Bipolar Fuzzy BRK-ideals in BRK-algebras

Abstract

In this paper, we investigated bipolar fuzzy BRK-ideals in BRK-algebras and discussed related properties. We presented some results on images and pre-images of bipolar fuzzy BRK-ideals in BRK-algebras. Finally, we introduced translation, extension and multiplications of bipolar fuzzy BRK-ideals in BRK-algebras and discussed related results.

1 Introduction

The fundamental concept of fuzzy set, popularized by Zadeh [1], was used to generalize several basic concepts of algebra. Fuzzy sets are extremely useful to deal with the many problems in applied mathematics, control engineering, information sciences, expert systems and theory of automata etc. Although, there are many generalizations of fuzzy sets but none of these deal with the problems related to the contrary characteristics of the members having membership degree 0. Lee [2] handled this problem by introducing the concept of bipolar fuzzy (BF) sets. The BF set theory has been widely applied to solve real life problems. The sweet taste of foodstuffs is a BF set. Assuming that sweet taste of foodstuff as a positive membership value then bitter taste of foodstuffs as a negative membership value. The remaining foodstuffs of taste like acidic, saline, chilly etc. are extraneous to the sweet and bitter foodstuffs. Thus, these foodstuffs are accepted as zero membership values. Notice that every matter has two sides and bipolarity as well as fuzziness, is an inherent and internal part of human thinking [3, 4]. A BF set is a pair of fuzzy sets, namely a membership and a non-membership function, which represent positive and negative aspects of the given information.

Imai and Iseki investigated two classes of abstract algebras: BCI-algebras and BCK-algebras [5]. In 2002, Neggers et al. [6], presented B-algebra and discussed related properties. The generalization of B-algebra called BF-algebra and BG-algebra proposed by Walendziak [7], and Kim [8], respectively. Recently, Bandaru [9], investigated BRK-algebra which is a generalization of BCK/BCI/Q-algebras. In [10, 11], El-Gendy introduced fuzzy BRK-ideal of BRK-algebra and cubic BRK-ideal of BRK-algebra. Some properties of n-dimensional fuzzy subalgebra in BRK-algebras investigated by Zulfiqar [12]. Fuzzy translations and fuzzy multiplications of BCK/BCI-algebras presented in [13]. As a generalization of fuzzy set theory, BF set theory makes descriptions of the objective world more realistic, practical and very accurate in many cases, making it very promising. In the past few decades, BF set theory has been successfully applied to various algebraic structures. Lee [14], introduced BF ideals of BCK/BCI-algebras and Akram [15] introduced BF graphs. Recently, Hayat et al. [16, 17], characterized himi-rings by their BF h-ideals and BAF h-ideals.

The contents of the present paper are organized as follows: In Sect. 2, we presented some basic definitions and preliminaries. In Sect. 3, we investigated bipolar fuzzy BRK-ideals in BRK-algebras and discussed related properties. In Sect. 4, we presented some results on images and pre-images of bipolar fuzzy BRK-ideals in BRK-algebras. In Sect. 5, we introduced translation, extension and multiplications of bipolar fuzzy BRK-ideals in BRK-algebras and discussed related results. Finally, we presented some conclusions and future work.

2 Preliminaries

In this section, some elementary aspects that are necessary for this paper are included.

Definition 2.1

A BRK-algebra is a non-empty set X with a constant 0 and a binary operation “$*$” satisfying the following conditions:

• (BRK$_{1}$) $x*0=x$,

• (BRK$_{2}$) $\left( x*y\right) *x=0*y$ for all $x,y\in X$.

A partial ordered relation $\preceq$ can be defined by $x\preceq y$ if and only if $x*y=0$. Throughout this paper, X denotes BRK-algebra.

Definition 2.2

[9]. If $\left( X,*,0\right)$ is a BRK-algebra, the following conditions hold:

• (BRK$_{3}$) $x*x=0$,

• (BRK$_{4}$) $\left( x*y\right) =0$ implies $0*x=0*y$ for all $x,y\in X$,

• (BRK$_{5}$) $0*\left( a*b\right) =\left( 0*a\right) *\left( 0*b\right)$ for all $a,b\in X.$

Definition 2.3

[11, 12]. A subset S of a BRK-algebra X is said to be BRK-subalgebra of X, if $x,y\in S$, implies $x*y\in S$.

Definition 2.4

[11]. A subset S of a BRK-algebra X is said to be a BRK-ideal of X ( briefly $S _{BRK}X$ ) if it satisfies:

• (i) $0\in S$,

• (ii) $0*\left( x*y\right) \in S$ and $0*y\in S\implies 0*x\in S$ for all $x,y\in X$.

Example 1

Consider a set $X=\{0,a_{1},a_{2},a_{3}\}$. We define “$*$” on X as the following table:

Clearly, X is a BRK-algebra. Then $S=\{0,a_{1},a_{2}\}_{BRK}X.$

Definition 2.5

[11]. Let $(X_{1},*,0)$ and $(X_{2},*^{\prime },0^{\prime })$ be two BRK-algebras. A mapping $\psi :X_{1}\longrightarrow X_{2}$ is said to be a homomorphism if $\psi (x*y)=\psi (x)*^{\prime }\psi (y)$, for all $x,y\in X_{1}$.

Definition 2.6

[3]. A bipolar fuzzy set is a pair $(\lambda ^{+},\lambda ^{-})$, where $\lambda ^{+}:X\longrightarrow [0,1]$, and $\lambda ^{-}:X\longrightarrow [-1,0]$ are any mappings.

Definition 2.7

[3]. Let $B_{1}=(\lambda ^{+},\lambda ^{-})$ and $B_{2}=(\mu ^{+},\mu ^{-})$ be two BF sets in X. Thenfollowingconditionshold : 

• (i) $B_{1}\le B_{2}$ if and only if $\lambda ^{+}\le \mu ^{+}$ and $\lambda ^{-}\ge \mu ^{-}.$

• (ii) $\max \left\{ B_{1},B_{2}\right\} =\left( \max \left\{ \lambda ^{+},\mu ^{+}\right\} ,\min \left\{ \lambda ^{-},\mu ^{-}\right\} \right)$,

  $\min \left\{ B_{1},B_{2}\right\} =\left( \min \left\{ \lambda ^{+},\mu ^{+}\right\} ,\max \left\{ \lambda ^{-},\mu ^{-}\right\} \right) .$

3 Bipolar Fuzzy BRK-ideals in BRK-algebras

In this section, we introduced BF BRK-ideals in BRK-algebras and discussed related properties.

Definition 3.1

A BF set $B=\left( \lambda ^{+},\lambda ^{-}\right)$ of X is called BF BRK-ideal of X if it satisfies following conditions hold:

• (BF$_{1}$) $\lambda ^{+}\left( 0\right) \ge \lambda ^{+}\left( x\right) ,\lambda ^{-}\left( 0\right) \le \lambda ^{-}\left( x\right) ,$

• (BF$_{2}$) $\lambda ^{+}\left( 0*x\right) \ge \min \left\{ \lambda ^{+}\left( 0*\left( x*y\right) \right) ,\lambda ^{+}\left( 0*y\right) \right\} ,$

  $\lambda ^{-}\left( 0*x\right) \le \max \left\{ \lambda ^{-}\left( 0*\left( x*y\right) \right) ,\lambda ^{-}\left( 0*y\right) \right\} .$

Example 2

Consider a set $X=\{0,a_{1},a_{2},a_{3}\}$. We define “$*$” on X as the following table:

Clearly, X is a BRK-algebra. Define a bipolar fuzzy set,

where $t_{1},t_{2}\in [0,1]$ and $s_{1},s_{2},s_{3}\in [-1,0]$ with $t_{1}>t_{2}$ and $s_{1}<s_{2}<s_{3}$, routine calculation gives that $B=(\lambda ^{+},\lambda ^{-})$ is a BF BRK-ideal of X.

Lemma 3.1

Let B be a BF BRK-ideal of BRK-algebra X. If $y*x\le x$ holds in X, then $\lambda ^{+}\left( 0*y\right) \ge \lambda ^{+}\left( 0*x\right)$ and $\lambda ^{-}\left( 0*y\right) \le \lambda ^{-}\left( 0*x\right)$.

Proof

Assume that $y*x\le x$ holds in X. Then $\left( y*x\right) *x=0$. By (BRK$_{2}$),

Also,

$$\begin{aligned} \lambda ^{+}\left( 0*\left( y*x\right) \right)\ge & {} \min \left\{ \lambda ^{+}\left( \left( 0*\left( y*x\right) \right) *x\right) ,\lambda ^{+}\left( 0*x\right) \right\} \\= & {} \min \left\{ \lambda ^{+}\left( 0\right) ,\lambda ^{+}\left( 0*x\right) \right\} \\= & {} \lambda ^{+}\left( 0*x\right) , \end{aligned}$$

and

$$\begin{aligned} \lambda ^{-}\left( 0*\left( y*x\right) \right)\le & {} \max \left\{ \lambda ^{-}\left( \left( 0*\left( y*x\right) \right) *x\right) ,\lambda ^{-}\left( 0*x\right) \right\} \\= & {} \max \left\{ \lambda ^{-}\left( 0\right) ,\lambda ^{-}\left( 0*x\right) \right\} \\= & {} \lambda ^{-}\left( 0*x\right) . \end{aligned}$$

Hence $\lambda ^{+}\left( 0*y\right) \ge \lambda ^{+}\left( 0*x\right)$ and $\lambda ^{-}\left( 0*y\right) \le \lambda ^{-}\left( 0*x\right)$.

Lemma 3.2

Let B be a BF BRK-ideal of BRK-algebra X. If $x\le y$ holds in X, then $\lambda ^{+}\left( 0*x\right) \ge \lambda ^{+}\left( 0*y\right)$ and $\lambda ^{-}\left( 0*x\right) \le \lambda ^{-}\left( 0*y\right)$.

Theorem 3.1

Let $B_{i}=\{\left( \lambda _{i}^{+},\lambda _{i}^{-}\right) :i\in \Omega \}$ be a family of BF BRK-ideals in X. Then$\ B=\underset{i\in \Omega }{ \wedge }$ $B_{i}$ is also a BF BRK-ideal in X, where $B=\left( \lambda ^{+},\lambda ^{-}\right)$ that is $\lambda ^{+}=\underset{i\in \Omega }{ \wedge }$ $\lambda _{i}^{+}$ and $\lambda ^{-}=\underset{i\in \Omega }{\vee }$ $\lambda _{i}^{-}$ ( $\lambda ^{+}\le \lambda _{i}^{+},\lambda ^{-}\ge \lambda _{i}^{-}$ $\forall$ $i\in \Omega )$.

Definition 3.2

Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be BF set in BRK-algebra X and $\left( \alpha ,\beta \right)$ $\in [-1,0] \times [ 0,1]$, then

• (1) The set $\widetilde{B}_{\beta }^{+}=\left\{ x\in R:\mu ^{+}\left( x\right) \ge \beta \right\}$ is called positive $\beta$-cut of B.

• (2) The set $B_{\alpha }^{-}=\left\{ x\in R:\mu ^{-}\left( x\right) \le \alpha \right\}$ is called negative $\alpha$-cut of B.

• (3) The set $B_{\left( \alpha ,\beta \right) }=\left\{ x\in R:\mu ^{-}\left( x\right) \le \alpha \text { and }\mu ^{+}\left( x\right) \ge \beta \right\}$ is called $(\alpha ,\beta )$-cut of B.

For every $\gamma \in (0,1]$ and $B_{\gamma }^{+}\cap B_{-\gamma }^{-}$ is called $\gamma$-cut of B.

Theorem 3.2

A BF set $B=\left( \mu ^{+},\mu ^{-}\right)$ is a BF BRK-ideal in X iff the followings hold:

• (i) For all $\beta \in [0,1]$, $B_{\beta }^{+}$ is non-empty this implies $B_{\beta }^{+}$ is a BRK -ideal of X.

• (ii) For all $\alpha \in [-1,0]$, $B_{\alpha }^{-}$ is non-empty this implies $B_{\alpha }^{-}$ is a BRK -ideal of X.

Proof

Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be a BF BRK-ideal in X. For $x\in B_{\beta }^{+}$ so $\mu ^{+}\left( x\right) \ge \beta$ where $\beta \in [0,1]$. Now $\mu ^{+}\left( 0\right) \ge \mu ^{+}\left( x\right) \ge \beta$. This implies that $0\in$ $B_{\beta }^{+}$.

Next, let $0*\left( x*y\right) \in$ $B_{\beta }^{+}$ and $0*y\in$ $B_{\beta }^{+}$, this means that $\mu ^{+}\left( 0*\left( x*y\right) \right) \ge \beta$ and $\mu ^{+}\left( 0*y\right) \ge \beta .$ Then $\lambda ^{+}\left( 0*x\right) \ge \min \left\{ \lambda ^{+}\left( 0*\left( x*y\right) \right) ,\lambda ^{+}\left( 0*y\right) \right\} \ge \beta$. Hence $B_{\beta }^{+}$ is a BRK-ideal of X.

Analogously, we can prove that $B_{\alpha }^{-}$ is a BRK-ideal of X.

Corollary 3.1

If $B=\left( \mu ^{+},\mu ^{-}\right)$ is a BF BRK-ideal in X, then the sets $\widetilde{B}_{\mu ^{+}\left( 0\right) }^{+}$ and $\widetilde{B}_{\mu ^{-}\left( 0\right) }^{-}$ are BRK-ideals of X.

Corollary 3.2

Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be BF set in X. If $B=\left( \mu ^{+},\mu ^{-}\right)$ is a BF BRK-ideal in X, then for all $\gamma \in [0,1]$ the $\gamma$-cut of B is a BRK-ideal of X.

Proof

It is analogous to the proof of Theorem 3.2.

Corollary 3.3

If $B=\left( \mu ^{+},\mu ^{-}\right)$ is a BF BRK-ideal in X, then $\widetilde{B}_{\left( \alpha ,\beta \right) }$ is a BRK-ideal in X, $\forall \left( \alpha ,\beta \right)$ $\in [-1,0] \times [ 0,1]$. In particular, $\gamma$-cut of B is a BRK-ideal in X, for all $\gamma \in [0,1]$.

4 Images and Pre-images of a BF BRK-ideal

In this section, we introduced images and pre-images of BF BRK-ideals and discussed some theorems.

Definition 4.1

Let $\psi :X_{1}\rightarrow X_{2}$ be a mapping of BRK-algebras. If $B=\left( \mu ^{+},\mu ^{-}\right)$ and $V=\left( \upsilon ^{+},\upsilon ^{-}\right)$ are BF set of $X_{1}$ and $X_{2}$ respectively. Then

$\mu ^{+}\left( \psi ^{-1}(y)\right) =\upsilon ^{+}(y)=\left\{ \begin{array}{c} \underset{x\in \psi ^{-1}\left( y\right) }{\vee }\mu ^{+}\left( x\right) , { \ \ { if} \ }\psi ^{-1}\left( y\right) \ne \emptyset ,\\ 1,{ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \ { Otherwise,}} \end{array} \right. ,$

and

$\mu ^{-}\left( \psi ^{-1}(y)\right) =\upsilon ^{-}(y)=\left\{ \begin{array}{c} \underset{x\in \psi ^{-1}\left( y\right) }{\wedge }\mu ^{-}\left( x\right) , { \ \ { if }\ }\psi ^{-1}\left( y\right) \ne \emptyset , \\ -1,{ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \ { Otherwise,}} \end{array} \right.$

for all $x\in R_{2}$ is called image of $B=\left( \mu ^{+},\mu ^{-}\right)$ under $\psi$, where $\psi ^{-1}\left( y\right) =\{x\in X_{1} \psi \left( x\right) =y\}$. Also the pre-image $B=V\circ \psi$ in $X_{1}$ defined as, $\upsilon ^{+}\left( \psi (x)\right) =\mu ^{+}(x)$, and $\upsilon ^{-}\left( \psi (x)\right) =\mu ^{-}(x)$.

Theorem 4.1

An into homomorphic pre-image of a BF BRK-ideal is also a BF BRK-ideal.

Proof

Let $\psi :X_{1}\rightarrow X_{2}$ be an into homomorphism of BRK-algebras. Suppose that $V=\left( \upsilon ^{+},\upsilon ^{-}\right)$ be a BF BRK-ideal in $\left( X_{2},*^{\prime },0^{\prime }\right)$ and $B=\left( \mu ^{+},\mu ^{-}\right)$ be a BF BRK-ideal in $\left( X_{1},*,0\right)$. Then for all $x\in$ $X_{1}$, $\mu ^{+}\left( 0\right) =\upsilon ^{+}\left( \psi \left( 0\right) \right) \ge \upsilon ^{+}\left( \psi \left( x\right) \right) =\mu ^{+}(x)$ and $\mu ^{-}\left( 0\right) =\upsilon ^{-}\left( \psi \left( 0\right) \right) \le \upsilon ^{-}\left( \psi \left( x\right) \right) =\mu ^{-}(x)$.

Now, let $x,y\in$ $X_{1}$. Then

$$\begin{aligned} \begin{array}{lcl} \mu ^{+}\left( 0*x\right) &{}=&{}\upsilon ^{+}(\psi \left( 0*x\right) )\\ &{}=&{}\upsilon ^{+}\left( \psi \left( 0\right) *^{\prime }\psi \left( x\right) \right) \\ &{}\ge &{} \min \{\upsilon ^{+}(\psi \left( 0\right) *^{\prime }\left( \psi \left( x\right) *^{\prime }\psi \left( y\right) \right) ),\upsilon ^{+}\left( \psi \left( 0\right) *^{\prime }\psi \left( y\right) \right) \}\\ &{}=&{}\min \{\upsilon ^{+}(\psi \left( 0*\left( x*y\right) \right) ),\upsilon ^{+}\left( \psi \left( 0*y\right) \right) \}\\ &{}=&{}\min \{\mu ^{+}(0*\left( x*y\right) ),\mu ^{+}\left( 0*y\right) \} \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{lcl} \mu ^{-}\left( 0*x\right) &{}=&{}\upsilon ^{-}(\psi \left( 0*x\right) )\\ &{}=&{}\upsilon ^{-}\left( \psi \left( 0\right) *^{\prime }\psi \left( x\right) \right) \\ &{}\le &{} \max \{\upsilon ^{-}(\psi \left( 0\right) *^{\prime }\left( \psi \left( x\right) *^{\prime }\psi \left( y\right) \right) ),\upsilon ^{-}\left( \psi \left( 0\right) *^{\prime }\psi \left( y\right) \right) \}\\ &{}=&{}\max \{\upsilon ^{-}(\psi \left( 0*\left( x*y\right) \right) ),\upsilon ^{-}\left( \psi \left( 0*y\right) \right) \}\\ &{}=&{}\max \{\mu ^{-}(0*\left( x*y\right) ),\mu ^{-}\left( 0*y\right) \}. \end{array} \end{aligned}$$

Hence pre-image of a BF BRK-ideal is also a BF BRK-ideal.

Definition 4.2

Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be a BF set in X. Then for $K\subseteq X_{1}$ there exist $m,n\in K$ such that $\mu ^{+}\left( m\right) = \underset{m\in K}{\vee }\mu ^{+}\left( m\right)$ and $\mu ^{-}\left( n\right) =\underset{n\in K}{\wedge }\mu ^{-}\left( n\right)$.

Theorem 4.2

An onto homomorphic image of a BF BRK-ideal is also a BF BRK-ideal.

Proof

Let $\psi :X_{1}\rightarrow X_{2}$ be an onto homomorphism of BRK-algebras and $V=\left( \upsilon ^{+},\upsilon ^{-}\right)$ be a BF BRK-ideal in $\left( X_{2},*^{\prime },0^{\prime }\right)$. Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be a BF BRK-ideal in $\left( X_{1},*,0\right)$ with sup and inf properties. By Definition 4.2, we get $\upsilon ^{+}(y^{\prime })=\mu ^{+}\left( \psi ^{-1}(y^{\prime })\right) =\underset{ x\in \psi ^{-1}\left( y\right) }{\vee }\mu ^{+}\left( x\right)$ and $\upsilon ^{-}(y^{\prime })=\mu ^{-}\left( \psi ^{-1}(y^{\prime })\right) = \underset{x\in \psi ^{-1}\left( y\right) }{\wedge }\mu ^{-}\left( x\right)$ for all $y^{\prime }\in X_{2}$. Since $B=\left( \mu ^{+},\mu ^{-}\right)$ be BF BRK-ideal in $X_{1}$, we have $\mu ^{+}\left( 0\right) \ge \mu ^{+}\left( x\right)$ and $\mu ^{-}\left( 0\right) \le \mu ^{-}\left( x\right)$. Note that $0\in \psi ^{-1}(0^{\prime })$. Thus $\upsilon ^{+}(0^{\prime })=\mu ^{+}\left( \psi ^{-1}(0^{\prime })\right) =\underset{ a\in \psi ^{-1}\left( 0^{\prime }\right) }{\vee }\mu ^{+}\left( a\right) =$ $\mu ^{+}\left( 0\right) \ge \mu ^{+}\left( x\right)$ and $\upsilon ^{-}(0^{\prime })=\mu ^{-}\left( \psi ^{-1}(0^{\prime })\right) =\underset{ a\in \psi ^{-1}\left( 0^{\prime }\right) }{\wedge }\mu ^{-}\left( a\right)$ $=\mu ^{-}\left( 0\right) \le \mu ^{-}\left( x\right)$. This implies that $\upsilon ^{+}(0^{\prime })\ge \underset{a\in \psi ^{-1}\left( x^{\prime }\right) }{\vee }\mu ^{+}\left( a\right) =\upsilon ^{+}(x^{\prime })$ and $\upsilon ^{-}(0^{\prime })\le \underset{a\in \psi ^{-1}\left( x^{\prime }\right) }{\wedge }\mu ^{-}\left( a\right) =\upsilon ^{-}(x^{\prime })$for all $x^{\prime }\in X_{2}$.

Now, let $x^{\prime },y^{\prime },z^{\prime }\in X_{2}$ and $0_{0}\in \psi ^{-1}(0^{\prime })$, $x_{0}\in \psi ^{-1}(x^{\prime })$, $y_{0}\in \psi ^{-1}(y^{\prime })$ be such that $\mu ^{+}(0_{0}*x_{0})=\underset{a\in \psi ^{-1}\left( 0^{\prime }*x^{\prime }\right) }{\vee }\mu ^{+}\left( a\right)$ and $\mu ^{+}(0_{0}*y_{0})=\underset{a\in \psi ^{-1}\left( 0^{\prime }*y^{\prime }\right) }{\vee }\mu ^{+}\left( a\right)$ and $\mu ^{+}(0_{0}*(x_{0}*y_{0}))=\upsilon ^{+}(\psi (0_{0}*(x_{0}*y_{0})))=\upsilon ^{+}(0^{\prime }*(x^{\prime }*y^{\prime }))$ $=\underset{0_{0}*\left( x_{0}*y_{0}\right) \in \psi ^{-1}\left( 0^{\prime }*\left( x^{\prime }*y^{\prime }\right) \right) }{\vee } \mu ^{+}\left( 0_{0}*\left( x_{0}*y_{0}\right) \right) =\underset{ a\in \psi ^{-1}\left( 0^{\prime }*\left( x^{\prime }*y^{\prime }\right) \right) }{\vee }\mu ^{+}\left( a\right) .$

Thus

$$\begin{aligned} \upsilon ^{+}(0^{\prime }*x^{\prime })= & {} \underset{a\in \psi ^{-1}\left( 0^{\prime }*x^{\prime }\right) }{\vee }\mu ^{+}\left( a\right) \\= & {} \mu ^{+}(0_{0}*x_{0})\\\ge & {} \min \{\mu ^{+}(0_{0}*\left( x_{0}*y_{0}\right) ),\mu ^{+}(0_{0}*y_{0})\}\\= & {} \min \{\underset{a\in \psi ^{-1}\left( 0^{\prime }*\left( x^{\prime }*y^{\prime }\right) \right) }{\vee }\mu ^{+}\left( a\right) ,\underset{a\in \psi ^{-1}\left( 0^{\prime }*x^{\prime }\right) }{\vee }\mu ^{+}\left( a\right) \}\\= & {} \min \{\upsilon ^{+}(0^{\prime }*\left( x^{\prime }*y^{\prime }\right) ),\upsilon ^{+}(0^{\prime }*y^{\prime })\}. \end{aligned}$$

On the other hand, we have $\mu ^{-}(0_{0}*x_{0})=\underset{a\in \psi ^{-1}\left( 0^{\prime }*x^{\prime }\right) }{\wedge }\mu ^{-}\left( a\right)$ and $\mu ^{-}(0_{0}*y_{0})=\underset{a\in \psi ^{-1}\left( 0^{\prime }*y^{\prime }\right) }{\wedge }\mu ^{-}\left( a\right)$ and $\mu ^{-}(0_{0}*\left( x_{0}*y_{0}\right) )=\upsilon ^{-}\left( \psi (0_{0}*\left( x_{0}*y_{0}\right) )\right) =\upsilon ^{-}(0^{\prime }*\left( x^{\prime }*y^{\prime }\right) )$ $=\underset{0_{0}*\left( x_{0}*y_{0}\right) \in \psi ^{-1}\left( 0^{\prime }*\left( x^{\prime }*y^{\prime }\right) \right) }{\wedge } \mu ^{-}\left( 0_{0}*\left( x_{0}*y_{0}\right) \right) =\underset{ a\in \psi ^{-1}\left( 0^{\prime }*\left( x^{\prime }*y^{\prime }\right) \right) }{\wedge }\mu ^{-}\left( a\right) .$

Thus,

$$\begin{aligned} \upsilon ^{-}(0^{\prime }*x^{\prime })= & {} \underset{a\in \psi ^{-1}\left( 0^{\prime }*x^{\prime }\right) }{\wedge }\mu ^{-}\left( a\right) \\= & {} \mu ^{-}(0_{0}*x_{0})\\\le & {} \max \{\mu ^{-}(0_{0}*\left( x_{0}*y_{0}\right) ),\mu ^{-}(0_{0}*y_{0})\}\\= & {} \max \{\underset{a\in \psi ^{-1}\left( 0^{\prime }*\left( x^{\prime }*y^{\prime }\right) \right) }{\wedge }\mu ^{-}\left( a\right) ,\underset{a\in \psi ^{-1}\left( 0^{\prime }*x^{\prime }\right) }{\wedge }\mu ^{-}\left( a\right) \}\\= & {} \max \{\upsilon ^{-}(0^{\prime }*\left( x^{\prime }*y^{\prime }\right) ),\upsilon ^{-}(0^{\prime }*y^{\prime })\}. \end{aligned}$$

Hence onto homomorphic image of a BF BRK-ideal is also a BF BRK-ideal.

5 BF Translation and BF Extension on BF BRK-Ideals

For any BF set $B=\left( \lambda ^{+},\lambda ^{-}\right)$ in X, we denote $\top =1-\sup \{\lambda ^{+}(x)\mid x\in X\}$ and $\bot =-1-\inf \{\lambda ^{-}(x)\mid x\in X\}$.

Definition 5.1

Let $B=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF set in X and $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$. By a BF $(\gamma$, $\delta )$-translation of B we mean a BF set $B_{(\gamma ,\delta )}^{T}=(\lambda _{(\gamma ,T)}^{+},\lambda _{(\delta ,T)}^{-})$ where

Theorem 5.1

If $B=\left( \lambda ^{+},\lambda ^{-}\right)$ is a BF BRK-ideal in X, then the BF $(\gamma$, $\delta )$-translation $B_{(\gamma ,\delta )}^{T}=(\lambda _{(\gamma ,T)}^{+},\lambda _{(\delta ,T)}^{-})$ of B is a BF BRK -ideal in X for all $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0].$

Proof

Let $a\in X$. Then $\lambda _{(\gamma ,T)}^{+}(0)=\lambda ^{+}(0)+\gamma \ge \lambda ^{+}(a)+\gamma =\lambda _{(\gamma ,T)}^{+}(a)$, and $\lambda _{(\delta ,T)}^{-}(0)=\lambda ^{-}(0)+\delta \le \lambda ^{-}(a)+\delta =\lambda _{(\delta ,T)}^{-}(a)$. Now, let

$$\begin{aligned} \lambda _{(\gamma ,T)}^{+}(0*x)= & {} \lambda ^{+}(0*x)+\gamma \\\ge & {} \min \{\lambda ^{+}(0*\left( x*y\right) ),\lambda ^{+}(0*y)\}+\gamma \\= & {} \min \{\lambda ^{+}(0*\left( x*y\right) )+\gamma ,\lambda ^{+}(0*y)+\gamma \}\\= & {} \min \{\lambda _{(\gamma ,T)}^{+}\left( 0*\left( x*y\right) \right) ,\lambda _{(\gamma ,T)}^{+}\left( 0*y\right) \}, \end{aligned}$$

and

$$\begin{aligned} \lambda _{(\delta ,T)}^{-}(0*x)= & {} \lambda ^{-}(0*x)+\delta \\\le & {} \max \{\lambda ^{-}(0*\left( x*y\right) ),\lambda ^{-}(0*y)\}+\delta \\= & {} \max \{\lambda ^{-}(0*\left( x*y\right) )+\delta ,\lambda ^{-}(0*y)+\delta \}\\= & {} \max \{\lambda _{(\delta ,T)}^{-}\left( 0*\left( x*y\right) \right) ,\lambda _{(\delta ,T)}^{-}\left( 0*y\right) \}. \end{aligned}$$

Hence $B=\left( \lambda ^{+},\lambda ^{-}\right)$ is a BF BRK-ideal in X.

Theorem 5.2

Let $B=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF set in X such that the BF $(\gamma$, $\delta )$-translation $B_{(\gamma ,\delta )}^{T}=(\lambda _{(\gamma ,T)}^{+},\lambda _{(\delta ,T)}^{-})$ of B is a BF BRK-ideal in X for some $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$. Then $B=\left( \lambda ^{+},\lambda ^{-}\right)$ is a BF BRK-ideal in X.

Proof

Let $a\in X$. Then $\lambda ^{+}(0)+\gamma =\lambda _{(\gamma ,T)}^{+}(0)\ge \lambda _{(\gamma ,T)}^{+}(a)=\lambda ^{+}(a)+\gamma$, and $\lambda ^{-}(0)+\delta =\lambda _{(\delta ,T)}^{-}(0)\le \lambda _{(\delta ,T)}^{-}(a)=\lambda ^{-}(a)+\delta$. Thus $\lambda ^{+}(0)\ge \lambda ^{+}(a)$ and $\lambda ^{-}(0)\le \lambda ^{-}(a)$. Now, let

$$\begin{aligned} \lambda ^{+}(0*x)+\gamma= & {} \lambda _{(\gamma ,T)}^{+}(0*x)\\\ge & {} \min \{\lambda _{(\gamma ,T)}^{+}\left( 0*\left( x*y\right) \right) ,\lambda _{(\gamma ,T)}^{+}\left( 0*y\right) \}\\= & {} \min \{\lambda ^{+}(0*\left( x*y\right) )+\gamma ,\lambda ^{+}(0*y)+\gamma \}\\= & {} \min \{\lambda ^{+}(0*\left( x*y\right) ),\lambda ^{+}(0*y)\}+\gamma , \end{aligned}$$

and

$$\begin{aligned} \lambda ^{-}(0*x)+\delta= & {} \lambda _{(\delta ,T)}^{-}(0*x)\\\le & {} \max \{\lambda _{(\delta ,T)}^{-}\left( 0*\left( x*y\right) \right) ,\lambda _{(\delta ,T)}^{-}\left( 0*y\right) \}\\= & {} \max \{\lambda ^{-}(0*\left( x*y\right) )+\delta ,\lambda ^{-}(0*y)+\delta \}\\= & {} \max \{\lambda ^{-}(0*\left( x*y\right) ),\lambda ^{-}(0*y)\}+\delta . \end{aligned}$$

Thus $\lambda ^{+}(0*x)\ge \min \{\lambda ^{+}(0*\left( x*y\right) ),\lambda ^{+}(0*y)\}$ and $\lambda ^{-}(0*x)\le \max \{\lambda ^{-}(0*\left( x*y\right) ),\lambda ^{-}(0*y)\}$. Hence $B=\left( \lambda ^{+},\lambda ^{-}\right)$ is a BF BRK-ideal in X.

Definition 5.2

Let $B_{1}=\left( \lambda ^{+},\lambda ^{-}\right)$ and $B_{2}=\left( \mu ^{+},\mu ^{-}\right)$ be two BF sets in X. If $\lambda ^{+}\left( x\right) \le \mu ^{+}\left( x\right)$ and $\lambda ^{-}\left( x\right) \ge \mu ^{-}\left( x\right)$ for all $x\in X$, then we say that $B_{2}$ is a BF extension of $B_{1}$.

Definition 5.3

Let $B_{1}=\left( \lambda ^{+},\lambda ^{-}\right)$ and $B_{2}=\left( \mu ^{+},\mu ^{-}\right)$ be two BF sets in X. Then $B_{2}$ is called a BF BRK-ideal extension of $B_{1}$ if following conditions are holds:

• (i) $B_{2}$ is called a BF extension of $B_{1}$,

• (ii) If $B_{2}$ is a BF BRK-ideal extension of X then $B_{1}$ is also a BF BRK-ideal extension of X.

By definition $\lambda _{(\gamma ,T)}^{+}\ge \lambda ^{+}\left( x\right)$ and $\lambda _{(\delta ,T)}^{-}\le \lambda ^{-}\left( x\right)$ for all $x\in X$. Therefore, we have following theorem.

Theorem 5.3

If $B=\left( \lambda ^{+},\lambda ^{-}\right)$ is a BF BRK-ideal of X, then the BF $(\gamma$, $\delta )$-translation $B_{(\gamma ,\delta )}^{T}=(\lambda _{(\gamma ,T)}^{+},\lambda _{(\delta ,T)}^{-})$ of B is a BF extension of X for all $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$.

Converse of above theorem is not true in general as seen in following Example.

Example 3

Consider BRK-algebra $X=\{0,a_{1},a_{2},a_{3}\}$ defined in Example 2. Let $B_{1}=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF set in X defined by

Clearly, B is a BF BRK-ideal of X. Let $B_{2}=\left( \mu ^{+},\mu ^{-}\right)$ be a BF set of X defined by

Then $B_{2}$ is a BF BRK-ideal extension of $B_{1}$ but it is not BF $(\gamma$, $\delta )$-translation of $B_{1}$, for all $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$.

Definition 5.4

Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be a BF set in X, $\left( \alpha , \beta \right)$ $\in \left[ -1,0\right] \times \left[ 0,1\right]$, $\gamma \in [0,\top ]$ and $\delta \in [\bot ,0]$. We define $B_{\left( \beta ,\gamma \right) }^{+T}=\left\{ x\in X:\mu ^{+}\left( x\right) \ge \beta -\gamma \right\}$, $B_{\left( \alpha ,\delta \right) }^{-T}=\left\{ x\in X:\mu ^{-}\left( x\right) \le \alpha -\delta \right\}$ and $B_{\left( \left( \alpha ,\beta \right) ,\left( \gamma ,\delta \right) \right) }^{T}=\left\{ x\in X:\mu ^{-}\left( x\right) \le \alpha -\gamma \right. \left. \text {and }\mu ^{+}\left( x\right) \ge \beta -\delta \right\}$.

Theorem 5.4

If $B=\left( \mu ^{+},\mu ^{-}\right)$ is BF BRK-ideal of X, then $B_{\left( \beta ,\gamma \right) }^{+T}$ and $B_{\left( \alpha ,\delta \right) }^{-T}$ are BRK-ideal of X for all $\alpha \in Im(\mu ^{-})$ and $\beta \in Im(\mu ^{+})$ with $\beta \ge \gamma$ and $\alpha \le \delta$.

If we do not take the condition that $B=\left( \mu ^{+},\mu ^{-}\right)$ is BF BRK-ideal of X, then either both $B_{\left( \beta ,\gamma \right) }^{+T}$ and $B_{\left( \alpha ,\delta \right) }^{-T}$ are not BRK-ideals of X, or one of them is not a BRK-ideal of X.

Example 4

Consider BRK-algebra $X=\{0,1,2\}$ with the following operation

Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be a BF set of X defined by

Clearly, B is a BF BRK-ideal of X. Take $\beta =0.32$, $\alpha =-0.35$, $\gamma =0.05$ and $\delta =-0.04$. Then $B_{\left( \beta ,\gamma \right) }^{+T}=\{0,2\}$ and $B_{\left( \alpha ,\delta \right) }^{-T}=\{0,1,2\}$ are BRK-ideals of X. On the other hand, BF set $B=\left( \mu ^{\prime +},\mu ^{\prime -}\right)$ defined by

is not a BF BRK-ideal of X. Take $\beta =0.55$, $\alpha =-0.8$, $\gamma =0.03$ and $\delta =-0.07$. Then $B_{\left( \beta ,\gamma \right) }^{+T}=\{0,1\}$ is not a BRK-ideal of X, but $B_{\left( \alpha ,\delta \right) }^{-T}=\{0\}$ is a BRK-ideal of X.

Corollary 5.1

If $B=\left( \mu ^{+},\mu ^{-}\right)$ is BF BRK-ideal of X, then $B_{\left( \left( \alpha ,\beta \right) ,\left( \gamma ,\delta \right) \right) }^{T}$ is a BRK-ideal of X, for all $\left( \alpha ,\beta \right)$ $\in \left[ -1,0\right] \times \left[ 0,1\right]$ and $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$.

Theorem 5.5

Let $B=\left( \mu ^{+},\mu ^{-}\right)$ be a BF set of X. Then the BF $(\gamma$, $\delta )$-translation of B is a BF BRK-ideal of X if and only if $B_{\left( \beta ,\gamma \right) }^{+T}$ and $B_{\left( \alpha ,\delta \right) }^{-T}$ are BRK -ideal of X for all $\beta \in Im(\mu ^{+})$, $\alpha \in Im(\mu ^{-})$ and $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$ with $\alpha <\delta$ and $\beta >\gamma$.

Proof

Assume that $B_{(\gamma ,\delta )}^{T}=(\mu _{(\gamma ,T)}^{+},\mu _{(\delta ,T)}^{-})$ is a BF BRK-ideal of X. Let $x\in B_{\left( \beta ,\gamma \right) }^{+T}$. Then $\mu ^{+}\left( x\right) \ge \beta -\gamma$ such that $\mu _{(\gamma ,T)}^{+}\left( 0\right) \ge \mu _{(\gamma ,T)}^{+}\left( x\right) \ge \beta$. This shows that $\mu ^{+}\left( 0\right) +\gamma \ge \beta$. Thus $\mu ^{+}\left( 0\right) \ge \beta -\gamma$, which implies that $0\in B_{\left( \beta ,\gamma \right) }^{+T}$. Now, let $0*\left( a*b\right) \in B_{\left( \beta ,\gamma \right) }^{+T}$ and $0*b\in B_{\left( \beta ,\gamma \right) }^{+T}$ for all $a,b\in X$. Since $\mu ^{+}\left( 0*a\right) \ge \min \{\mu _{(\gamma ,T)}^{+}\left( 0*\left( a*b\right) \right) ,\mu _{(\gamma ,T)}^{+}\left( 0*b\right) \}\ge \beta$. This shows that $\mu ^{+}\left( 0*a\right) +\gamma \ge \beta$. Thus $\mu ^{+}\left( 0*a\right) \ge \beta -\gamma$, which implies that $0*a\in B_{\left( \beta ,\gamma \right) }^{+T}$. Hence $B_{\left( \beta ,\gamma \right) }^{+T}$ is a BRK-ideal of X. Analogously, we can prove that $B_{\left( \alpha ,\delta \right) }^{-T}$ is a BRK-ideal of X.

Conversely, suppose $B_{\left( \beta ,\gamma \right) }^{+T}$ and $B_{\left( \alpha ,\delta \right) }^{-T}$ are BRK-ideals of X for all $\beta \in Im(\mu ^{+})$, $\alpha \in Im(\mu ^{-})$ and $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$ with $\alpha <\delta$ and $\beta >\gamma$. Assume that there exist $a\in X$, such that $\mu _{(\gamma ,T)}^{+}\left( 0\right)<\beta ^{\prime }<\mu _{(\gamma ,T)}^{+}\left( a\right)$ and $\mu _{\delta ,T)}^{-}\left( 0\right)>\alpha ^{\prime }>\mu _{\delta ,T)}^{-}\left( a\right)$. Then $\mu ^{+}\left( 0\right) >\beta ^{\prime }-\gamma$, $\mu ^{+}\left( 0\right) >\beta ^{\prime }-\gamma$, $\mu ^{-}\left( 0\right) <\alpha ^{\prime }-\delta$ and $\mu ^{-}\left( 0\right) <\alpha ^{\prime }-\delta$. This shows that $a\in B_{\left( \beta ,\gamma \right) }^{+T}$ and $a\in B_{\left( \alpha ,\delta \right) }^{-T}$, but $0\notin B_{\left( \beta ,\gamma \right) }^{+T}$ and $0\notin B_{\left( \alpha ,\delta \right) }^{-T}$ which is contradiction. Hence $\mu _{(\gamma ,T)}^{+}\left( 0\right) \ge \mu _{(\gamma ,T)}^{+}\left( a\right)$ and $\mu _{(\delta ,T)}^{-}\left( 0\right) \le \mu _{(\delta ,T)}^{-}\left( a\right)$. Similarly, we can prove that $\mu _{(\gamma ,T)}^{+}\left( 0*a\right) \ge \min \{\mu _{(\gamma ,T)}^{+}\left( 0*\left( a*b\right) \right) ,\mu _{(\gamma ,T)}^{+}\left( 0*b\right) \}$, $\mu _{(\delta ,T)}^{-}\left( 0*a\right) \le \max \{\mu _{(\delta ,T)}^{-}\left( 0*\left( a*b\right) \right) ,\mu _{(\delta ,T)}^{-}\left( 0*b\right) \}$. Hence BF $(\gamma$, $\delta )$-translation of B is a BF BRK-ideal of X.

Theorem 5.6

Let $B=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF BRK-ideal in X, $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$ and $(\gamma ^{\prime }$, $\delta ^{\prime })\in [0,\top ]\times [\bot ,0]$. If $(\gamma$, $\delta )\ge (\gamma ^{\prime }$, $\delta ^{\prime })$, then BF $(\gamma$, $\delta )$-translation $B_{(\gamma ,\delta )}^{T}$ of B is a BF BRK-ideal extension of $(\gamma ^{\prime }$, $\delta ^{\prime })$-translation $B_{(\gamma ^{\prime },\delta ^{\prime })}^{T}$ of B.

Theorem 5.7

Let $B=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF BRK-ideal in X and $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$. For every BF BRK -ideals extension $B^{\prime }=\left( \nu ^{+},\nu ^{-}\right)$ of the BF $(\gamma$, $\delta )$ -translation $B_{(\gamma ,\delta )}^{T}$ of B, there exist $(\gamma ^{\prime }$, $\delta ^{\prime })\in [0,\top ]\times [\bot ,0]$ such that $(\gamma$, $\delta )\ge (\gamma ^{\prime }$, $\delta ^{\prime })$ and $B^{\prime }$ is a BF BRK -ideal extension of the BF $(\gamma ^{\prime }$, $\delta ^{\prime })$ -translation $B_{(\gamma ^{\prime },\delta ^{\prime })}^{T}$ of B.

Proof

Assume that for every BF BRK-ideal extension $B^{\prime }=\left( \nu ^{+},\nu ^{-}\right)$ of $B_{(\gamma ,\delta )}^{T}$, there does not exist $(\gamma ^{\prime }$, $\delta ^{\prime })\in [0,\top ]\times [\bot ,0]$ $(\gamma$, $\delta )\ge (\gamma ^{\prime }$, $\delta ^{\prime })$ such that $B^{\prime }$ is not BF BRK-ideal extension of $B_{(\gamma ^{\prime },\delta ^{\prime })}^{T}$. Then $\nu ^{+}\left( x\right) >\mu _{(\gamma ^{\prime },T)}^{+}\left( x\right) =\mu ^{+}\left( x\right) +\gamma ^{\prime }$ and $\nu ^{-}\left( x\right) <\mu _{(\delta ^{\prime },T)}^{-}\left( x\right) =\mu ^{-}\left( x\right) +\delta ^{\prime }$ for $x\in R$. Since, $(\gamma$, $\delta )<(\gamma ^{\prime }$, $\delta ^{\prime })$ so that $\nu ^{+}\left( x\right)>\mu ^{+}\left( x\right) +\gamma ^{\prime }>\mu ^{+}\left( x\right) +\gamma$ and $\nu ^{-}\left( x\right)<\mu ^{-}\left( x\right) +\delta ^{\prime }<\mu ^{-}\left( x\right) +\delta$. This shows that $\nu ^{+}\left( x\right) >\mu _{(\gamma ,T)}^{+}\left( x\right)$ and $\nu ^{-}\left( x\right) <\mu _{(\delta ,T)}^{-}\left( x\right)$ which is contradiction. Thus, there exist $(\gamma ^{\prime }$, $\delta ^{\prime })\in [0,\top ]\times [\bot ,0]$ such that $(\gamma$, $\delta )\ge (\gamma ^{\prime }$, $\delta ^{\prime })$ and $B^{\prime }$ is a BF BRK-ideal extension of $B_{(\gamma ^{\prime },\delta ^{\prime })}^{T}$.

Definition 5.5

Let $B=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF set of X and $\zeta ,\eta \in [0,1]$. By a BF $(\zeta$, $\eta )$ -multiplication of B we mean a BF set $B_{(\zeta ,\eta )}^{m}=(\lambda _{\zeta }^{+m},\lambda _{\eta }^{-m})$ express as

For a BF BRK-ideal, BF (0, 0)-multiplication $B_{(0,0)}^{m}$ is a BF BRK-ideal of X.

Theorem 5.8

If $B=\left( \lambda ^{+},\lambda ^{-}\right)$ is a BF BRK-ideal of X, then BF $(\zeta$, $\eta )$-multiplication $B_{(\zeta ,\eta )}^{m}$ of B is a BF BRK-ideal of X.

Theorem 5.9

Let $B=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF set of X. Then BF $(\zeta$, $\eta )$-multiplication $B_{(\zeta ,\eta )}^{m}$ of B is a BF BRK-ideal of X if and only if B is a BF BRK-ideal of X for some $\zeta ,\eta \in [0,1]$.

Proof

Necessity obtains from Theorem 5.8. Let $\zeta ,\eta \in [0,1]$ such that BF $(\zeta$, $\eta )$-multiplication $B_{(\zeta ,\eta )}^{m}$ of B is a BF BRK-ideal of X. Let $x\in X$. Then $\lambda ^{+}\left( 0\right) \zeta =\lambda _{\zeta }^{+m}\left( 0\right) \ge \lambda _{\zeta }^{+m}\left( x\right) \ge \lambda ^{+}\left( x\right) \zeta$, $\lambda ^{-}\left( 0\right) \eta =\lambda _{\eta }^{+m}\left( 0\right) \le \lambda _{\eta }^{+m}\left( x\right) \le \lambda ^{+}\left( x\right) \eta$ which implies that $\lambda ^{+}\left( 0\right) \ge \lambda ^{+}\left( x\right)$. Analogously, $\lambda ^{-}\left( 0\right) \le \lambda ^{-}\left( x\right)$. Next, let for $x,y\in X$

which implies that $\lambda ^{+}\left( x\right) \ge \min \{\lambda ^{+}\left( 0*\left( x*y\right) \right) ,\lambda ^{+}\left( 0*y\right) \}$. And,

which implies that $\lambda ^{-}\left( x\right) \le \max \{\lambda ^{-}\left( 0*\left( x*y\right) \right) ,\lambda ^{-}\left( 0*y\right) \}$ for $x,y\in X$. Hence B is a BF BRK-ideal of X.

Theorem 5.10

Let $B=\left( \lambda ^{+},\lambda ^{-}\right)$ be a BF set of X, $(\gamma$, $\delta )\in [0,\top ]\times [\bot ,0]$ and $\zeta$, $\eta \in (0,1]$. Then every BF $(\gamma$, $\delta )$ -translation $B_{(\gamma ,\delta )}^{T}$ of B is a BF BRK -ideal extension of the $(\zeta$, $\eta )$-multiplication $B_{(\zeta ,\eta )}^{m}$ of B.

Proof

For all $a\in R$, we have $\mu _{(\gamma ,T)}^{+}\left( a\right) =\mu ^{+}\left( a\right) +\gamma \ge \mu ^{+}\left( a\right) \ge \mu ^{+}\left( a\right) \zeta =\mu _{\zeta }^{+m}\left( a\right)$ and $\mu _{(\delta ,T)}^{-}\left( a\right) =\mu ^{-}\left( a\right) +\delta \le \mu ^{-}\left( a\right) \le \mu ^{-}\left( a\right) \eta =\mu _{\eta }^{-m}\left( a\right)$. Thus $B_{(\gamma ,\delta )}^{T}$ is a BF extension of the $(\zeta$, $\eta )$-multiplication $B_{(\zeta , \eta )}^{m}$ of B. Assume that BF $(\zeta$, $\eta )$-multiplication $B_{(\zeta ,\eta )}^{m}$ of B is a BF BRK-ideal of X. Then by Theorem 5.9, B is a BF BRK-ideal of X. By Theorem 5.1, $B_{(\gamma ,\delta )}^{T}$ is a BF BRK-ideal of X. Thus every BF $(\gamma$, $\delta )$-translation is a BF BRK-ideal extension of the $(\zeta$, $\eta )$-multiplication of B.

6 Conclusion

In this study, we investigated BF BRK-ideals in BRK-algebras and discussed related properties. We introduced translation, extension and multiplications of BF BRK-ideals in BRK-algebras and discussed related results. As an extension of above results, one could study bipolar anti fuzzy BRK-algebras and BF ideals in other algebraic structures. Applications of the method to consider the related problems in machines learning, decision makings, information sciences, cognitive science, intelligent decision-making system, and so on.