New types of fuzzy ( m,n )-ideals in ordered semigroups

Abstract

In this paper, we introduce new types of fuzzy (m, n)-ideals in ordered semigroups. In fact, the notion of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals of the ordered semigroups is introduced. Further, we present the characterzations of this notion in different ways. Then the (κ^*, κ)-lower part of the (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals is defined and its associated properties are investigated. After that, (m, n)-regular ordered semigroups are characterized in terms of its (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals and their (κ^*, κ)-lower parts.

Keywords

Ordered semigroups fuzzy sets

1 Introduction

Since Zadeh introduced the idea of fuzzy sets in 1965 [20], fuzzy sets and fuzzy systems theories quickly developed. In the pioneering paper of Rosenfeld [18] in 1971, the study of the fuzzy algebraic structures began. Kuroki [10] applied the theory of fuzzy sets to semigroups and fuzzy sets to ordered semigroups were first considered by Kehayopulu and Tsingelis in [6]. Jun et al. [2] introduced the concept of (α, β)-fuzzy generalized bi-ideals in ordered semigroups and provided several characterizations of regular and intra-regular ordered semigroups in terms of (∈ , ∈ ∨ q)-fuzzy generalized bi-ideals. To generalize the concept of (α, β)-fuzzy generalized bi-ideals of ordered semigroups, Khan et al. introduced the concept of (∈ , ∈ ∨ q_k)-fuzzy generalized bi-ideals in [8] of ordered semigroups. By using the concept of (∈ , ∈ ∨ q_k)-fuzzy ideals, Shabir et al. [19] characterized regular semigroups. The concept of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy generalized bi-ideals and its (κ^*, κ)-lowe part in ordered semigroups introduced and charactized by Khan et al. [7]. Also, different notions of fuzzy set have been applied in semigroups (see for e.g., [3–5 , 15–17]). For related notions of the present study, the reader is referred to [12 –14].

In this paper, we establishe new types of fuzzy ideals in ordered semigroups. In fact, on ordered semigroups, the notion of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals is defined and some vital properties are investigated. In addition, (m, n)-regular ordered semigroups are described in terms of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal and their (κ^*, κ)-lower parts. The structure of paper is organized as: Section 2 summarizes some concepts and properties relating to ordered semigroups, ideals, fuzzy subsets, fuzzy subsemigroups that are required to produce our key results. The notion of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal of ordered semigroups is discussed in Section 3. Section 4 is dedicated to the analysis of (κ^*, κ)-lower part of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal. In Section 5, directions for some potential work are given. The paper finishes with a list of references.

2 Preliminaries

A semigroup (S, ·) is said to be an ordered semigroup if (S, ≤) is a partially ordered set such that a ≤ b ⇒ ac ≤ bc and ca ≤ cb, ∀a, b, c ∈ S.

For A ⊆ S, we denote (A] = {t ∈ S | t ≤ a forsome a ∈ A} . For subsets ∅ ≠ A, B ⊆ S, the product of A and B is define as AB = {ab | a ∈ A and b ∈ B}.

A subset ∅ ≠ Ω of S is called a subsemigroup if ΩΩ ⊆ Ω. Further, Ω is called a left (resp. right) ideal of S if SΩ ⊆ Ω (resp . ΩS ⊆ Ω) and (Ω] ⊆ Ω. It is called an ideal of S if it is both left and right ideal of S. A subsemigroup A of S is called a bi-ideal of S if ASA ⊆ A and (A] ⊆ A. A subsemigroup A of S is called an (m, n)-ideal of S [1] if A^mSAⁿ ⊆ A and (A] ⊆ A, where m, n are non-negative integers. Here A⁰S = SA⁰ = S.

For (∅ ≠) A ⊆ S, the principal (m, n)-ideal [A] _(m,n) generated by A is defined [1] as follows: $[A]_{(m, n)} = (⋃_{i = 1}^{m + n} A^{i} \cup A^{m} S A^{n}] .$

Throughout this paper, $\hat{I_{(m, n)}}$ will stand for the set of all (m, n)-ideals of S.

A mapping η : S → [0, 1] is called the fuzzy subset of S. For Ω ⊆ S, the characteristic function χ_Ω is defined as: $χ_{Ω} (x) = {\begin{matrix} 1 & if x \in Ω, \\ 0 & if x \notin Ω . \end{matrix}$ For any fuzzy subsets η and ξ of S, the η ∩ ξ, η ∪ ξ and η ∘ ξ are defined as: $\begin{matrix} (η \cap ξ) (x) & = \min {η (x), ξ (x)} \\ (η \cup ξ) (x) & = \max {η (x), ξ (x)} \end{matrix}$ and $(η \circ ξ) (x) = {\begin{matrix} \underset{(y, z) \in Ω_{x}}{⋁} {η (y) \land ξ (z)} & if Ω_{x} \neq \emptyset, \\ 0 & if Ω_{x} = \emptyset, \end{matrix}$ where Ω_x = {(r, s) ∈ S × S | x ≤ rs}. Define an order relation ⪯ on the set $F (S)$ of all fuzzy subsets of S by $η ⪯ ξ \Leftrightarrow η (x) \leq ξ (x) \forall x \in S .$ If $η, ξ \in F (S)$ such that η ⪯ ξ, then $\forall λ \in F (S)$ , η ∘ λ ⪯ ξ ∘ λ and λ ∘ η ⪯ λ ∘ ξ. We denote by 1 the fuzzy subset of S defined by 1 : S → [0, 1] |x ↦ 1 (x) =1.

Let A, B ⊆ S. Then A ⊆ B ⇔ η_A ⪯ η_B; η_A ∩ η_B = η_A∩B; η_A ∘ η_B = η_(AB].

$F (S) ∋ η$ is called a fuzzy subsemigroup [11] of S if η (ℏ ℓ) ≥ min {η (ℏ) , η (ℓ)} ∀ ℏ , ℓ ∈ S and it is called a fuzzy left (resp. right) ideal [11] of S if ℏ ≤ ℓ ⇒ η (ℏ) ≥ η (ℓ) , η (ℏ ℓ) ≥ η (ℓ) (resp. η (ℏ ℓ) ≥ η (ℏ)) ∀ ℏ , ℓ ∈ S. If η is both fuzzy left and right ideal of S, then it is called a fuzzy ideal of S. Fuzzy subset η is a strongly convex fuzzy subset ⇔ ℏ ≤ ℓ ⇒ η (ℏ) ≥ η (ℓ) ∀ ℏ , ℓ ∈ S . A fuzzy subsemigroup η of S is called a fuzzy bi-ideal [11] of S if ℏ ≤ ℓ ⇒ η (ℏ) ≥ η (ℓ) and η (ℏ ℓ z) ≥ min {η (ℏ) , η (ı)} ∀ ℏ , ℓ , ı ∈ S.

Let a ∈ S and ω ∈ (0, 1]. An ordered fuzzy point (OFP) a_ω of S is defined by $a_{ω} (x) = {\begin{matrix} ω, & if x \in (a], \\ 0, & if x \notin (a] . \end{matrix}$ For $η \in F (S)$ , a_ω ∈ η will stand for a_ω ⊆ η. Thus a_ω ∈ η ⇔ η (a) ≥ ω.

An OFP a_ω of S, for any κ^* ∈ (0, 1], is said to be (κ^*, q)-quasi-coincident with a fuzzy subset η of S, written as a_ω (κ^*, q) η, if η (a) + ω > κ^* .

Let 0 ≤ κ < κ^* ≤ 1. For any OFP ℏ_ω, we say that

ℏ_ω (κ^*, q_κ) η if η (ℏ) + ω + κ > κ^*;

ℏ_ω ∈ ∨ (κ^*, q_κ) η if ℏ_ω ∈ η or ℏ_ω (κ^*, q_κ) η;

$ℏ_{ω} \bar{α} η$ if ℏ_ωαη does not hold for α ∈ {(κ^*, q_κ), ∈ ∨ (κ^*, q_κ)}.

$F (S) ∋ η$ is called an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S if ℏ_ω, ℓ _ɛ ∈ η ⇒ (ℏ ℓ) _min{ω,ɛ} ∈ ∨ (κ^*, q_κ) η ∀ ω, ɛ ∈ (0, 1] , ℏ , ℓ ∈ S. It is called an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy left (resp. right) ideal of S if ℏ≤ ℓ, ℓ_ω ∈ η ⇒ ℏ _ω ∈ ∨ (κ^*, q_κ) η, and ℏ ∈ S, ℓ_ω ∈ η ⇒ (ℏ ℓ) _ω ∈ ∨ (κ^*, q_κ) η (resp. (ℓ ℏ) _ω ∈ ∨ (κ^*, q_κ) η). If it is both an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy left and right ideal of S, then it is called an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy ideal. An (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S is called an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy bi-ideal if ℏ≤ ℓ, ℓ_ω ∈ η ⇒ ℏ _ω ∈ ∨ (κ^*, q_κ) η and ℏ_ω ∈ η, ı_ɛ ∈ η ⇒ (ℏ ℓ ı) _min{ω,ɛ} ∈ ∨ (κ^*, q_κ) η ∀ ω, ɛ ∈ (0, 1] , ℏ , ℓ , ı ∈ S.

Theorem 2.1. [7] Let η be a fuzzy subset of S. Then η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S if and only if $η (xy) \geq min {η (x), η (y), \frac{κ^{*} - κ}{2}}$ for all x, y ∈ S.

Theorem 2.2. [7] Let η be a fuzzy subset of S. Then η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy bi-ideal of S if and only if

x ≤ y $\Rightarrow η (x) \geq min {η (y), \frac{κ^{*} - κ}{2}}$ , and

$η (xy) \geq min {η (x), η (y), \frac{κ^{*} - κ}{2}}$ for all x, y ∈ S, and

$η (xyz) \geq min {η (x), η (z), \frac{κ^{*} - κ}{2}}$ for all x, y, z ∈ S.

3 (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals

Definition 3.1. A fuzzy subset η of S is called an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal of S if

x¹ ≤ z¹, $z_{ω_{1}}^{1} \in η \Rightarrow x_{ω_{1}}^{1} \in \lor (κ^{*}, q_{κ}) η$ ,

$x_{ω_{1}}^{1} \in η$ , $z_{ɛ_{1}}^{1} \in η \Rightarrow (x^{1} z^{1})_{min {ω_{1}, ɛ_{1}}} \in \lor (κ^{*}, q_{κ}) η$ , and

$x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η$ , $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η \Rightarrow$ (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) η,

for all ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n ∈ (0, 1] and x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S

Example 3.2. Define a binary operation ^′ · ′ and an order ^′ ≤ ′ on the set S = {ℏ , ı , 𝚥 , ℓ} as follows:

·	ℏ	ı	𝚥	ℓ
ℏ	ℏ	ℏ	ℏ	ℏ
ı	ı	ı	ı	ı
𝚥	𝚥	𝚥	𝚥	𝚥
ℓ	ℏ	ℏ	ı	ℏ

\leq : = {(ℏ, ℏ), (ı, ı), (𝚥, 𝚥), (ℓ, ℓ), (ℏ, ı), (ℏ, ℓ)} .

Then S is an ordered semigroup. Now define a fuzzy subset η of S as follows:

η (a) = {\begin{matrix} 0.4 & if a \in {ℏ, ı}, \\ 0 & if a \in {𝚥, ℓ} . \end{matrix}

Take κ^* = 0.9 and κ = 0.1. It is easy to verify that η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal of S for each

m, n \in ℤ^{+}

Throughout the paper, the set of all (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals and fuzzy bi-ideals of S will be denoted by $\hat{F_{(m, n)}}$ and $\hat{F_{B}}$ .

Lemma 3.3. Let $η \in F (S)$ . Then $η \in \hat{F_{B}} \Rightarrow η \in \hat{F_{(m, n)}}$ $\forall m, n \in ℤ^{+}$ .

Proof. Let $η \in \hat{F_{B}}$ . Then, first two conditions of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal hold. Suppose that $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η$ , $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η$ for any x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S and ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n ∈ (0, 1]. By Theorem 2.2, we have η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) $\geq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}} .$

Since $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η$ and $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η$ , we have η (x¹) ≥ ω₁, η (x²) ≥ ω₂, …, η (x^m) ≥ ω_m and η (z¹) ≥ ɛ₁, η (z²) ≥ ɛ₂, …, η (zⁿ) ≥ ɛ_n. So η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) ≥ min {η (x¹) , η (x²) , … , η (x^m) , η (z¹) , η (z²) , … $, η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}, \frac{κ^{*} - κ}{2}} .$ If $min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} \leq \frac{κ^{*} - κ}{2}$ , then $\begin{matrix} η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) \\ \geq min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} . \end{matrix}$ So, (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ η. If $min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} > \frac{κ^{*} - κ}{2}$ , then $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) \geq \frac{κ^{*} - κ}{2} .$ So η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) + min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, $ɛ_{n}} > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ,$ which implies that (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} (κ^*, q_κ) η. So (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) η. Hence $η \in \hat{F_{(m, n)}}$ . □

Remark 3.4. In general, $η \in \hat{F_{(m, n)}} ≱ η \in \hat{F_{B}}$ .

Example 3.5. Let S be an ordered semigroup of Example 3.2. Define a fuzzy subset η of S as follows: $η (x) = {\begin{matrix} 0.3 & if x \in {ℏ, ℓ}, \\ 0 & if x \in {ı, 𝚥} . \end{matrix}$ Take κ^* = 0.7 and κ = 0.1. Then $η \in \hat{F_{(m, n)}}$ ∀ m, n ≥ 2, but $η \notin \hat{F_{B}}$ because $0 = η (ı) = η (ℓ 𝚥 ℓ) ≱ min {η (ℓ), η (ℓ), \frac{κ^{*} - κ}{2}} = 0.3 .$

Theorem 3.6. Let $η \in F (S)$ . Then $η \in \hat{F_{(m, n)}}$ ⇔

$x^{1} \leq z^{1} \Rightarrow η (x^{1}) \geq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ ,

$η (x^{1} z^{1}) \geq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ ∀ x¹, z¹ ∈ S, and

η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) ≥ min {η (x¹) , η (x²) , …, η (x^m) , η (z¹) , η (z²) , …, η (zⁿ) , $\frac{κ^{*} - κ}{2}}$ ∀ x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S.

Proof. (⇒) Let x¹, z¹ ∈ S such that x¹ ≤ z¹. Now on contrary suppose that $η (x^{1}) < min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ for some x¹, z¹ ∈ S. So ∃ ω ∈ (0, 1] such that $η (x^{1}) < ω \leq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ . Then $z_{ω}^{1} \in η$ but $x_{ω}^{1} \bar{\in \lor (κ^{*}, q_{κ})} η$ , which is a contradiction. Hence $η (x^{1}) \geq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ . Thus (1) holds.

Next, suppose to the contrary that $η (x^{1} z^{1}) < min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ for some x¹, z¹ ∈ S. So ∃ ω ∈ (0, 1] such that $η (x^{1} z^{1}) < ω \leq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ . Then $x_{ω}^{1} \in η$ and $z_{ω}^{1} \in η$ , but $(x^{1} z^{1})_{ω} \bar{\in \lor (κ^{*}, q_{κ})} η$ , which is a contradiction. Thus $η (x^{1} z^{1}) \geq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ which proves (2).

Finally, to show that (3) holds, suppose to the contrary that $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) < min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ for some x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S. So ∃ ω ∈ (0, 1] such that η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) < ω and $ω \leq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ . Then $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η$ and $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η$ , but $(x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n})_{ω} \bar{\in \lor (κ^{*}, q_{κ})} η$ , which is a contradiction. Hence η (x¹x²⋯ x^myz¹z² ⋯ zⁿ) ≥ min {η (x¹) , η (x²) , …, η (x^m) , η (z¹) , η (z²) , … $, η (z^{n}), \frac{κ^{*} - κ}{2}} .$ Thus (3) holds.

(⇐) Assume that the conditions (1), (2) and (3) hold. By (1), we have $η (x^{1}) \geq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ for any x¹, z¹ ∈ S with x¹ ≤ z¹. Let $z_{ω}^{1} \in η$ (ω ∈ (0, 1]). Then η (z¹) ≥ ω. Thus $η (x^{1}) \geq min {ω, \frac{κ^{*} - κ}{2}}$ . Now, if $ω \leq \frac{κ^{*} - κ}{2}$ , then η (x¹) ≥ ω. So $x_{ω}^{1} \in η$ . Again, if $ω > \frac{κ^{*} - κ}{2}$ , then $η (x^{1}) \geq \frac{κ^{*} - κ}{2}$ . So $η (x^{1}) + ω > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ$ , which implies that $x_{ω}^{1} (κ^{*}, q_{κ}) η$ . Hence $x_{ω}^{1} \in \lor (κ^{*}, q_{κ}) η$ .

Next, assume that $x_{ω_{1}}^{1} \in η$ and $z_{ɛ_{1}}^{1} \in η$ (ω₁, ɛ₁ ∈ (0, 1]). Then η (x¹) ≥ ω₁ and η (z¹) ≥ ɛ₁. Now, as $η (x^{1} z^{1}) \geq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ , we have $η (x^{1} z^{1}) \geq min {ω_{1}, ɛ_{1}, \frac{κ^{*} - κ}{2}}$ . If $min {ω_{1}, ɛ_{1}} \leq \frac{κ^{*} - κ}{2}$ , then η (x¹z¹) ≥ min {ω, ɛ} implies that (x¹z¹) _{min{ω₁,ɛ₁}} ∈ η . In the other case when $min {ω_{1}, ɛ_{1}} > \frac{κ^{*} - κ}{2}$ , we have $η (x^{1} z^{1}) \geq \frac{κ^{*} - κ}{2}$ . So $η (x^{1} z^{1}) + min {ω_{1}, ɛ_{1}} > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ$ implies (x¹z¹) _min{ω,ɛ} (κ^*, q_κ) η. Thus (x¹z¹) _min{ω,ɛ} ∈ ∨ (κ^*, q_κ) η.

Finally, take any x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S and ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n in (0, 1] such that $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η$ , $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η$ . By (3), η (x¹x² ⋯ x^myz¹z² $\dots z^{n}) \geq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ . As $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η$ and $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η$ for all ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n ∈ (0, 1], we have η (x¹) ≥ω₁, η (x²) ≥ ω₂, …, η (x^m) ≥ ω_m and η (z¹) ≥ ɛ₁, η (z²) ≥ ɛ₂, …, η (zⁿ) ≥ ɛ_n. So η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) $\geq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}} \geq min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}, \frac{κ^{*} - κ}{2}} .$ If $min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} \leq \frac{κ^{*} - κ}{2}$ , then η (x¹x² ⋯ x^myz¹z², ⋯ zⁿ) ≥ min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n} implies that $(x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n})_{min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}}} \in η .$ Again, if $min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} > \frac{κ^{*} - κ}{2}$ , then $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) \geq \frac{κ^{*} - κ}{2} .$ So $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) + min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ$ , which implies that (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} (κ^*, q_κ) η . Therefore (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) η. Hence $η \in \hat{F_{(m, n)}}$ .□

Definition 3.7. S is said to be (m, n)-regular if ∀ a ∈ S ∃ x ∈ S such that a ≤ a^mxaⁿ.

Lemma 3.8 If S is (m, n)-regular, then $η \in \hat{F_{(m, n)}} \Rightarrow η \in \hat{F_{B}}$ .

Proof. Suppose that $η \in \hat{F_{(m, n)}}$ and x, a, y ∈ S . Since S is (m, n)-regular, there exist p, q ∈ S such that xay ≤ x^mpxⁿay^mqyⁿ. Now η (xay) = η (x^mpxⁿay^mqyⁿ) = η (x^m (pxⁿay^mq) yⁿ) $\geq min {η (x), η (x), \dots, η (x), η (y), η (y), \dots, η (y), \frac{κ^{*} - κ}{2}}$ $= min {η (x), η (y), \frac{κ^{*} - κ}{2}} .$

Hence by Theorem 2.2, $η \in \hat{F_{B}}$ .□

Lemma 3.9 Let ${η_{i} \in \hat{F_{(m, n)}} | i \in I}$ . Then $⋂_{i \in I} η_{i} \in \hat{F_{(m, n)}}$ .

Proof. Take any x, y ∈ S such that x ≤ y and y_ω ∈ ⋂ _i∈Iη_i. Then y_ω ∈ η_i for each i ∈ I. As each $η_{i} \in \hat{F_{(m, n)}}$ , x_ω ∈ η_i for each i ∈ I. Thus x_ω ∈ ⋂ _i∈Iη_i.

Next take any x, y ∈ S and ω, ɛ ∈ (0, 1] such that x_ω, y_ɛ ∈ ⋂ _i∈Iη_i. Then x_ω, y_ɛ ∈ η_i for each i ∈ I. So η_i (x) ≥ ω, η_i (y) ≥ ɛ. Thus, we have ⋂_i∈Iη_i (xy) = ⋀ _i∈Iη_i (xy) ≥ ⋀ _i∈I min {η_i (x) , η_i (y)} ≥ min {ω, ɛ} . So (xy) _min{ω,ɛ} ∈ ⋂ _i∈Iη_i (xy).

Finally, take any $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in ⋂_{i \in I} η_{i}$ and $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in ⋂_{i \in I} η_{i}$ for each x₁, x₂, ⋯ , x_m, z, y₁, y₂, ⋯ , y_n ∈ S and ω₁, ω₂, ⋯ , ω_m, ɛ₁, ɛ₂, ⋯ , ɛ_n ∈ (0, 1]. Therefore $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η_{i}$ and $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η_{i}$ for i ∈ I. As each $η_{i} \in \hat{F_{(m, n)}}$ , so (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) η_i ∀ i ∈ I. Thus (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ η_i or η_i (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) + min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n} + κ ≥ κ^*. If (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,} ɛ₂, …, ɛ_n} ∈ η_i, then

⋂_i∈Iη_i (x₁x₂ ⋯ x_mzy₁y₂ ⋯ y_n) = ⋀ _i∈Iη_i (x₁x₂ ⋯ x_mzy₁y₂ ⋯ y_n) ≥ ⋀ _i∈I min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n} = min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n} .

Therefore (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ⋂ _i∈Iη_i, which implies that (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) ⋂ _i∈Iη_i. Similarly, if η_i (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) + min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n} + κ ≥ κ^*, then (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) ⋂ _i∈Iη_i . Hence $⋂_{i \in I} η_{i} \in \hat{F_{(m, n)}}$ . □

Remark 3.10. Let ${η_{i} \in \hat{F_{(m, n)}} | i \in I}$ . Then $⋃_{i \in I} η_{i} \notin \hat{F_{(m, n)}}$ in general. The following example validates the above claim:

Example 3.11. Let S = {ℏ , ı , 𝚥 , ℓ}. Define a binary operation ^′ · ′ and an order ≤ on S as follows:

·	ℏ	ı	𝚥	ℓ
ℏ	ℏ	ℏ	ℏ	ℏ
ı	ℏ	ℏ	ℓ	ℏ
𝚥	ℏ	ℏ	ℏ	ℏ
ℓ	ℏ	ℏ	ℏ	ℏ

\leq : = {(ℏ, ℏ), (ı, ı), (𝚥, 𝚥), (ℓ, ℓ), (ℏ, ı)} .

Then S is an ordered semigroup. Define $η_{1}, η_{2} \in F (S)$ by $\begin{matrix} η_{1} (ℏ) & = 0.4, η_{1} (ı) = 0.4, η_{1} (𝚥) = 0, η_{1} (ℓ) = 0; \\ η_{2} (ℏ) & = 0.4, η_{2} (ı) = 0, η_{2} (𝚥) = 0.4, η_{2} (ℓ) = 0 . \end{matrix}$ Then $η_{1}, η_{2} \in \hat{F_{(m, n)}}$ , but $η_{1} \cup η_{2} \notin \hat{F_{(m, n)}}$ because $0 = η_{1} (ℓ) \lor η_{1} (ℓ) = (η_{1} \cup η_{2}) (ℓ) = (η_{1} \cup η_{2}) (ı 𝚥) < \min {(η_{1} \cup η_{2}) (ı), (η_{1} \cup η_{2}) (𝚥), \frac{κ^{*} - κ}{2}}$ .

Definition 3.12. [9] Let $η \in F (S)$ . Then the set $[η]_{ω} = {x \in S | x_{ω} \in \lor (κ^{*}, q_{κ}) η},$ where ω ∈ (0, 1] , is called an (∈ ∨ (κ^*, q_κ))-level subset of η.

Theorem 3.13. Let η be any strongly convex fuzzy subset of S. Then $η \in \hat{F_{(m, n)}}$ ⇔ the (∈ ∨ (κ^*, q_κ))-level subset ${[η]}_{ω} \in \hat{I_{(m, n)}}$ , ∀ ω ∈ (0, 1].

Proof. (⇒) Take any x ∈ S and y ∈ [η] _ω such that x ≤ y. As y ∈ [η] _ω, we have y_ω ∈ ∨ (κ^*, q_κ) η; that is η (y) ≥ ω or η (y) + ω + κ > κ^*. As η is strongly convex, η (x) ≥ η (y) ≥ ω or η (x) ≥ η (y) ≥ κ^* - ω - κ; that is x_ω ∈ ∨ (κ^*, q_κ) η. So x ∈ [η] _ω.

Next, take any x¹, x², …, x^m, z¹, z², …, zⁿ ∈ [η] _ω for ω ∈ (0, 1]. Then $x_{ω}^{i} \in \lor (κ^{*}, q_{κ}) η$ and $z_{ω}^{j} \in \lor (κ^{*}, q_{κ}) η$ ; that is η (xⁱ) ≥ ω or η (xⁱ) + ω + κ > κ^* and η (z^j) ≥ ω or η (z^j) + ω + κ > κ^* for each i ∈ {1, 2, …, m} and j ∈ {1, 2, …, n}. Since $η \in \hat{F_{(m, n)}}$ , by Theorem 3.6, $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) \geq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ .

Case (1). Let η (x^ı) ≥ ω, η (z^𝚥) ≥ ω for each ı ∈ {1, 2, . . , m} , 𝚥 ∈ {1, 2, …, n}. If $ω > \frac{κ^{*} - κ}{2}$ , then η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) $\geq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {ω, ω, . . ., ω, ω, ω, \dots, ω, \frac{κ^{*} - κ}{2}}$ $= \frac{κ^{*} - κ}{2},$ and, thus, (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω (κ^*, q_κ) η. If $ω \leq \frac{κ^{*} - κ}{2}$ , then η (xy) ≥ min {η (x¹) , η (x²) , …, η (x^m) , η (z¹) , η (z²) , …, $η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {ω, ω, . ., ω, ω, ω, \dots, ω, \frac{κ^{*} - κ}{2}}$ = ω and, so, (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω ∈ η. Hence (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω ∈ ∨ (κ^*, q_κ) η.

Case (2). Let η (x^ı) ≥ ω for each ı ∈ {1, 2, . . , m}, η (z^𝚥) ≥ ω for each 𝚥 ∈ {1, 2, . . , n - 1}, and η (zⁿ) + ω + κ > κ^*. If $ω > \frac{κ^{*} - κ}{2}$ , then

η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) ≥ min {η (x¹) , η (x²) , …, η (x^m) , η (z¹) , η (z²) , $\dots η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq {ω, ω, \dots, ω, ω, ω, \dots ω, η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq {ω, η (z^{n}), \frac{κ^{*} - κ}{2}}$ $> min {ω, (κ^{*} - ω - κ), \frac{κ^{*} - κ}{2}}$ = κ^* - ω - κ ; i.e., η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) + ω + κ > κ^*, and, so, (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω (κ^*, q_κ) η. Again, if $ω \leq \frac{κ^{*} - κ}{2}$ , then

η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) ≥ min {η (x¹) , η (x²) , …, η (x^m) , η (z¹) , η (z²) , …, $η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq {ω, ω, \dots, ω, ω, ω, \dots ω, η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq {ω, η (z^{n}), \frac{κ^{*} - κ}{2}}$ $> min {ω, (κ^{*} - ω - κ), \frac{κ^{*} - κ}{2}} = ω$ and, thus, (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω ∈ η. Hence (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω ∈ ∨ (κ^*, q_κ) η.

Case (3). Let η (x^ı) ≥ ω for each ı ∈ {1, 2, …, m}, η (z^𝚥) ≥ ω for each 𝚥 ∈ {1, 2, …, n - 2}, and η (z^κ) + ω + κ > κ^* for κ ∈ {n - 1, n}. In this case, the proof is similar to the proof of case (2).

Continuing in this way, we may show that (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω ∈ ∨ (κ^*, q_κ) η.

Thus in each case, we have (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _ω ∈ ∨ (κ^*, q_κ) η, and so x¹x² ⋯ x^myz¹ z² ⋯ zⁿ ∈ [η] _ω. Hence $[η]_{ω} \in \hat{I_{(m, n)}}$ . (⇐) Let x¹, z¹ ∈ S such that x¹ ≤ z¹. Suppose to the contrary that $η (x^{1}) < min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ . Then $η (x^{1}) < ω \leq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ , for some $ω \in (0, \frac{κ^{*} - κ}{2}]$ , implies that z¹ ∈ [η] _ω, but x¹ ∉ [η] _ω, a contradiction. Thus $η (x^{1}) \geq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ .

Next, we assume that if, $η (x^{1} z^{1}) < min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ for some x¹, y¹ ∈ S. Then $\exists ω \in (0, \frac{κ^{*} - κ}{2}]$ such that $η (x^{1} z^{1}) < ω \leq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ . Now x¹, z¹ ∈ [η] _ω which implies that x¹z¹ ∈ [η] _ω. Then η (x¹z¹) ≥ ω or η (x¹z¹) + ω + κ > κ^* which is not possible. Therefore $η (x^{1} z^{1}) \geq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ for all x¹, z¹ ∈ S. Using the similar argument, we get $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) \geq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ for all x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S. Hence by Theorem 3.6, $η \in \hat{F_{(m, n)}}$ .□

Theorem 3.14. Let $B \in \hat{I_{(m, n)}}$ and let η be defined by $η (b) = {\begin{matrix} \frac{κ^{*} - κ}{2} & if b \in B, \\ 0 & if b \notin B . \end{matrix}$ Then

η is an ((k^*, q) , ∈ ∨ (κ^*, q_κ))-fuzzy (m,n)-ideal of S; and

η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m,n)-ideal of S.

Proof. (1). Let x¹, z¹ ∈ S with x¹ ≤ z¹ and let ω ∈ (0, 1] be such that $z_{ω}^{1} (k^{*}, q) η$ . Then z¹ ∈ B, η (z¹) + ω > k^*. Since $B \in \hat{I_{(m, n)}}$ and x¹ ≤ z¹ ∈ B, we have x¹ ∈ B. Thus $η (x^{1}) \geq \frac{κ^{*} - κ}{2}$ . If $ω \leq \frac{κ^{*} - κ}{2}$ , then η (x¹) ≥ ω. So $x_{ω}^{1} \in η$ . If $ω > \frac{κ^{*} - κ}{2}$ , then $η (x^{1}) + ω > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ$ . Therefore $x_{ω}^{1} (κ^{*}, q_{κ}) η$ .

Next, we assume that x¹, z¹ ∈ S and ω₁, ɛ₁ ∈ (0, 1] be such that $x_{ω_{1}}^{1} (k^{*}, q) η$ and $z_{ɛ_{1}}^{1} (k^{*}, q) η$ . Then x¹, z¹ ∈ B, η (x¹) + ω₁ > k^* and η (z¹) + ɛ₁ > k^*. As $B \in \hat{I_{(m, n)}}$ , we have x¹z¹ ∈ B. Thus $η (x^{1} z^{1}) \geq \frac{κ^{*} - κ}{2}$ . If $min {ω_{1}, ɛ_{1}} \leq \frac{κ^{*} - κ}{2}$ , then η (x¹z¹) ≥ min {ω₁, ɛ₁} and, so, (x¹z¹) _{min{ω₁,ɛ₁}} ∈ η. Again, if $min {ω_{1}, ɛ_{1}} > \frac{κ^{*} - κ}{2}$ , then

$η (x^{1} z^{1}) + min {ω_{1}, ɛ_{1}} > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ .$

So (x¹z¹) _{min{ω₁,ɛ₁}} (κ^*, q_κ) η. Therefore (x¹z¹) _{min{ω₁,ɛ₁}} ∈ ∨ (κ^*, q_κ) η .

Finally, assume that x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S and ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n ∈ (0, 1] be such that $x_{ω_{1}}^{1} (k^{*}, q) η, x_{ω_{2}}^{2} (k^{*}, q) η, \dots, x_{ω_{m}}^{m} (k^{*}, q) η$ and $z_{ɛ_{1}}^{1} (k^{*}, q) η,$ $z_{ɛ_{2}}^{2} (k^{*}, q) η, \dots, z_{ɛ_{n}}^{n} (k^{*}, q) η$ . Then x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ B, η (x¹) + ω₁ > k^*, η (x²) + ω₂ > k^*, …, η (x^m) + ω_m > k^* and η (z¹) + ɛ₁ > k^*, η (z²) + ɛ₂ > k^*, …, η (zⁿ) + ɛ_n > k^*. As B is an (m, n)-ideal, x¹x² ⋯ x^myz¹z² ⋯ zⁿ ∈ B. Thus $η (x^{1} x^{2} \dots x^{m} y z^{1} z^{2} \dots z^{n}) \geq \frac{κ^{*} - κ}{2}$ . If $min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} \leq \frac{κ^{*} - κ}{2}$ , then η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) ≥ min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n} , and, so, (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ η. Again, if min {ω₁, ω₂, …, ω_m, $ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} > \frac{κ^{*} - κ}{2}$ , then $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) + min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ .$ Thus (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) η. Hence η is a ((k^*, q) , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal of S.

(2). Let x¹, z¹ ∈ S with x¹ ≤ z¹ and let u ∈ (0, 1] be such that $z_{ω}^{1} \in η$ . Then z¹ ∈ B and η (z¹) ≥ ω. As $B \in \hat{I_{(m, n)}}$ and x¹ ≤ z¹ ∈ B, we have x¹ ∈ B. Thus $η (x^{1}) \geq \frac{κ^{*} - κ}{2}$ . If $ω \leq \frac{κ^{*} - κ}{2}$ , then η (x¹) ≥ ω. So $x_{ω}^{1} \in η$ . If $ω > \frac{κ^{*} - κ}{2}$ , then $η (x^{1}) + ω > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ$ . So $x_{ω}^{1} (κ^{*}, q_{κ}) η$ . Therefore $x_{ω}^{1} \in \lor (κ^{*}, q_{κ}) η$ .

Next, assume that x¹, z¹ ∈ S and ω₁, ɛ₁ ∈ (0, 1] be such that $x_{ω_{1}}^{1} \in η$ and $z_{ɛ_{1}}^{1} \in η$ . Then x¹, z¹ ∈ B, η (x¹) > ω₁ and η (z¹) > ɛ₁. Since $B \in \hat{I_{(m, n)}}$ , x¹z¹ ∈ B. Thus $η (x^{1} z^{1}) \geq \frac{κ^{*} - κ}{2}$ . If $min {ω_{1}, ɛ_{1}} \leq \frac{κ^{*} - κ}{2}$ , then η (x¹z¹) ≥ min {ω₁, ɛ₁} and, so, (x¹z¹) _{min{ω₁,ɛ₁}} ∈ η. Again, if $min {ω_{1}, ɛ_{1}} > \frac{κ^{*} - κ}{2}$ , then $η (x^{1} z^{1}) + min {ω_{1}, ɛ_{1}} > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ .$ So (x¹z¹) _{min{ω₁,ɛ₁}} (κ^*, q_κ) η. Hence (x¹z¹) _{min{ω₁,ɛ₁}} ∈ ∨ (κ^*, q_κ) η.

Finally, assume that x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S and ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n ∈ (0, 1] be such that $x_{ω_{1}}^{1}, x_{ω_{2}}^{2}, \dots, x_{ω_{m}}^{m} \in η$ and $z_{ɛ_{1}}^{1}, z_{ɛ_{2}}^{2}, \dots, z_{ɛ_{n}}^{n} \in η$ . Then x¹, x², …, x^m, z¹, z², …, zⁿ ∈ B, η (x¹) > ω₁, η (x²) > ω₂, …, η (x^m) > ω_m and η (z¹) > ɛ₁, η (z²) > ɛ₂, …, η (zⁿ) > ɛ_n. As $B \in \hat{I_{(m, n)}}$ , x¹x² ⋯ x^myz¹z² ⋯ zⁿ ∈ B. Thus $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) \geq \frac{κ^{*} - κ}{2}$ . If $min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} \leq \frac{κ^{*} - κ}{2}$ , then η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) ≥ min {ω₁, ω₂, …, ω_m, ɛ₁, ɛ₂, …, ɛ_n} . Therefore (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁}, ɛ₂, …, ɛ_n} ∈ η . Again, if $min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} > \frac{κ^{*} - κ}{2}$ , then $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) + min {ω_{1}, ω_{2}, \dots, ω_{m}, ɛ_{1}, ɛ_{2}, \dots, ɛ_{n}} > \frac{κ^{*} - κ}{2} + \frac{κ^{*} - κ}{2} = κ^{*} - κ .$ Therefore (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) _{min{ω₁,ω₂,…,ω_m,ɛ₁,ɛ₂,…,ɛ_n}} ∈ ∨ (κ^*, q_κ) η, as required. □

Theorem 3.15. Let (∅ ≠) B ⊆ S. Then $B \in \hat{I_{(m, n)}}$ ⇔ $η_{B} \in F_{(m, n)}$ , where $η_{B} (b) = {\begin{matrix} \frac{κ^{*} - κ}{2} & if b \in B, \\ 0 & if b \notin B . \end{matrix}$ .

Proof. (⇒) Let $B \in \hat{I_{(m, n)}}$ . Then, by Theorem 3.14, $η_{B} \in \hat{F_{(m, n)}}$ .

(⇐) Let x, y ∈ S with x ≤ y. If y ∈ B, then $η_{B} (y) = \frac{κ^{*} - κ}{2}$ . Since $η_{B} \in \hat{F_{(m, n)}}$ and x ≤ y, we have $η_{B} (x) \geq min {η_{B} (y), \frac{κ^{*} - κ}{2}} = \frac{κ^{*} - κ}{2}$ . Thus $η_{B} (x) = \frac{κ^{*} - κ}{2}$ , and, so, x ∈ B.

Next, we show that B is subsemigroup of S. To show this, take any x, y ∈ B. Then $η_{B} (x) = \frac{κ^{*} - κ}{2}$ and $η_{B} (y) = \frac{κ^{*} - κ}{2}$ . Now $η_{B} (xy) \geq min {η_{B} (x), η_{B} (u), \frac{κ^{*} - κ}{2}} = \frac{κ^{*} - κ}{2} .$ Therefore $η_{B} (xy) = \frac{κ^{*} - κ}{2}$ and, so, xy ∈ B. Thus B is a subsemigroup of S.

Finally, take any x₁, x₂, …, x_m, z₁, z₂, …, z_n ∈ B and y ∈ S. Then $η_{B} (x_{1}) = η_{B} (x_{2}) = \dots = η_{B} (x_{m}) = η_{B} (z_{1}) = η_{B} (z_{2}) = \dots = η_{B} (z_{n}) = \frac{κ^{*} - κ}{2}$ . Now

η_B (x₁x₂ ⋯ x_myz₁z₂ ⋯ z_n) ≥ min {η_B (x₁) , η_B (x₂) , …, η_B (x_m) , η_B (z₁) , $η_{B} (z_{2}), \dots, η_{B} (z_{n}), \frac{κ^{*} - κ}{2}}$ $= \frac{κ^{*} - κ}{2} .$ Therefore $η_{B} (x_{1} x_{2} \dots x_{m} {yz}_{1} z_{2} \dots z_{n}) = \frac{κ^{*} - κ}{2}$ and, so, x₁x₂ ⋯ x_myz₁z₂ ⋯ z_n ∈ B. Hence $B \in \hat{I_{(m, n)}}$ . □

Theorem 3.16. Let $η \in F (S)$ . Then $η \in \hat{F_{(m, n)}}$ ⇔ the level subset U (η ; ω) of η which is defined as $U (η; ω) = {x \in S | η (x) \geq ω},$ for $ω \in (0, \frac{κ^{*} - κ}{2}]$ is an (m,n)-ideal of S.

Proof. (⇒) Suppose that x, y ∈ S such that x ≤ y ∈ U (η ; ω), where $ω \in (0, \frac{κ^{*} - κ}{2}]$ . Then η (y) ≥ ω. By Theorem 3.6, $η (x) \geq min {η (y), \frac{κ^{*} - κ}{2}} \geq min {ω, \frac{κ^{*} - κ}{2}} = ω$ . Therefore x ∈ U (η ; ω).

Next, take any x, y ∈ U (η ; ω). Then η (x) ≥ ω and η (y) ≥ ω. So, by Theorem 3.6, $η (xy) \geq min {η (x), η (y), \frac{κ^{*} - κ}{2}} \geq min {ω, \frac{κ^{*} - κ}{2}} = ω .$ Therefore xy ∈ U (η ; ω). Thus U (η ; ω) is a subsemigroup of S.

Finally, take any x¹, x², …, x^m, z¹, z², …, zⁿ ∈ U (η ; ω) and y ∈ S. Then min {η (x¹) , η (x²) , …, η (x^m) , η (z¹) , η (z²) , …, η (zⁿ)} ≥ ω. Now, by Theorem 3.6, we have

η (x¹x² ⋯ x^myz¹z² ⋯ zⁿ) $\geq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {ω, \frac{κ^{*} - κ}{2}} = ω .$

Thus x¹x² ⋯ x^myz¹z² ⋯ zⁿ ∈ U (η ; ω). Hence $U (η; ω) \in \hat{I_{(m, n)}}$ .

(⇐) Let x¹, z¹ ∈ S be such that x¹ ≤ z¹. Suppose to the contrary that $η (x^{1}) < min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ . Then $η (x^{1}) < ω \leq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ for some $ω \in (0, \frac{κ^{*} - κ}{2}]$ . This implies that z¹ is in U (η ; ω), but x¹ ∉ U (η ; ω), a contradiction. Thus $η (x^{1}) \geq min {η (z^{1}), \frac{κ^{*} - κ}{2}}$ .

Suppose next to the contrary that $η (x^{1} z^{1}) < min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ for some x¹, z¹ ∈ S. Then there exist $ω \in (0, \frac{κ^{*} - κ}{2}]$ such that $η (x^{1} z^{1}) < ω \leq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ . This implies that x¹ ∈ U (η ; ω) and z¹ ∈ U (η ; ω), but x¹z¹ ∉ U (η ; ω), a contradiction. Therefore $η (x^{1} z^{1}) \geq min {η (x^{1}), η (z^{1}), \frac{κ^{*} - κ}{2}}$ for all x¹, z¹ ∈ S.

Finally, assume to the contrary that $η (x^{1} x^{2} \dots x^{m} y z^{1} z^{2} \dots z^{n}) < min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ for some x¹, x², … x^m, y, z¹, z², …, zⁿ ∈ S. Then $\exists t \in (0, \frac{κ^{*} - κ}{2}]$ such that $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) < t \leq min {η (x^{1}), η (x^{2}), \dots, η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ . Thus x¹, x², …, x^m ∈ U (η ; ω) and z¹, z², …, zⁿ ∈ U (η ; ω), but x¹x² ⋯ x^myz¹z² ⋯ zⁿ ∉ U (η ; ω), which is a contradiction. Therefore $η (x^{1} x^{2} \dots x^{m} {yz}^{1} z^{2} \dots z^{n}) \geq min {η (x^{1}), η (x^{2}), \dots η (x^{m}), η (z^{1}), η (z^{2}), \dots, η (z^{n}), \frac{κ^{*} - κ}{2}}$ for all x¹, x², …, x^m, y, z¹, z², …, zⁿ ∈ S. Hence, by Theorem 3.6, $η \in \hat{F_{(m, n)}}$ . □

Example 3.17. Define a binary operation ^′ · ′ and an order ^′ ≤ ′ on the set S = {0, ℏ , ℓ , 𝚥} as follows:

·	ℏ	ℓ	𝚥
0	0	0	0
ℏ	ℏ	ℏ	ℏ
ℓ	ℏ	ℏ	ℏ
𝚥	ℏ	ℏ	ℏ

≤ : = {(0, 0) , (ℏ , ℏ) , (ℓ , ℓ) , (0, ℏ) , (ℏ , ℓ) , (ℓ , 𝚥)} Then (S, · , ≤) is an ordered semigroup. Now define a fuzzy set η of S as: η (0) =0.5, η (ℏ) =0.4, η (ℓ) =0.1 and η (𝚥) =0.3. Therefore $U (η; ω) = {\begin{matrix} S, & if 0 < ω \leq 0.1; \\ {0, ℏ, 𝚥}, & if 0.1 < ω \leq 0.3; \\ {0, ℏ}, & if 0.3 < ω \leq 0.4; \\ {0}, & if 0.4 < ω \leq 0.5; \\ \emptyset, & if 0.5 < ω \leq 1 . \end{matrix}$ Since for all $ω \in (0, \frac{κ^{*} - κ}{2}]$ , where κ^* = 1 and κ = 0, U (η ; ω) is bi-ideal of S. Therefore by Theorem 3.16, η is (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy bi-ideal of S.

4 (κ^, κ)-lower parts of (∈ , ∈ ∨ (κ^, q_κ))-fuzzy (m, n)-ideals

Definition 4.1. The (κ^*, κ)-lower part $\underline{η_{κ}^{κ^{*}}}$ of fuzzy subset η is defined [7] as follows: $\underline{η_{κ}^{κ^{*}}} (x) = min {η (x), \frac{κ^{*} - κ}{2}}$ ∀ x ∈ S and 0 ≤ κ < κ^* ≤ 1. For any (∅ ≠) A ⊆ S and $η \in F (S)$ , the (κ^*, κ)-lower part of the characteristic function η_A, will be denoted by $(\underline{η_{κ}^{κ^{*}}})_{A}$ .

Definition 4.2. [7] Let $η, ξ \in F (S)$ . Define $η (\cap)_{κ}^{κ^{*}} ξ$ , $η (\cup)_{κ}^{κ^{*}} ξ$ and $η (\circ)_{κ}^{κ^{*}} ξ$ as follows: $\begin{matrix} (η (\cap)_{κ}^{κ^{*}} ξ) (x) & = min {(η \cap ξ) (x), \frac{κ^{*} - κ}{2}}; \\ (η (\cup)_{κ}^{κ^{*}} ξ) (x) & = min {(η \cup ξ) (x), \frac{κ^{*} - κ}{2}}; \\ (η (\circ)_{κ}^{κ^{*}} ξ) (x) & = min {(η \circ ξ) (x), \frac{κ^{*} - κ}{2}} \end{matrix}$ ∀ x ∈ S and 0 ≤ k < k^* ≤ 1.

Lemma 4.3. Let l ∈ Z⁺ and x ∈ S . Then $\underline{η_{κ}^{κ^{*}}} (x) \leq (\underline{η_{κ}^{κ^{*}}})^{l} (x^{l})$ for any $η \in F (S)$ .

Proof. Now

$(\underline{η_{κ}^{κ^{*}}})^{l} (x^{l})$ $= ((\underline{η_{κ}^{κ^{*}}})^{l - 1} (\circ)_{κ}^{κ^{*}} \underline{η_{κ}^{κ^{*}}}) (x^{l})$ $= min {((\underline{η_{κ}^{κ^{*}}})^{l - 1} \circ \underline{η_{κ}^{κ^{*}}}) (x^{l}), \frac{κ^{*} - κ}{2}}$ $= min {⋁_{(u, v) \in A_{x^{l}}} min {(\underline{η_{κ}^{κ^{*}}})^{l - 1} (u), \underline{η_{κ}^{κ^{*}}} (v)}, \frac{κ^{*} - κ}{2}}$ $\geq min {min {(\underline{η_{κ}^{κ^{*}}})^{l - 1} (x^{l - 1}), \underline{η_{κ}^{κ^{*}}} (x)}, \frac{κ^{*} - κ}{2}}$ $\geq min {min {min {min {(\underline{η_{κ}^{κ^{*}}})^{l - 2} (x^{l - 2}), \underline{η_{κ}^{κ^{*}}} (x)}, \frac{κ^{*} - κ}{2}}, \underline{η_{x}^{k^{*}}} (x)}, \frac{κ^{*} - κ}{2}}$ $= min {(\underline{η_{κ}^{κ^{*}}})^{l - 2} (x^{l - 2}), \underline{η_{κ}^{κ^{*}}} (x), \underline{η_{κ}^{κ^{*}}} (x), \frac{κ^{*} - κ}{2}}$ ⋮ $= min {\underline{η_{κ}^{κ^{*}}} (x), \frac{κ^{*} - κ}{2}}$ $= \underline{η_{κ}^{κ^{*}}} (x) .$ □

Lemma 4.4. Let (∅ ≠) B ⊆ S. Then $(\underline{η_{κ}^{κ^{*}}})_{B} \in \hat{F_{(m, n)}}$ ⇔ $B \in \hat{I_{(m, n)}}$ .

Proof. Follows from Theorem 3.15. □

Lemma 4.5. Let $η \in F (S)$ . Then

η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S ⇔ $η (\circ)_{κ}^{κ^{*}} η ⪯ \underline{η_{κ}^{κ^{*}}}$ .

η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S ⇔ $\underline{η_{κ}^{κ^{*}}} (\circ)_{κ}^{κ^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}}$ .

Proof. (1). (⇒) Let a ∈ S. If A_a =∅, then $η (\circ)_{κ}^{κ^{*}} η ⪯ η$ . Suppose that (u, v) ∈ A_a. Then, we have $(η (\circ)_{κ}^{κ^{*}} η) (a)$ $= min {(η \circ η) (a), \frac{κ^{*} - κ}{2}}$ $= min {⋁_{(u, v) \in A_{a}} min {η (u), η (v)}, \frac{κ^{*} - κ}{2}}$ $= min {⋁_{(u, v) \in A_{a}} {min {η (u), η (v), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}}$ $\leq ⋁_{(u, v) \in A_{a}} min {η (uv), \frac{κ^{*} - κ}{2}}$ $\leq ⋁_{(u, v) \in A_{a}} min {η (a), \frac{κ^{*} - κ}{2}}$ $\leq \underline{η_{κ}^{κ^{*}}} (a) .$

(⇐) Assume that $η (\circ)_{κ}^{κ^{*}} η ⪯ \underline{η_{κ}^{κ^{*}}}$ . Now, for any x, y ∈ S, we have

$\underline{η_{κ}^{κ^{*}}} (xy) \geq (η (\circ)_{κ}^{κ^{*}} η) (xy)$ $= min {⋁_{(u, v) \in A_{xy}} min {η (u), η (v)}, \frac{κ^{*} - κ}{2}}$ $\geq min {η (x), η (y), \frac{κ^{*} - κ}{2}} .$ Hence η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S.

(2). (⇒) By (1). we have $η (\circ)_{κ}^{κ^{*}} η ⪯ \underline{η_{κ}^{κ^{*}}}$ . Therefore $\underline{η_{κ}^{κ^{*}}} (\circ)_{κ}^{κ^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ η (\circ)_{κ}^{κ^{*}} η ⪯ \underline{η_{κ}^{κ^{*}}} .$ Thus $\underline{η_{κ}^{κ^{*}}} (\circ)_{κ}^{κ^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}} .$

(⇐) For any x, y ∈ S, we have η (xy) $\geq \underline{η_{κ}^{κ^{*}}} (xy)$ $\geq (\underline{η_{κ}^{κ^{*}}} (\circ)_{κ}^{κ^{*}} \underline{η_{κ}^{κ^{*}}}) (xy)$ $= min {⋁_{(p, q) \in A_{xy}} min {\underline{η_{κ}^{κ^{*}}} (p), \underline{η_{κ}^{κ^{*}}} (q)}, \frac{κ^{*} - κ}{2}}$ $= min {⋁_{(p, q) \in A_{xy}} min {min {η (p), \frac{κ^{*} - κ}{2}}, min {η (q), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}$ $= min {⋁_{(p, q) \in A_{xy}} min {η (p), η (q), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}$ $\geq min {η (x), η (y), \frac{κ^{*} - κ}{2}} .$

Hence η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S. □

Theorem 4.6. Let $η \in F (S)$ . Then $η \in \hat{F_{(m, n)}}$ ⇔

If x ≤ y, then $η (x) \geq min {η (y), \frac{κ^{*} - κ}{2})}$ ;

$η (\circ)_{κ}^{κ^{*}} η ⪯ η$ ; and

$η^{m} (\circ)_{κ}^{κ^{*}} 1 (\circ)_{κ}^{κ^{*}} η^{n} ⪯ \underline{η_{κ}^{κ^{*}}}$ .

Proof. (⇒) Let $η \in \hat{F_{(m, n)}}$ . Then condition (1) follows by Lemma 3.6. If A_a =∅, then $η (\circ)_{κ}^{κ^{*}} η ⪯ η$ . Suppose that (u, v) ∈ A_a. Then, we have

$(η (\circ)_{κ}^{κ^{*}} η) (a)$ $= min {(η \circ η) (a), \frac{κ^{*} - κ}{2}}$ $= min {⋁_{(u, v) \in A_{a}} min {η (u), η (v)}, \frac{κ^{*} - κ}{2}}$ $= min {⋁_{(u, v) \in A_{a}} {min {η (u), η (v), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}}$ $\leq ⋁_{(u, v) \in A_{a}} min {η (uv), \frac{κ^{*} - κ}{2}}$ ≤η (a) . Thus (2) holds.

Let a ∈ S. If $(η^{m} (\circ)_{κ}^{κ^{*}} 1 (\circ)_{κ}^{κ^{*}} η^{n}) (a) = 0$ , then (3) is obvious. So assume that $(η^{m} (\circ)_{κ}^{κ^{*}} 1 (\circ)_{κ}^{κ^{*}}$ ηⁿ) (a) ≠0. Then ∃ (x, y) ∈ A_a such that (η^m ∘ 1) (x) ≠0 and ηⁿ (y) ≠0. Now the following cases arise:

Case 1. When (η^m ∘ 1) (x) ≠0. In this case ∃ (u₁, v₁) ∈ A_x such that η^m (u₁) ≠0, ⇒ ∃ (u₂, v₂) ∈ A_{u
₁} such that η^m-1 (u₂) ≠0 and η (v₂) ≠0. ⇒ ∃ (u₃, v₃) ∈ A_{u
₁} such that η^m-2 (u₃) ≠0 and η (v₃) ≠0. ⋮ ⇒ ∃ (u_m-1, v_m-1) ∈ A_{u
_m-2} such that η² (u_m-1) = η^m-m-2 (u_m-1) ≠0 and η (v_m-1) ≠0. ⇒ ∃ (u_m, v_m) ∈ A_{u
_m-1} such that η (u_m) ≠0 and η (v_m) ≠0.

Case 2. When ηⁿ (y) ≠0. Then $\exists (u_{1}^{'}, v_{1}^{'}) \in A_{y}$ such that $η^{n - 1} (u_{1}^{'}) \neq 0$ and $η (v_{1}^{'}) \neq 0$ . ⇒ $\exists (u_{2}^{'}, v_{2}^{'}) \in A_{u_{1}^{'}}$ such that $η^{n - 2} (u_{2}^{'}) \neq 0$ and $η (v_{2}^{'}) \neq 0$ . ⇒ $\exists (u_{3}^{'}, v_{3}^{'}) \in A_{u_{2}^{'}}$ such that $η^{n - 3} (u_{3}^{'}) \neq 0$ and $η (v_{3}^{'}) \neq 0$ . ⋮ ⇒ $\exists (u_{n - 2}^{'}, v_{n - 2}^{'}) \in A_{u_{n - 3}^{'}}$ such that $η^{n - (n - 2)} (u_{n - 2}^{'}) \neq 0$ and $η (v_{n - 2}^{'}) \neq 0$ . ⇒ $\exists (u_{n - 1}^{'}, v_{n - 1}^{'}) \in A_{u_{n - 2}^{'}}$ such that $η (u_{n - 1}^{'}) \neq 0$ and $η (v_{n - 1}^{'}) \neq 0$ . So $(η^{m} (\circ)_{κ}^{κ^{*}} 1 (\circ)_{κ}^{κ^{*}} η^{n}) (a)$ $= ⋁_{(x, y) \in A_{a}} min {min {(η^{m} \circ 1) (x), \frac{κ^{*} - κ}{2}}, η^{n} (y), \frac{κ^{*} - κ}{2}}$ $= ⋁_{(x, y) \in A_{a}} min {min {⋁_{(u_{1}, v_{1}) \in A_{x}} min {η^{m} (u_{1}), 1 (v_{1}), \frac{κ^{*} - κ}{2}}, {⋁_{(u_{1}^{'}, v_{1}^{'}) \in A_{y}} min {{η^{n - 1} (u_{1}^{'}), η (v_{1}^{'})}, \frac{κ^{*} - κ}{2}}}, \frac{κ^{*} - κ}{2}}$ $= ⋁_{(x, y) \in A_{a}} ⋁_{(u_{1}, v_{1}) \in A_{x}} ⋁_{(u_{1}^{'}, v_{1}^{'}) \in A_{y}} min {η^{m} (u_{1}), 1 (v_{1}), η^{n - 1} (u_{1}^{'}), η (v_{1}^{'}), \frac{κ^{*} - κ}{2}}$ $= ⋁_{(x, y) \in A_{a}} ⋁_{(u_{1}, v_{1}) \in A_{x}} ⋁_{(u_{1}^{'}, v_{1}^{'}) \in A_{y}} min {η^{m} (u_{1}), η^{n - 1} (u_{1}^{'}), η (v_{1}^{'}), \frac{κ^{*} - κ}{2}} = ⋁_{(x, y) \in A_{a}} ⋁_{(u_{1}, v_{1}) \in A_{x}} ⋁_{(u_{1}^{'}, v_{1}^{'}) \in A_{y}} min {min {⋁_{(u_{2}, v_{2}) \in A_{u_{1}}} min {η^{m - 1} (u_{2}), η (v_{2})}, \frac{κ^{*} - κ}{2}} min {⋁_{(u_{2}^{'}, v_{2}^{'}) \in A_{u_{1}^{'}}} min {η^{n - 2} (u_{2}^{'}), η (v_{2}^{'})}, \frac{κ^{*} - κ}{2}}, η (v_{1}^{'}), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}$ $= ⋁_{(x, y) \in A_{a}} ⋁_{(u_{1}, v_{1}) \in A_{x}} ⋁_{(u_{1}^{'}, v_{1}^{'}) \in A_{y}} ⋁_{(u_{2}, v_{2}) \in A_{u_{1}}} ⋁_{(u_{2}^{'}, v_{2}^{'}) \in A_{u_{1}^{'}}} min {{η^{m - 1} (u_{2}), η (v_{2}), η^{n - 2} (u_{2}^{'}), η (v_{2}^{'}), η (v_{1}^{'}), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}$ ⋮ $= ⋁_{(x, y) \in A_{a}} ⋁_{(u_{1}, v_{1}) \in A_{x}} ⋁_{(u_{1}^{'}, v_{1}^{'}) \in A_{y}} ⋁_{(u_{2}, v_{2}) \in A_{u_{1}}} ⋁_{(u_{2}^{'}, v_{2}^{'}) \in A_{u_{1}^{'}}} \dots ⋁_{(u_{m}, v_{m}) \in A_{u_{m - 1}}} ⋁_{(u_{n - 1}^{'}, v_{n - 1}^{'}) \in A_{u_{n - 2}^{'}}} min {min {η (u_{m - 1}), η (v_{m - 1}), \dots, η (v_{2}), η (u_{n - 1}^{'}), η (v_{n - 1}^{'}), \dots, η (v_{2}^{'}), η (v_{1}^{'}), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}$ $\leq ⋁_{(x, y) \in A_{a}} min {min {η (u_{m - 1}, η (v_{m - 1}), \dots, η (v_{2}), η (v_{1}), η (u_{n - 1}^{'}) η (v_{n - 1}^{'}), \dots, η (v_{2}^{'}), η (v_{1}^{'}), \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}$ $\leq min {η (a), \frac{κ^{*} - κ}{2}} (since a \leq u_{m - 1} v_{m - 1} \dots v_{2} v_{1} u_{n - 1}^{'} v_{n - 1}^{'} \dots v_{2}^{'} v_{1}^{'})$ $= \underline{η_{κ}^{κ^{*}}} (a) .$

(⇐) Assume that (1) , (2) and (3) hold in S. By condition (2), it follows from Lemma 4.5 that η is an (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy subsemigroup of S. Therefore, by Theorem 2.1, $η (xy) \geq {η (x), η (y), \frac{k - k^{*}}{2}}$ .

Next, take any x₁, x₂, …, x_m, z, y₁, y₂, …, y_n ∈ S and let a = x₁x₂ ⋯ x_mzy₁y₂ ⋯ y_n. Now

η (x₁x₂ ⋯ x_mzy₁y₂ ⋯ y_n) = η (a) $\geq \underline{η_{κ}^{κ^{*}}} (a)$ $\geq (η^{m} \circ_{k}^{k^{*}} 1 \circ_{k}^{k^{*}} η^{n}) (a)$ $= min {⋁_{(p, q) \in A_{a}} min {(η^{m} \circ_{k}^{k^{*}} 1) (p), η^{n} (q)}, \frac{κ^{*} - κ}{2}}$ $\geq min {(η^{m} \circ_{k}^{k^{*}} 1) (x_{1} x_{2} \dots x_{m} z), η^{n} (y_{1} y_{2} \dots y_{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {min {⋁_{(u, v) \in A_{x_{1} x_{2} \dots x_{m} z}} min {η^{m} (u), 1 (u)}, \frac{κ^{*} - κ}{2}}, min {⋁_{(u^{'}, v^{'}) \in A_{y_{1} y_{2} \dots y_{n}}} min {η^{n - 1} (u^{'}), η (v^{'})}, \frac{κ^{*} - κ}{2}}, \frac{κ^{*} - κ}{2}}$ $\geq min {min {η^{m} (x_{1} x_{2} \dots x_{m}), 1 (z)}, min {η^{n - 1} (y_{1} y_{2} \dots y_{n - 1}), η (y_{n})}, \frac{κ^{*} - κ}{2}}$ $\geq min {η^{m} (x_{1} x_{2} \dots x_{m}), η^{n - 1} (y_{1} y_{2} \dots y_{n - 1}), η (y_{n})}, \frac{κ^{*} - κ}{2}}$ ⋮ $\geq min {η (x_{1}), η (x_{2}), \dots, η (x_{m}), η (y_{1}), η (y_{2}), \dots, η (y_{n}), \frac{κ^{*} - κ}{2}} .$ Hence $η \in \hat{F_{(m, n)}}$ . □

Corollary 4.7. Let $η \in F (S)$ . Then $η \in \hat{F_{(m, n)}}$ ⇔

If x ≤ y, then $η (x) \geq min {η (y), \frac{κ^{*} - κ}{2})}$ ;

$\underline{η_{κ}^{κ^{*}}} (\circ)_{κ}^{κ^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}}$ ; and

$(\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{κ}^{κ^{*}} 1 (\circ)_{κ}^{κ^{*}} (\underline{η_{κ}^{κ^{*}}})^{n} ⪯ \underline{η_{κ}^{κ^{*}}}$ .

Theorem 4.8. The following assertions hold in S:

S is (m, n)-regular ⇔ $\underline{η_{κ}^{κ^{*}}} ⪯ η^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} η^{n}$ $\forall η \in F (S)$ .

S is (m, n)-regular ⇔ $\underline{η_{κ}^{κ^{*}}} ⪯ (\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{n}$ $\forall η \in F (S)$ .

Proof. (⇒) Let a ∈ S . Then a ≤ a^mxaⁿ, it follows that (a^mx, aⁿ) ∈ A_a. Now, we have

$(η^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} η^{n}) (a)$ $= ⋁_{(r, s) \in A_{a}} min {(\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1) (r), (\underline{η_{κ}^{κ^{*}}})^{n} (s), \frac{κ^{*} - κ}{2}}$ $\geq min {((\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1) (a^{m} x), ((\underline{η_{κ}^{κ^{*}}})^{n}) (a^{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {min {⋁_{(p, q) \in A_{a^{m} x}} min {((\underline{η_{κ}^{κ^{*}}})^{m}) (p), (1) (q)}, \frac{κ^{*} - κ}{2}}, (\underline{η_{κ}^{κ^{*}}})^{n} (a^{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {min {(\underline{η_{κ}^{κ^{*}}})^{m} (a^{m}), 1 (x), \frac{κ^{*} - κ}{2}}, (\underline{η_{κ}^{κ^{*}}})^{n} (a^{n}), \frac{κ^{*} - κ}{2}}$ $= min {(\underline{η_{κ}^{κ^{*}}})^{m} (a^{m}), (\underline{η_{κ}^{κ^{*}}})^{n} (a^{n}), \frac{κ^{*} - κ}{2}}$ $\geq min {\underline{η_{κ}^{κ^{*}}} (a), \underline{η_{κ}^{κ^{*}}} (a), \frac{κ^{*} - κ}{2}} (by Lemma 4)$ $= min {\underline{η_{κ}^{κ^{*}}} (a), \frac{κ^{*} - κ}{2}}$ $= \underline{η_{κ}^{κ^{*}}} (a) .$

Therefore $\underline{η_{κ}^{κ^{*}}} ⪯ η^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} η^{n}$ .

(⇐) Take any a ∈ S. Since $(\underline{η_{κ}^{κ^{*}}})_{a}$ is fuzzy subset of S, by hypothesis, $\frac{κ^{*} - κ}{2} = (\underline{η_{κ}^{κ^{*}}})_{a} (a) = ((\underline{η_{κ}^{κ^{*}}})_{a})_{k}^{k^{*}} (a) \leq (((\underline{η_{κ}^{κ^{*}}})_{a})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} ((\underline{η_{κ}^{κ^{*}}})_{a})^{n}) (a)$ implies $(((\underline{η_{κ}^{κ^{*}}})_{a})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} ((\underline{η_{κ}^{κ^{*}}})_{a})^{n}) (a) \neq 0 .$ Therefore ∃ (x, y) ∈ A_a such that $(((\underline{η_{κ}^{κ^{*}}})_{a})^{m} (\circ)_{k}^{k^{*}} 1) (x) \neq 0$ and $(((\underline{η_{κ}^{κ^{*}}})_{a})^{n}) (y) \neq 0$ . We have following cases:

Case 1. When $(((\underline{η_{κ}^{κ^{*}}})_{a})^{m} \circ 1) (x) \neq 0$ . Then ∃ (u₁, v₁) ∈ A_x such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{m} (u_{1}) \neq 0$ . ⇒ ∃ (u₂, v₂) ∈ A_{u
₁} such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{m - 1} (u_{2}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{2}) \neq 0$ implies v₂ = a . ⇒ ∃ (u₃, v₃) ∈ A_{u
₂} such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{m - 2} (u_{3}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{3}) \neq 0$ implies v₃ = a . ⋮ ⇒ ∃ (u_m-1, v_m-1) ∈ A_{u
_m-2} such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{2} (u_{m - 1}) = ((\underline{η_{κ}^{κ^{*}}})_{a})^{m - (m - 2)} (u_{m - 1}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{m - 1}) \neq 0$ implies v_m-1 = a . ⇒ ∃ (u_m, v_m) ∈ A_{u
_m-1} such that $((\underline{η_{κ}^{κ^{*}}})_{a}) (u_{m}) \neq 0$ and $((\underline{η_{κ}^{κ^{*}}})_{a}) (v_{m}) \neq 0$ implies u_m = a and v_m = a .

Case 2. When $((\underline{η_{κ}^{κ^{*}}})_{a})^{n} (y) \neq 0$ . Then $\exists (u_{1}^{'}, v_{1}^{'}) \in A_{y}$ such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{n - 1} (u_{1}^{'}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{1}^{'}) \neq 0$ implies $v_{1}^{'} = a$ . ⇒ $\exists (u_{2}^{'}, v_{2}^{'}) \in A_{u_{1}^{'}}$ such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{n - 2} (u_{2}^{'}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{2}^{'}) \neq 0$ implies $v_{2}^{'} = a$ . ⇒ $\exists (u_{3}^{'}, v_{3}^{'}) \in A_{u_{2}^{'}}$ such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{n - 3} (u_{3}^{'}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{3}^{'}) \neq 0$ implies $v_{3}^{'} = a$ . ⋮ ⇒ $\exists (u_{n - 2}^{'}, v_{n - 2}^{'}) \in A_{u_{n - 3}^{'}}$ such that $((\underline{η_{κ}^{κ^{*}}})_{a})^{n - (n - 2)} (u_{n - 2}^{'}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{n - 2}^{'}) \neq 0$ implies $v_{n - 2}^{'} = a$ . ⇒ $\exists (u_{n - 1}^{'}, v_{n - 1}^{'}) \in A_{u_{n - 2}^{'}}$ such that $((\underline{η_{κ}^{κ^{*}}})_{a}) (u_{n - 1}^{'}) \neq 0$ and $(\underline{η_{κ}^{κ^{*}}})_{a} (v_{n - 1}^{'}) \neq 0$ implies $u_{n - 1}^{'} = a$ and $v_{n - 1}^{'} = a$ . Therefore by above two cases we have $a \leq xy \leq u_{1} v_{1} u_{1}^{'} v_{1}^{'} \dots \leq u_{m} v_{m} v_{m - 1} \dots v_{2} v_{1} u_{n - 1}^{'} v_{n - 1}^{'} v_{n - 2}^{'} \dots v_{2}^{'} v_{1}^{'} \leq aa \dots a v_{1} aa \dots a = a^{m} v_{1} a^{n}$ , as required. □

Theorem 4.9 The following assertions hold in S:

S is (m, n)-regular ⇔ $\underline{η_{κ}^{κ^{*}}} = η^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} η^{n}$ $\forall η \in \hat{F_{(m, n)}}$ .

S is (m, n)-regular ⇔ $\underline{η_{κ}^{κ^{*}}} = (\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{n}$ $\forall η \in \hat{F_{(m, n)}}$ .

Proof. (1) . (⇒) Let $η \in \hat{F_{(m, n)}}$ . Then by Theorems 4.6 and 4.8, $η^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} η^{n} ⪯ \underline{η_{κ}^{κ^{*}}}$ and $\underline{η_{κ}^{κ^{*}}} ⪯ η^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} η^{n}$ . Hence $\underline{η_{κ}^{κ^{*}}} = η^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} η^{n}$ .

(⇐) Let a ∈ S. Then $[a]_{(m, n)} \in \hat{I_{(m, n)}}$ . By hypothesis, we have $((χ_{[a]_{(m, n)}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (χ_{[a]_{(m, n)}})^{n}) (a) = (\underline{χ_{κ}^{κ^{*}}})_{[a]_{(m, n)}} (a) = \frac{κ^{*} - κ}{2}$ implies $((χ_{[a]_{(m, n)}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (χ_{[a]_{(m, n)}})^{n}) (a) \neq 0$ . Therefore ∃ (x, y) ∈ A_a such that $((χ_{[a]_{(m, n)}})^{m} (\circ)_{k}^{k^{*}} 1) (x) \neq 0$ and (χ_{[a]_(m,n)}) ⁿ (y) ≠0. Again, ∃ (u, v) ∈ A_x such that (χ_{[a]_(m,n)}) ^m (u) ≠0. Now we have (χ_{[a]_(m,n)}) ⁿ (y) ≠0 and (χ_{[a]_(m,n)}) ^m (u) ≠0, it follows that y ∈ (([a] _(m,n)) ⁿ] and u ∈ (([a] _(m,n)) ^m]. Since (x, y) ∈ A_a and (u, v) ∈ A_x, therefore a ≤ xy ≤ uvy ∈ (([a] _(m,n)) ^m] S (([a] _(m,n)) ⁿ] ⊆ (([a] _(m,n)]) ^mS (([a] _(m,n)]) ⁿ ⊆ (a^mSaⁿ]. Thus [a] _(m,n) ⊆ (a^mSaⁿ]. Since (a^mSaⁿ] ⊆ [a] _(m,n). Therefore [a] _(m,n) = (a^mSaⁿ], and hence by Theorem [1, Theorem 2.4] S is (m, n)-regular. □

Lemma 4.10. Let $m, n \in ℤ^{+}$ with m ≥ 2 or n ≥ 2. Then S is (m, n)-regular ⇔ B = (B²] $\forall B \in \hat{I_{(m, n)}}$ .

Proof. (⇒) Let $B \in \hat{I_{(m, n)}}$ . Then B = (B^mSBⁿ]. Now, we have B = (B^mSBⁿ] = (B^mS ((B^mSBⁿ]) ⁿ] ⊆ (B^mSBⁿB^mSBⁿ] = (BB] .

Also, (BB] ⊆ B. Hence B = (B²].

(⇐) Take any x ∈ S . Then, as $[x]_{m, n} \in \hat{I_{(m, n)}}$ , we have [x] _m,n = ([x] _m,n [x] _m,n] = ([x] _m,n [x] _m,n [x] _m,n] = … = (([x] _m,n) ^m+n+1] ⊆ (x^mSxⁿ] .

Since x ∈ [x] _m,n, we have x ∈ (x^mSxⁿ]. Hence S is (m, n)-regular. □

The conditions m ≥ 2 or n ≥ 2 in Lemma 4.10 is necessary, we show it by the following example:

Example 4.11. Let S = {ℏ , ı , 𝚥 , ℓ , ϑ}. Define operation ^′ · ′ and ordered ≤ on S as follows:

·	ℏ	ı	𝚥	ℓ	ϑ
ℏ	ℏ	ℏ	ℏ	ℏ	ℏ
ı	ℏ	ı	ℏ	ℓ	ℏ
𝚥	ℏ	ϑ	𝚥	𝚥	ϑ
ℓ	ℏ	ı	ℓ	ℓ	ı
ϑ	ℏ	ϑ	ℏ	𝚥	ℏ

≤ : = {(ℏ , ℏ) , (ı , ı) , (𝚥 , 𝚥) , (ℓ , ℓ) , (ϑ, ϑ) , (ℏ , ı) , (ℏ , 𝚥) , (ℏ , ℓ) , (ℏ , ϑ)} . Then S is regular ordered semigroup. It is a routine to verify that B = {ℏ , ϑ} is a bi-ideal of S, but B ≠ (B²].

Theorem 4.12. Let $m, n \in ℤ^{+}$ with m ≥ 2 or n ≥ 2. Then

S is (m, n)-regular ⇔ $\underline{η_{κ}^{κ^{*}}} = η (\circ)_{k}^{k^{*}} η$ $\forall η \in \hat{F_{(m, n)}}$ .

S is (m, n)-regular ⇔ $\underline{η_{κ}^{κ^{*}}} = \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}}$ $\forall η \in \hat{F_{(m, n)}}$ .

Proof. (1). (⇒) Let $η \in \hat{F_{(m, n)}}$ . Then by Theorem 4.9(2), $\underline{η_{κ}^{κ^{*}}} = (\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{n}$ , which implies $\underline{η_{κ}^{κ^{*}}}$ $= (\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{n}$ $= (\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} ((\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{n})^{n}$ $⪯ (\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{n} (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}})^{n}$ $= \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}}$ $⪯ η (\circ)_{k}^{k^{*}} η .$

Thus $\underline{η_{κ}^{κ^{*}}} ⪯ η (\circ)_{k}^{k^{*}} η$ . Also, $η (\circ)_{k}^{k^{*}} η ⪯ η$ because $η \in \hat{F_{(m, n)}}$ . Hence $\underline{η_{κ}^{κ^{*}}} = η (\circ)_{k}^{k^{*}} η$ .

(⇐) Let $B \in \hat{I_{(m, n)}}$ and b ∈ B. Then by hypothesis $(\underline{χ_{κ}^{κ^{*}}})_{B} (b) = (χ_{B} (\circ)_{k}^{k^{*}} χ_{B}) (b) = ((\underline{χ_{κ}^{κ^{*}}})_{(B^{2}]}) (b) .$ Since $((\underline{χ_{κ}^{κ^{*}}})_{B}) (b) = \frac{κ^{*} - κ}{2}$ , $((\underline{χ_{κ}^{κ^{*}}})_{(B^{2}]}) (b) = \frac{κ^{*} - κ}{2}$ implies b ∈ (B²]. Therefore B ⊆ (B²]. Also, (BB] ⊆ B. Therefore B = (B²], and hence by Lemma 4.10, S is (m, n)-regular. □

Theorem 4.13. Let $m, n \in ℤ^{+}$ with m ≥ 2 or n ≥ 2. Then S is (m, n)-regular ⇔ $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} = \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}}$ $\forall η, ξ \in \hat{F_{(m, n)}}$ .

Proof. (⇒) Let $η, ξ \in \hat{F_{(m, n)}}$ . Then $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} \in \hat{F_{(m, n)}}$ . Since $(\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}})_{k}^{k^{*}} = \underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}$ , by Theorem 4.12, $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} = (\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) ⪯ \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}$ . Similarly $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} ⪯ \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}}$ . Therefore, $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} .$ Now

$(\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}})^{n} = ((\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (\circ)_{k}^{k^{*}} \dots (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}})) (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} ((\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (\circ)_{k}^{k^{*}} \dots (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}})) = (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \dots (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \dots (\circ)_{k}^{k^{*}} \underline{η_{k}^{k^{*}}}) (\circ)_{k}^{k^{*}} \underline{ξ_{k}^{k^{*}}}$ $⪯ \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}$ $= \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} (\underline{ξ_{κ}^{κ^{*}}})^{m} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} (\underline{ξ_{κ}^{κ^{*}}})^{n}$ $⪯ \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}},$ and similarly $(\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (\circ)_{k}^{k^{*}} (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) ⪯ \underline{η_{κ}^{κ^{*}}}$ . Also for any x, y ∈ S such that x ≤ y implies $(\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (x) \geq min {(\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}}) (y), \frac{κ^{*} - κ}{2}}$ . Hence $\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} \in \hat{F_{(m, n)}}$ .

Similarly $\underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} \in \hat{F_{(m, n)}}$ . Therefore $\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} \cap \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} \in \hat{F_{(m, n)}}$ . Again, by Theorem 4, $\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} \cap \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} = (\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} \cap \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}})^{2} ⪯ \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} 1 (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}} .$

Similarly $\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} \cap \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{ξ_{κ}^{κ^{*}}} .$ Therefore $\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} \cap \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} .$ Hence $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} = \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}}$ .

Converse is obvious. □

Lemma 4.14. Let $m, n \in ℤ^{+}$ with m ≥ 2 or n ≥ 2. Then S is (m, n)-regular ⇔ A ∩ B = (AB] ∩ (BA] $\forall A, B \in \hat{F_{(m, n)}}$ .

Proof. (⇒) Let $A, B \in \hat{I_{(m, n)}}$ . Then $A \cap B \in \hat{I_{(m, n)}}$ . So, by Lemma 4.10, A ∩ B = ((A ∩ B) ²] = ((A ∩ B) (A ∩ B)] ⊆ (AB]. Similarly A ∩ B ⊆ (BA]. Therefore A ∩ B ⊆ (AB] ∩ (BA] . Now ((AB]) ^mS ((AB]) ⁿ $= \underset{m - times}{\underset{︸}{(AB] (AB] \dots (AB]}} S \underset{n - times}{\underset{︸}{(AB] (AB] \dots (AB]}}$ = ((AB) (ABAB ⋯ ABSABAB ⋯ A) B] ⊆ (ABSB] = (AB^mSBⁿ] ⊆ (AB] . Thus ((AB]) ^mS ((AB]) ⁿ ⊆ (AB], and similarly (AB] (AB] ⊆ (AB]. Therefore $(AB] \in \hat{I_{(m, n)}}$ . Similarly $(BA] \in \hat{I_{(m, n)}}$ . So $(AB] \cap (BA] \in \hat{I_{(m, n)}}$ . Again, by Lemma 4.10, (AB] ∩ (BA] = (((AB] ∩ (BA]) ²] ⊆ ((AB] (BA] ⊆ ((ASA] ⊆ A. Similarly ((AB] (BA] ⊆ B . Therefore (AB] ∩ (BA] ⊆ A ∩ B . Hence A ∩ B = (AB] ∩ (BA] .

Converse is obvious. □

Lemma 4.15. Let $m, n \in ℤ^{+}$ such that m ≥ 2 or n ≥ 2. Then for any $A, B, C \in \hat{I_{(m, n)}}$ such that (AB] ∩ (BA] ⊆ C implies A ⊆ C or B ⊆ C ⇔ S is (m, n)-regular and the set of all (m, n)-ideals $\hat{I_{(m, n)}}$ of S forms a chain.

Proof. (⇒) Take x ∈ S. Since $[x]_{m, n} \in \hat{I_{(m, n)}}$ of S and (([x] _m,n) ²] ⊆ (([x] _m,n) ²], by hypothesis, [x] _m,n ⊆ (([x] _m,n) ²]. Now [x] _m,n ⊆ (([x] _m,n) ²] ⊆ (([x] _m,n) ³] ⊆ … ⊆ ((([x] _m,n) ^m] H (([x] _m,n) ⁿ]] ⊆ (x^mSxⁿ].

Since x ∈ [x] _m,n, x ∈ (x^mSxⁿ] . Hence S is (m, n)-regular. Next, we show that $\hat{I_{(m, n)}}$ a chain. Let $A, B \in \hat{I_{(m, n)}}$ . By Lemma 4.14, A ∩ B = (AB] ∩ (BA]. Since $A \cap B \in \hat{I_{(m, n)}}$ , by hypothesis, A ⊆ A ∩ B or B ⊆ A ∩ B. If A ⊆ A ∩ B, then A ⊆ B. In the other case if B ⊆ A ∩ B, then B ⊆ A, as required.

(⇐) Assume that $A, B, C \in \hat{F_{(m, n)}}$ such that (AB] ∩ (BA] ⊆ C. As S is (m, n)-regular, by Lemma 4.14, A ∩ B = (AB] ∩ (BA] ⊆ C. Now, by hypothesis, either A ⊆ B or B ⊆ A. Therefore either A ∩ B = A or A ∩ B = B. Hence, either A ⊆ C or B ⊆ C. □

Theorem 4.16. Let $m, n \in ℤ^{+}$ either m ≥ 2 or n ≥ 2. Then for any $η, ξ, λ \in F_{(m, n)}$ such that $\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{λ_{κ}^{κ^{*}}}$ implies $\underline{η_{κ}^{κ^{*}}} ⪯ \underline{λ_{κ}^{κ^{*}}}$ or $\underline{ξ_{κ}^{κ^{*}}} ⪯ \underline{λ_{κ}^{κ^{*}}}$ ⇔ S is (m, n)-regular and the set $F_{(m, n)}$ forms a chain.

Proof. (⇒) Let $A, B, C \in \hat{I_{(m, n)}}$ such that (AB] ∩ (BA] ⊆ C. By Lemma 4.4, $(\underline{χ_{κ}^{κ^{*}}})_{A}, (\underline{χ_{κ}^{κ^{*}}})_{B}, (\underline{χ_{κ}^{κ^{*}}})_{C} \in \hat{F_{(m, n)}}$ . Let x ∈ (AB] ∩ (BA]. Then x ∈ (AB] , x ∈ (BA]. Now $((\underline{χ_{κ}^{κ^{*}}})_{A} (\circ)_{k}^{k^{*}} (\underline{χ_{κ}^{κ^{*}}})_{B} (\land)_{k}^{k^{*}} (\underline{χ_{κ}^{κ^{*}}})_{B} (\circ)_{k}^{k^{*}} (\underline{χ_{κ}^{κ^{*}}})_{A}) (x) = ((\underline{χ_{κ}^{κ^{*}}})_{A} (\circ)_{k}^{k^{*}} (\underline{χ_{κ}^{κ^{*}}})_{B}) (x) \cap ((\underline{χ_{κ}^{κ^{*}}})_{B} (\circ)_{k}^{k^{*}} (\underline{χ_{κ}^{κ^{*}}})_{A}) (x) = (\underline{χ_{κ}^{κ^{*}}})_{(AB]} (x) \cap (\underline{χ_{κ}^{κ^{*}}})_{(BA]} (x) = \frac{κ^{*} - κ}{2} .$ Since x ∈ (AB] ∩ (BA] and (AB] ∩ (BA] ⊆ C, $(\underline{χ_{κ}^{κ^{*}}})_{C} (x) = \frac{κ^{*} - κ}{2}$ .

(⇐) Assume that for any $η, ξ, λ \in \hat{F_{(m, n)}}$ such that $\underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{λ_{κ}^{κ^{*}}}$ . As S is (m, n)-regular, by Theorem 4.13, $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} = \underline{η_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} (\circ)_{k}^{k^{*}} \underline{η_{κ}^{κ^{*}}} ⪯ \underline{λ_{κ}^{κ^{*}}} .$ Now, by hypothesis, either $\underline{η_{κ}^{κ^{*}}} ⪯ \underline{ξ_{κ}^{κ^{*}}}$ or $\underline{ξ_{κ}^{κ^{*}}} ⪯ \underline{η_{κ}^{κ^{*}}}$ . Therefore either $\underline{η_{κ}^{κ^{*}}} (\cap)_{k}^{k^{*}} \underline{ξ_{κ}^{κ^{*}}} = \underline{η_{κ}^{κ^{*}}}$ or $\underline{η_{κ}^{κ^{*}}} \land \underline{ξ_{κ}^{κ^{*}}} = \underline{ξ_{κ}^{κ^{*}}}$ . Hence either $\underline{η_{κ}^{κ^{*}}} ⪯ \underline{λ_{κ}^{κ^{*}}}$ or $\underline{ξ_{κ}^{κ^{*}}} ⪯ \underline{λ_{κ}^{κ^{*}}}$ . □

5 Conclusion

This paper’s primary objective is to introduce the notion of (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideal in the ordered semigroups, and to improve the concept of (m, n)-regular ordered semigroups by considering the (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals. In this article, the special cases are as follows:

(1). By taking m = 1 = n, all results of [7] will be special cases of this article results.

(2). If we take m = 1 = n, κ^* = 1, then most of the results from [8] are obtained as direct corollaries of the present paper results.

(3). If we put m = 1 = n, κ^* = 1 and κ = 0, then most of the results of the paper [2] are deduced as direct corollaries of the results of this paper.

Conflict of interest

The authors have no conflict of interest.

Footnotes

Acknowledgments

The authors are grateful to the anonymous referees for a careful checking of the details and for helpful comments that improved this paper. This work was supported by the Taif University Researchers Supporting Project (TURSP-2020/246), Taif University, Taif, Saudi Arabia.

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New types of fuzzy ( m,n )-ideals in ordered semigroups

Abstract

Keywords

1 Introduction

2 Preliminaries

3 (∈ , ∈ ∨ (κ*, q κ ))-fuzzy (m, n)-ideals

4 (κ*, κ)-lower parts of (∈ , ∈ ∨ (κ*, q κ ))-fuzzy (m, n)-ideals

5 Conclusion

Conflict of interest

Footnotes

Acknowledgments

References

3 (∈ , ∈ ∨ (κ^*, q_κ))-fuzzy (m, n)-ideals

4 (κ^, κ)-lower parts of (∈ , ∈ ∨ (κ^, q_κ))-fuzzy (m, n)-ideals