. We first show that φ^-1 (f) is a fuzzy hyperideal of S . Indeed, let x, y ∈ S and x ≤ y . Then, since φ is a strong homomorphism of ordered semihypergroups from S to T, we have φ (x) ≤ ₁φ (y) . Since f is a fuzzy hyperideal of T, we get f (φ (x)) ≥ f (φ (y)) , i.e., φ^-1 (f) (x) ≥ φ^-1 (f) (y) . Furthermore, for any x, y ∈ S, we have $\begin{matrix} \underset{z \in x \circ y}{⋀} φ^{- 1} (f) (z) \\ = \underset{z \in x \circ y}{⋀} f (φ (z)) = \underset{φ (z) \in φ (x) \circ_{1} φ (y)}{⋀} f (φ (z)) \\ (Since φ is a strong homomorphism) \\ \geq f (φ (x)) (Since f is a fuzzy hyperideal of T) \\ = φ^{- 1} (f) (x) . \end{matrix}$

Thus φ^-1 (f) is a fuzzy right hyperideal of S . Similarly, we can prove that φ^-1 (f) is also a fuzzy left hyperideal of S .

Moreover, φ^-1 (f) is semiprime. In fact, for any a ∈ S, we have $\begin{matrix} \underset{x \in a \circ a}{⋀} φ^{- 1} (f) (x) \\ = \underset{x \in a \circ a}{⋀} f (φ (x)) = \underset{φ (x) \in φ (a) \circ_{1} φ (a)}{⋀} f (φ (x)) \\ (Since φ is a strong homomorphism) \\ = f (φ (a)) (By Theorem 4.4) \\ = φ^{- 1} (f) (a) . \end{matrix}$

Therefore, φ^-1 (f) is a semiprime fuzzy hyperideal of S by Theorem 4.4.

5 Weakly prime and weakly semiprime fuzzy hyperideals of ordered semihypergroups

In what follows, we denote by Z⁺ the set of positive integers. In this section, we introduce the concepts of weakly prime and weakly semiprime fuzzy hyperideals of an ordered semihypergroup, and give some characterizations of them.

Definition 5.1. A fuzzy hyperideal f of an ordered semihypergroup S is called weakly prime if for any two fuzzy hyperideals g and h of S, g ∗ h ⊆ f implies g ⊆ f or h ⊆ f .

Theorem 5.2. Let A be a nonempty subset of an ordered semihypergroup S . Then A is a weakly prime hyperideal of S if and only if the characteristic function f_A of A is a weakly prime fuzzy hyperideal of S .

Proof. Let A be a weakly prime hyperideal of S . Then, by Corollary 3.3, f_A is a fuzzy hyperideal of S . Suppose that f and g are fuzzy hyperideals of S such that f ∗ g ⊆ f_A . If f⊈f_A, then there exists an ordered fuzzy point x_λ ∈ f (λ > 0) such that x_λ ∉ f_A. For any y_μ ∈ g (μ > 0) , since I (x_λ) ∗ I (y_μ) ⊆ f ∗ g ⊆ f_A, and notice that $(I (x_{λ}) * I (y_{μ})) (z) = {\begin{matrix} λ \land μ, & if z \in (I (x) \circ I (y)], \\ 0, & otherwise, \end{matrix}$ for any z ∈ S . We deduce that (I (x) ∘ I (y)] ⊆ A . Since A is weakly prime and I (x) ∘ I (y) ⊆ (I (x) ∘ I (y)] ⊆ A, we have I (x) ⊆ A or I (y) ⊆ A . Also, since x_λ ∉ f_A, we have x ∉ A, and thus I (x) ⊈A . Hence I (y) ⊆ A, which implies that y_μ ∈ f_A . Therefore, $g = ⋃_{y_{μ} \in g} y_{μ} \subseteq f_{A} .$

Conversely, let f_A be a weakly prime fuzzy hyperideal of S . Then, by Corollary 3.3, A is a hyperideal of S . To prove that A is weakly prime, let B, C be two hyperideals of S such that B ∘ C ⊆ A . Then, by Corollary 3.3 and Lemma 2.9, f_B and f_C are fuzzy hyperideals of S, and f_B ∗ f_C = f_(B∘C] ⊆ f_(A] = f_A . Since f_A is weakly prime, we have f_B ⊆ f_A or f_C ⊆ f_A, which means B ⊆ A or C ⊆ A . Hence A is a semiprime hyperideal of S .

Lemma 5.3. Let S be an ordered semihypergroup. If f is a nonconstant weakly prime fuzzy hyperideal of S, then |Im (f) |=2 .

Proof. Since f is a nonconstant weakly prime fuzzy hyperideal of S, we have |Im (f) |≥2 . Suppose |Im (f) |≥3 . Then there exist x, y, z ∈ S such that f (x) , f (y) and f (z) are different from each other. Without loss of generality, it can be assumed that $f (x) < f (y) < f (z) .$

Thus there exist r, t ∈ (0, 1) such that $f (x) < r < f (y) < t < f (z) .$ Then, for any u ∈ S, we have $(I (x_{r}) * I (y_{t})) (u) = {\begin{matrix} r \land t = r, & u \in (I (x) \circ I (y)], \\ 0, & otherwise . \end{matrix}$

If u ∈ (I (x) ∘ I (y)] , then there exist a ∈ I (x) , b ∈ I (y) such that u ∈ (a ∘ b] , and there exists c ∈ a ∘ b such that u ≤ c . Since f is a fuzzy hyperideal of S, we have $f (u) \geq \underset{\binom{c \in a \circ b}{u \leq c}}{⋀} f (c) \geq \underset{c \in a \circ b}{⋀} f (c) \geq f (b) .$

Since b ∈ (y ∪ S ∘ y ∪ y ∘ S ∪ S ∘ y ∘ S] , we have b ∈ (y] or b ∈ (S ∘ y] or b ∈ (y ∘ S] or b ∈ (S ∘ y ∘ S] . Similar to the above proof, we can obtain that f (b) ≥ f (y) . Hence f (u) ≥ f (y) > r . It follows that I (x_r) ∗ I (y_t) ⊆ f . Thus I (x_r) ⊆ f or I (y_t) ⊆ f because f is a weakly prime fuzzy hyperideal of S . Let I (x_r) ⊆ f . Then we have f (x) ≥ I (x_r) (x) = r, which is impossible. From I (y_t) ⊆ f, similarly, we get a contradiction. This completes the proof.

Theorem 5.4. Let S be an ordered semihypergroup. If f is a nonconstant weakly prime fuzzy hyperideal of S, then there exists x₀ ∈ S such that f (x₀) =1 .

Proof. By Lemma 5.3, |Im (f) |=2 . If f (x) <1 for all x ∈ S, then Im (f) = {s, t} , s < t < 1 . Hence there exist x, y ∈ S and m ∈ (0, 1] such that $f (x) = s < t = f (y) < m \leq 1 .$

Let t₁, t₂ ∈ (0, 1) such that s < t₁ < t < t₂ < m . Then by the similar way of the proof in Lemma 5.3, we have I (x_{t
₁}) ∗ I (y_{t
₂}) ⊆ f . Since f is a weakly prime fuzzy hyperideal of S, we have I (x_{t
₁}) ⊆ f or I (y_{t
₂}) ⊆ f, that is, f (x) ≥ t₁ or f (y) ≥ t₂, which is impossible. Thus there exists x₀ ∈ S such that f (x₀) =1 .

Theorem 5.5. Let S be an ordered semihypergroup. If f is a weakly prime fuzzy hyperideal of S, then the level subset f_t (t ∈ (0, 1]) of f is a weakly prime hyperideal of S for f_t ≠ ∅ .

Proof. By Proposition 3.4, for any t ∈ (0, 1] , f_t is a hyperideal of S for f_t ≠ ∅ . To prove that f_t is weakly prime, let A and B be hyperideals of S such that A ∘ B ⊆ f_t, and let g = tf_A and h = tf_B . Then, by Proposition 3.2, g and h are fuzzy hyperideals of S . Furthermore, we have g ∗ h ⊆ f, that is, (g ∗ h) (x) ≤ f (x) for all x ∈ S . Indeed, if (g ∗ h) (x) =0, then it is obvious. If (g ∗ h) (x) ≠0, then H_x ≠ ∅ , and there exist y, z ∈ S such that x ⪯ y ∘ z, 0 < g (y) ∧ h (z) ≤ t . Thus y ∈ A and z ∈ B, and so x ∈ (A ∘ B] ⊆ (f_t] = f_t, and f (x) ≥ t . Consequently, $(g * h) (x) = \underset{(y, z) \in H_{x}}{⋁} [g (y) \land h (z)}] \leq t \leq f (x) .$

Hence it can be obtained that g ∗ h ⊆ f . Since f is a weakly prime fuzzy hyperideal of S, it can be followed that g ⊆ f or h ⊆ f . Say g ⊆ f, then for any x ∈ A, g (x) = t ≤ f (x) , and x ∈ f_t . Thus A ⊆ f_t . Similarly, say h ⊆ f, we have B ⊆ f_t . Therefore, f_t is a weakly prime hyperideal of S for f_t ≠ ∅ .

By Theorems 5.4 and 5.5, we immediately obtain the following corollary:

Corollary 5.6. Let S be an ordered semihypergroup. If f is a nonconstant weakly prime fuzzy hyperideal of S, then f₁ is a weakly prime hyperideal of S .

Remark 5.7. The inverse of Theorem 5.5 is not true, for example, let A be a weakly prime hyperideal of S, A ≠ S, and let $f (x) = {\begin{matrix} λ, & if x \in A, \\ 0, & if x \notin A, \end{matrix}$ for any x ∈ S, where 0 < λ < 1 . Then f is a fuzzy hyperideal of S . For any t ∈ (0, 1], if f_t ≠ ∅ , then f_t = A, which is a weakly prime hyperideal of S . But f is not weakly prime since f₁ = ∅ .

Now, weakly prime fuzzy hyperideals of ordered semihypergroups can be characterized.

Theorem 5.8. Let f be a nonconstant fuzzy subset of an ordered semihypergroup S . Then f is a weakly prime fuzzy hyperideal of S if and only if f satisfies the following conditions:

|Im (f) |=2 .

f₁ ≠ ∅ , and f₁ is a weakly prime hyperidealof S .

Proof. Suppose that f is a nonconstant weakly prime fuzzy hyperideal of S . Then, by Lemma 5.3, Theorem 5.4 and Corollary 5.6, the conditions (1) and (2) hold.

Conversely, assume that the conditions (1) and (2) hold. Since |Im (f) |=2, by hypothesis we have Im (f) = {t, 1} (t < 1) . Thus

(A) f is a fuzzy hyperideal of S . To prove this assertion, we consider the following four cases:

Case 1. If x, y ∈ f₁, then f (x) = f (y) =1, and by (2), we have x ∘ y ⊆ f₁, which implies that f (z) =1for any z ∈ x ∘ y . Hence $\underset{z \in x \circ y}{⋀} f (z) = 1 = f (x) \lor f (y) .$

Case 2. If x, y ∉ f₁, then f (x) = f (y) = t . Thus, by hypothesis, $\underset{z \in x \circ y}{⋀} f (z) \geq f (x) \lor f (y) .$

Case 3. If x ∉ f₁, y ∈ f₁, then f (x) = t, f (y) =1, and by (2), we have x ∘ y ⊆ f₁ . Consequently, $\underset{z \in x \circ y}{⋀} f (z) = 1 = f (x) \lor f (y) .$

Case 4. If x ∈ f₁, y ∉ f₁, then, it is similar to Case 3, we obtain that $\underset{z \in x \circ y}{⋀} f (z) = 1 = f (x) \lor f (y) .$

Thus, in any case, $\underset{z \in x \circ y}{⋀} f (z) \geq f (x) \lor f (y)$ for all x, y ∈ S . Furthermore, let x, y ∈ S such that x ≤ y . Then f (x) ≥ f (y) . In fact, if y ∉ f₁, then f (y) = t ≤ f (x) . If y ∈ f₁, then, since f₁ is a hyperideal of S, we have x ∈ f₁ . Thus f (x) =1 = f (y) .

(B) f is weakly prime. In fact, let g and h be fuzzy hyperideals of S such that g ∗ h ⊆ f . We claim that g ⊆ f or h ⊆ f . If g⊈f and h⊈f, then there exist x, y ∈ S such that g (x) > f (x) and h (y) > f (y) . Hence x, y ∉ f₁, which implies (x ∘ S ∘ y] ⊈f₁ . Otherwise, by Lemma 2.5, we have

$\begin{matrix} (S \circ x \circ S] \circ (S \circ y \circ S] \subseteq (S \circ (x \circ S \circ y] \circ S] \\ \subseteq (S \circ f_{1} \circ S] \subseteq (f_{1}] \subseteq f_{1} . \end{matrix}$

Since f₁ is a weakly prime hyperideal of S, by Lemma 2.5(7) we have (S ∘ x ∘ S] ⊆ f₁ or (S ∘ y ∘ S] ⊆ f₁ . Say (S ∘ x ∘ S] ⊆ f₁, we can deduce that (I (x)) ³ ⊆ (S ∘ x ∘ S] ⊆ f₁ . It follows that x ∈ I (x) ⊆ f₁ because f₁ is weakly prime. Impossible. Say (S ∘ y ∘ S] ⊆ f₁, similarly, we get a contradiction. Thus (x ∘ S ∘ y] ⊈f₁, and there exists a ∈ (x ∘ S ∘ y] such that a ∉ f₁ . Then f (a) = t and there exists s ∈ S such that a ⪯ x ∘ s ∘ y . Thus there exists b ∈ s ∘ y such that a ⪯ x ∘ b . Since f₁ is a hyperideal of S, by Lemma 2.5(8) we have x ∘ b⊈f₁, which implies that there exists c ∈ x ∘ b such that c ∉ f₁, that is, f (c) = t . By (A), f is a fuzzy hyperideal of S, and we have

$\underset{z \in x \circ b}{⋀} f (z) = t \geq f (x),$ which implies that f (x) = t . In a similar way, we can show that f (y) = t . Consequently,

$\begin{matrix} (g * h) (a) & = & \underset{(u, v) \in H_{a}}{⋁} [g (u) \land h (v)] \geq g (x) \land h (b) \\ \geq & g (x) \land (\underset{b \in s \circ y}{⋀} h (b)) \geq g (x) \land h (y) \end{matrix}$

$\begin{matrix} (Since h is a fuzzy hyperideal of S) \\ > & t = f (a), \end{matrix}$ which contradicts the fact that g ∗ h ⊆ f . Therefore, f is a weakly prime fuzzy hyperideal of S .

Definition 5.9. Let S be an ordered semihypergroup. A fuzzy hyperideal f of S is called proper if f ≠ 1 .

Theorem 5.10. Let S be an ordered semihypergroup. If f is a nonconstant weakly prime fuzzy hyperideal of S, then there exists a proper weakly prime fuzzy hyperideal g of S such that f ⊂ g .

Proof. Let f be a nonconstant weakly prime fuzzy hyperideal of S. By Theorem 5.8, there exists x₀ ∈ S such that f (x₀) =1, and Im (f) = {t, 1} , where t < 1 . Let g be a fuzzy subset of S defined by $(\forall x \in S) g (x) = \frac{1}{2} f (x) + \frac{1}{2} .$

Then, it is easy to prove that g is a fuzzy hyperideal of S, and |Im (g) |=2 .

On the other hand, since g₁ = f₁, by Theorem 5.8, g₁ is a weakly prime hyperideal of S and g is a weakly prime fuzzy hyperideal of S . Let y ∈ S such thatf (y) = t . Then $f (y) < \frac{1}{2} (f (y) + 1) = g (y) < 1,$ which implies that g is a proper weakly prime fuzzy hyperideal of S and f ⊂ g . The proof is completed.

We now characterize the weakly prime fuzzy hyperideals by ordered fuzzy points.

Theorem 5.11. Let f be a fuzzy hyperideal of an ordered semihypergroup S . Then the following statements are equivalent:

f is weakly prime.

For any ordered fuzzy point x_r, y_s of S, if x_r ∗ 1 ∗ y_s ⊆ f, then x_r ∈ f or y_s ∈ f .

For any ordered fuzzy point x_r, y_s of S, if I (x_r) ∗ I (y_s) ⊆ f, then x_r ∈ f or y_s ∈ f .

If g and h are fuzzy right hyperideals of S such that g ∗ h ⊆ f, then g ⊆ f or h ⊆ f .

If g and h are fuzzy left hyperideals of S such that g ∗ h ⊆ f, then g ⊆ f or h ⊆ f .

If g is a fuzzy right hyperideal of S and h is a fuzzy left hyperideal of S such that g ∗ h ⊆ f, then g ⊆ f or h ⊆ f .

Proof. (1) ⇒ (2) . Let f be a weakly prime fuzzy hyperideal of S . For any ordered fuzzy point x_r, y_s of S, if x_r ∗ 1 ∗ y_s ⊆ f, then, by Lemmas 2.6 and 3.8, we have

(1 ∗ x_r ∗ 1) ∗ (1 ∗ y_s ∗ 1) ⊆1 ∗ (x_r ∗ 1 ∗ y_s) ∗1 ⊆ f .

By Proposition 3.17(1), 1 ∗ x_r ∗ 1 and 1 ∗ y_s ∗ 1 are fuzzy hyperideals of S . Then, by hypothesis, we have 1 ∗ x_r ∗ 1 ⊆ f or 1 ∗ y_s ∗ 1 ⊆ f . Say 1 ∗ x_r ∗ 1 ⊆ f, then, by Proposition 3.17(6), we have $(I (x_{r}))^{3} \subseteq 1 * x_{r} * 1 \subseteq f .$

Since f is weakly prime, we have x_r ∈ I (x_r) ⊆ f . Similarly, say 1 ∗ y_s ∗ 1 ⊆ f, we have y_s ∈ I (y_s) ⊆ f .

(2) ⇒ (3) . Let I (x_r) ∗ I (y_s) ⊆ f . By Lemma 2.6 and Proposition 3.17(5), we have $x_{r} * 1 * y_{s} \subseteq x_{r} * I (y_{s}) \subseteq I (x_{r}) * I (y_{s}) \subseteq f .$

Thus, by hypothesis, x_r ∈ f or y_s ∈ f .

(3) ⇒ (4) . Let g and h be fuzzy right hyperideals of S such that g ∗ h ⊆ f . If g⊈f, then there exists x ∈ S such that g (x) > f (x) . Consequently, x_g(x) ∉ f . For any y_s ∈ h, by Proposition 3.17(5), Lemmas 2.7 and 3.8, we have $\begin{matrix} I (x_{g (x)}) * I (y_{s}) & \subseteq & (g * h) \cup (1 * g * h) \\ \subseteq & f \cup (1 * f) \subseteq f . \end{matrix}$

By assumption, y_s ∈ f . Thus $h = ⋃_{y_{s} \in h} y_{s} \subseteq f .$

(3) ⇒ (5) . Similar to the proof of (3) ⇒ (4) .

(5) ⇒ (1), (4) ⇒ (1) and (6) ⇒ (1) are obvious.

(3) ⇒ (6) . Let g be a fuzzy right hyperideal of S and h a fuzzy left hyperideal of S such that g ∗ h ⊆ f . If g⊈f, then there exists an ordered fuzzy point x_r ∈ g such that x_r ∉ f . For any y_s ∈ h, by Proposition 3.17(5), Lemmas 2.7, 3.6 and 3.7, we have $\begin{matrix} I (x_{r}) * I (y_{s}) \\ = (x_{r} \cup 1 * x_{r} \cup x_{r} * 1 \cup 1 * x_{r} * 1) * I (y_{s}) \\ \subseteq (g \cup 1 * g) * (y_{s} \cup y_{s} * 1 \cup 1 * y_{s} \cup 1 * y_{s} * 1) \\ \subseteq (g \cup 1 * g) * (h \cup h * 1) \\ \subseteq (g * h) \cup (g * h * 1) \cup (1 * g * h) \cup (1 * g * h * 1) . \end{matrix}$

Since g ∗ h ⊆ f, by Lemma 3.8 we have I (x_r) ∗ I (y_s) ⊆ f . Also, since x_r ∉ f, we have I (x_r) ⊈f . Hence by hypothesis, y_s ∈ I (y_s) ⊆ f . Thus $h = ⋃_{y_{s} \in h} y_{s} \subseteq f .$

In the following we shall give the relationships among the prime fuzzy hyperideals, weakly prime fuzzy hyperideals and semiprime fuzzy hyperideals in ordered semihypergroups.

Theorem 5.12. Let S be an ordered semihypergroup and f a fuzzy hyperideal of S . Then f is prime if and only if f is weakly prime and semiprime. In commutative ordered semihypergroups the prime fuzzy hyperideals and weakly prime fuzzy hyperidealscoincide.

Proof. Let f be a prime fuzzy hyperideal of S . Clearly, f is semiprime. By Theorem 4.2 and Definition 5.1, f is also weakly prime.

Conversely, assume that f is weakly prime and semiprime. Let a_λ, b_μ be ordered fuzzy points of S such that a_λ ∗ b_μ ∈ f. Then $\begin{matrix} (b_{μ} * 1 * a_{λ}) * (b_{μ} * 1 * a_{λ}) \\ \subseteq b_{μ} * 1 * (a_{λ} * b_{μ}) * 1 * a_{λ} \\ \subseteq 1 * (a_{λ} * b_{μ}) * 1 \subseteq 1 * f * 1 \subseteq f . \end{matrix}$

Since f is semiprime, then, by Theorem 4.3, we have b_μ ∗ 1 ∗ a_λ ⊆ f . Thus $\begin{matrix} (1 * b_{μ} * 1) * (1 * a_{λ} * 1) \\ \subseteq 1 * (b_{μ} * 1 * a_{λ}) * 1 \subseteq 1 * f * 1 \subseteq f . \end{matrix}$

By Proposition 3.17(1), 1 ∗ a_λ ∗ 1 and 1 ∗ b_μ ∗ 1 are fuzzy hyperideals of S . Since f is weakly prime, we have 1 ∗ a_λ ∗ 1 ⊆ f or 1 ∗ b_μ ∗ 1 ⊆ f . Say 1 ∗ a_λ ∗ 1 ⊆ f . Then, by Proposition 3.17(6), we have $(I (a_{λ}))^{3} \subseteq 1 * a_{λ} * 1 \subseteq f .$

Consequently, (I (a_λ)) ² ⊆ f or I (a_λ) ⊆ f . If I (a_λ) ⊆ f, then a_λ ∈ I (a_λ) ⊆ f . If (I (a_λ)) ² ⊆ f, then, since f is weakly prime, I (a_λ) ⊆ f . Say 1 ∗ b_μ ∗ 1 ⊆ f, similarly, it can be obtained that b_μ ∈ f .

In particular, if S is commutative, suppose that f is a weakly prime fuzzy hyperideal of S . Let a_λ, b_μ be ordered fuzzy points of S such that a_λ ∗ b_μ ∈ f. Then, by Proposition 3.17(5), we have $\begin{matrix} I (a_{λ}) * I (b_{μ}) & = & (a_{λ} \cup 1 * a_{λ} \cup a_{λ} * 1 \cup 1 * a_{λ} * 1) \\ * (b_{μ} \cup 1 * b_{μ} \cup b_{μ} * 1 \cup 1 * b_{μ} * 1) \\ \subseteq & (a_{λ} \cup 1 * a_{λ}) * (b_{μ} \cup 1 * b_{μ}) \\ \subseteq & (a_{λ} * b_{μ}) \cup (1 * a_{λ} * b_{μ}) \\ \subseteq & f \cup (1 * f) \subseteq f . \end{matrix}$

By hypothesis, we have $a_{λ} \in I (a_{λ}) \subseteq f or b_{μ} \in I (b_{μ}) \subseteq f,$ which implies that f is a prime fuzzy hyperideal of S . This completes the proof.

In the following we shall define and study the weakly semiprime fuzzy hyperideals of ordered semihypergroups.

Definition 5.13. A fuzzy subset f of an ordered semihypergroup S is called weakly semiprime if f is not a constant function and for any fuzzy hyperideal g of S, g ∗ g ⊆ f implies g ⊆ f .

Theorem 5.14. Let S be an ordered semihypergroup and f a fuzzy hyperideal of S . Then f is weakly semiprime if and only if for any fuzzy hyperideal g of S, gⁿ ⊆ f, n ∈ Z⁺ implies that g ⊆ f .

Proof. ⟸ . This is obvious.

⇒ . Let f be a weakly semiprime fuzzy hyperideal of S . Here we prove the result by induction. Clearly the result holds for n = 2 . Let k ≥ 2 be any positive integer and let the result holds for every positive integer n, 1 ≤ n ≤ k . We claim that g^k+1 ⊆ f implies g ⊆ f . We consider the following two cases:

Case 1. If k is odd, let k = 2m + 1 . Then g^k+1 = g^2(m+1) = (g^m+1) ² .

Case 2. If k is even, let k = 2m . Then, by Proposition 3.9, Lemmas 2.7 and 3.8, we have $\begin{matrix} g^{k + 1} & = & g^{2 m + 1} \supseteq g^{2 m + 1} * 1 \\ \supseteq & g^{2 m + 1} * g = g^{2 m + 2} = (g^{m + 1})^{2} . \end{matrix}$

Thus, in both cases, if g^k+1 ⊆ f, then g^m+1 ⊆ f . Since m + 1 ≤ k, the induction hypothesis insures that g ⊆ f . The proof is completed.

Remark 5.15. In the above theorem, we have characterized weakly semiprime fuzzy hyperideals of an ordered semihypergroup S . The characterization, however, make no reference to the grade of membership of an element of S . The purpose of following Theorem 5.16 is to characterize weakly semiprime fuzzy hyperideal in terms of its effect on the elements of S . We shall see that the following theorem is simpler to use.

Theorem 5.16. Let S be an ordered semihypergroup and f a fuzzy hyperideal of S . Then f is weakly semiprime if and only if $f (a) = \underset{b \in (a \circ S \circ a]}{⋀} f (b)$ for all a ∈ S .

Proof. Assume that $f (a) = \underset{b \in (a \circ S \circ a]}{⋀} f (b)$ for any a ∈ S . Let g be any fuzzy hyperideal of S such that g ∗ g ⊆ f . If g⊈f, then there exists a ∈ S such that g (a) > f (a) . Since $f (a) = \underset{b \in (a \circ S \circ a]}{⋀} f (b),$ there exists t ∈ S such that b ⪯ a ∘ t ∘ a and f (a) = f (b) . Then there exists c ∈ a ∘ t ∘ a such that b ≤ c, and there exists x ∈ a ∘ t such that c ∈ x ∘ a . Since f is a fuzzy hyperideal of S, we have $f (c) \leq f (b) = f (a) < g (a) .$

Furthermore, according to g ∗ g ⊆ f, we have $\begin{matrix} g (a) & > & f (c) \geq (g * g) (c) \\ = & \underset{(u, v) \in H_{c}}{⋁} [g (u) \land g (v)] \geq g (x) \land g (a) \\ \geq & (\underset{x \in a \circ t}{⋀} g (x)) \land g (a) \\ \geq & g (a) \land g (a) = g (a), \end{matrix}$ which is a contradiction.

Conversely, suppose that f is a weakly semiprime fuzzy hyperideal of S . If $f (a) \neq \underset{b \in (a \circ S \circ a]}{⋀} f (b)$ for some a ∈ S, then $f (a) < \underset{b \in (a \circ S \circ a]}{⋀} f (b) .$ In fact, for any b ∈ (a ∘ S ∘ a] , there exists t ∈ S such that b ⪯ a ∘ t ∘ a . Then there exists c ∈ a ∘ t ∘ a such that b ≤ c, and there exists x ∈ a ∘ t such that c ∈ x ∘ a . Since f is a fuzzy hyperideal of S, we have $f (b) \geq f (c) \geq \underset{c \in x \circ a}{⋀} f (c) \geq f (a) .$

Let $\underset{b \in (a \circ S \circ a]}{⋀} f (b) = m .$ Define a fuzzy subset g of S as follows: $(\forall x \in S) g (x) = {\begin{matrix} m & if x \in (S \circ a \circ S], \\ 0 & if x \notin (S \circ a \circ S] . \end{matrix}$

Then, by Proposition 3.17(1), g is a fuzzy hyperideal of S . Furthermore, we can show that g ∗ g ⊆ f . It is enough to prove that (g ∗ g) (x) ≤ f (x) for all x ∈ S . Indeed, if (g ∗ g) (x) =0, then it is obvious that (g ∗ g) (x) ≤ f (x) . Let (g ∗ g) (x) = m . Then we have $\underset{(y, z) \in H_{x}}{⋁} [g (y) \land g (z)] = m,$ which means there exist u, v ∈ (S ∘ a ∘ S] such that x ⪯ u ∘ v . Put u ⪯ s ∘ a ∘ t, v ⪯ p ∘ a ∘ q for some s, t, p, q ∈ S . Then, by Lemma 2.5(9), we have $\begin{matrix} x & ⪯ & u \circ v ⪯ (s \circ a \circ t) \circ (p \circ a \circ q) \\ = & s \circ (a \circ t \circ p \circ a) \circ q, \end{matrix}$ and there exists y ∈ a ∘ t ∘ p ∘ a such that x ⪯ s ∘ y ∘ q . Then there exists z ∈ s ∘ y such that x ⪯ z ∘ q, and x ≤ w for some w ∈ z ∘ q . Since f is a fuzzy hyperideal of S, we have $\begin{matrix} f (x) & \geq & \underset{\binom{x \leq w}{w \in z \circ q}}{⋀} f (w) \geq \underset{w \in z \circ q}{⋀} f (w) \geq f (z) \\ \geq & \underset{z \in s \circ y}{⋀} f (z) \geq f (y) \geq \underset{y \in (a \circ t \circ p \circ a]}{⋀} f (y) \\ \geq & \underset{y \in (a \circ S \circ a]}{⋀} f (y) = m = (g * g) (x) . \end{matrix}$

It implies that g ∗ g ⊆ f . By hypothesis, g ⊆ f . Again define a fuzzy subset h of S as follows: $(\forall x \in S) h (x) = {\begin{matrix} m & if x \in I (a), \\ 0 & if x \notin I (a) . \end{matrix}$

Clearly, h = mf_I(a) . Then, by Proposition 3.2, h is a fuzzy hyperideal of S . Moreover, h⁴ ⊆ f . Indeed, since $h^{4} (x) = \underset{x ⪯ x_{1} \circ x_{2} \circ x_{3} \circ x_{4}}{⋁} [h (x_{1}) \land h (x_{2}) \land h (x_{3}) \land h (x_{4})] = m$ only if x₁, x₂, x₃, x₄ ∈ I (a) . We can easily verify that $x ⪯ x_{1} \circ x_{2} \circ x_{3} \circ x_{4} \subseteq S \circ a \circ S,$ which implies that x ∈ (S ∘ a ∘ S] . Thus h⁴ (x) = m implies g (x) = m . Consequently, h⁴ ⊆ g ⊆ f . Since f is weakly semiprime, and so, by Proposition 3.9, we have h ∗ h ⊆ f and h ⊆ f . Thus m = h (a) ≤ f (a) , which contradicts the fact that f (a) < m . This completes the proof.

Corollary 5.17. Let S be an ordered semihypergroup. If f is a semiprime fuzzy hyperideal of S, then f is weakly semiprime.

Proof. Let f be a semiprime fuzzy hyperideal of S and a ∈ S . Then, by Theorem 4.4, we have $\begin{matrix} f (a) & = & \underset{x \in a \circ a}{⋀} f (x) = \underset{x \in a \circ a}{⋀} (\underset{b \in x \circ x}{⋀} f (b)) \\ \geq & \underset{b \in a \circ a \circ a \circ a}{⋀} f (b) \geq \underset{b \in (a \circ S \circ a]}{⋀} f (b) . \end{matrix}$

On the other hand, since f is a fuzzy hyperideal of S, we have $\begin{matrix} \underset{b \in (a \circ S \circ a]}{⋀} f (b) = ⋀ {f (b) | b ⪯ a \circ t \circ a, t \in S} \\ = & ⋀ {f (b) | b ⪯ a \circ x, x \in t \circ a, t \in S} \\ = & ⋀ {f (b) | b \leq c, c \in a \circ x, x \in t \circ a, t \in S} \\ \geq & ⋀ {f (c) | c \in a \circ x, x \in t \circ a, t \in S} \\ \geq & \underset{c \in a \circ x}{⋀} f (c) \geq f (a) . \end{matrix}$

Hence $f (a) = \underset{b \in (a \circ S \circ a]}{⋀} f (b) .$ It thus follows, byTheorem 5.16, that f is a weakly semiprime fuzzy hyperideal of S .

The following Corollary 5.18 shows that the converse of Corollary 5.17 also holds on commutative ordered semihypergroups.

Corollary 5.18. Let S be a commutative ordered semihypergroup and f a weakly semiprime fuzzy hyperideal of S . Then f is semiprime.

Proof. Suppose that f is a weakly prime fuzzy hyperideal of a commutative ordered semihypergroup S . If $f (a) \neq \underset{x \in a \circ a}{⋀} f (x)$ for some a ∈ S, then we have $\underset{x \in a \circ a}{⋀} f (x) > f (a) = \underset{b \in (a \circ S \circ a]}{⋀} f (b) .$

Thus there exists t ∈ S such that b ⪯ a ∘ t ∘ a and f (a) = f (b) . Since S is commutative, we have b ⪯ a ∘ a ∘ t . Then there exists y ∈ a ∘ a such that b ⪯ y ∘ t, that is, b ≤ c for some c ∈ y ∘ t . Thus $\begin{matrix} \underset{x \in a \circ a}{⋀} f (x) & > & f (a) = f (b) \geq \underset{b ⪯ a \circ a \circ t}{⋀} f (b) \\ = & \underset{y \in a \circ a}{⋀} (\underset{b ⪯ y \circ t}{⋀} f (b)) \geq \underset{y \in a \circ a}{⋀} (\underset{\binom{b \leq c}{c \in y \circ t}}{⋀} f (c)) \\ \geq & \underset{y \in a \circ a}{⋀} (\underset{c \in y \circ t}{⋀} f (c)) \geq \underset{y \in a \circ a}{⋀} f (y), \end{matrix}$ which is a contradiction. We have thus shown, byTheorem 4.4, that f is semiprime.

6 Characterizations of semisimple ordered semihypergroups

In this section, we investigate mainly the properties of semisimple ordered semihypergroups. In particular, we give some characterizations of semisimple ordered semihypergroups by fuzzy hyperideals generated by ordered fuzzy points.

Definition 6.1. An ordered semihypergroup S is called semisimple if (I²] = I holds for every hyperideal I of S .

Lemma 6.2. Let S be an ordered semihypergroup. Then the following statements are equivalent:

S is semisimple.

a ∈ (S ∘ a ∘ S ∘ a ∘ S] for all a ∈ S .

Proof. The proof is straightforward verification, and we omit it.

Theorem 6.3. Let S be an ordered semihypergroup. Then the following statements are equivalent:

S is semisimple.

f ∗ f = f for every fuzzy hyperideal f of S .

f ∗ g = f ∩ g for all fuzzy hyperideals f and g of S .

Proof. (1) ⇒ (3) . Let S be a semisimple ordered semihypergroup and a ∈ S . Then, by Lemma 6.2, we have a ∈ (S ∘ a ∘ S ∘ a ∘ S] , and there exist x, y, z ∈ S such that a ⪯ x ∘ a ∘ y ∘ a ∘ z . Then there exist b ∈ x ∘ a, c ∈ y ∘ a ∘ z such that a ⪯ b ∘ c, and there exists d ∈ y ∘ a such that c ∈ d ∘ z . For any two fuzzy hyperideals f and g of S, we have $\begin{matrix} (f * g) (a) & = & \underset{(u, v) \in H_{a}}{⋁} [f (u) \land g (v)] \geq f (b) \land g (c) \\ \geq & (\underset{b \in x \circ a}{⋀} f (b)) \land (\underset{\binom{c \in d \circ z}{d \in y \circ a}}{⋀} g (c)) \\ \geq & f (a) \land (\underset{c \in d \circ z}{⋀} g (c)) \geq f (a) \land g (d) \\ \geq & f (a) \land (\underset{d \in y \circ a}{⋀} g (d)) \\ \geq & f (a) \land g (a) = (f \cap g) (a), \end{matrix}$ which means that f ∩ g ⊆ f ∗ g. On the other hand, since f, g are fuzzy hyperideals of S, by Lemma 3.8 we have f ∗ g ⊆ f ∩ g . Therefore, f ∗ g = f ∩ g .

(3) ⇒ (2) . Obviously.

(2) ⇒ (1) . Let I be any hyperideal of S . Then, by Corollary 3.3, the characteristic function f_I of I is a fuzzy hyperideal of S . Thus, by (2), we have f_I ∗ f_I = f_I . By Lemma 2.9(2), we have f_(I²] = f_I, and thus (I²] = I . Hence S is a semisimple ordered semihypergroup.

Now, we give characterizations of a semisimple ordered semihypergroup S by fuzzy hyperideals generated by ordered fuzzy points of S .

Theorem 6.4. Let S be an ordered semihypergroup, λ > 0, μ > 0 . Then the following statements are equivalent:

S is semisimple.

I (a_λ) ∗ I (b_μ) = I (a_λ) ∩ I (b_μ) for all ordered fuzzy points a_λ and b_μ of S .

(I (a_λ)) ² = I (a_λ) for every ordered fuzzy point a_λ of S .

a_λ ∈ 1 ∗ a_λ ∗ 1 ∗ a_λ ∗ 1 for every ordered fuzzy point a_λ of S .

Proof. (1) ⇒ (2) . It is obvious by Theorem 6.3.

(2) ⇒ (3) . Clearly.

(3) ⇒ (4) . Let a_λ be any ordered fuzzy point of S . By (3), (I (a_λ)) ² = I (a_λ) . Then we have $a_{λ} \in I (a_{λ}) = (I (a_{λ}))^{5} = (I (a_{λ}))^{4} * I (a_{λ}) .$ By Proposition 3.17(6), (I (a_λ)) ³ ⊆ 1 ∗ a_λ ∗ 1 . Then $\begin{matrix} (I (a_{λ}))^{4} \\ \subseteq (1 * a_{λ} * 1) * (a_{λ} \cup 1 * a_{λ} \cup a_{λ} * 1 \cup 1 * a_{λ} * 1) \\ \subseteq 1 * a_{λ} * 1 * a_{λ} \cup 1 * a_{λ} * 1 * a_{λ} * 1 . \end{matrix}$

Thus we have $\begin{matrix} (I (a_{λ}))^{5} & \subseteq & (1 * a_{λ} * 1 * a_{λ} \cup 1 * a_{λ} * 1 * a_{λ} * 1) * \\ (a_{λ} \cup 1 * a_{λ} \cup a_{λ} * 1 \cup 1 * a_{λ} * 1) \\ \subseteq & 1 * a_{λ} * 1 * a_{λ} * 1 . \end{matrix}$

Therefore, a_λ ∈ 1 ∗ a_λ ∗ 1 ∗ a_λ ∗ 1 .

(4) ⇒ (1) . Let a_λ ∈ 1 ∗ a_λ ∗ 1 ∗ a_λ ∗ 1 for every ordered fuzzy point a_λ of S . Then (1 ∗ a_λ ∗ 1 ∗ a_λ ∗ 1) (a) ≥ a_λ (a) = λ > 0, and we have $\underset{a ⪯ b \circ c}{⋁} [(1 * a_{λ} * 1) (b) \land (a_{λ} * 1) (c)] > 0 .$

Thus, by Proposition 3.17, there exist b ∈ (S ∘ a ∘ S] , c ∈ (a ∘ S] such that a ⪯ b ∘ c, i.e., a ∈ (b ∘ c] . Hence we have $\begin{matrix} a & \in & (b \circ c] \subseteq ((S \circ a \circ S] \circ (a \circ S]] \\ = & (S \circ a \circ S \circ a \circ S] . \end{matrix}$

By Lemma 6.2, S is semisimple.

Theorem 6.5. Let S be an ordered semihypergroup. Then the fuzzy hyperideals of S are weakly prime if and only if they form a chain and S is semisimple.

Proof. Let g and h are fuzzy hyperideals of S . Then, by Proposition 3.9, g ∗ h is a fuzzy hyperideal of S. By hypothesis, g ∗ h is weakly prime. Since g ∗ h ⊆ g ∗ h, we have g ⊆ g ∗ h ⊆ 1 ∗ h ⊆ h or h ⊆ g ∗ h ⊆ g ∗ 1 ⊆ g . Hence the fuzzy hyperideals of S form a chain. Moreover, for any fuzzy hyperideal f of S, obviously, f ∗ f ⊆ f. On the other hand, since f ∗ f ⊆ f ∗ f and f ∗ f is a weakly prime fuzzy hyperideal of S, we have f ⊆ f ∗ f . Thus f ∗ f = f . Therefore, S is semisimpie by Theorem 6.3.

Conversely, suppose that f is a fuzzy hyperideal of S . Let g, h be any fuzzy hyperideals of S such that g ∗ h ⊆ f . By hypothesis, we have g ⊆ h or h ⊆ g . Say g ⊆ h, then, by Theorem 6.3, g = g ∗ g ⊆ g ∗ h ⊆ f . Similarly, say h ⊆ g, we have h ⊆ f . Hence f is weakly prime.

Lemma 6.6. Let S be a semisimple ordered semihypergroup and f a fuzzy hyperideal of S . If f (x) = α, x ∈ S, α ∈ [0, 1] , then there exists a weakly prime fuzzy hyperideal g of S such that f ⊆ g and g (x) = α .

Proof. Let $B = {h | h hyperideal of S such that f \subseteq h, h (x) = α} .$ Then $B \neq \emptyset$ because $f \in B .$ Thus $(B, \subseteq)$ is an ordered set. Let $C$ be a chain in $B .$ Then, by Theorem 3.11, the set $⋃_{h \in C} h$ is a fuzzy hyperideal of S and $f \subseteq ⋃_{h \in C} h .$ Since for any $h \in C,$ h (x) = α, we have $(⋃_{h \in C} h) (x) = α .$ Thus the fuzzy hyperideal $⋃_{h \in C} h$ is an upper bound of $C$ in $B .$ By Zorn’s lemma, $B$ has a maximal element, say g . Then f ⊆ g, g (x) = α .

First, we shall prove the fact that for any fuzzy hyperideals h₁, h₂ of S such that g = h₁ ∩ h₂, it implies that g = h₁ or g = h₂ . Indeed, if g ≠ h₁ and g ≠ h₂, then, since g is maximal with respect to the property that f ⊆ g, g (x) = α, we have h₁ (x) ≠ α, h₂ (x) ≠ α . Thus g (x) = (h₁ ∩ h₂) (x) ≠ α, which is a contradiction.

We now prove that g is a weakly prime fuzzy hyperideal of S . Let h₁, h₂ be fuzzy hyperideals of S such that h₁ ∗ h₂ ⊆ g . Thus (h₁ ∗ h₂) ∪ g = g . Since S is semisimple, by Theorem 6.3, we have $\begin{matrix} g & = & (h_{1} * h_{2}) \cup g = (h_{1} \cap h_{2}) \cup g \\ = & (h_{1} \cup g) \cap (h_{2} \cup g) . \end{matrix}$

Thus, by the above fact, g = h₁ ∪ g or g = h₂ ∪ g, that is, h₁ ⊆ g or h₂ ⊆ g . Consequently, g is a weakly prime fuzzy hyperideal of S .

In the following we shall give a characterization of semisimple ordered semihypergroups by weakly prime fuzzy hyperideals.

Theorem 6.7. Let S be an ordered semihypergroup. Then S is semisimple if and only if every fuzzy hyperideal of S is the intersection of all weakly prime fuzzy hyperideals of S containing it.

Proof. Suppose that f is a fuzzy hyperideal of a semisimple ordered semihypergroup S . Let $M = {g_{α} | g_{α} is a weakly prime fuzzy hyperideal of S such that f \subseteq g_{α}} .$ Then, obviously, $f \subseteq ⋂_{g_{α} \in M} g_{α} .$ On the other hand, let x ∈ S . By Lemma 6.6, there exists a weakly prime fuzzy hyperideal g_β such that f ⊆ g_β and g_β (x) = f (x) . Hence $g_{β} \in M$ and $(⋂_{g_{α} \in M} g_{α}) (x) \leq g_{β} (x) = f (x) .$ It thus follows that $⋂_{g_{α} \in M} g_{α} \subseteq f .$ Consequently, $f = ⋂_{g_{α} \in M} g_{α} .$

Conversely, assume that f is any fuzzy hyperideal of S . Then, by Proposition 3.9, f ∗ f is also a fuzzy hyperideal of S . By hypothesis, we have

$f * f = ⋂_{g \in N} g,$ where $N$ is the set of all weakly prime fuzzy hyperideals of S containing f ∗ f . Furthermore, we prove that f ∗ f = f . In fact, for any $g \in N,$ clearly, f ∗ f ⊆ g . Since g is weakly prime, it can be obtained that f ⊆ g . Then we have $f \subseteq ⋂_{g \in N} g = f * f .$ On the other hand, since f is a fuzzy hyperideal of S, by Lemma 3.8, we have f ∗ f ⊆ f ∗ 1 ⊆ f . Thus f ∗ f = f . Therefore, S is semisimple by Theorem 6.3.

Acknowledgments

The authors are extremely grateful to the Associate Editor Professor Bijan Davvaz and the referees for their valuable comments and helpful suggestions which help to improve the presentation of this paper. This work was supported by the Natural Science Foundation of China (No. 11371177) and the University Natural Science Project of Anhui Province (No. KJ2015A161, 2015KJ003).

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