SBE-Algebras via Intuitionistic Fuzzy Structures

Abstract

The study introduces the concept of intuitionistic fuzzy SBE-subalgebras, ideals, and filters, along with level sets of intuitionistic fuzzy sets within the framework of Sheffer stroke BE-algebras. These concepts are shown to be crucial for understanding the behavior of intuitionistic fuzzy logic in this algebraic structure. The study further establishes a bidirectional relationship between subalgebras, ideals, filters, and their respective level sets on Sheffer stroke BE-algebras, demonstrating that the level set of an intuitionistic fuzzy SBE-subalgebra, ideal, or filter is itself a subalgebra, ideal, or filter on this algebra, and vice versa.

1. Introduction

The Sheffer operation, also known as the Sheffer stroke or NAND operator, was first introduced by H. M. Sheffer [1]. This operation is significant because it alone can be used to construct an entire logical system, allowing any axiom of the system to be restated solely in terms of the Sheffer operation. This unique property facilitates control over certain aspects of the logical system. Notably, even the axioms of Boolean algebra, the algebraic counterpart of classical propositional calculus, can be expressed using only the Sheffer operation, highlighting its fundamental role and versatility in both logical and algebraic systems. The Sheffer stroke has been applied to several algebraic structures, for example, Boolean algebra, BCK-algebra, basic algebras, etc. (see [2,3,4]).

BE-algebras, defined by H. S. Kim and Y. H. Kim [5] as a generalization of BCK-algebras (see [6,7]), have been extensively studied by many researchers [8,9,10,11,12,13,14,15,16]. The theory of intuitionistic fuzzy sets, introduced by K. T. Atanassov [17] as an extension of L. A. Zadeh’s fuzzy set theory [18], has been widely applied to solve problems in various mathematical fields, particularly in algebraic structures. The fuzzification of BE-algebras and their properties has been investigated in [19,20,21,22], while intuitionistic fuzzy BE-algebras have been explored in [23,24,25].

Recently, T. Katican et al. [26] introduced the concept of a Sheffer stroke BE-algebra (SBE-algebra) and identified relationships among SBE-filters, upper sets, and SBE-subalgebras. Additionally, T. Katican [27] investigated obstinate SBE-filters and characterized the branches of a Sheffer stroke BE-algebra using a tile-based approach, describing branchwise commutative and self-distributive properties. Further developments include the work of T. Oner et al. [28], who explored fuzzy Sheffer stroke BE-algebras and defined fuzzy SBE-subalgebras, (weak) fuzzy SBE-filters, Cartesian products of fuzzy subsets, and fuzzy points on Sheffer stroke BE-algebras. N. Chunsee et al. [29] applied fuzzy set theory to SBE-algebras, defining fuzzy and anti-fuzzy SBE-ideals and investigating their properties, especially by characterizing these ideals via level subsets.

In this paper, we introduced and examined the concepts of intuitionistic fuzzy SBE-subalgebras, ideals, filters, and level sets within the framework of Sheffer stroke BE-algebras. We demonstrated the significant interplay among these notions, proving that the level set of an intuitionistic fuzzy SBE-subalgebra, ideal, or filter within this algebraic framework is itself a subalgebra, ideal, or filter, respectively. Moreover, we highlighted the deep interconnections between subalgebras and level sets, revealing their intricate relationships. These findings provide valuable insights into the behavior of intuitionistic fuzzy logic in algebraic contexts, establishing a solid foundation for further exploration of intuitionistic fuzzy systems within Sheffer stroke BE-algebras.

Future research could broaden these findings to encompass other algebraic systems, such as residuated lattices or Brouwerian algebras, extending the applicability of intuitionistic fuzzy logic. Investigating computational aspects of intuitionistic fuzzy SBE-subalgebras, ideals, and filters might also enable practical applications, particularly in decision-making and optimization processes. Additionally, exploring the topological properties of intuitionistic fuzzy level sets could uncover new applications in artificial intelligence and fuzzy logic systems. Advancing these research directions has the potential to significantly enhance the theoretical and practical development of intuitionistic fuzzy logic in solving complex real-world problems.

2. Preliminaries

In this section, we review some fundamental concepts and results related to Sheffer stroke BE-algebras that will be used in the following sections.

Definition 1

([5]). Let $L=\langle L,|\rangle$ be a groupoid. The operation | is called a Sheffer stroke operation if it satisfies the following conditions:

$$\begin{array}{c}\hfill \begin{array}{cc}\left(S1\right)\hfill & x|y=y|x,\hfill \\ \left(S2\right)\hfill & \left(x\right|x\left)\right|\left(x\right|y)=x,\hfill \\ \left(S3\right)\hfill & x\left|\right(\left(y\right|z\left)\right|\left(y\right|z\left)\right)=\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right|z,\hfill \\ \left(S4\right)\hfill & \left(x\right|\left(\right(x\left|x\right)\left|\right(y\left|y\right)\left)\right)\left|\right(x\left|\right(\left(x\right|x\left)\right|\left(y\right|y\left)\right))=x.\hfill \end{array}\end{array}$$

Definition 2

([5]). An algebra $(L,\to ,1)$ of type $(2,0)$ is called a BE-algebra, if it satisfies the following axioms—for all $x,y,z\in L$:

$$\begin{array}{c}\hfill \begin{array}{cc}\left(BE1\right)\hfill & x\to x=1,\hfill \\ \left(BE2\right)\hfill & x\to 1=1,\hfill \\ \left(BE3\right)\hfill & 1\to x=x,\hfill \\ \left(BE4\right)\hfill & x\to (y\to z)=y\to (x\to z).\hfill \end{array}\end{array}$$

Definition 3

([26]). A Sheffer stroke BE-algebra(briefly, SBE-algebra)is a structure $\langle L,|\rangle$ of type $\left(2\right)$, where 1 is a fixed element in L. For all $x,y,z\in L$, the following conditions are satisfied:

$$\begin{array}{c}\hfill \begin{array}{cc}\left(\mathrm{SBE}1\right)\hfill & \left(x\right|\left(x\right|x\left)\right)\left|\right(x\left|\right(x\left|x\right))=1,\hfill \\ \left(\mathrm{SBE}2\right)\hfill & \left(0\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right)=x|x.\hfill \end{array}\end{array}$$

Proposition 1

([26]). Let $\langle L,|\rangle$ be an SBE-algebra. Then, the binary relation $x\le y$ if and only if $\left(y\right|\left(x\right|x\left)\right)\left|\right(y\left|\right(x\left|x\right))=1$ defines a partial order on L.

Definition 4

([26]). A nonempty subset G of a Sheffer stroke BE-algebra L is called an SBE-subalgebra of L if $\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right))\in G$ for all $x,y\in G$.

Definition 5

([26]). A nonempty subset G of a Sheffer stroke BE-algebra L is called an SBE-ideal of L if, for all $x,y\in G$:

• $1\in G$;
• $\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right))\in G$ and $y\in G\Rightarrow x\in G$.

Definition 6

([28]). A fuzzy set μ in an SBE-algebra L is called a fuzzy SBE-subalgebra of L if:

$$\begin{array}{c}\hfill (\forall x,y\in L)\mu \left(x\right|\left(y\right|y\left)\right)\ge min\left\{\mu \right(x),\mu (y\left)\right\}.\end{array}$$

(1)

Definition 7

([29]). A fuzzy set μ in an SBE-algebra L is called a fuzzy SBE-ideal of L if:

$$\begin{array}{c}\hfill (\forall x,y,z\in L)\left\{\begin{array}{c}\mu \left(x\right|\left(y\right|y\left)\right)\ge \mu \left(y\right),\hfill \\ \mu \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\ge min\left\{\mu \right(x),\mu (y\left)\right\}.\hfill \end{array}\right.\end{array}$$

(2)

Definition 8

([28]). A fuzzy set μ in an SBE-algebra L is called a fuzzy SBE-filter of L if:

$$\begin{array}{c}\hfill (\forall x,y\in L)\mu \left(1\right)\ge \mu \left(y\right)\ge min\left\{\mu \right(x\left|\right(y\left|y\right)),\mu (x\left)\right\}.\end{array}$$

(3)

Lemma 1

([26]). Let $\langle L,|,1\rangle$ be an SBE-algebra. Then, the following properties hold:

• $x\left|\right(1\left|1\right)=1$;
• $1\left|\right(x\left|x\right)=x$;
• $x\left|\right(\left(y\right|\left(x\right|x\left)\right)\left|\right(y\left|\right(x\left|x\right)\left)\right)=1$;
• $x\left|\right(\left(\right(x\left|\right(y\left|y\right)\left)\right|\left(y\right|y\left)\right)\left|\right(\left(x\right|\left(y\right|y\left)\right)\left|\right(y\left|y\right)\left)\right)=1$;
• $\left(x\right|1\left)\right|\left(x\right|1)=x$;
• $\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right|\left(x\right|x)=1$ and $\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right|\left(y\right|y)=1$;
• $x\left|\right(\left(x\right|y\left)\right|\left(x\right|y\left)\right)=x|y=(\left(x\right|y\left)\right|\left(x\right|y\left)\right)|y$.

Definition 9

([26]). An SBE-algebra L is called:

• Transitive if $b\left|\right(c\left|c\right)\le \left(a\right|\left(b\right|b\left)\right)\left|\right(\left(a\right|\left(c\right|c\left)\right)\left|\right(a\left|\right(c\left|c\right)\left)\right)$ for all $a,b,c\in L$;
• Commutative if $\left(a\right|\left(b\right|b\left)\right)\left|\right(b\left|b\right)=\left(b\right|\left(a\right|a\left)\right)\left|\right(a\left|a\right)$ for all $a,b\in L$;
• Self-distributive if $a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)=\left(a\right|\left(b\right|b\left)\right)\left|\right(\left(a\right|\left(c\right|c\left)\right)\left|\right(a\left|\right(c\left|c\right)\left)\right)$ for all $a,b,c\in L$.

Lemma 2

([26]). Let $\langle L,|,1\rangle$ be an SBE-algebra. Then, the following statements hold:

• If $x\le y$, then $y|y\le x|x$;
• $x\le y\left|\right(x\left|x\right)$;
• $y\le \left(y\right|\left(x\right|x\left)\right)\left|\right(x\left|x\right)$;
• If S is self-distributive, then $x\le y$ implies $y|z\le x|z$;
• If S is self-distributive, then $y\left|\right(z\left|z\right)\le \left(z\right|\left(x\right|x\left)\right)\left|\right(\left(y\right|\left(x\right|x\left)\right)\left|\right(y\left|\right(x\left|x\right)\left)\right)$.

Definition 10

([17]). Let H be a nonempty set. The intuitionistic fuzzy set on A is defined to be a structure:

$$\begin{array}{c}\hfill A:=\{\langle x,\alpha (x),\beta (x)\mid x\in L\},\end{array}$$

(4)

where $\alpha :H\to [0,1]$ is the degree of membership of x to A and $\beta :H\to [0,1]$ is the degree of nonmembership of x to A such that $0\le \alpha \left(x\right)+\beta \left(x\right)\le 1$.

Lemma 3

([30]). Let $a,b,c\in \mathbb{R}$. Then, the following statements hold:

• $a-min\{b,c\}=max\{a-b,a-c\}$;
• $a-max\{b,c\}=min\{a-b,a-c\}$.

3. Intuitionistic Fuzzy Sets in SBE-Algebras

In this section, we introduce the concept of intuitionistic fuzzy SBE-subalgebras within the framework of Sheffer stroke BE-algebras. This exploration aims to enhance the understanding of how intuitionistic fuzzy logic operates in this specific algebraic context. It is important to clarify that, unless explicitly stated otherwise, L will refer to a Sheffer stroke BE-algebra throughout this discussion.

Definition 11.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ in an SBE-algebra L is called an intuitionistic-valued fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$ if it satisfies the following conditions for all $x,y\in L$:

$$\begin{array}{cc}\hfill & \left(\forall x,y\in L\right)\left(\begin{array}{c}\alpha \left(x\right|\left(y\right|y\left)\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}\hfill \\ \beta \left(x\right|\left(y\right|y\left)\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}\hfill \end{array}\right).\hfill \end{array}$$

(5)

Example 1.

Given a structure $\langle L,|,1\rangle$ where $L=\{0,u,v,1\}$ and a binary operation | with Cayley table as below:

	|	0	u	v	1
	0	1	1	1	1
|	u	1	v	1	v
	v	1	1	u	u
	1	1	v	u	0

Then, $\langle L,|,1\rangle$ is a SBE-algebra (see [26]). We analyze the relationships between membership and nonmembership functions and ensure that they satisfy the condition $0\le \alpha \left(x\right)+\beta \left(x\right)\le 1$, where $\alpha \left(x\right)$ represents the membership function, and $\beta \left(x\right)$ represents the nonmembership function. The chosen values for the membership and nonmembership functions are as follows:

$$\alpha \left(0\right)=0.1,\beta \left(0\right)=0.9,\alpha \left(u\right)=0.5,\beta \left(u\right)=0.5,$$

$$\alpha \left(v\right)=0.7,\beta \left(v\right)=0.3,\alpha \left(1\right)=0.9,\beta \left(1\right)=0.1.$$

It is routine to verify that the conditions for an intuitionistic fuzzy set to be an intuitionistic-valued fuzzy SBE-subalgebra are satisfied.

Definition 12.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set on an SBE-algebra L, and let $t\in [0,1]$. We define the following sets in L:

$$\begin{array}{cc}\hfill U(\alpha ,t)& =\{x\in L:\alpha (x)\ge t\},\hfill \\ \hfill U^{+}(\alpha ,t)& =\{x\in L:\alpha (x)>t\},\hfill \\ \hfill L(\alpha ,t)& =\{x\in L:\alpha (x)\le t\},\hfill \\ \hfill L^{+}(\alpha ,t)& =\{x\in L:\alpha (x)<t\}.\hfill \end{array}$$

Theorem 1.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of L if and only if, for all $s,t\in [0,1]$, the sets $U(\alpha ,t)$ and $L(\beta ,s)$ are either empty or SBE-subalgebras of L.

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$ and $L(\beta ,s)\ne \varnothing \ne U(\alpha ,t)$ for all $s,t\in [0,1]$. Let $x,y,a,b\in L$ be such that $(x,a)\in L(\beta ,s)\times U(\alpha ,t)$ and $(y,b)\in L(\beta ,s)\times U(\alpha ,t)$. Then, $\beta \left(x\right)\le s$, $\beta \left(y\right)\le s$, $\alpha \left(a\right)\ge t$ and $\alpha \left(b\right)\ge t$. Then, we get $\beta \left(x\right|\left(y\right|y\left)\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}\le s$ and $\alpha \left(a\right|\left(b\right|b\left)\right)\ge min\left\{\alpha \right(a),\alpha (b\left)\right\}\ge t$, and so, $x\left|\right(y\left|y\right),a\left|\right(b\left|b\right)\in L(\beta ,s)\times U(\alpha ,t)$. Therefore, $L(\beta ,s)$ and $U(\alpha ,t)$ are SBE-subalgebras of $\mathcal{L}=(L,|)$.

Conversely, let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set in L for which its negative s-cut and positive t-cut are SBE-subalgebras of $\mathcal{L}=(L,|)$ whenever they are nonempty for all $s,t\in [0,1]$. Suppose that $\beta \left(a\right|\left(b\right|b\left)\right)>max\left\{\beta \right(a),\beta (b\left)\right\}$ or $\alpha \left(x\right|\left(y\right|y\left)\right)<min\left\{\alpha \right(x),\alpha (y\left)\right\}$ for some $a,b,x,y\in L$. Then, $a,b\in L\left(\beta s\right)$ or $x,y\in U(\alpha ,t)$ where $s=max\left\{\beta \right(a),\beta (b\left)\right\}$ and $t=min\left\{\alpha \right(x),\alpha (y\left)\right\}$. However, $a\left|\right(b\left|b\right)\notin L(\beta ,s)$ or $\left(x\right|\left(y\right|y\left)\right)\notin U(\alpha ,t)$, which is a contradiction. Therefore, $\beta \left(a\right|\left(b\right|b\left)\right)\le max\left\{\beta \right(a),\beta (b\left)\right\}$ and $\alpha \left(x\right|\left(y\right|y\left)\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}$ for all $a,b,x,y\in L$. Consequently, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$. □

Theorem 2.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ in L is an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$ if and only if the fuzzy sets $\overline{\beta }$ and α are fuzzy SBE-subalgebras of $\mathcal{L}=(L,|)$.

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$. It is clear that $\alpha$ is an intuitionistic fuzzy subalgebra of $\mathcal{L}=(L,|)$. For every $x,y\in L$:

$$\begin{array}{ccc}\overline{\beta }\left(x\right|\left(y\right|y))& =& 1-\beta \left(x\right|\left(y\right|y\left)\right)\hfill \\ & \ge & 1-max\left\{\beta \right(x),\beta (y\left)\right\}\hfill \\ & =& min\{1-\beta (x),1-\beta (y\left)\right\}\hfill \\ & =& min\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}.\hfill \end{array}$$

Hemce, $\overline{\beta }$ is an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$. Conversely, let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set of $\mathcal{L}=(L,|)$ for which $\overline{\beta }$ and $\alpha$ are fuzzy SBE-subalgebras of $\mathcal{L}=(L,|)$. Let $x,y\in L$. Then:

$$\begin{array}{ccc}1-\beta \left(x\right|\left(y\right|y\left)\right)& =& \overline{\beta }\left(x\right|\left(y\right|y))\hfill \\ & \ge & min\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}\hfill \\ & =& min\{1-\beta (x),1-\beta (y\left)\right\}\hfill \\ & =& 1-max\left\{\beta \right(x),\beta (y\left)\right\}\hfill \\ \beta \left(x\right|\left(y\right|y\left)\right)& \le & max\left\{\beta \right(x),\beta (y\left)\right\}.\hfill \end{array}$$

Hence, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$. □

Theorem 3.

$\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of L if and only if both $(L,\alpha ,\overline{\alpha })$ and $(L,\beta ,\overline{\beta })$ are intuitionistic fuzzy SBE-subalgebras of L.

Proof. 

If $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of L, then $\alpha =\overline{\overline{\alpha }}$ and $\beta$ are fuzzy SBE-subalgebra of L from Theorem 2, and hence, $(L,\alpha ,\overline{\alpha })$ and $(L,\overline{\beta },\beta )$ are intuitionistic fuzzy SBE-subalgebras of L.

Conversely, if $(L,\alpha ,\overline{\alpha })$ and $(L,\beta ,\overline{\beta })$ are intuitionistic fuzzy SBE-subalgebras of L, then $\alpha$ and $\overline{\beta }$ are fuzzy SBE-subalgebras of L, and hence, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of L. □

Corollary 1.

Let ${\chi }_M$ be the characteristic function of an intuitionistic fuzzy SBE-subalgebra of L. Then, $\overline{M}=(L,{\chi }_M,\overline{{\chi }_M})$ is intutionstic fuzzy SBE-subalgebra of L.

Lemma 4.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$. Then:

$$\begin{array}{c}\hfill (\forall x,\in L)\left(\begin{array}{c}\alpha \left(1\right)\ge \alpha \left(x\right)\hfill \\ \beta \left(1\right)\le \beta \left(x\right)\hfill \end{array}\right).\end{array}$$

(6)

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-subalgebra of $\mathcal{L}=(L,|)$. Then:

$$\alpha \left(1\right)=\alpha \left(x\right|\left(x\right|x\left)\right)\ge min\left\{\alpha \right(x),\alpha (x\left)\right\}=\alpha \left(x\right),$$

$$\beta \left(1\right)=\beta \left(x\right|\left(x\right|x\left)\right)\le max\left\{\beta \right(x),\beta (x\left)\right\}=\beta \left(x\right),$$

for all $x\in L$. □

Theorem 4.

An intuitionistic fuzzy SBE-subalgebra $\mathcal{L}=(L,\alpha ,\beta )$ of a SBE-algebra L satisfies $\alpha \left(x\right)\le \alpha \left(x\right|\left(y\right|y\left)\right)$ and $\beta \left(x\right|\left(y\right|y\left)\right)\le \beta \left(x\right)$ for all $x,y\in L$ if and only if α and β are constants.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-subalgebra of L satisfying $\alpha \left(x\right)\le \alpha \left(x\right|\left(y\right|y\left)\right)$, $\beta \left(x\right|\left(y\right|y\left)\right)\le \beta \left(x\right)$ for all $x,y\in L$. Since $\alpha \left(1\right)\le \alpha \left(1\right|\left(x\right|x\left)\right)=\alpha \left(x\right)$ and $\beta \left(x\right)=\beta \left(1\right|\left(x\right|x\left)\right)\le \beta \left(1\right)$ from Lemma 1 (2), it follows from Lemma 4 that $\alpha \left(x\right)=\alpha \left(1\right)$ and $\beta \left(x\right)=\beta \left(1\right)$ for all $x\in L$. Hence, $\alpha$ and $\beta$ are constants.

Conversely, it is obvious by the fact that $\alpha$ and $\beta$ are constants. □

4. Intuitionistic Fuzzy Filters in SBE-Algebras

In this section, we introduce the concept of intuitionistic fuzzy SBE-filters in the context of Sheffer stroke BE-algebras.

Definition 13.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ of L is called an intuitionistic fuzzy SBE-filter of L if it satisfies the following conditions:

$$\begin{array}{c}\hfill (\forall x,y\in L)\left(\begin{array}{c}\alpha \left(1\right)\ge \alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\},\hfill \\ \beta \left(1\right)\le \beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}\hfill \end{array}\right).\end{array}$$

(7)

Example 2.

To construct an intuitionistic fuzzy SBE-filer in the SBE-algebra $\langle L,|,1\rangle$ with $L=\{0,u,v,1\}$ and the given Cayley table in Example 1, we define a fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ with $\alpha \left(0\right)=0.1,\beta \left(0\right)=0.9,\alpha \left(u\right)=0.5,\beta \left(u\right)=0.5,\alpha \left(v\right)=0.7,\beta \left(v\right)=0.3,\alpha \left(1\right)=0.9,$ and $\beta \left(1\right)=0.1.$ For each combination of x and y, the conditions in $\left(4.1\right)$ are satisfied, confirming that L is an intuitionistic fuzzy SBE-filter.

Lemma 5.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ in L is an intuitionistic fuzzy SBE-filter of $\mathcal{L}=(L,|)$ if and only if the following conditions hold:

$$\begin{array}{cc}\hfill & (\forall x,y\in L)\left(\begin{array}{c}x⪯y\Rightarrow \left\{\begin{array}{c}\alpha \left(x\right)\ge \alpha \left(y\right)\hfill \\ \beta \left(y\right)\le \beta \left(x\right)\hfill \end{array}\right.\hfill \end{array}\right),\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & (\forall x,y\in L)\left(\begin{array}{c}\alpha \left(\right(x\left|y\right)\left|\right(x\left|y\right))\ge min\{\alpha \left(x\right),\alpha \left(y\right)\}\hfill \\ \beta \left(\right(x\left|y\right)\left|\right(x\left|y\right))\le max\{\beta \left(x\right),\beta \left(y\right)\}\hfill \end{array}\right).\hfill \end{array}$$

(9)

Proof. 

Assume that $x⪯y$. Then, $x\left|\right(y\left|y\right)=1$. Thus:

$$\alpha \left(x\right)=min\left\{\alpha \right(1),\alpha (x\left)\right\}=min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}\le \alpha \left(y\right)$$

and

$$\beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}=max\left\{\beta \right(1),\beta (x\left)\right\}\le \beta \left(x\right).$$

Since $y⪯y\left|\right(\left(\right(x\left|x\right)\left|\right(x\left|x\right)\left)\right|\left(\right(x\left|x\right)\left|\right(x\left|x\right)\left)\right)=x\left|\right(x\left|y\right)$:

$$\begin{array}{ccc}\alpha \left(\right(x\left|y\right)\left|\right(x\left|y\right))& \ge & min\left\{\alpha \right(x),\alpha (x\left|\right(\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right|\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right)\left)\right\}\hfill \\ & =& min\left\{\alpha \right(x),\alpha (x\left|\right(x\left|y\right)\left)\right\}\hfill \\ & \ge & min\left\{\alpha \right(x),\alpha (y\left)\right\}\hfill \end{array}$$

$$\begin{array}{ccc}\beta \left(\right(x\left|y\right)\left|\right(x\left|y\right))& \le & max\left\{\beta \right(x),\beta (x\left|\right(\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right|\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right)\left)\right\}\hfill \\ & =& max\left\{\beta \right(x),\beta (x\left|\right(x\left|y\right)\left)\right\}\hfill \\ & \le & max\left\{\beta \right(x),\beta (y\left)\right\}.\hfill \end{array}$$

Conversely, let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set on L satisfying (8) and (9). Since $x\le 1$, $\alpha \left(1\right)\ge \alpha \left(x\right)$ and $\beta \left(1\right)\le \beta \left(x\right)$, for all $x\in S$. Also, $\left(\right(x\left|\right(x\left|\right(y\left|y\right)\left)\right)\left|\right(x\left|\right(x\left|\right(y\left|y\right)\left)\right)\left)\right|\left(y\right|y)=(x\left|\right(y\left|y\right)\left)\right|\left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))=1$, and so, $\left(x\right|\left(x\right|\left(y\right|y\left)\right)\left)\right|\left(x\right|\left(x\right|\left(y\right|y\left)\right))\le y$. Then:

$$\alpha \left(y\right)\ge \alpha \left(\right(x\left|\right(x\left|\right(y\left|y\right)\left)\right)\left|\right(x\left|\right(x\left|\right(y\left|y\right)\left)\right))\ge min\{\alpha \left(x\right|\left(y\right|y\left)\right),\alpha \left(x\right)\}$$

and

$$\beta \left(y\right)\le \beta \left(\right(x\left|\right(x\left|\right(y\left|y\right)\left)\right)\left|\right(x\left|\right(x\left|\right(y\left|y\right)\left)\right))\le max\{\beta \left(x\right|\left(y\right|y\left)\right),\beta \left(x\right)\}.$$

Therefore, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L. □

Lemma 6.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-filter of L. Then, for all $x,y,z\in L$, the following statements hold:

• $\alpha \left(y\right|\left(x\right|x\left)\right)\ge \alpha \left(x\right)$ and $\beta \left(y\right|\left(x\right|x\left)\right)\le \beta \left(x\right)$,
• $\alpha \left(x\right|\left(y\right|y\left)\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}$ and $\beta \left(x\right|\left(y\right|y\left)\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}$,
• $\alpha \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(y\right|y\left)\right)\ge \alpha \left(x\right)$ and $\beta \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(y\right|y\left)\right)\le \beta \left(x\right)$,
• $\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}$ and
  $\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}$.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-filter of L. Then:

1. It is proved from Lemma 2 (2) and (8).

2. It follows from (1) that $\alpha \left(x\right|\left(y\right|y\left)\right)\ge \alpha \left(y\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}$ and $\beta \left(x\right|\left(y\right|y\left)\right)\le \beta \left(y\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}$ for all $x,y\in L$.

3. We get from Lemma 2 (3) and (8).

4. It is obtained from (3) and (SBE-2) that: $\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)$

$$\begin{array}{ccc}& \ge & min\left\{\alpha \right(x),\alpha (x\left|\right(\left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\left|\right(\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\left)\right)\left)\right\}\hfill \\ & =& min\left\{\alpha \right(x),\alpha (\left(y\right|\left(\right(x\left|\right(z\left|z\right)\left)\right|\left(x\right|\left(z\right|z\left)\right)\left)\right)\left|\right(\left(x\right|\left(z\right|z\left)\right)\left|\right(x\left|\right(z\left|z\right)\left)\right)\left)\right\}\hfill \\ & \ge & min\left\{\alpha \right(x),\alpha (y\left)\right\},\hfill \end{array}$$

$\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)$

$$\begin{array}{ccc}& \le & max\left\{\beta \right(x),\beta (x\left|\right(\left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\left|\right(\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\left)\right)\left)\right\}\hfill \\ & =& max\left\{\beta \right(x),\beta (\left(y\right|\left(\right(x\left|\right(z\left|z\right)\left)\right|\left(x\right|\left(z\right|z\left)\right)\left)\right)\left|\right(\left(x\right|\left(z\right|z\left)\right)\left|\right(x\left|\right(z\left|z\right)\left)\right)\left)\right\}\hfill \\ & \le & max\left\{\beta \right(x),\beta (y\left)\right\},\hfill \end{array}$$

for all $x,y,z\in L$. □

Theorem 5.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set on L. Then, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L if and only if:

$$\begin{array}{cc}\hfill & (\forall x,y,z\in L)\left(z\in U(x,y)\Rightarrow \left\{\begin{array}{c}\alpha \left(z\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\},\hfill \\ \beta \left(z\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}\hfill \end{array}\right.\right).\hfill \end{array}$$

(10)

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L and $z\in U(x,y)$. Since $x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)=1$, we have $x\le y\left|\right(z\left|z\right)$. By Equation (8), it follows that:

$$\alpha \left(z\right)\ge min\left\{\alpha \right(y\left|\right(z\left|z\right)),\alpha (y\left)\right\}\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}$$

and

$$\beta \left(z\right)\le max\left\{\beta \right(y\left|\right(z\left|z\right)),\beta (y\left)\right\}\le max\left\{\beta \right(x),\beta (y\left)\right\}.$$

Conversely, let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set on L satisfying condition (10). Since $x\left|\right(\left(x\right|\left(1\right|1\left)\right)\left|\right(x\left|\right(1\left|1\right)\left)\right)=1$ from Lemma 1 (1), it follows that $1\in U(x,x)$ for all $x\in L$. Consequently, we have:

$$\alpha \left(1\right)\ge min\left\{\alpha \right(x),\alpha (x\left)\right\}=\alpha \left(x\right)$$

and

$$\beta \left(1\right)\le max\left\{\beta \right(x),\beta (x\left)\right\}=\beta \left(x\right).$$

Next, since $x\left|\right(\left(\right(x\left|\right(y\left|y\right)\left)\right|\left(y\right|y\left)\right)\left|\right(\left(x\right|\left(y\right|y\left)\right)\left|\right(y\left|y\right)\left)\right)=1$, it follows that $y\in U(x,x|\left(y\right|y\left)\right)$. By applying condition (10), we get:

$$\alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}$$

and

$$\beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}$$

for all $x,y\in L$. Therefore, $\mathcal{L}=(L,\alpha ,\beta )$ is indeed an intuitionistic fuzzy SBE-filter of L. □

Definition 14.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ of L is called an implicative intuitionistic fuzzy SBE-filter of L if it satisfies the following conditions:

$$\begin{array}{cc}\hfill & (\forall x\in L)\left(\begin{array}{c}\alpha \left(1\right)\ge \alpha \left(x\right)\hfill \\ \beta \left(1\right)\le \beta \left(x\right)\hfill \end{array}\right),\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill & (\forall x,y,z\in L)\left(\begin{array}{c}\alpha \left(x\right|\left(z\right|z\left)\right)\ge min\left\{\alpha \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\alpha (x\left|\right(y\left|y\right)\left)\right\}\hfill \\ \beta \left(x\right|\left(z\right|z\left)\right)\le max\left\{\beta \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\beta (x\left|\right(y\left|y\right)\left)\right\}\hfill \end{array}\right).\hfill \end{array}$$

(12)

Lemma 7.

Every implicative intuitionistic fuzzy SBE-filter of L is also an intuitionistic fuzzy SBE-filter of L.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an implicative intuitionistic fuzzy SBE-filter of L. Then, $\alpha \left(1\right)\ge \alpha \left(x\right)$ and $\beta \left(1\right)\ge \beta \left(x\right)$ for all $x\in L$. Since:

$$\alpha \left(y\right)=\alpha \left(1\right|\left(y\right|y\left)\right)\ge min\left\{\alpha \right(1\left|\right(\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right)\left)\right)),\alpha (1\left|\right(x\left|x\right)\left)\right\}$$

and

$$\beta \left(y\right)=\beta \left(1\right|\left(y\right|y\left)\right)\le max\left\{\beta \right(1\left|\right(\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right)\left)\right)),\beta (1\left|\right(x\left|x\right)\left)\right\}$$

for all $x,y,z\in L$, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L. □

Example 3.

Let $\langle L,|,1\rangle$ with $L=\{0,u,v,1\}$ be an SBE-algebra having Cayley table in Example 1. We define intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ of L with the following membership functions:

$$\alpha \left(0\right)=0,\beta \left(0\right)=1,\alpha \left(u\right)=1,\beta \left(u\right)=0,$$

$$\alpha \left(v\right)=0.5,\beta \left(v\right)=0.5,\alpha \left(1\right)=1,\beta \left(1\right)=0.$$

This shows that L forms an intuitionistic fuzzy SBE-filter but not an implicative intuitionistic fuzzy SBE-filter.

Theorem 6.

Every intuitionistic fuzzy SBE-filter of L is also an intuitionistic fuzzy SBE-subalgebra of L.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-filter of L. Since $\left(\right(x\left|y\right)\left|\right(x\left|y\right)\left)\right|\left(y\right|y)=1$, $\left(x\right|y\left)\right|\left(x\right|y)\le y$ for all $x,y\in L$. Then:

$$\alpha \left(x\right|\left(y\right|y\left)\right)\ge \alpha \left(y\right)\ge \alpha \left(\right(x\left|y\right)\left|\right(x\left|y\right))\ge min\{\alpha \left(x\right),\alpha \left(y\right)\}$$

and

$$\beta \left(x\right|\left(y\right|y\left)\right)\le \beta \left(y\right)\le \beta \left(\right(x\left|y\right)\left|\right(x\left|y\right))\le max\{\beta \left(x\right),\beta \left(y\right)\}$$

for all $x,y\in L$. Thereby, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of L. □

Lemma 8.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-subalgebra of L, satisfying:

$$\begin{array}{cc}\hfill & (\forall x,y,z\in L)\left(\begin{array}{c}\alpha \left(x\right|\left(z\right|z\left)\right)\ge min\left\{\alpha \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\alpha (x\left|\right(y\left|y\right)\left)\right\},\hfill \\ \beta \left(x\right|\left(z\right|z\left)\right)\le max\left\{\beta \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\beta (x\left|\right(y\left|y\right)\left)\right\}.\hfill \end{array}\right)\hfill \end{array}$$

(13)

Then, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L.

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-subalgebra of L satisfying the condition (13). By Lemma 4, $\alpha \left(1\right)\ge \alpha \left(x\right)$ and $\beta \left(1\right)\le \beta \left(x\right)$ for all $x\in L$. Then, it is obtained from Lemma 1 (2) that:

$$\begin{array}{ccc}\hfill \alpha \left(y\right)& =& \alpha \left(1\right|\left(y\right|y\left)\right)\hfill \\ & \ge & min\left\{\alpha \right(1\left|\right(\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right)\left)\right)),\alpha (1\left|\right(x\left|x\right)\left)\right\}\hfill \\ & =& min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \beta \left(y\right)& =& \beta \left(1\right|\left(y\right|y\left)\right)\hfill \\ & \le & max\left\{\beta \right(1\left|\right(\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right)\left)\right)),\beta (1\left|\right(x\left|x\right)\left)\right\}\hfill \\ & =& max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}\hfill \end{array}$$

for all $x,y\in L$. Hence, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L. □

Theorem 7.

Let L be a self-distributive SBE-algebra. Then, every intuitionistic fuzzy SBE-filter of L is an implicative intuitionistic fuzzy SBE-filter of L.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-filter of a self-distributive SBE-algebra L. Since $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L, $\alpha \left(1\right)\ge \alpha \left(x\right)$ and $\beta \left(1\right)\le \beta \left(x\right)$ and for all $x\in L$. Then:

$min\left\{\alpha \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\alpha (x\left|\right(y\left|y\right)\left)\right\}$

$$\begin{array}{ccc}\alpha \left(x\right|\left(z\right|z\left)\right)& \ge & min\left\{\alpha \right(\left(x\right|\left(y\right|y\left)\right)\left|\right(\left(x\right|\left(z\right|z\left)\right)\left|\right(x\left|\right(z\left|z\right)\left)\right)),\alpha (x\left|\right(y\left|y\right)\left)\right\}\hfill \\ & =& min\left\{\alpha \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\alpha (x\left|\right(y\left|y\right)\left)\right\}\hfill \end{array}$$

and

$$\begin{array}{ccc}\beta \left(x\right|\left(z\right|z\left)\right)& \le & max\left\{\beta \right(\left(x\right|\left(y\right|y\left)\right)\left|\right(\left(x\right|\left(z\right|z\left)\right)\left|\right(x\left|\right(z\left|z\right)\left)\right)),\beta (x\left|\right(y\left|y\right)\left)\right\}\hfill \\ & =& max\left\{\beta \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\beta (x\left|\right(y\left|y\right)\left)\right\}\hfill \end{array}$$

for all $x,y,z\in L$. Thus, $\mathcal{L}=(L,\alpha ,\beta )$ is an implicative intuitionistic fuzzy SBE-filter of L. □

Lemma 9.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an(implicative)intuitionistic fuzzy SBE-filter of L. Then, the following subsets:

$$L_{\alpha }^1=\{x\in L:\alpha \left(x\right)=\alpha \left(1\right)\}$$

and

$$L_{\beta }^1=\{x\in L:\beta \left(x\right)=\beta \left(1\right)\}$$

are(implicative)SBE-filters of L.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-filter of L. Then, it is obvious that $1\in L_{\alpha }^1,L_{\beta }^1$. Assume that $x,x|\left(y\right|y)\in L_{\alpha }^1,L_{\beta }^1$. Since $\alpha \left(x\right)=\alpha \left(1\right)=\alpha \left(x\right|\left(y\right|y\left)\right)$, $\beta \left(x\right)=\beta \left(1\right)=\beta \left(x\right|\left(y\right|y\left)\right)$, $\alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}=min\left\{\alpha \right(1),\alpha (1\left)\right\}=\beta \left(1\right)$ and $\beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}=max\left\{\beta \right(1),\beta (1\left)\right\}=\beta \left(1\right)$, which imply that $\alpha \left(y\right)=\alpha \left(1\right)$ and $\beta \left(y\right)=\beta \left(1\right)$. Then, $y\in L_{\alpha }^1,L_{\beta }^1$. Hence, $L_{\alpha }^1,L_{\beta }^1$ are SBE-filters of S. Let $\mathcal{L}=(L,\alpha ,\beta )$ be an implicative intuitionistic fuzzy SBE-filter of L. Suppose that $x|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))),x|\left(y\right|y)\in L_{\alpha }^1,L_{\beta }^1$. Since:

$$\alpha \left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)=\alpha \left(1\right)=\alpha \left(x\right|\left(y\right|y\left)\right),$$

$$\beta \left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)=\beta \left(1\right)=\beta \left(x\right|\left(y\right|y\left)\right),$$

$$\begin{array}{ccc}\alpha \left(x\right|\left(z\right|z\left)\right)& \ge & min\left\{\alpha \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\alpha (x\left|\right(y\left|y\right)\left)\right\}\hfill \\ & =& min\left\{\alpha \right(1),\alpha (1\left)\right\}\hfill \\ & =& \alpha \left(1\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\beta \left(x\right|\left(z\right|z\left)\right)& \le & max\left\{\beta \right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)),\beta (x\left|\right(y\left|y\right)\left)\right\}\hfill \\ & =& max\left\{\beta \right(1),\beta (1\left)\right\}\hfill \\ & =& \beta \left(1\right),\hfill \end{array}$$

which imply that $\alpha \left(x\right|\left(z\right|z\left)\right)=\alpha \left(1\right)$, $\beta \left(x\right|\left(z\right|z\left)\right)=\beta \left(1\right)$. Thus, $x|\left(z\right|z)\in L_{\alpha }^1,L_{\beta }^1$. Therefore, $L_{\alpha }^1,L_{\beta }^1$ are implicative SBE-filters of L. □

Lemma 10.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L if and only if α and $\overline{\beta }$ are fuzzy SBE-filters of L.

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intutionstic fuzzy SBE-filter of L. Then, $\alpha$ is a fuzzy SBE-filter of L. Consider for every $x,y\in L$ we have, $\overline{\beta }\left(1\right)=1-\beta \left(1\right)\ge 1-\beta \left(x\right)=\overline{\beta }\left(x\right)$. Also:

$$\begin{array}{ccc}\overline{\beta }\left(y\right)& =& 1-\beta \left(y\right)\hfill \\ & \ge & 1-max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}\hfill \\ & =& min\{1-\beta (x\left|\right(y\left|y\right)),1-\beta (x\left)\right\}\hfill \\ & =& min\{\overline{\beta }\left(x\right|\left(y\right|y)),\overline{\beta }\left(x\right)\}.\hfill \end{array}$$

Hence, $\overline{\beta }$ is a fuzzy SBE-filter of L.

Conversely, let us take $\alpha$ and $\overline{\beta }$ are fuzzy SBE-filters of L. Then, obviously, for every $x\in L$, we have $\alpha \left(1\right)\ge \alpha \left(x\right)$, $1-\beta \left(1\right)=\overline{\beta }\left(1\right)\ge \overline{\beta }\left(x\right)=1-\beta \left(x\right)$, that is, $\beta \left(1\right)\le \beta \left(x\right)$. Moreover, $\alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}$ and:

$$\begin{array}{ccc}1-\beta \left(y\right)& =& \overline{\beta }\left(y\right)\hfill \\ & \ge & min\{\overline{\beta }\left(x\right|\left(y\right|y)),\overline{\beta }\left(x\right)\}\hfill \\ & =& min\{1-\beta (x\left|\right(y\left|y\right)),1-\beta (x\left)\right\}\hfill \\ & =& 1-max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}.\hfill \end{array}$$

Hemce, $\beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}$. Thus, $\mathcal{L}=(L,\alpha ,\beta )$ is an intutionstic fuzzy SBE-filter of L. □

Theorem 8.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L if and only if $(L,\alpha ,\overline{\alpha })$ and $(L,\beta ,\overline{\beta })$ are fuzzy SBE-filters of L.

Proof. 

If an intutionstic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L, then $\alpha =\overline{\overline{\alpha }}$ and $\beta$ are intuitionistic fuzzy SBE-filter of L by Lemma 10, and hence, $(L,\alpha ,\overline{\alpha })$ and $(L,\overline{\beta },\beta )$ are intuitionistic fuzzy SBE-filters of L.

Conversely, if $(L,\alpha ,\overline{\alpha })$ and $(L,\beta ,\overline{\beta })$ are intuitionistic fuzzy SBE-filters of L, then $\alpha$ and $\overline{\beta }$ are SBE-filters of L, and hence, the intutionstic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L. □

Theorem 9.

Let A be a nonempty subset of L and let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set defined by:

$$\alpha \left(x\right)=\left\{\begin{array}{cc}{\alpha }_0\hfill & \mathrm{if}x\in A,\hfill \\ {\alpha }_1\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

$$\beta \left(x\right)=\left\{\begin{array}{cc}{\beta }_0\hfill & \mathrm{if}x\in A,\hfill \\ {\beta }_1\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

for all $x\in L$, where ${\alpha }_i,{\beta }_i\in [0,1]$ such that ${\alpha }_0>{\alpha }_1$, ${\beta }_0<{\beta }_1$, and ${\alpha }_i+{\beta }_i\le 1$ for $i=0,1$.

Then, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L, and ${\mu }_{{\alpha }_0}=A={\gamma }_{{\beta }_0}$.

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L. Since $\alpha \left(1\right)\ge \alpha \left(x\right)$ and $\beta \left(1\right)\le \beta \left(x\right)$ for all $x\in L$, $\alpha \left(1\right)={\alpha }_1$ and $\beta \left(1\right)={\beta }_1$, and so, $1\in A$. Let $x,y\in L$ such that $x\left|\right(y\left|y\right),x\in A$, we get $\alpha \left(y\right)\ge min\{\alpha \left(x\right|\left(y\right|y)),\alpha \left(x\right)\}={\alpha }_1$ and $\beta \left(y\right)\le max\{\beta \left(x\right|\left(y\right|y)),\beta \left(x\right)\}={\beta }_1$, which implies that $\alpha \left(y\right)={\alpha }_1$ and $\beta \left(y\right)={\beta }_1$. It follows that $y\in A$. Therefore, A is an SBE-filter of L.

Conversely, suppose that A is an SBE-filter of L. Since $1\in A$, $\alpha \left(1\right)={\alpha }_1\ge \alpha \left(x\right)$ for all $x\in L$. Let $x,y\in L$. If $x\left|\right(y\left|y\right)\in L\setminus A$ or $x\in L\setminus A$, then $\alpha \left(x\right|\left(y\right|y))={\alpha }_2$ or $\alpha \left(x\right)={\alpha }_2$. It follows that $\alpha \left(y\right)\ge {\alpha }_2=min\{\alpha \left(x\right|\left(y\right|y)),\alpha \left(x\right)\}$. Also, if $x\left|\right(y\left|y\right)\in L\setminus A$ or $x\in L\setminus A$, then $\beta \left(x\right|\left(y\right|y))={\beta }_2$ or $\beta \left(x\right)={\beta }_2$. It follows that $\beta \left(y\right)\le {\beta }_2=max\{\beta \left(x\right|\left(y\right|y)),\beta \left(x\right)\}$. Assume that $x\left|\right(y\left|y\right),x\in A$. Then, $y\in A$ and thus $\alpha \left(y\right)={\alpha }_1=min\{\alpha \left(x\right|\left(y\right|y)),\alpha \left(x\right)\}$ and $\beta \left(y\right)={\beta }_1=max\{\beta \left(x\right|\left(y\right|y)),\beta \left(x\right)\}$. Hence, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L. □

Theorem 10.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-filter of L. Then, for every $t\in Im\left(\alpha \right)\cap Im\left(\beta \right)\cap [0,0.5]$, the subsets $U(\alpha ,t)$ and $L(\beta ,t)$ of L are SBE-filters of L.

Proof. 

Suppose that $(L,\alpha ,\beta )$ be an intuitionistic SBE-filter of L and $t\in Im\left(\alpha \right)\cap Im\left(\beta \right)\cap [0,0.5]$. Let $x\in U(\alpha ,t)$. Then, $\alpha \left(x\right)\ge t$. Since $(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L, $\alpha \left(1\right)\ge \alpha \left(x\right)\ge t$. Hence, $1\in U(\alpha ,t)$. Let $x\in L(\beta ,t)$. Then, $\beta \left(x\right)\le t$. Since $(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L, $\beta \left(1\right)\le \beta \left(x\right)\le t$. Hence, $1\in L(\beta ,t)$. Let $x,y\in L$ and $x\left|\right(y\left|y\right),x\in U(\alpha ,t)$. Then, $\alpha \left(x\right|\left(y\right|y\left)\right)\ge t$ and $\alpha \left(x\right)\ge t$. Since $(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L, $\alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}\ge t$. Hence, $y\in U(\alpha ,t)$. Let $x^{\prime }|\left(y^{\prime }\right|y^{\prime })\in L(\beta ,t)$. Then, $\beta \left(x^{\prime }\right|\left(y^{\prime }\right|y^{\prime }))\le t$ and $\beta \left(x^{\prime }\right)\le t$. Since $(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L, $\beta \left(y^{\prime }\right)\le max\{\beta \left(x^{\prime }\right|\left(y^{\prime }\right|y^{\prime })),\beta \left(x^{\prime }\right)\}\le t$. Hence, $x^{\prime }|\left(y^{\prime }\right|y^{\prime })\in L(\beta ,t)$. Thus, the sets $U(\alpha ,t)$ and $L(\beta ,t)$ are SBE-filters of L for every $t\in Im\left(\alpha \right)\cap Im\left(\beta \right)\cap [0,0.5]$. □

Corollary 2.

Let ${\chi }_M$ be the characteristic function of an intuitionistic fuzzy SBE-filter of L. Then, the intuitionistic fuzzy set $\overline{M}=(L,{\chi }_M,\overline{{\chi }_M})$ is an intuitionistic fuzzy SBE-filter of L.

Theorem 11.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of L if and only if, for all $s,t\in [0,1]$, the sets $U(\alpha ,t)$ and $L(\beta ,s)$ are either empty or SBE-filters of L.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intutionstic fuzzy SBE-filter of L and let $s,t\in [0,1]$ be such that $U(\alpha ,t)$ and $L(\beta ,s)$ are nonempty sets of L. It is clear that $1\in U(\alpha ,t)\cap L(\beta ,s)$, since $\alpha \left(1\right)\ge t$ and $\beta \left(1\right)\le s$. Let $x,y\in L$ and $x\left|\right(y\left|y\right),x\in U(\alpha ,t)$. Then, $\alpha \left(x\right|\left(y\right|y\left)\right)\ge t$ and $\alpha \left(x\right)\ge t$. Hence, $\alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}\ge t$ so that $y\in U(\alpha ,t)$. Hence, $U(\alpha ,t)$ is an SBE-filter of L. Let $x,y\in L$ and $x\left|\right(y\left|y\right),x\in L(\beta ,s)$. Then, $\beta \left(x\right|\left(y\right|y\left)\right)\le s$ and $\beta \left(x\right)\le s$. Hence, $\beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}\le s$ so that $y\in L(\beta ,s)$. Hence, $L(\beta ,s)$ is an SBE-filter of L. Assume now that every nonempty set $U(\alpha ,t)$ and $L(\beta ,s)$ are SBE-filter in L. If $\alpha \left(1\right)\ge \alpha \left(x\right)$ is not true for all $x\in L$, then there exists $x_0\in L$ such that $\alpha \left(1\right)<\alpha \left(x_0\right)$. However, in this case, for $s={\displaystyle \frac{1}{2}}(\alpha \left(1\right)+\alpha \left(x_0\right))$. Then, $x_0\in U(\alpha ,s)$, that is $U(\alpha ,s)\ne \varnothing$. Since by the assumption, $U(\alpha ,s)$ is an SBE-filter of L, then $\alpha \left(1\right)\ge s$, which is impossible. Hence, $\alpha \left(1\right)\ge \alpha \left(x\right)$. If $\beta \left(1\right)\le \beta \left(x\right)$ is not true, then there exists $y_0\in L$ such that $\beta \left(1\right)<\beta \left(y_0\right)$. However, in this case, for $s_0={\displaystyle \frac{1}{2}}(\beta \left(1\right)+\beta \left(y_0\right))$. Then, $y_0\in L(\beta ,s_0)$, that is $L(\beta ,s_0)\ne \varnothing$. Since by the assumption, $L(\beta ,s_0)$ is an SBE-filter of L, then $\beta \left(1\right)\le s_0$, which is impossible. Hence, $\beta \left(1\right)\le \beta \left(x\right)$.

Suppose that $\alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}$ is not true for all $x,y\in L$. Then, there exist $x_0,y_0\in L$ such that $\alpha \left(y_0\right)<min\{\alpha \left(x_0\right|\left(y_0\right|y_0)),\alpha \left(x_0\right)\}$. Taking $p={\displaystyle \frac{1}{2}}(\alpha \left(y_0\right)+min\{\alpha \left(x_0\right|\left(y_0\right|y_0)),\alpha \left(x_0\right)\})$. Then, we have $\alpha \left(y_0\right)<p<min\{\alpha \left(x_0\right|\left(y_0\right|y_0)),\alpha \left(x_0\right)\}$, which prove that $x_0,y_0\in U(\alpha ,p)$. Since $U(\alpha ,p)$ is an SBE-filter of L, $y_0\in U(\alpha ,p)$, which is a contradiction. Thus, $\alpha \left(y\right)\ge min\left\{\alpha \right(x\left|\right(y\left|y\right)),\alpha (x\left)\right\}$ is true for all $x,y\in L$. Suppose that $\beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}$ is not true for all $x,y\in L$. Then, there exist $x_0,y_0\in L$ such that $\beta \left(y_0\right)>max\{\beta \left(x_0\right|\left(y_0\right|y_0)),\beta \left(x_0\right)\}$. Taking $p_0={\displaystyle \frac{1}{2}}(\beta \left(y_0\right)+max\{\beta \left(x_0\right|\left(y_0\right|y_0)),\beta \left(x_0\right)\})$. Then, $\beta \left(y_0\right)>p_0>max\{\beta \left(x_0\right|\left(y_0\right|y_0)),\beta \left(x_0\right)\}$, which proves that $x_0,y_0\in L(\beta ,p_0)$. Since $L(\beta ,p_0)$ is an SBE-filter of L, $y_0\in L(\beta ,p_0)$, which is a contradiction. Thus, $\beta \left(y\right)\le max\left\{\beta \right(x\left|\right(y\left|y\right)),\beta (x\left)\right\}$ is true for all $x,y\in L$. Hence, $\mathcal{L}=(L,\alpha ,\beta )$ is an intutionstic fuzzy SBE-filter of L. □

Theorem 12.

Let $\{I_t:t\in \Delta \subset [0,1]\}$ be a collection of SBE-filters of L such that $L=\bigcup _{t\in \Delta }{I}_t$, and for all $s,t\in \Delta$, $s>t$ if and only if $I_s\subset I_t$. Then, the intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ defined by:

$$\alpha \left(x\right)=sup\{t\in \Delta :x\in I_t\}$$

and

$$\beta \left(x\right)=inf\{t\in \Delta :x\in I_t\}$$

for all $x\in L$ is an intuitionistic fuzzy SBE-filter of L.

Proof. 

According to Theorem 11, it is sufficient to show that the nonempty sets $U(\alpha ,t)$ and $L(\beta ,t)$ are SBE-filters of L. In order to prove that $U(\alpha ,t)$ is an SBE-filter of L, we divide the proof into the following two cases:

• $t=sup\{q\in \Delta :q<t\}$;
• $t\ne sup\{q\in \Delta :q<t\}$.

Case (1) implies that $x\in U(\alpha ,t)$⇔$x\in I_q$, $\forall q<t$⇔$x\in \bigcap _{q<t}{I}_q$, so that $U(\alpha ,t)=\bigcap _{q<t}{I}_q$, which is an SBE-filter of L. For case (2), we claim that $U(\alpha ,t)=\bigcup _{q\ge t}{I}_q$. If $x\in \bigcup _{q\ge t}{I}_q$, then $x\in I_q$ for some $q\ge t$. It follows that $\alpha \left(x\right)\ge q\ge t$ so that $x\in U(\alpha ,t)$. This shows that $\bigcup _{q\ge t}{I}_q\subset U(\alpha ,t)$. Now, assume that $x\notin \bigcup _{q\ge t}{I}_q$. Then, $x\notin I_q$ for all $q\ge t$. Since $t=sup\{q\in \Delta :q<t\}$, there exists $ϵ>0$ such that $(t-ϵ,t)\cap \Delta =\varnothing$. Hence, $x\notin I_q$ for all $q>t-ϵ$, which means that $x\in I_q$, then $q\le t-ϵ$. Thus, $\alpha \left(x\right)\le t-ϵ<t$, and so, $x\notin U(\alpha ,t)$. Therefore, $U(\alpha ,t)\subset \bigcup _{q\ge t}{I}_q$, and thus, $U(\alpha ,t)=\bigcup _{q\ge t}{I}_q$, which is an SBE-filter of L.

Next, we prove that $L(\beta ,t)$ is an SBE-filter of L. We consider the following two cases:

• $s=inf\{r\in \Delta :s<r\}$;
• $s\ne inf\{r\in \Delta :s<r\}$.

For case (3), we have $x\in L(\beta ,s)$⇔$x\in I_r$, $\forall s<r$⇔$x\in \bigcap _{s<r}{I}_r$, and hence, $L(\beta ,s)=\bigcap _{s<r}{I}_r$, which is an SBE-filter of L. For case (4), there exists $ϵ>0$ such that $(s,s+ϵ)\cap \Delta =\varnothing$. We will show that $L(\beta ,s)=\bigcup _{s\ge r}{I}_r$. If $x\in \bigcup _{s\ge r}{I}_r$, then $x\in I_r$ for some $r\le s$. It follows that $\beta \left(x\right)\le r\le s$ so that $x\in L(\beta ,s)$. Hence, $\bigcup _{s\ge r}{I}_r\subset L(\beta ,s)$.

Conversely, if $x\notin \bigcup _{s\ge r}{I}_r$, then $x\notin I_r$ for all $r\le s$, which implies that $x\notin I_r$ for all $r<s+ϵ$, that is, if $x\in I_r$, then $r\ge s+ϵ$. Thus, $\beta \left(x\right)\ge s+ϵ>s$, that is, $x\notin L(\beta ,s)$. Therefore, $L(\beta ,s)\subset \bigcup _{s\ge r}{I}_r$, and consequently, $L(\beta ,s)=\bigcup _{s\ge r}{I}_r$, which is an SBE-filter of L. □

5. Intuitionistic Fuzzy Ideals in SBE-Algebras

In this section, we introduce the concept of intuitionistic fuzzy SBE-ideals within the framework of Sheffer stroke BE-algebras. This study aims to explore how these ideals function in this specific algebraic context, enhancing our understanding of their properties and applications.

Definition 15.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ of L is called an intuitionistic fuzzy SBE-ideal of L if it satisfies the following conditions:

$$\begin{array}{cc}\hfill & (\forall x,y\in L)\left\{\begin{array}{c}\alpha \left(x\right|\left(y\right|y\left)\right)\ge \alpha \left(y\right)\hfill \\ \beta \left(x\right|\left(y\right|y\left)\right)\le \beta \left(y\right)\hfill \end{array}\right.,\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill & (\forall x,y,z\in L)\left\{\begin{array}{c}\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}\hfill \\ \beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}\hfill \end{array}\right..\hfill \end{array}$$

(15)

Example 4.

To provide an example of an intuitionistic fuzzy SBE-ideal in the SBE-algebra $\langle L,|,1\rangle$ with $L=\{0,u,v,1\}$ and the given Cayley table in Example 1, we define the fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$. Let $\alpha \left(0\right)=0.1,\beta \left(0\right)=0.9,\alpha \left(u\right)=0.5,\beta \left(u\right)=0.5,\alpha \left(v\right)=0.7,\beta \left(v\right)=0.3,\alpha \left(1\right)=0.9,$ and $\beta \left(1\right)=0.1.$ From the Cayley table, we compute the values of $y|y$ for each $y\in L$: $0|0=1,u|u=v,v|v=u$ and $1|1=0$.

We check the condition in (14) for all pairs $x,y\in L$. For instance, let $x=0$ and $y=0$. Then, $\alpha \left(0\right|\left(0\right|0\left)\right)=\alpha \left(1\right)=0.9\ge \alpha \left(0\right)=0.1$ and $\beta \left(0\right|\left(0\right|0\left)\right)=\beta \left(1\right)=0.1\le \alpha \left(0\right)=0.9.$ the condition in (14). We repeat this process for all possible pairs $x,y\in L$.

Let $x=0,y=u$ and $z=v$. We compute $z|z=u,y|\left(z\right|z)=u|u=v,x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)=0|u=1$. We check $\alpha \left(1\right)=0.9\ge min\left\{\alpha \right(0),\alpha (u\left)\right\}=min\{0.1,0.5\}=0.1$ and $\beta \left(1\right)=0.1\le max\left\{\beta \right(0),\beta (u\left)\right\}=max\{0.9,0.5\}=0.9.$ the condition in (15). We repeat this for all combinations of $x,y,z\in L$. Therefore, the fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ defined above satisfies both conditions for an intuitionistic fuzzy SBE-ideal in SBE-algebra $\langle L,|,1\rangle$.

Theorem 13.

Every intuitionistic fuzzy SBE-ideal of L is an intuitionistic fuzzy SBE-subalgebra of L.

Proof. 

It follows from (14). □

We concluded that every intuitionistic fuzzy SBE-ideal of L is an intuitionistic fuzzy SBE-subalgebra. However, the converse of this conclusion, that every intuitionistic fuzzy SBE-subalgebra is an intuitionistic fuzzy SBE-ideal, is not always true. We now provide an example to demonstrate this.

Example 5.

Consider an SBE-algebra $\langle L,|,1\rangle$, where $L=\{0,u,v,1\}$ and the Cayley table is given in Example 1. Let $S=\{0,u,v\}$ be a subset of L, and define the corresponding membership and nonmembership functions as follows:

$$\alpha \left(0\right)=1,\beta \left(0\right)=0,\alpha \left(u\right)=0.7,\beta \left(u\right)=0.3,\alpha \left(v\right)=0.6,\beta \left(v\right)=0.4.$$

The set S is an intuitionistic fuzzy SBE-subalgebra of L but not necessarily an an intuitionistic fuzzy SBE-ideal.

Theorem 14.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ in L is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$ if and only if the subsets $U(\alpha ,t)$ and $L(\beta ,s)$ are SBE-ideals of $\mathcal{L}=(L,|)$ whenever they are nonempty for all $s,t\in [0,1]$.

Proof. 

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$ and $L(\alpha ,t)\ne \varnothing \ne U(\beta ,s)$ for all $t,s\in [0,1]$. Let $x,y,a,b\in L$ be such that $(y,b)\in L(\beta ,s)\times U(\alpha ,t)$. This shows that $\alpha \left(b\right)\ge t$ and $\beta \left(y\right)\le s$. Then, $\alpha \left(a\right|\left(b\right|b\left)\right)\ge \alpha \left(b\right)\ge t$ and $\beta \left(x\right|\left(y\right|y\left)\right)\le \beta \left(y\right)\le s$. Thus, $\left(x\right|\left(y\right|y),a|\left(b\right|b\left)\right)\in L(\beta ,s)\times U(\alpha ,t)$. Let $x,y,z,a,b,c\in L$ be such that $(x,a)\in L(\beta ,s)\times U(\alpha ,t)$ and $(y,b)\in L(\beta ,s)\times U(\alpha ,t)$. Then: $\alpha \left(a\right)\ge t$, $\alpha \left(b\right)\ge t$, $\beta \left(x\right)\le s$ and $\beta \left(y\right)\le s$. Then,

$$\alpha \left(\right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)\left)\right|\left(c\right|c\left)\right)\ge min\left\{\alpha \right(a),\alpha (b\left)\right\}\ge t$$

and

$$\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}\le s$$

and so:

$$\left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z),(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)\left)\right|\left(c\right|c\left)\right)\in L(\beta ,s)\times U(\alpha ,t).$$

Therefore, $L(\beta ,s)$ and $U(\alpha ,t)$ are SBE-ideals of $\mathcal{L}=(L,|)$.

Conversely, let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set in L for which $L(\beta ,s)$ and $U(\alpha ,t)$ are SBE-ideals of $\mathcal{L}=(L,|)$ whenever they are nonempty for all $s,t\in [0,1]$. Suppose that $\beta \left(a\right|\left(b\right|b\left)\right)>\beta \left(b\right)$ for some $b\in L$. Then, $b\in L(\beta ,\beta (b\left)\right)$ but $a\left|\right(b\left|b\right)\notin L(\beta ,\beta (b\left)\right)$, which is a contradiction. Hence, $\beta \left(x\right|\left(y\right|y\left)\right)\le \beta \left(y\right)$ for all $y\in L$. Suppose that $\alpha \left(x\right|\left(y\right|y\left)\right)<\alpha \left(y\right)$ for some $y\in L$. Then, $y\in U(\alpha ,t(y\left)\right)$ but $x\left|\right(y\left|y\right)\notin U(\alpha ,t(y\left)\right)$, which is a contradiction. Hence, $\beta \left(x\right|\left(y\right|y\left)\right)\ge \beta \left(y\right)$ for all $y\in L$. Suppose that $\beta \left(\right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)\left)\right|\left(c\right|c\left)\right)>max\left\{\beta \right(a),\beta (b\left)\right\}$ or $\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)<min\left\{\alpha \right(x),\alpha (y\left)\right\}$ for some $a,b,c,x,y,z\in L$. Then, $a,b\in L(\beta ,s)$ or $x,y\in U(\alpha ,t)$ where $s=max\left\{\beta \right(a),\beta (b\left)\right\}$ and $t=min\left\{\alpha \right(x),\alpha (y\left)\right\}$. However:

$$\left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right)\left|\right(c\left|c\right)\notin L(\beta ,s)$$

or

$$\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\notin U(\alpha ,t),$$

which are contradictions. Therefore, $\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\ge min\left\{\alpha \right(x),\alpha (y\left)\right\}$ and $\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\le max\left\{\beta \right(x),\beta (y\left)\right\}$ for all $x,y\in L$. Consequently, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$. □

Theorem 15.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ in L is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$ if and only if the fuzzy sets α and $\overline{\beta }$ are fuzzy SBE-ideals of $\mathcal{L}=(L,|)$.

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$. It is clear that $\alpha$ is a fuzzy SBE-ideal of $\mathcal{L}=(L,|)$. For every $x,y\in L$:

$$\begin{array}{ccc}\overline{\beta }\left(x\right|\left(y\right|y))& =& 1-\beta \left(x\right|\left(y\right|y\left)\right)\hfill \\ & \ge & 1-\beta \left(y\right)\hfill \\ & =& \overline{\beta }\left(y\right),\hfill \end{array}$$

$$\begin{array}{ccc}\overline{\beta }\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))& =& 1-\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\hfill \\ & \ge & 1-max\left\{\beta \right(x),\beta (y\left)\right\}\hfill \\ & =& min\{1-\beta (x),1-\beta (y\left)\right\}\hfill \\ & =& min\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}.\hfill \end{array}$$

Hence, $\overline{\beta }$ is a fuzzy SBE-ideal of $\mathcal{L}=(L,|)$.

Conversely, let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set of $\mathcal{L}=(L,|)$ for which $\overline{\beta }$ and $\alpha$ are fuzzy SBE-ideals of $\mathcal{L}=(L,|)$. Let $x,y\in L$. Then:

$$\begin{array}{ccc}1-\beta \left(x\right|\left(y\right|y\left)\right)& =& \overline{\beta }\left(x\right|\left(y\right|y))\hfill \\ & \ge & \overline{\beta }\left(y\right)\hfill \\ & =& 1-\beta \left(y\right)\hfill \\ \beta \left(x\right|\left(y\right|y\left)\right)& \le & \beta \left(y\right),\hfill \end{array}$$

$$\begin{array}{ccc}1-\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& =& \overline{\beta }\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))\hfill \\ & \ge & min\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}\hfill \\ & =& min\{1-\beta (x),1-\beta (y\left)\right\}\hfill \\ & =& 1-max\left\{\beta \right(x),\beta (y\left)\right\}\hfill \\ \beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& \le & max\left\{\beta \right(x),\beta (y\left)\right\}.\hfill \end{array}$$

Hence, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$. □

Theorem 16.

Given a nonempty subset F of L, let ${\mathcal{L}}_F=(L,{\alpha }_F,{\beta }_F)$ be an intuitionistic fuzzy set in L defined as follows:

$$\alpha \left(x\right)=\left\{\begin{array}{cc}{\alpha }_0\hfill & \mathrm{if}x\in F,\hfill \\ {\alpha }_1\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

$$\beta \left(x\right)=\left\{\begin{array}{cc}{\beta }_0\hfill & \mathrm{if}x\in F,\hfill \\ {\beta }_1\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

for all $x\in L$ and ${\alpha }_i,{\beta }_i\in [0,1]$ such that ${\alpha }_0>{\alpha }_1$, ${\beta }_0<{\beta }_1$, and ${\alpha }_i+{\beta }_i\le 1$ for $i=0,1$.

Then, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$ if and only if F is an SBE-ideal of $\mathcal{L}=(L,|)$.

Proof. 

Assume that ${\mathcal{L}}_F=(L,{\alpha }_F,{\beta }_F)$ is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$. Let $x,y\in L$ be such that $x,y\in F$. Then, $\alpha \left(x\right|\left(y\right|y))\ge \alpha \left(y\right)={\alpha }_0,$$\beta \left(x\right|\left(y\right|y))\le \beta \left(y\right)={\beta }_0,$ and so $\alpha \left(x\right|\left(y\right|y))={\alpha }_0$ and $\beta \left(x\right|\left(y\right|y))={\beta }_0$. This shows that $x\left|\right(y\left|y\right)\in F$. Also:

$$\alpha \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))\ge min\{\alpha \left(x\right),\alpha \left(y\right)\}={\alpha }_0,$$

and

$$\beta \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))\le max\{\beta \left(x\right),\beta \left(y\right)\}={\beta }_0.$$

Thus:

$$\alpha \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))={\alpha }_0$$

and

$$\beta \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))={\beta }_0.$$

This shows that $\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\in F$. Therefore, F is a SBE-ideal of $\mathcal{L}=(L,|)$. Conversely, let F be a SBE-ideal of $\mathcal{L}=(L,|)$. For every $x,y\in L$, if $y\in F$, then $x\left|\right(y\left|y\right)\in F$, which implies that $\alpha \left(x\right|\left(y\right|y))={\alpha }_0=\alpha \left(y\right)$ and $\beta \left(x\right|\left(y\right|y))={\beta }_0=\beta \left(y\right)$. If $x\left|\right(y\left|y\right)\notin F$, then $\alpha \left(x\right|\left(y\right|y))={\alpha }_1<\alpha \left(y\right)$ and $\beta \left(x\right|\left(y\right|y))={\beta }_1>\beta \left(y\right)$. For every $x,y,z\in L$, if $x,y\in F$, then $\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\in F$ which implies that:

$$\alpha \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))={\alpha }_0=min\{\alpha \left(x\right),\alpha \left(y\right)\}$$

and

$$\beta \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))={\beta }_0=max\{\beta \left(x\right),\beta \left(y\right)\}.$$

If $x\notin F$ or $y\notin F$, then $\alpha \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))\ge {\alpha }_1=min\{\alpha \left(x\right),\alpha \left(y\right)\}$ and $\beta \left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))\le {\beta }_1=max\{\beta \left(x\right),\beta \left(y\right)\}$. Therefore, ${\mathcal{L}}_F=(L,{\alpha }_F,{\beta }_F)$ is an intuitionistic fuzzy SBE-ideal of $\mathcal{L}=(L,|)$. □

Proposition 2.

If ${\mathcal{L}}_i=\{(L,{\alpha }_i,{\beta }_i):i\in \Delta \}$ is a family of intuitionistic fuzzy SBE-ideals of L, then ${\displaystyle \underset{i\in \Delta }{\bigwedge }}{\mathcal{L}}_i$ is an intuitionistic fuzzy SBE-ideal of L.

Proof. 

Let ${\mathcal{L}}_i=\{(L,{\alpha }_i,{\beta }_i):i\in \Delta \}$ be a family of intuitionistic fuzzy SBE-ideals of L. Let $x,y\in L$, we have:

$$\left(\underset{i\in \Delta }{\bigwedge }{\alpha }_i\right)\left(x\right|\left(y\right|y))=\underset{i\in \Delta }{inf}\left\{{\alpha }_i\left(x\right|\left(y\right|y))\right\}\ge \underset{i\in \Delta }{inf}\left\{{\alpha }_i\left(y\right)\right\}=\left(\underset{i\in \Delta }{\bigwedge }{\alpha }_i\right)\left(y\right),$$

$$\left(\underset{i\in \Delta }{\bigwedge }{\beta }_i\right)\left(x\right|\left(y\right|y))=\underset{i\in \Delta }{sup}\left\{{\beta }_i\left(x\right|\left(y\right|y))\right\}\le \underset{i\in \Delta }{sup}\left\{{\beta }_i\left(y\right)\right\}=\left(\underset{i\in \Delta }{\bigwedge }{\beta }_i\right)\left(y\right).$$

Let $x,y,z\in L$, we have:

$\left(\underset{i\in \Delta }{\bigwedge }{\alpha }_i\right)\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))$

$$\begin{array}{ccc}& =& \underset{i\in \Delta }{inf}\left\{{\alpha }_i\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))\right\}\hfill \\ & \ge & \underset{i\in \Delta }{inf}\{min\{{\alpha }_i\left(x\right),{\alpha }_i\left(y\right)\}\}\hfill \\ & =& min\{\underset{i\in \Delta }{inf}{\alpha }_i\left(x\right),\underset{i\in \Delta }{inf}{\alpha }_i\left(y\right)\}\hfill \\ & =& min\{\left(\underset{i\in \Delta }{\bigwedge }{\alpha }_i\right)\left(x\right),\left(\underset{i\in \Delta }{\bigwedge }{\alpha }_i\right)\left(y\right)\},\hfill \end{array}$$

and

$\left(\underset{i\in \Delta }{\bigwedge }{\beta }_i\right)\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))$

$$\begin{array}{ccc}& =& \underset{i\in \Delta }{sup}\left\{{\beta }_i\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))\right\}\hfill \\ & \le & \underset{i\in \Delta }{sup}\{max\{{\beta }_i\left(x\right),{\beta }_i\left(y\right)\}\}\hfill \\ & =& max\{\underset{i\in \Delta }{sup}{\beta }_i\left(x\right),\underset{i\in \Delta }{sup}{\beta }_i\left(y\right)\}\hfill \\ & =& max\{\left(\underset{i\in \Delta }{\bigwedge }{\beta }_i\right)\left(x\right),\left(\underset{i\in \Delta }{\bigwedge }{\beta }_i\right)\left(y\right)\}.\hfill \end{array}$$

Hence, $\underset{i\in \Delta }{\bigwedge }{\mathcal{L}}_i$ is an intuitionistic fuzzy SBE-ideal of a SBE-algebra L. □

Proposition 3.

If $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of L and $a,b\in L$, then $\beta \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)\le \beta \left(a\right)$ and $\alpha \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)\ge \alpha \left(a\right)$.

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of L and $a,b\in L$. Using Lemma 1 (2) and (15), we get:

$$\begin{array}{ccc}\alpha \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)& =& \alpha \left(\right(a\left|\right(\left(1\right|\left(b\right|b\left)\right)\left|\right(1\left|\right(b\left|b\right)\left)\right)\left)\right|\left(b\right|b\left)\right)\hfill \\ & \ge & min\left\{\alpha \right(a),\alpha (1\left)\right\}\hfill \\ & =& min\left\{\alpha \right(a),\alpha (a\left|\right(a\left|a\right)\left)\right\}\hfill \\ & \ge & \alpha \left(a\right),\hfill \end{array}$$

$$\begin{array}{ccc}\beta \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)& =& \beta \left(\right(a\left|\right(\left(1\right|\left(b\right|b\left)\right)\left|\right(1\left|\right(b\left|b\right)\left)\right)\left)\right|\left(b\right|b\left)\right)\hfill \\ & \le & max\left\{\beta \right(a),\beta (1\left)\right\}\hfill \\ & =& max\left\{\beta \right(a),\beta (a\left|\right(a\left|a\right)\left)\right\}\hfill \\ & \le & \beta \left(a\right).\hfill \end{array}$$

□

Proposition 4.

Every fuzzy SBE-ideal $\mathcal{L}=(L,\alpha ,\beta )$ of L is order-preserving.

Proof. 

Let $a,b\in L$ be such that $a\preccurlyeq b$. Then, $a\left|\right(b\left|b\right)=1$. From Lemma 1 (2) and Proposition 3, we have $\beta \left(b\right)=\beta \left(1\right|\left(b\right|b\left)\right)=\beta \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)\le \beta \left(a\right)$ and $\alpha \left(b\right)=\alpha \left(1\right|\left(b\right|b\left)\right)=\alpha \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)\ge \alpha \left(a\right)$. Hence, $\mathcal{L}=(L,\alpha ,\beta )$ is order-preserving. □

Proposition 5.

If $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of L, then:

$$\begin{array}{c}\hfill (\forall a\in L)\left(\begin{array}{c}\alpha \left(1\right)\ge \alpha \left(a\right)\hfill \\ \beta \left(1\right)\le \beta \left(a\right)\hfill \end{array}\right).\end{array}$$

(16)

Proof. 

Let $a\in L$. By using (SBE1) and (14), we have $\beta \left(1\right)=\beta \left(a\right|\left(a\right|a\left)\right)\le \beta \left(a\right)$ and $\alpha \left(1\right)=\alpha \left(a\right|\left(a\right|a\left)\right)\ge \alpha \left(a\right)$. □

Proposition 6.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set of L, which satisfies:

$$\begin{array}{cc}\hfill & (\forall a,b,c\in L)\left(\begin{array}{c}\alpha \left(1\right)\ge \alpha \left(a\right)\hfill \\ \beta \left(1\right)\le \beta \left(a\right)\hfill \\ \alpha \left(a\right|\left(c\right|c\left)\right)\ge min\left\{\alpha \right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)),\alpha (b\left)\right\}\hfill \\ \beta \left(a\right|\left(c\right|c\left)\right)\le max\left\{\beta \right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)),\beta (b\left)\right\}\hfill \end{array}\right).\hfill \end{array}$$

(17)

Then, $\mathcal{L}=(L,\alpha ,\beta )$ is order-preserving.

Proof. 

Let $a,b\in L$ be such that $a\preccurlyeq b$. Then:

$$\begin{array}{ccc}\alpha \left(b\right)& =& \alpha \left(1\right|\left(b\right|b\left)\right)\hfill \\ & \ge & min\left\{\alpha \right(1\left|\right(\left(a\right|\left(b\right|b\left)\right)\left|\right(a\left|\right(b\left|b\right)\left)\right)),\alpha (a\left)\right\}\hfill \\ & =& min\left\{\alpha \right(1\left|\right(1\left|1\right)),\alpha (a\left)\right\}\hfill \\ & =& min\left\{\alpha \right(1),\alpha (a\left)\right\}\hfill \\ & =& \alpha \left(a\right).\hfill \end{array}$$

Hence, $\mathcal{L}=(L,\alpha ,\beta )$ is order-preserving. □

Theorem 17.

An intuitionistic fuzzy set $\mathcal{L}=(L,\alpha ,\beta )$ in a transitive SBE-algebra L is an intuitionistic fuzzy SBE-ideal of L if and only if it satisfies (17).

Proof. 

Assume that $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of L. By Proposition 5, we have $\alpha \left(1\right)\ge \alpha \left(a\right)$ and $\beta \left(1\right)\le \beta \left(a\right)$ for all $a\in L$. Since L is transitive, $\left(b\right|\left(c\right|c\left)\right)\left|\right(c\left|c\right)\preccurlyeq \left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right))$ for all $a,b,c\in L$. Thus, $\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(c\right|c\left)\right)\left|\right(\left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right|\left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right)=1$ for all $a,b,c\in L$. We consider:

$$\begin{array}{ccc}\alpha \left(a\right|\left(c\right|c\left)\right)& =& \alpha \left(1\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right)\hfill \\ & =& \alpha \left(\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(c\left|c\right)\left|\right(\left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right|\hfill \\ & & \left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right))\hfill \\ & \ge & min\left\{\alpha \right(\left(b\right|\left(c\right|c\left)\right)\left|\right(c\left|c\right)),\alpha (a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)\left)\right\}\hfill \\ & \ge & min\left\{\alpha \right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)),\alpha (b\left)\right\}\hfill \end{array}$$

and

$$\begin{array}{ccc}\beta \left(a\right|\left(c\right|c\left)\right)& =& \beta \left(1\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right)\hfill \\ & =& \beta \left(\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(c\left|c\right)\left|\right(\left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right|\hfill \\ & & \left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right))\hfill \\ & \le & max\left\{\beta \right(\left(b\right|\left(c\right|c\left)\right)\left|\right(c\left|c\right)),\beta (a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)\left)\right\}\hfill \\ & \le & max\left\{\beta \right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)),\beta (b\left)\right\},\hfill \end{array}$$

for all $a,b,c\in L$. Hence, the conditions (17) are true.

Conversely, using (17), (SBE1), and Lemma 1 (2), we have:

$$\begin{array}{ccc}\alpha \left(a\right|\left(b\right|b\left)\right)& \ge & min\left\{\alpha \right(a\left|\right(\left(b\right|\left(b\right|b\left)\right)\left|\right(b\left|\right(b\left|b\right)\left)\right),\alpha \left(b\right)\}\hfill \\ & =& min\left\{\alpha \right(a\left|\right(a\left|1\right)),\alpha (b\left)\right\}\hfill \\ & =& min\left\{\alpha \right(1),\alpha (b\left)\right\}\hfill \\ & =& \alpha \left(b\right),\hfill \end{array}$$

$$\begin{array}{ccc}\beta \left(a\right|\left(b\right|b\left)\right)& \le & max\left\{\beta \right(a\left|\right(\left(b\right|\left(b\right|b\left)\right)\left|\right(b\left|\right(b\left|b\right)\left)\right),\beta \left(b\right)\}\hfill \\ & =& max\left\{\beta \right(a\left|\right(a\left|1\right)),\beta (b\left)\right\}\hfill \\ & =& max\left\{\beta \right(1),\beta (b\left)\right\}\hfill \\ & =& \beta \left(b\right),\hfill \end{array}$$

$$\begin{array}{ccc}\alpha \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)& \ge & min\left\{\alpha \right(\left(a\right|\left(b\right|b\left)\right)\left|\right(\left(a\right|\left(b\right|b\left)\right)\left|\right(a\left|\right(b\left|b\right)\left)\right),\alpha \left(a\right)\}\hfill \\ & =& min\left\{\alpha \right(1),\alpha (a\left)\right\}\hfill \\ & =& \alpha \left(a\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\beta \left(\right(a\left|\right(b\left|b\right)\left)\right|\left(b\right|b\left)\right)& \le & max\left\{\beta \right(\left(a\right|\left(b\right|b\left)\right)\left|\right(\left(a\right|\left(b\right|b\left)\right)\left|\right(a\left|\right(b\left|b\right)\left)\right),\beta \left(a\right)\}\hfill \\ & =& max\left\{\beta \right(1),\beta (a\left)\right\}\hfill \\ & =& \beta \left(a\right),\hfill \end{array}$$

for all $a,b\in L$. Since $\mathcal{L}=(L,\alpha ,\beta )$ satisfies (17) and by Proposition 6, we have that $\mathcal{L}=(L,\alpha ,\beta )$ is order-preserving. Since L is transitive, we see that:

$$\beta \left(\right(b\left|\right(c\left|c\right)\left)\right|\left(c\right|c\left)\right)\ge \beta \left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right)$$

and

$$\alpha \left(\right(b\left|\right(c\left|c\right)\left)\right|\left(c\right|c\left)\right)\le \alpha \left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right|\left(\right(a\left|\right(c\left|c\right)\left)\right|\left(a\right|\left(c\right|c\left)\right)\left)\right),$$

for all $a,b,c\in L$. Hence, for all $a,b,c\in L$, we have:

$\alpha \left(\right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)\left)\right|\left(c\right|c\left)\right)$

$$\begin{array}{ccc}& \ge & min\left\{\alpha \right(\left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right)\left|\right(\left(a\right|\left(c\right|c\left)\right)\left|\right(a\left|\right(c\left|c\right)\left)\right)),\alpha (a\left)\right\}\hfill \\ & \ge & min\left\{\alpha \right(\left(b\right|\left(c\right|c\left)\right)\left|\right(c\left|c\right)),\alpha (a\left)\right\}\hfill \\ & \ge & min\left\{\alpha \right(a),\alpha (b\left)\right\},\hfill \end{array}$$

and

$\beta \left(\right(a\left|\right(\left(b\right|\left(c\right|c\left)\right)\left|\right(b\left|\right(c\left|c\right)\left)\right)\left)\right|\left(c\right|c\left)\right)$

$$\begin{array}{ccc}& \le & max\left\{\alpha \right(\left(a\right|\left(\right(b\left|\right(c\left|c\right)\left)\right|\left(b\right|\left(c\right|c\left)\right)\left)\right)\left|\right(\left(a\right|\left(c\right|c\left)\right)\left|\right(a\left|\right(c\left|c\right)\left)\right)),\beta (a\left)\right\}\hfill \\ & \le & max\left\{\beta \right(\left(b\right|\left(c\right|c\left)\right)\left|\right(c\left|c\right)),\beta (a\left)\right\}\hfill \\ & \le & max\left\{\beta \right(a),\beta (b\left)\right\}.\hfill \end{array}$$

Therefore, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of L. □

Corollary 3.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set in a self-distributive SBE-algebra L. Then, $\mathcal{L}=(L,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of L if and only if it satisfies conditions (17).

Proof. 

Straightforward. □

Definition 16.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set of L. The intuitionistic fuzzy set $\overline{\mathcal{L}}=(L,\overline{\alpha },\overline{\beta })$ defined by: for all $x\in L$, $\overline{\alpha }=1-\alpha \left(x\right),$ $\overline{\beta }=1-\beta \left(x\right),$ is called the complement of $\mathcal{L}=(L,\beta ,\alpha )$ in L.

Theorem 18.

Let $\mathcal{L}=(L,\alpha ,\beta )$ be an intuitionistic fuzzy set in L. Then, $\overline{\mathcal{L}}=(L,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-subalgebra of L if and only if for all $s,t\in [0,1]$, $U(\beta ,s)$ and $L(\alpha ,t)$ are SBE-subalgebras of L if $U(\beta ,s)$ and $L(\alpha ,t)$ are nonempty.

Proof. 

Assume that $\overline{\mathcal{L}}=(L,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-subalgebra of L. Let $s,t\in [0,1]$ be such that $U(\beta ,s)$ and $L(\alpha ,t)$ are nonempty. (i) Let $x,y\in U(\beta ,s)$. Then, $\beta \left(x\right)\ge s$ and $\beta \left(y\right)\ge s$, so s is a lower bound of $\left\{\beta \right(x),\beta (y\left)\right\}$. Since $\overline{\mathcal{L}}=(L,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-subalgebra of L, $\overline{\beta }\left(\left(x\right|\left(y\right|y))\right|\left(x\right|\left(y\right|y)))\le max\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}$. By Lemma 3 (1), we have $1-\beta \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\le max\{1-\beta \left(x\right),1-\beta \left(y\right)\}=1-min\{\beta \left(x\right),\beta \left(y\right)\}.$ Thus, $\beta \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\ge min\{\beta \left(x\right),\beta \left(y\right)\}\ge s$, and so, $\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right))\in U(\beta ,s)$. Therefore, $U(\beta ,s)$ is a SBE-subalgebra of L. Let $x,y\in L(\alpha ,t)$. Then, $\alpha \left(x\right)\le t$ and $\alpha \left(y\right)\le t$, so t is an upper bound of $\left\{\alpha \right(x),\alpha (y\left)\right\}$. Since $\overline{\mathcal{L}}=(L,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-subalgebra of L, $\overline{\alpha }\left(\left(x\right|\left(y\right|y))\right|\left(x\right|\left(y\right|y)))\ge min\{\overline{\alpha }\left(x\right),\overline{\alpha }\left(y\right)\}$. By Lemma 3 (2), we have $1-\alpha \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\ge min\{1-\alpha \left(x\right),1-\alpha \left(y\right)\}=1-max\{\alpha \left(x\right),\alpha \left(y\right)\}.$ Thus, $\alpha \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\le max\{\alpha \left(x\right),\alpha \left(y\right)\}\le t$, and so, $\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right))\in L(\alpha ,t)$. Therefore, $L(\alpha ,t)$ is a SBE-subalgebra of L.

Conversely, assume that for all $s,t\in [0,1]$, $U(\beta ,s)$ and $L(\alpha ,t)$ are SBE-subalgebras of L if $U(\beta ,s)$ and $L(\alpha ,t)$ are nonempty.

(i). Let $x,y\in L$. Then, $\beta \left(x\right),\beta \left(y\right)\in [0,1]$. Choose $s=min\left\{\beta \right(x),\beta (y\left)\right\}$. Thus, $\beta \left(x\right)\ge s$ and $\beta \left(y\right)\ge s$, so $x,y\in U(\beta ,s)\ne \varnothing$. By assumption, we have that $U(\beta ,s)$ is a SBE-subalgebra of L, and so, $\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right))\in U(f,s)$. Thus $\beta \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\ge s=min\{\beta \left(x\right),\beta \left(y\right)\}$. By Lemma 3 (1), we have:

$$\begin{array}{ccc}\overline{\beta }\left(\left(x\right|\left(y\right|y))\right|\left(x\right|\left(y\right|y)))& =& 1-\beta \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\hfill \\ & \le & 1-min\{f-(x),\beta (y\left)\right\}\hfill \\ & =& max\{1-\beta (x),1-\beta (y\left)\right\}\hfill \\ & =& max\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}.\hfill \end{array}$$

(ii). Let $x,y\in L$. Then, $\alpha \left(x\right),\alpha \left(y\right)\in [0,1]$. Choose $t=max\left\{\alpha \right(x),\alpha (y\left)\right\}$. Thus, $\alpha \left(x\right)\le t$ and $\alpha \left(y\right)\le t$, so $x,y\in L(\alpha ,t)\ne \varnothing$. By assumption, we have $L(\alpha ,t)$ is a SBE-subalgebra of L, and so, $\left(x\right|\left(y\right|y\left)\right)\left|\right(x\left|\right(y\left|y\right))\in L(\alpha ,t)$. Thus, $\alpha \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\le t=max\{\alpha \left(x\right),\alpha \left(y\right)\}.$ By Lemma 3 (2), we have:

$$\begin{array}{ccc}\overline{\alpha }\left(\left(x\right|\left(y\right|y))\right|\left(x\right|\left(y\right|y)))& =& 1-\alpha \left(\right(x\left|\right(y\left|y\right)\left)\right|\left(x\right|\left(y\right|y\left)\right))\hfill \\ & \ge & 1-max\left\{\alpha \right(x),\alpha (y\left)\right\}\hfill \\ & =& min\{1-\alpha (x),1-\alpha (y\left)\right\}\hfill \\ & =& min\{\overline{\alpha }\left(x\right),\overline{\alpha }\left(y\right)\}.\hfill \end{array}$$

Hence, $\overline{\mathcal{L}}=(L,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-subalgebra of L. □

Theorem 19.

Let $\overline{\mathcal{L}}=(L,\overline{\beta },\overline{\alpha })$ be an intuitionistic fuzzy set in L. Then, $\mathcal{L}=(L,\beta ,\alpha )$ is an intuitionistic fuzzy SBE-ideal of L if and only if for all $s,t\in [0,1]$, $U(\beta ,s)$ and $L(\alpha ,t)$ are SBE-ideals of L if $U(\beta ,s)$ and $L(\alpha ,t)$ are nonempty.

Proof. 

Assume that $\overline{\mathcal{L}}=(A,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-ideal of L. Let $s,t\in [0,1]$ be such that $U(\beta ,s)$ and $L(\alpha ,t)$ are nonempty. Let $x\in L$ be such that $x\in U(\beta ,s)$. Then, $\beta \left(y\right)\ge s$. Since $\mathcal{L}=(A,\beta ,\alpha )$ is an intuitionistic fuzzy SBE-ideal of L, we have:

$$\begin{array}{ccc}\overline{\beta }\left(x\right|\left(y\right|y))& \le & \overline{\beta }\left(y\right)\hfill \\ 1-\alpha \left(x\right|\left(y\right|y\left)\right)& \le & 1-\alpha \left(y\right)\hfill \\ \alpha \left(x\right|\left(y\right|y\left)\right)& \ge & \alpha \left(y\right)\hfill \\ & \ge & s.\hfill \end{array}$$

Hence, $x\left|\right(y\left|y\right)\in U(\beta ,s)$. Next, let $x,y,z\in L$ be such that $x,y\in U(f,s)$. Then, $\beta \left(x\right)\ge s$ and $\beta \left(y\right)\ge s$. Since $\overline{\mathcal{L}}=(A,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-ideal of L, we have:

$$\begin{array}{ccc}\overline{\beta }\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))& \le & max\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}\hfill \\ 1-\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& \le & max\{1-\beta (x),1-\beta (y\left)\right\}\hfill \\ 1-\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& \le & 1-min\left\{\beta \right(x),\beta (y\left)\right\}\hfill \\ \beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& \ge & min\left\{\beta \right(x),\beta (y\left)\right\}\hfill \\ & \ge & s.\hfill \end{array}$$

Hence, $\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\in U(\beta ,s)$. Therefore, $U(f,t)$ is an SBE-ideal of L. Let $x,y\in L$ be such that $x\in L(\alpha ,t)$. Then, $\alpha \left(x\right)\ge t$. Since $\mathcal{L}=(A,\beta ,\alpha )$ is an intuitionistic fuzzy SBE-ideal of L, we have:

$$\begin{array}{ccc}\overline{\alpha }\left(x\right|\left(y\right|y))& \ge & \overline{\alpha }\left(y\right)\hfill \\ 1-\alpha \left(x\right|\left(y\right|y\left)\right)& \ge & 1-\alpha \left(y\right)\hfill \\ \alpha \left(x\right|\left(y\right|y\left)\right)& \le & \alpha \left(y\right)\hfill \\ & \le & s.\hfill \end{array}$$

Hence, $x\left|\right(y\left|y\right)\in L(f,s)$. Next, let $x,y,z\in L$ be such that $x,y\in L(\alpha ,t)$. Then, $\alpha \left(x\right)\ge t$ and $\alpha \left(y\right)\ge t$. Since $\overline{\mathcal{L}}=(A,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-ideal of L, we have:

$$\begin{array}{ccc}\overline{\alpha }\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))& \le & max\{\overline{\alpha }\left(x\right),\overline{\alpha }\left(y\right)\}\hfill \\ 1-\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& \le & max\{1-\alpha (x),1-\alpha (y\left)\right\}\hfill \\ 1-\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& \le & 1-min\left\{\alpha \right(x),\alpha (y\left)\right\}\hfill \\ \alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)& \ge & min\left\{\alpha \right(x),\alpha (y\left)\right\}\hfill \\ & \ge & t.\hfill \end{array}$$

Hence $\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\in L(\alpha ,t)$. Therefore, $L(\alpha ,t)$ is an SBE-ideal of L.

Conversely, assume that for all $s,t\in [0,1]$, $U(\beta ,s)$ and $L(\alpha ,t)$ are SBE-ideals of L if $U(\beta ,s)$ and $L(\alpha ,t)$ are nonempty. Let $x\in L$. Then, $\beta \left(x\right)\in [0,1]$. Choose $s=\beta \left(x\right)$. Thus, $\beta \left(x\right|\left(y\right|y\left)\right)\ge \beta \left(y\right)=s$, so $x\left|\right(y\left|y\right)\in U(\beta ,s)\ne \varnothing$. By assumption, we have that $U(\beta ,s)$ is an SBE-ideal of L, and so, $x\left|\right(y\left|y\right)\in U(\beta ,s)$. Thus, $\beta \left(x\right|\left(y\right|y\left)\right)\ge s=\beta \left(y\right)$, and so, $\overline{\beta }\left(x\right|\left(y\right|y))=1-\beta \left(x\right|\left(y\right|y))\le 1-\beta \left(y\right)=\overline{\beta }\left(y\right)$. Let $x,y,z\in L$. Then, $\beta \left(x\right),\beta \left(y\right)\in [0,1]$. Choose $s=min\left\{\beta \right(x),\beta (y\left)\right\}$. Thus, $\beta \left(x\right)\ge s$ and $\beta \left(y\right)\ge s$, so $x,y\in U(\beta ,s)\ne \varnothing$. By assumption, we have $U(\beta ,s)$ is an SBE-ideal of L and so $\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\in U(\beta ,s)$. Thus, $\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\ge s=min\left\{\beta \right(x),\beta (y\left)\right\}$. By Lemma 3 (1), we have:

$$\begin{array}{ccc}\overline{\beta }\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))& =& 1-\beta \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\hfill \\ & \le & 1-min\left\{\beta \right(x),\beta (y\left)\right\}\hfill \\ & =& max\{1-\beta (x),1-\beta (y\left)\right\}\hfill \\ & =& max\{\overline{\beta }\left(x\right),\overline{\beta }\left(y\right)\}.\hfill \end{array}$$

Let $x,y\in L$. Then, $\alpha \left(x\right)\in [0,1]$. Choose $t=\alpha \left(x\right)$. Thus, $\alpha \left(x\right|\left(y\right|y\left)\right)\le \alpha \left(y\right)=t$, so $y\in L(\alpha ,t)\ne \varnothing$. By assumption, we have $L(\alpha ,t)$ is an SBE-ideal of L and so $x\left|\right(y\left|y\right)\in L(\alpha ,t)$. Thus, $\alpha \left(x\right|\left(y\right|y\left)\right)\ge t=\alpha \left(y\right)$, and so, $\overline{\alpha }\left(x\right|\left(y\right|y))=1-\alpha \left(x\right|\left(y\right|y))\ge 1-\alpha \left(y\right)=\overline{\alpha }\left(y\right)$. Let $x,y,z\in L$. Then, $\alpha \left(x\right),\alpha \left(y\right)\in [0,1]$. Choose $t=max\left\{\alpha \right(x),\alpha (y\left)\right\}$. Thus, $\alpha \left(x\right)\le t$ and $\alpha \left(y\right)\le t$, so $x,y\in L(\alpha ,t)\ne \varnothing$. By assumption, we have $L(\alpha ,t)$ is an SBE-ideal of L and so $\left(x\right|\left(\right(y\left|\right(z\left|z\right)\left)\right|\left(y\right|\left(z\right|z\left)\right)\left)\right)\left|\right(z\left|z\right)\in L(\alpha ,t)$. Thus, $\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\ge t=max\left\{\alpha \right(x),\alpha (y\left)\right\}$. By Lemma 3 (1), we have:

$$\begin{array}{ccc}\overline{\alpha }\left(\left(x\right|\left(\left(y\right|\left(z\right|z))\right|\left(y\right|\left(z\right|z))))\right|\left(z\right|z))& =& 1-\alpha \left(\right(x\left|\right(\left(y\right|\left(z\right|z\left)\right)\left|\right(y\left|\right(z\left|z\right)\left)\right)\left)\right|\left(z\right|z\left)\right)\hfill \\ & \le & 1-max\left\{\alpha \right(x),\alpha (y\left)\right\}\hfill \\ & =& min\{1-\alpha (x),1-\alpha (y\left)\right\}\hfill \\ & =& min\{\overline{\alpha }\left(x\right),\overline{\alpha }\left(y\right)\}.\hfill \end{array}$$

Hence, $\overline{\mathcal{L}}=(A,\overline{\beta },\overline{\alpha })$ is an intuitionistic fuzzy SBE-ideal of L. □

The proof of the following theorem is similar to the proof of Theorem 19.

Theorem 20.

Let $\overline{\mathcal{L}}=(A,\overline{\beta },\overline{\alpha })$ be an intuitionistic fuzzy set in L. Then, $\mathcal{L}=(A,\beta ,\alpha )$ is an intuitionistic fuzzy SBE-filter of L if and only if for all $s,t\in [0,1]$, $U(\beta ,s)$ and $L(\alpha ,t)$ are SBE-filters of L if $U(\beta ,s)$ and $L(\alpha ,t)$ are nonempty.

Definition 17

([26]). Let $\langle A,{|}_A,1_A\rangle$ and $\langle B,{|}_B,1_B\rangle$ be SBE-algebras. Then, a mapping $f:A\to B$ is called a homomorphism if $f\left(x{|}_{A}y\right)=f\left(x\right){|}_{B}f\left(y\right)$ for all $x,y\in L$ and $f\left(1_A\right)=1_B$.

Theorem 21.

Let $\langle A,{|}_A,1_A\rangle$ and $\langle B,{|}_B,1_B\rangle$ be SBE-algebras, $f:A\to B$ be a surjective homomorphism and $\mathcal{B}=(B,\alpha ,\beta )$ be an intuitionistic fuzzy set on B. Then, $\mathcal{B}=(B,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of B if and only if ${\mathcal{B}}^f=(A,{\alpha }^f,{\beta }^f)$ is an intuitionistic fuzzy SBE-subalgebra of L, where ${\alpha }^f\to [0,1]$ on L are defined by ${\alpha }^f=\alpha \left(f\left(x\right)\right)$ for all $x\in L$, respectively.

Proof. 

Let $\langle A,{|}_A,1_A\rangle$ and $\langle B,{|}_B,1_B\rangle$ be SBE-algebras, $f:A\to B$ be a surjective homomorphism, and $\mathcal{B}=(B,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-subalgebra of B. Let $x_1,x_2\in A$. Then:

${\alpha }^f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)$

$$\begin{array}{ccc}& =& \alpha \left(f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\right)\hfill \\ & =& \alpha \left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right){|}_B\left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right)\right)\right)\hfill \\ & \ge & min\{\alpha \left(f\left(x_1\right)\right),\alpha \left(f\left(x_2\right)\right)\}\hfill \\ & =& min\{{\alpha }^f\left(x_1\right),{\alpha }^f\left(x_2\right)\},\hfill \end{array}$$

and

${\beta }^f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)$

$$\begin{array}{ccc}& =& \beta \left(f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\right)\hfill \\ & =& \beta \left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right){|}_B\left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right)\right)\right)\hfill \\ & \le & max\{\beta \left(f\left(x_1\right)\right),\beta \left(f\left(x_2\right)\right)\}\hfill \\ & =& max\{{\beta }^f\left(x_1\right),{\beta }^f\left(x_2\right)\}.\hfill \end{array}$$

Hence, ${\mathcal{B}}^f=(A,{\alpha }^f,{\beta }^f)$ is an intuitionistic fuzzy SBE-subalgebra of L.

Conversely, let ${\mathcal{B}}^f=(A,{\alpha }^f,{\beta }^f)$ be an intuitionistic fuzzy SBE-subalgebra of L. Let $y_1,y_2\in B$ such that $f\left(x_1\right)=y_1$ and $f\left(x_2\right)=y_2$ for $x_1,x_2\in A$. Then:

$\alpha \left(\left(y_1{|}_B\left(y_2{|}_B{y}_2\right)\right){|}_B\left(y_1{|}_B\left(y_2{|}_B{y}_2\right)\right)\right)$

$$\begin{array}{ccc}& =& \alpha \left(\left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right)\right){|}_B\left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right)\right)\right)\hfill \\ & =& \alpha \left(f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\right)\hfill \\ & =& {\alpha }^f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\hfill \\ & \ge & min\{{\alpha }^f\left(x_1\right),{\alpha }^f\left(x_2\right)\}\hfill \\ & =& min\{\alpha \left(f\left(x_1\right)\right),\alpha \left(f\left(x_2\right)\right)\}\hfill \\ & =& min\{\alpha \left(y_1\right),\alpha \left(y_2\right)\}\hfill \end{array}$$

and

$\beta \left(\left(y_1{|}_B\left(y_2{|}_B{y}_2\right)\right){|}_B\left(y_1{|}_B\left(y_2{|}_B{y}_2\right)\right)\right)$

$$\begin{array}{ccc}& =& \beta \left(\left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right)\right){|}_B\left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right)\right)\right)\hfill \\ & =& \beta \left(f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\right)\hfill \\ & =& {\beta }^f\left(\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right){|}_A\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\hfill \\ & \le & max\{{\beta }^f\left(x_1\right),{\beta }^f\left(x_2\right)\}\hfill \\ & =& max\{\beta \left(f\left(x_1\right)\right),\beta \left(f\left(x_2\right)\right)\}\hfill \\ & =& max\{\beta \left(y_1\right),\beta \left(y_2\right)\}.\hfill \end{array}$$

Hence, $\mathcal{B}=(B,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-subalgebra of B. □

Theorem 22.

Let $\langle A,{|}_A,1_A\rangle$ and $\langle B,{|}_B,1_B\rangle$ be SBE-algebras, and let $f:A\to B$ be a surjective homomorphism. If $\mathcal{B}=(B,\alpha ,\beta )$ is an intuitionistic fuzzy set on B, then $\mathcal{B}=(B,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of B if and only if ${\mathcal{B}}^f=(A,{\alpha }^f,{\beta }^f)$ is an intuitionistic fuzzy SBE-ideal of A.

Proof. 

Let $\langle A,{|}_A,1_A\rangle$ and $\langle B,{|}_B,1_B\rangle$ be SBE-algebras, $f:A\to B$ be a surjective homomorphism, and $\mathcal{B}=(B,\alpha ,\beta )$ be an intuitionistic fuzzy SBE-ideal of B. Let $x_1,x_2\in A$. Then:

$$\begin{array}{ccc}{\alpha }^f\left(x_2\right)& =& \alpha \left(f\left(x_2\right)\right)\hfill \\ & \ge & \alpha \left(f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\hfill \\ & =& {\alpha }^f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right),\hfill \end{array}$$

${\alpha }^f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)$

$$\begin{array}{ccc}& =& \alpha \left(f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)\right)\hfill \\ & \ge & min\{\alpha \left(f\left(x_2\right)\right),\alpha \left(f\left(x_1\right)\right)\}\hfill \\ & =& min\{{\alpha }^f\left(x_2\right),{\alpha }^f\left(x_2\right)\},\hfill \end{array}$$

$$\begin{array}{ccc}{\beta }^f\left(x_2\right)& =& \beta \left(f\left(x_2\right)\right)\hfill \\ & \le & \beta \left(f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\hfill \\ & =& {\beta }^f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right),\hfill \end{array}$$

and

${\beta }^f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)$

$$\begin{array}{ccc}& =& \beta \left(f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)\right)\hfill \\ & \le & max\{\beta \left(f\left(x_1\right)\right),\beta \left(f\left(x_2\right)\right)\}\hfill \\ & =& max\{{\beta }^f\left(x_1\right),{\beta }^f\left(x_2\right)\}.\hfill \end{array}$$

Hence, ${\mathcal{B}}^f=(A,{\alpha }^f,{\beta }^f)$ is an intuitionistic fuzzy SBE-ideal of L.

Conversely, let ${\mathcal{B}}^f=(A,{\alpha }^f,{\beta }^f)$ be an intuitionistic fuzzy SBE-ideal of L. Let $y_1,y_2\in B$ such that $f\left(x_1\right)=y_1$ and $f\left(x_2\right)=y_2$ for $x_1,x_2\in A$. Then:

$$\begin{array}{ccc}\alpha \left(x_2\right)& =& \alpha \left(f\left(x_2\right)\right)\hfill \\ & \ge & {\alpha }^f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\hfill \\ & =& \alpha \left(f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\hfill \\ & =& \alpha \left(f\left(x_1\right)\right|\left(f\left(x_2\right)\right|f\left(x_2\right)))\hfill \\ & =& \alpha \left(y_1\right|\left(y_2\right|y_2)),\hfill \end{array}$$

$\alpha \left(\left(y_1{|}_B\left(\left(y_2{|}_B\left(y_3{|}_B{y}_3\right)\right){|}_B\left(y_2{|}_B\left(y_3{|}_B{y}_3\right)\right)\right)\right)\right)$

$$\begin{array}{ccc}& =& \alpha \left(\left(f\left(x_1\right){|}_B\left(\left(f\left(x_2\right){|}_B\left(f\left(x_3\right){|}_{B}f\left(x_3\right)\right)\right){|}_B\left(f\left(x_2\right){|}_B\left(f\left(x_3\right){|}_{B}f\left(x_3\right)\right)\right)\right)\right)\right)\hfill \\ & =& \alpha \left(f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)\right)\hfill \\ & =& {\alpha }^f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)\hfill \\ & \ge & min\{{\alpha }^f\left(x_1\right),{\alpha }^f\left(x_2\right)\}\hfill \\ & =& min\{\alpha \left(f\left(x_1\right)\right),\alpha \left(f\left(x_2\right)\right)\}\hfill \\ & =& min\{\alpha \left(y_1\right),\alpha \left(y_2\right)\},\hfill \end{array}$$

$$\begin{array}{ccc}\beta \left(x_2\right)& =& \beta \left(f\left(x_2\right)\right)\hfill \\ & \le & {\beta }^f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\hfill \\ & =& \beta \left(f\left(x_1{|}_A\left(x_2{|}_A{x}_2\right)\right)\right)\hfill \\ & =& \beta \left(f\left(x_1\right){|}_B\left(f\left(x_2\right){|}_{B}f\left(x_2\right)\right)\right)\hfill \\ & =& \beta \left(y_1{|}_B\left(y_2{|}_B{y}_2\right)\right)\hfill \end{array}$$

and

$\beta \left(\left(y_1{|}_B\left(\left(y_2{|}_B\left(y_3{|}_B{y}_3\right)\right){|}_B\left(y_2{|}_B\left(y_3{|}_B{y}_3\right)\right)\right)\right)\right)$

$$\begin{array}{ccc}& =& \beta \left(\left(f\left(x_1\right){|}_B\left(\left(f\left(x_2\right){|}_B\left(f\left(x_3\right){|}_{B}f\left(x_3\right)\right)\right){|}_B\left(f\left(x_2\right){|}_B\left(f\left(x_3\right){|}_{B}f\left(x_3\right)\right)\right)\right)\right)\right)\hfill \\ & =& \beta \left(f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)\right)\hfill \\ & =& {\beta }^f\left(\left(x_1{|}_A\left(\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right){|}_A\left(x_2{|}_A\left(x_3{|}_A{x}_3\right)\right)\right)\right)\right)\hfill \\ & \le & max\{{\beta }^f\left(x_1\right),{\beta }^f\left(x_2\right)\}\hfill \\ & =& max\{\beta \left(f\left(x_1\right)\right),\beta \left(f\left(x_2\right)\right)\}\hfill \\ & =& max\{\beta \left(y_1\right),\beta \left(y_2\right)\}.\hfill \end{array}$$

Hence, $\mathcal{B}=(B,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-ideal of B. □

The proof of the following theorem is similar to the proof of Theorem 22.

Theorem 23.

Let $\langle A,{|}_A,1_A\rangle$ and $\langle B,{|}_B,1_B\rangle$ be SBE-algebras, and let $f:A\to B$ be a surjective homomorphism. If $\mathcal{B}=(B,\alpha ,\beta )$ is an intuitionistic fuzzy set on B, then $\mathcal{B}=(B,\alpha ,\beta )$ is an intuitionistic fuzzy SBE-filter of B if and only if ${\mathcal{B}}^f=(A,{\alpha }^f,{\beta }^f)$ is an intuitionistic fuzzy SBE-filter of A.

6. Conclusions and Future Work

In conclusion, the Sheffer stroke operation’s capacity to construct entire logical systems underscores its foundational role across various branches of logic and algebra. Our work extends this versatility into the realm of intuitionistic fuzzy SBE-algebras, establishing a structured relationship between subalgebras, ideals, filters, and level sets within this context. By defining intuitionistic fuzzy SBE-subalgebras and exploring their properties, we have demonstrated that the level set of a fuzzy SBE-subalgebra, ideal, or filter retains its structural identity within Sheffer stroke BE-algebras. This not only highlights the intricate connections between these concepts but also opens new avenues for further research in the interplay between intuitionistic fuzzy logic and algebraic structures. The results here contribute to a deeper understanding of intuitionistic fuzzy logic’s application in Sheffer stroke BE-algebras, reinforcing the Sheffer operation’s relevance in both theoretical and applied algebra.

That said, we recognize the potential applications of our results and are committed to pursuing these directions in future research. Specifically, the structures introduced in this paper could find use in the following areas, which we plan to explore in subsequent studies:

• Decision-making systems: SBE-algebras can model multi-criteria decision-making problems by leveraging intuitionistic preferences, enabling reasoning in scenarios with uncertainty and partial truth.
• Optimization problems: Intuitionistic fuzzy filters could aid in solving optimization problems, such as resource allocation and scheduling, where conflicting criteria are present.
• Artificial intelligence: These structures may enhance reasoning systems in AI, especially in knowledge representation, natural language processing, and adaptive learning models.
• Cryptography and secure systems: The algebraic structure of Sheffer stroke BE-algebras could contribute to designing secure encoding schemes.
• Network analysis: Fuzzy logic principles within these algebras could support the optimization of routing protocols in network systems.

By continuing to explore these potential applications, we aim to bridge the gap between abstract algebraic theory and practical, real-world problems, further enriching the utility of intuitionistic fuzzy logic in various domains.