Some Results of Fermatean Fuzzy Set on Subalgebras and Ideals of Bn-Algebras

Definition 3.1:

Let F be a fermatean fuzzy set (FFS) in $X$ . then f is called a fermatean fuzzy BN-subalgebra of x if the following conditions hold for all $x,y\in X:$

Example 3.2:

Let x = {0, 1, 2, 3, 4} be a set, and let * be defined by Table 1. The operation table defining the BN-algebra structure used in the following example is given in Table 1.

Then $(X,\ast ,0)$ is a BN-algebra. Define ${\delta }_F:X\to [0,1]$ by

Define ${\theta }_F\left(x\right)=1-{\delta }_F\left(x\right).\mathrm{\text{Then}}{\theta }_F\left(0\right)=0.1,{\theta }_F\left(1\right)=0.5,{\theta }_F\left(2\right)=0.4,{\theta }_F\left(3\right)=0.5,{\theta }_F\left(4\right)=0.4.$

Let $S=\{0,1\}\subseteq X$ . Then $S$ is a subalgebra of the BN-algebra $X$ .

We have ${\delta }_F(0\ast 1)={\delta }_F\left(1\right)\ge 0.9\wedge 0.5={\delta }_F\left(0\right)\wedge {\delta }_F\left(1\right)$ implies ${\delta }_F(0\ast 1)\ge {\delta }_F\left(0\right)\wedge {\delta }_F\left(1\right).$

Also ${\theta }_F(0\ast 1)={\theta }_F\left(1\right)\le 0.1\vee 0.5={\theta }_F\left(0\right)\vee {\theta }_F\left(1\right).$

Hence ${\delta }_F(x\ast y)\ge {\delta }_F\left(x\right)\wedge {\delta }_F\left(y\right)$ and ${\theta }_F(x\ast y)\le {\theta }_F\left(x\right)\vee {\theta }_F\left(y\right)$ for all $x,y\in X.$

Thus, $F=\{x\in S:{\delta }_F\left(x\right),{\theta }_F\left(x\right)\}$ is a fermatean fuzzy subalgebra of $X.$

Proposition 3.3:

Let X be a BN-algebra and s be a subalgebra of x. Then

Proof:

Let $S$ be a subalgebra of $X$ . Then $x,y\in S$ imply $x\ast y\in S$ . Since ${\theta }_F\left(0\right)=0$ and ${\delta }_F\left(0\right)=$ 1, it follows that $0\in F$ . Hence $F$ is nonempty.

Let $x,y\in F$ . Then ${\theta }_F\left(x\right)=0,{\delta }_F\left(x\right)=1$ and ${\theta }_F\left(y\right)=0,{\delta }_F\left(y\right)=1$ . Since $x,y\in F$ implies $x,y\in S$ , it follows that $x\ast y\in S$ . We get ${\theta }_F(x\ast y)=0,{\delta }_F(x\ast y)=1.$

Hence $F$ is a fermatean fuzzy subalgebra of x. Moreover $0\le {\left({\theta }_F\left(x\right)\right)}^{³}+{\left({\delta }_F\left(x\right)\right)}^{³}\le 1$ for all $x,y\in F.$

Definition 3.4:

Let $X$ be a BN-algebra and $S_{\mathit{FF}}\left(X\right)$ be the set of all fermatean fuzzy subalgebras of $X$ . Then for ${\delta }_F,{\theta }_F\in S_{\mathit{FF}}\left(X\right)$ we have:

are called level cuts of s. Here $U({\delta }_F,t)$ and $L({\theta }_F,k)$ are called the upper-level cut and lower-level cut of s respectively.

Theorem 3.5:

$F=\{x\in X,{\delta }_F\left(x\right),{\theta }_F\left(x\right)\}$ is a fermatean subalgebra of x if and only if $U({\delta }_F,t)$ and $L({\theta }_F,k)$ for $k,t\in [0,1]$ are subalgebra of $X$ .

Proof:

Assume f is a fermatean fuzzy subalgebra of x. We need to prove $U({\delta }_F,t)$ and $L({\theta }_F,k)$ are subalgebra of $X$ for $k,t\in [0,1].$

Let $x_0\in X$ such that ${\delta }_F\left(x_0\right)\ge t.$ Since

$\left(x_0\right)\ge t$ , implies ${\delta }_F\left(0\right)\ge t$ and hence $0\in U({\delta }_F,t).$ Therefore, $U({\delta }_F,t)$ is nonempty.

Let $x,y\in U({\delta }_F,t).$ Then ${\delta }_F\left(x\right)\ge t$ and ${\delta }_F\left(y\right)\ge t$ . Put ${\delta }_F\left(x\right)=t$ and ${\delta }_F\left(y\right)=t$ . Now ${\delta }_F(x\ast y)\ge {\delta }_F\left(x\right)\wedge {\delta }_F\left(y\right)=t\wedge t=t.$ It follows that ${\delta }_F(x\ast y)\ge t.$ Hence $x\ast y\in U({\delta }_F,t),$ which implies $U({\delta }_F,t)$ is a subalgebra of $X$ . Similar result holds for $0<t_1<t_2\le 1$ or $0<t_2<t_1\le 1$ .

Again let $x_0\in X$ such that ${\theta }_F(x\_0)\le k$ , for ${\delta }_F(x\_0)=1-{\theta }_F(x\_0),k\in [0,1].{\theta }_F\left(0\right)={\theta }_F(x_0\ast x_0)\le {\theta }_F(x\_0)\vee {\theta }_F(x\_0)={\theta }_F(x\_0)\le k.$ Imply that $0\in L({\theta }_F,k)$ which implies $L({\theta }_F,k)$ is non-empty.

Let $x,y\in L({\theta }_F,k)$ then ${\theta }_F\left(x\right)\le k$ and ${\theta }_F\left(y\right)\le k$ . Put ${\theta }_F\left(x\right)=k={\theta }_F\left(y\right)$ now ${\theta }_F(x\ast y)\le {\theta }_F\left(x\right)\vee {\theta }_F\left(y\right)=k\vee k=k$ which imply ${\theta }_F(x\ast y)\le k$ we get $x\ast y\in L({\theta }_F,k)$ therefore $L({\theta }_F,k)$ is a subalgebra of x a similar result holds for $0<k_1<k_2\le 1\mathrm{\text{or}}0<k_2<k_1\le 1.$

Conversely, assume $U({\delta }_F,t$ ) and $L({\theta }_F,k)$ are subalgebras of $X$ . We must prove $F=\{x\in X,{\delta }_F\left(x\right),{\theta }_F\left(x\right)\}$ is a fermatean fuzzy subalgebra of $X$ .

Let $t,k\in [0,1]$ and because $U({\delta }_F,t)$ and $L({\theta }_F,k)$ are subalgebras, we have $U({\delta }_F,t)$ and $L({\theta }_F,k)$ nonempty. Let $x_0\in U({\delta }_F,t)$ and $x_0\in L({\theta }_F,k)$ then ${\delta }_F\left(x_0\right)\ge t$ and ${\theta }_F\left(x_0\right)\le k$ .

Suppose ${\delta }_F(x\ast y)<{\delta }_F\left(x\right)\wedge {\delta }_F\left(y\right)$ then put $\beta =\mathit{½}\{{\delta }_F(x\ast y)+{\delta }_F\left(x\right)\wedge {\delta }_F\left(y\right)\}$ so that ${\delta }_F(x\ast y)<\beta <{\delta }_F\left(x\right)\wedge {\delta }_F\left(y\right)\le {\delta }_F(x\ast y)$ implies that ${\delta }_F(x\ast y)<{\delta }_F(x\ast y)$ which is a contradiction. Hence ${\delta }_F(x\ast y)\ge {\delta }_F\left(x\right)\wedge {\delta }_F\left(y\right)$ for $x,y\in X$ the cases $x_0\in U({\delta }_F,t)$ and $y_0\in L({\theta }_F,k)$ with $x_0\ne y_0$ followed by a similar argument.

Again, let $x,y\in X$ such that ${\theta }_F(x\ast y)>{\theta }_F\left(x\right)\vee {\theta }_F\left(y\right)$ . Then put $\gamma =\mathit{½}\{{\theta }_F(x\ast y)+{\theta }_F\left(x\right)\vee {\theta }_F\left(y\right)\}$ then we have ${\theta }_F(x\ast y)>\gamma >{\theta }_F\left(x\right)\vee {\theta }_F\left(y\right)\ge {\theta }_F(x\ast y)$ which implies that ${\theta }_F(x\ast y)>{\theta }_F(x\ast y)$ which is not correct. Hence ${\theta }_F(x\ast y)\le {\theta }_F\left(x\right)\vee {\theta }_F\left(y\right)$ for all $x,y\in X$ the cases $x_0\in U({\delta }_F,t)$ and $y_0\in L({\theta }_F,k)$ with $x_0\ne y_0$ follow by a similar argument thus $F=\{x\in X,{\delta }_F\left(x\right),{\theta }_F\left(x\right)\}$ is a fermatean fuzzy subalgebra of $X.$

Definition 3.6:

Let $X$ be a BN-algebra and let s be a subalgebra of $X$ . Then

Proposition 3.7:

Let x be a BN-algebra and let s be a subalgebra of x then

Is a fermatean fuzzy subalgebra of $X$ with $0\le {\left({\theta }_F\left(x\right)\right)}^{³}+{\left({\delta }_F\left(x\right)\right)}^{³}\le 1,{\theta }_F:X\to [0,1],{\delta }_F:X\to [0,1]$ and ${\theta }_F\left(x\right)=1-{\delta }_F\left(x\right)$ for $x\in X.$

Proof:

Let S be a subalgebra of $X$ then $x,y\in S$ imply $x\ast y\in S$ since ${\theta }_F\left(0\right)=0$ and ${\delta }_F\left(0\right)=1$ , it follows that $0\in F$ hence $F$ is nonempty.

Let $x,y\in F$ then ${\theta }_F\left(x\right)=0$ and ${\delta }_F\left(x\right)=1$ and ${\theta }_F\left(y\right)=0$ and ${\delta }_F\left(y\right)=1$ since $x,y\in F$ imply $x,y\in S$ , it follows that $x\ast y\in S$ we get ${\theta }_F(x\ast y)=0,{\delta }_F(x\ast y)=1.$

Hence F is a fermatean fuzzy subalgebra of x moreover $0\le {\left({\theta }_F\left(x\right)\right)}^{³}+{\left({\delta }_F\left(x\right)\right)}^{³}\le 1$ for all $x,y\in F$ .

Theorem 3.8:

Let S be a subalgebra of $X$ then ${({\delta }_F,{\theta }_F)}_{\left\{\right(t,k\left)\right\}}$ is a fermatean level subset of s if and only if

Is a fermatean fuzzy subalgebra of $X$ here $0\le {\left({\delta }_F\left(x\right)\right)}^{³}+{\left({\theta }_F\left(x\right)\right)}^{³}\le 1$ for all $x\in S$ .

Proof:

Let ${({\delta }_F,{\theta }_F)}_{\left\{\right(t,k\left)\right\}}$ be a level subset of S in $X$ .

Claim:

F is a fermatean fuzzy subalgebra.

Hence f is a fermatean fuzzy subalgebra of $X.$

Definition 3.9:

Let X be a BN Algebra. then fermatean fuzzy ideal of x is defined by:

Example 3.10:

Let X = {0, a, b, c} and * be defined by Table 2.

Let $0\le t_3<t_2<t_1<1.$ Define a fuzzy subset ${\delta }_F:X\to [0,1]$ by

If $x=a$ , then ${\theta }_F\left(a\right)=1-t_3\le (1-t_3)\vee (1-t_1)={\theta }_F\left(a\right)\vee {\theta }_F(a\ast a)$ .

Proposition 3.11:

For a fermatean fuzzy ideal $I_{\mathit{FF}}$ of x and any $x,y\in X$ . if $x\le y$ then

Proof:

Let $x,y\in X$ such that $x\le y$ implies $x\ast y=0$ . it follows $y\ast x=0.$

Proposition 3.12:

Let I_FF be a fermatean fuzzy ideal of a BN-algebra x then ${\theta }_F\left(0\right)\le {\theta }_F\left(x\right)\mathrm{\text{for}}\mathrm{all}x\in X,{\delta }_F:X\to [0,1]\mathrm{\text{and}}{\theta }_F\left(x\right)=1-{\delta }_F\left(x\right)$ with the property $0\le {\left({\theta }_F\left(x\right)\right)}^{³}+{\left({\delta }_F\left(x\right)\right)}^{³}\le 1.$

Proof:

Let $I_{\mathit{FF}}$ be a fermatean fuzzy ideal of a BN-algebra $X$ and ${\delta }_F:X\to [0,1]$ be a degree membership function and ${\theta }_F\left(x\right)=1-{\delta }_F\left(x\right)$ for all $x\in X.$

${\theta }_F\left(0\right)=1-{\delta }_F\left(0\right)\le 1-{\delta }_F\left(x\right)={\theta }_F\left(x\right)$ for all $x\in X$ hence ${\theta }_F\left(0\right)\le {\theta }_F\left(x\right)$ for all $x\in X.$

Proposition 3.13:

Let $I_{\mathit{FF}}$ be a fermatean fuzzy ideal of $X$ and $J=I_{\mathit{FF}}$ then

Proof:

Let j be a fermatean fuzzy ideal of x and

Definition 3.14:

Let $X$ be a BN-algebra and let $I_{\mathit{FF}}\left(X\right)$ be the set of fermatean fuzzy ideals of $X$ then for ${\delta }_F\in I_{\mathit{FF}}\left(X\right)$ for ${\delta }_F\& {\theta }_F\in I_{\mathit{FF}}\left(X\right)$ we have:

Theorem 3.15:

Let $X$ be a BN-algebra and ${\delta }_F,{\theta }_F\in I_{\mathit{FF}}\left(X\right),{\theta }_F\left(x\right)=1-{\delta }_F\left(x\right)$ if and only if the non-empty level subsets $U({\delta }_F,t)$ and $L({\theta }_F,k),t,k\in [0,1]$ are ideals of $X$ .

Proof:

Let ${\delta }_F,{\theta }_F\in I_{\mathit{FF}}\left(X\right)$ and $t,k\in [0,1]$ with ${\theta }_F\left(x\right)=1-{\delta }_F\left(x\right).$ we need to prove $U({\delta }_F,t)$ and $L({\theta }_F,k)$ are ideals of $X.$

Let $x_0\in X$ such that ${\delta }_F\left(x_0\right)\ge t,t\in [0,1]$ since ${\delta }_F,{\theta }_F\in I_{\mathit{FF}}\left(X\right)$ we have ${\delta }_F\left(0\right)\ge {\delta }_F\left(x_0\right)\ge t$ implies $0\in U({\delta }_F,t)$ .

Again, for $x_0\in$ x, such that ${\theta }_F\left(x_0\right)\le k$ and ${\theta }_F\left(0\right)\le {\theta }_F\left(x_0\right)\le k$ it follows that ${\theta }_F\left(0\right)\le k$ hence $0\in L({\theta }_F,k).$

Let $x,y\in X$ and $x\ast y,y\in U({\delta }_F,t),t\in [0,1]$ , imply ${\delta }_F(x\ast y)\ge t$ and ${\delta }_F\left(y\right)\ge t$ since ${\delta }_F\in I_{\mathit{FF}}\left(X\right),{\delta }_F\left(x\right)\ge {\delta }_F(x\ast y)\wedge {\delta }_F\left(y\right)\ge t.{\delta }_F\left(x\right)\ge t$ it follows that $x\in U({\delta }_F,t).$

Again let $x,y\in X$ such that $x\ast y,y\in L({\theta }_F,k)$ then we have ${\theta }_F(x\ast y)\le k$ and ${\theta }_F\left(y\right)\le k$ since $0\le {\left({\theta }_F\left(x\right)\right)}^{³}+{\left({\delta }_F\left(x\right)\right)}^{³}\le 1$ for all $x\in X$ and ${\theta }_F\in I_{\mathit{FF}}\left(X\right)$ we get ${\theta }_F\left(x\right)\le {\theta }_F(x\ast y)\vee {\theta }_F\left(y\right)\le k$ it follows that ${\theta }_F\left(x\right)\le k$ hence $x\in L({\theta }_F,k)$ therefore $U({\delta }_F,t)$ and $L({\theta }_F,k),t,k\in [0,1]$ are ideals of $X.$