Characterization of Derived Nilpotent (Engel) Lie Ring of Fuzzy Hyperrings by Using Fuzzy Strongly Regular Relations

Abstract

In this paper, we determined a new characterisation of the derived nilpotent (Engel) Lie ring of fuzzy hyperrings by fuzzy strongly regular relation ${\zeta }_n^{\ast }$(${\nu }_{n,s}^{\ast }$). Moreover, we proved that for a fuzzy hyperring S, the quotient $S/{\zeta }_n^{\ast }$($S/{\nu }_{n,s}^{\ast }$) was a nilpotent (Engel) Lie ring. Also, we introduced the notion of an ζ-role of a fuzzy hyperring and investigated its essential properties. Basically, we stated a necessary and sufficient condition for transitivity of ζ. Also, we studied the relationship between the strongly regular relation and ζ-role of a given fuzzy hyperring.

Keywords:

• (Nilpotent) Lie ring
• Engel Lie ring
• fuzzy hyperring
• strongly regular relation

MSC (2010):

• 17B62
• 20N20

1. Introduction

The notion of nilpotency is the most critical concepts in the study of groups [1]. Nilpotent groups arise in Galois theory, as well as in the classification of groups. By Galois theory, specific problems in field theory replace with problems in group theory, that is in some sense better understood. The certain generalisation of nilpotent groups is Engel groups [2–4].

Hyperstructure theory introduced by Marty 1934, he defined hypergroups, investigated their properties, and applied them to groups and relational algebraic functions (see [5]). In 1971, Rosenfeld used the concept of fuzzy sets to introduce fuzzy groups(see [6]). Since then, many researchers extended the concepts of abstract algebra to the fuzzy sets (see [6–14, 19–21]). The main tool in the definition of fuzzy hyperstructures is the ‘fuzzy hyperoperation’, where the non-empty set replaces the nonzero fuzzy set. Vougiouklis studied the fundamental relations on hyperrings [15]. In [16], Nozari and Fahimi introduced fundamental relation and commutative fundamental relation on fuzzy hyperrings.

In this paper, we define the ${\zeta }^{\ast }$ and ${\nu }_{(n,s)}$ relations, which are defined in fuzzy hyperrings. These relations connect the fuzzy hyperrins with (Engel)nilpotent-Lie rings. Especially, we introduce the notion of an ζ-role of a fuzzy hyperring and investigate its basic properties. Moreover, we study the relationship between the strongly regular relation and ζ-role of a given fuzzy hyperring. Also, we obtain Engel Lie rings from fuzzy hyperrings, by a FSR.

2. Preliminaries

First we recall some definitions and theorems.

Definition 2.1

Let H be a non-empty set and $\mathcal{F}(H)$ be the set of all fuzzy subsets of H. Suppose ${\mathcal{F}}^{\ast }(H)=\mathcal{F}(H)\setminus \{0\}$, where 0 is the zero fuzzy set. Then H equieped with a fuzzy hyperoperation $\circ :H\times H\mapsto {\mathcal{F}}^{\ast }(H)$, is called a fuzzy semihypergroup if for any $x,y,z\in H,$ $(x\circ y)\odot z=x\odot (y\circ z)$, where for any $\alpha \in F^{\ast }(H)$, and $r\in H$, $\begin{array}{rl}(x\odot \alpha )(r)& =\{\begin{array}{ll}{\displaystyle \underset{u\in H}{\bigvee }((x\circ u)(r)\wedge \alpha (u)),}& \alpha \ne 0\\ 0,& \alpha =0\end{array}\\ (\alpha \odot x)(r)& =\{\begin{array}{ll}{\displaystyle \underset{u\in H}{\bigvee }(\alpha (u)\wedge (u\circ x)(r)),}& \alpha \ne 0\\ 0,& \alpha =0\end{array}\end{array}$

If A, B are non-empty subsets of H and $x\in H$, then for any $u\in H$ we have $(x\circ A)(u)=\underset{a\in A}{\bigvee }(x\circ a)(u),(A\circ x)(u)=\underset{a\in A}{\bigvee }(a\circ x)(u)\mathrm{and}(B\circ A)(u)=\underset{a\in A,b\in B}{\bigvee }(b\circ a)(u).$ If a fuzzy semihypergroup $(H,\circ )$ satisfies in the condition $a\circ H=H\circ a={\chi }_H$, for any $a\in H$, where ${\chi }_H$ is the characteristic function of H, then the pair $(H,\circ )$ is called a fuzzy hypergroup.

Definition 2.2

[8]

A non-empty set $(S,+,\cdot )$, where + and · are two fuzzy hyperoperations, is called a fuzzy hyperring (we write by $\mathbf{F}\mathbf{H}\mathbf{R}$) if for any $a,b,c\in S$, the following axioms are valid:

1. $a\oplus (b+c)=(a+b)\oplus c$,

2. $a+S=S+a={\chi }_S$,

3. $a\odot (b\cdot c)=(a\cdot b)\odot c$,

4. a + b = b + a,

5. $a\odot (b+c)=(a\cdot b)⊞(a\cdot c)$ and $(a+b)\odot c=(a\cdot c)⊞(b\cdot c)$,

where for fuzzy subsets $\alpha ,\nu$ of a fuzzy hypergroupoid $(H,+)$ we have $\begin{array}{rl}(x\oplus \alpha )(r)& =\{\begin{array}{ll}{\displaystyle \underset{u\in H}{\bigvee }((x+u)(r)\wedge \alpha (u)),}& \alpha \ne 0\\ 0;& \alpha =0\end{array}\\ (\alpha \oplus x)(r)& =\{\begin{array}{ll}{\displaystyle \underset{u\in H}{\bigvee }(\alpha (u)\wedge (u+x)(r)),}& \alpha \ne 0\\ 0;& \alpha =0\end{array}\\ (\alpha ⊞\nu )(u)& =\underset{p,q\in H}{\bigvee }(\alpha (p)\wedge (p+q)(u)\wedge \nu (q));\\ (x\odot \alpha )(r)& =\{\begin{array}{ll}{\displaystyle \underset{u\in H}{\bigvee }((x\cdot u)(r)\wedge \alpha (u)),}& \alpha \ne 0\\ 0;& \alpha =0\end{array}\\ (\alpha \odot x)(r)& =\{\begin{array}{ll}{\displaystyle \underset{u\in H}{\bigvee }(\alpha (u)\wedge (u\cdot x)(r)),}& \alpha \ne 0\\ 0.& \alpha =0\end{array}\end{array}$ The fuzzy hyperring $(S,+,\cdot )$ is called commutative if for any $a,b\in S$, $a\cdot b=b\cdot a$. Also, $S^{\mathrm{\prime }}\subseteq S$ is called a subfuzzy hyperring of S if for any $s_1,s_2\in S^{\mathrm{\prime }}$ and $x\in S$, the following axioms are valid:

1. if $(s_1+s_2)(x)>0$, then $x\in S^{\mathrm{\prime }}$,

2. $s_1+S^{\mathrm{\prime }}=S^{\mathrm{\prime }}+s_1={\chi }_{S^{\mathrm{\prime }}}$,

3. if $(s_1\cdot s_2)(x)>0$, then $x\in S^{\mathrm{\prime }}$.

Definition 2.3

[14]

Let ρ be an equivalence relation on a fuzzy semihypergroup $(H,\circ )$. For any $\alpha ,\vartheta \in {\mathcal{F}}^{\ast }(H)$, we define two relations $\overline{\rho }$ and $\overline{\overline{\rho }}$ on ${\mathcal{F}}^{\ast }(H)$ as follows:

1. $\alpha \overline{\rho }\nu$ if for any $a\in H$, there exists $b\in H$ such that $a\rho b$ and if $\alpha (a)>0$, then $\nu (b)>0$ and if $\nu (a)>0$, then $\alpha (b)>0$.

2. $\alpha \overline{\overline{\rho }}\nu$ if every element $x\in H$ such that $\alpha (x)>0$ is ρ equivalent to every $y\in H$ such that $\nu (y)>0$.

Definition 2.4

[14]

An equivalence relation ρ on a fuzzy semihypergroup $(H,\circ )$ is called a fuzzy (strongly) regular (we write by F(S)R) if $a\rho b$ and $a^{\mathrm{\prime }}\rho b^{\mathrm{\prime }}$, then $(a\circ a^{\mathrm{\prime }})\overline{\rho }(b\circ b^{\mathrm{\prime }})((a\circ a^{\mathrm{\prime }})\overline{\overline{\rho }}(b\circ b^{\mathrm{\prime }})).$

Definition 2.5

[16]

Consider $(S,+,\cdot )$ be a FHR and φ be a F(S)R on both $(S,+)$ and $(S,\cdot )$. Then φ is called a F(S)R on S.

Let $(S,+,\cdot )$ and $(S^{\prime },+,\cdot )$ be two FHR and g be a map from S to $S^{\prime }$. Then for any $x,y,t\in S$ we have $g(x+y)=\{g(t)|(x+y)(t)>0\}\mathrm{and}g(x\cdot y)=\{g(t)|(x\cdot y)(t)>0\}.$

Theorem 2.6

[16]

Let $(S,+,\cdot )$ be a FHR and φ be an equivalence relation on S. If for every $\frac{a}{\phi },\frac{b}{\phi }\in S/\phi$, $\begin{array}{rl}\frac{a}{\phi }⊛\frac{b}{\phi }& =\{\frac{c}{\phi }:(a^{\prime }+b^{\prime })(c)>0,a\phi a^{\prime },b\phi b^{\prime }\},\\ \frac{a}{\phi }⊚\frac{b}{\phi }& =\{\frac{c}{\phi }:(a^{\prime }\cdot b^{\prime })(c)>0,a\phi a^{\prime },b\phi b^{\prime }\},\end{array}$ then

(i)	the relation φ is a FR on $(S,+,\cdot )$ iff $(S/\phi ,⊛,⊚)$ is a hyperring,
(ii)	the relation φ is a FSR on $(S,+,\cdot )$ iff $(S/\phi ,⊛,⊚)$ is a ring.

Definition 2.7

[16]

Let $(S,+,\cdot )$ be a FHR. A (commutative) fundamental relation on S is the smallest equivalence relations such that the quotient structure $(S/\phi ,⊛,⊚)$ is a (coommutative) ring.

Assume $(S,+,\cdot )$ is a FHR and relation γ on S is defined as follows: $\begin{array}{rl}& a\gamma b\iff \mathrm{\exists }n\in \mathbb{N},\mathrm{\exists }{k}_1,\dots ,k_n\in \mathbb{N},\mathrm{\exists }(z_{i1},\dots ,z_{i{k}_i})\in S^{k_i},(i=1,\dots ,n)\\ & \text{ such that }\left(\underset{i=1}{\overset{n}{⨄}}\prod _{j=1}^{k_i}{z}_{ij}\right)(a)>0\\ & \mathrm{and}\left(\underset{i=1}{\overset{n}{⨄}}\prod _{j=1}^{k_i}{z}_{ij}\right)(b)>0.\end{array}$ Then ${\gamma }^{\ast }$, the transitive closure of γ, is a fundamental relation on S (See [16]).

We recall that for a set X, a one to one function from X onto X is called a permutation on X. We denote the set of all permutations of X by ${\mathbb{S}}_X$. If $X=\{x_1,x_2,\dots ,x_n\}$, then we write ${\mathbb{S}}_n$ instead of ${\mathbb{S}}_X$. ${\mathbb{S}}_X$ forms a group. ${\mathbb{S}}_X$ is the symmetric group on X (see [1]).

Consider S be a FHR. Nozari and Fahimi in [16], introduced the relation ϵ on S as follows: $\begin{array}{rl}& aϵb\iff \mathrm{\exists }n\in \mathbb{N},\mathrm{\exists }(k_1,\dots ,k_n)\in {\mathbb{N}}^n,\mathrm{\exists }\varrho \in {\mathbb{S}}_n\mathrm{and}\\ & \times [(z_{i1},z_{i2},\dots ,z_{i{k}_i})\in S^{k_i},\mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,\dots ,n)]\\ & \text{such that}\left(\underset{i=1}{\overset{n}{⨄}}\prod _{j=1}^{k_i}{z}_{ij}\right)(a)>0\text{and}\left(\underset{i=1}{\overset{n}{⨄}}{A}_{\varrho (i)}\right)(b)>0,\end{array}$ where $A_i=\prod _{j=1}^{k_i}{z}_{i{\sigma }_i(j)}$. The quotient $S/{ϵ}^{\ast }$, where ${ϵ}^{\ast }$ is the transitive closure of ϵ, is a commutative ring (see [14]).

Definition 2.8

[17]

Let $(L,+,\cdot )$ be a vector space over a field F. Consider the operation $[-,-]:L\times L⟶L$ defined by $(x,y)⟶[x,y]$. Then L is called a Lie algebra if the following axioms are satisfied:

1. $[x,x]=0,$

2. $[x+y,z]=[x,z]+[y,z],$

3. $[[x,y],z]+[[y,z],x]+[[z,x],y]=0$.

Also, $\varphi \ne I\subseteq L$ is called a Lie ideal if $[x,r]\in I$ for any $x\in I$ and $r\in L$.

Let $(R,+,\cdot )$ be a ring. We can introduce the Lie structure on R by defining the Lie product $[x,y]=x\cdot y-y\cdot x$ for $x,y\in R$. This Lie structure, denoted by $R_l$, is called the associated Lie ring of R. Also, the Lie ring $R_l$ is called nilpotent of class r if we have $R_l=l_1(R_l)\supset l_2(R_l)\supset \cdots \supset l_r(R_l)\supset l_{r+1}(R_l)=0$ where for $k\ne 1$, $l_k(R_l)=[l_{k-1}(R_l),R_l]$ is the Lie ideal of $R_l$ generated by all elements of the form $[x,y]$ with $x\in l_{k-1}(R_l)$ and $y\in R_l$. R is called nilpotent of lenth r if its Lie ring is nilpotent of lenth r. Clearly $[l_k(R_l),l_n(R_l)]\subseteq l_{n+k}(R_l)$.

The n-th Lie centre of $R_l$ is defined by $U_n(R_l)=\{r\in R_l\mid [r,x_1,\dots ,x_n]=0,\mathrm{\forall }{x}_i\in R_l,1⩽i⩽n\},$ where commutator of weight n($n\in \mathbb{N}$) is defined by $[x_1,x_2,\dots ,x_n]=[[x_1,x_2,\dots ,x_{n-1}],x_n]$ (see [18]).

We can show that $U_n(R_l)$ is a (unitary) subring of $R_l$ and $\{0\}=U_0(R_l)\subseteq U_1(R_l)\subseteq U_2(R_l)\subseteq \cdots \subseteq U_n(R_l)\subseteq \dots .$ If $U_m(R_l)=R_l$ for some integer m, then $R_l$ is nilpotent. The smallest such integer is called the class of $R_l$ (see [18]).

Now, in this paper, we define two new relations ${\zeta }_n^{\ast }$ and ${\nu }_{n,s}^{\ast }$ on FHR S such that $S/{\zeta }_n^{\ast }$ and $S/{\vartheta }_{n,s}^{\ast }$ are nilpotent, and Engel Lie rings, respectively.

Note. From now one, let $(S,+,\cdot )$ or S be a fuzzy hyperring (i.e. FHR), unless otherwise stated.

3. Nilpotent Fundamental Relation ${\zeta }^{\ast }$

In this section, we present that combining a new FSR with the fuzzy hyperring is a nilpotent Lie ring.

Note. Let $x,y\in S$. For simplify, we use xy instead of $x\cdot y$ and $\mathrm{\exists }{I}_n$ instead of: $\begin{array}{rl}& \mathrm{\exists }(k_1,\dots ,k_m)\in {\mathbb{N}}^m,\mathrm{\exists }\varrho \in {\mathbb{S}}_m,\mathrm{\exists }(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},\mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m);\\ & \varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j\text{ if }{x}_{ij}\notin L_n(S).\end{array}$ Also, we use $\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^n{x}_{ij}$ instead of: $\begin{array}{rl}& \{x_{11}+x_{12},x_{11}\oplus (x_{12}{x}_{22}),x_{11}\oplus ((x_{12}{x}_{22})\odot x_{32}),(x_{11}x21)⊞(x_{12}{x}_{22}),\\ & \times (x_{11}{x}_{21})⊞((x_{12}{x}_{22})\odot x_{32}),\dots \}.\end{array}$

Definition 3.1

We consider $L_0(S)=S$ and for any $k\ge 0$, $L_{k+1}(S)=\{t\in S|\mathrm{\exists }r\in S\text{ s.t. }(xy)(r)>0,\text{ and }(t\oplus (yx))(r)>0,\text{ for some }x\in L_k(S)\text{ and }y\in S\}$

For any $n\in \mathbb{N}$, we let ${\zeta }_{1,n}=\{(x,x)|x\in S\}$ and for any $a,b\in S$ and $n,m\in \mathbb{N}$, we define ${\zeta }_{m,n}$ as follows: $\begin{array}{rl}& a{\zeta }_{m,n}b\iff \mathrm{\exists }(k_1,\dots ,k_m)\in {\mathbb{N}}^m,\mathrm{\exists }\varrho \in {\mathbb{S}}_m,\mathrm{\exists }(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},\\ & \mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m)\\ & \varrho (i)=i\text{ and }{\varrho }_{\varrho (i)}(j)=j\text{ if }{x}_{ij}\notin L_n(S),\text{ such that}a\in A_1\text{ and }b\in A_2.\end{array}$ where $A_1=\{z\in S|\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(z)>0\},A_2=\{z\in S|\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(z)>0\}.$ i.e. $a{\zeta }_{m,n}b\iff \mathrm{\exists }{I}_n\text{ such that}a\in A_1\text{and}b\in A_2.$ Now, suppose that ${\zeta }_n=\bigcup _{m\ge 1}{\zeta }_{m,n}$. If $a{\zeta }_{n}b$, then $a{\zeta }_{m,n}b$ for some $m\in \mathbb{N}$. Thus, $\mathrm{\exists }(k_1,\dots ,k_m)\in {\mathbb{N}}^m,\mathrm{\exists }\varrho \in {\mathbb{S}}_m,\mathrm{\exists }(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},\mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m);$ (I) $\varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j\text{ if }{x}_{ij}\notin L_n(S),$(I) such that $\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(a)>0,\text{and}\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(b)>0.$ Put $I=\varrho (i)$ and $J={\varrho }_{\varrho (i)}(j)$. So ${\varrho }^{-1}(I)=i$ and (II) $j=({\varrho }_{\varrho (i)}{)}^{-1}(J)=({\varrho }_{\varrho ({\varrho }^{-1}(I))}{)}^{-1}(J).$(II) Thus, by (I) and (II), for $(k_1,\dots ,k_m)\in {\mathbb{N}}^m,{\varrho }^{-1}\in {\mathbb{S}}_m$, $(x_{I1},x_{I2},\dots ,x_{I{k}_I})\in S^{k_I},{{\varrho }^{-1}}_I\in {\mathbb{S}}_{k_I},(I=1,2,\dots ,m)$. If $x_{IJ}\notin L_n(S),\text{then}{\varrho }^{-1}(I)=i=\varrho (i)=I\text{and}J={\varrho }_{\varrho (i)}(j)=j=({\varrho }_{\varrho ({\varrho }^{-1}(I))}{)}^{-1}(J)$ and so $(\underset{I=1}{\overset{m}{⨄}}\prod _{J=1}^{k_{\varrho (I)}}{x}_{{\varrho }^{-1}(I){{\varrho }^{-1}}_{\varrho ({\varrho }^{-1})(I)}}(J))(a)>0,\left(\underset{I=1}{\overset{m}{⨄}}\prod _{J=1}^{k_I}{x}_{IJ}\right)(b)>0.$ Hence, $b{\zeta }_{n}a$. Therefore, ${\zeta }_n$ is symmetric.

Define for any $a\in S$, $a(a)=({\chi }_a)(a)=1$. Then ${\zeta }_n$ is reflexive. Hence ${\zeta }_n^{\ast }$, the transitive closure of ${\zeta }_n$, is an equivalence relation.

Theorem 3.2

[8]

Let $(R,\dotplus ,\circ )$ be a ring. Then $(R,+,\cdot )$, where $+\text{and}\cdot$ are defined as follows is a fuzzy hyperring. $\text{For}\text{any}a,b\in R,a+b={\chi }_{\{a,b\}}\text{and}a\cdot b={\chi }_{a\circ b}.$

Example 3.3

Let $S=({\mathbb{Z}}_2,\dotplus ,\circ )$. By Theorem 3.2, $(S,+,\cdot )$ is a FHR. Then, $L_0(S)={\mathbb{Z}}_2$, and $L_1(S)=\{t\in {\mathbb{Z}}_2|\mathrm{\exists }r\in {\mathbb{Z}}_2;(xy)(r)>0and(t\oplus (yx))(r)>0,\text{ for some}x,y\in {\mathbb{Z}}_2\}.$ Let r = 0, x = 0 and $y=1$. Then $(xy)(r)={\chi }_0(0)=1>0$ and $\begin{array}{rl}(t\oplus (yx))(r)& =(t\oplus {\chi }_0)(0)\\ & =\underset{s\in {\mathbb{Z}}_2}{\bigvee }((t+s)(0))\wedge ({\chi }_0(s))\\ & =\underset{s\in {\mathbb{Z}}_2}{\bigvee }(({\chi }_{\{t,s\}})(0))\wedge ({\chi }_0(s))\\ & =((({\chi }_{\{t,0\}})(0))\wedge ({\chi }_0(0)))\vee ((({\chi }_{\{t,1\}})(0))\wedge ({\chi }_0(1)))\\ & =1\\ & >0.\end{array}$ Thus, $L_1(S)={\mathbb{Z}}_2$.

Lemma 3.4

Let ϱ be a permutation of ${\mathbb{S}}_k$, $k^{\mathrm{\prime }}=k+1$ and ${\varrho }^{\mathrm{\prime }}(j)=\{\begin{array}{ll}\varrho (i),& \mathrm{\forall }i\in \{1,2,\dots ,k\}\\ k+1,& i=k+1.\end{array}$ Thus ${\varrho }^{\mathrm{\prime }}$ is a permutation of ${\mathbb{S}}_{k^{\mathrm{\prime }}}$

Proof.

We show that ${\varrho }^{\mathrm{\prime }}$ is one to one. For this let $r,s\in \{1,2,\dots ,(k+1)\}$ and ${\varrho }^{\mathrm{\prime }}(r)={\varrho }^{\mathrm{\prime }}(s)$. By the definition of ${\varrho }^{\mathrm{\prime }}$ for the case $r\in \{1,2,\dots ,k\}$ and s = k + 1, we have $\varrho (r)=k+1$. Since $\varrho \in {\mathbb{S}}_k$, we have $\varrho (r)\le k$ and so $k+1=\varrho (r)\le k$ which is a contradiction. For $s\in \{1,2,\dots ,k\}$ and r = k + 1, we have $k+1=\varrho (s)\le k$, which is a contradiction. Finally, for $r,s\in \{1,2,\dots ,k\}$, we have $\varrho (r)=\varrho (s)$, and so r = s. Therefore, ${\varrho }^{\mathrm{\prime }}$ is one to one. It is clear that, ${\varrho }^{\mathrm{\prime }}$ is onto. Therefore, ${\varrho }^{\mathrm{\prime }}$ is a permutation of ${\mathbb{S}}_{k^{\mathrm{\prime }}}$.

Theorem 3.5

For each $x,y,z\in S$ if $x{\zeta }_n^{\ast }y$, then:

1. $(xz)\overline{\overline{{\zeta }_n^{\ast }}}(yz)$ and $(zx)\overline{\overline{{\zeta }_n^{\ast }}}(zy);$

2. $(x+z)\overline{\overline{{\zeta }_n^{\ast }}}(y+z)$ and $(z+x)\overline{\overline{{\zeta }_n^{\ast }}}(z+y).$

Proof.

(i) If $x{\zeta }_{n}y$, then there exists $m\in \mathbb{N}$ such that $x{\zeta }_{m,n}y$. Then there exist $(k_1,\dots ,k_m)\in {\mathbb{N}}^m$,  $\varrho \in {\mathbb{S}}_m$ and there exists $(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m)$; ($\varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j\text{ if }{x}_{ij}\notin L_n(S)$), such that $\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(x)>0\text{and}\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(y)>0.$ Let $z\in S$ and $r,s\in S$ such that $(xz)(r)>0$ and $(yz)(s)>0$. Then $(\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)\odot z)(r)=\underset{p\in S}{\bigvee }\{\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(p)\wedge (pz)(r)\}.$ Let p = x, then $(((\underset{i=1}{\overset{m}{⨄}})(\prod _{j=1}^{k_i}{x}_{ij})\odot z))(r)>0$. Also $(\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)\odot z)(s)=\underset{q\in S}{\bigvee }\{\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(q)\wedge (qz)(s)\}.$ Let q = y, then $(((\underset{i=1}{\overset{m}{⨄}})(\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)})\odot z))(s)>0$.

We set $k_i^{\prime }=k_i+1$, and we define ${\varrho }_i^{\prime }(j)=\{\begin{array}{ll}{\varrho }_i(j),& \mathrm{\forall }j\in \{1,2,\dots ,k_i\}\\ k_i+1,& j=k_i+1.\end{array}$ By Lemma 3.4, ${\varrho }_i^{\prime }$ is a permutation of ${\mathbb{S}}_{k_i^{\prime }}$. Thus for $z=x_{i{k}_i^{\prime }}$ and any $r,s\in S$; $\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i^{\prime }}{x}_{ij}\right)(r)>0,\text{and}(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}^{\prime }}{x}_{\varrho (i){\varrho }_{\varrho (i)}^{\prime }}(s))>0$ such that $\varrho (i)=i$ and ${\varrho }_{\varrho (i)}^{\prime }(j)=j$ if $x_{ij}\notin L_n(S)$. Therefore, for any $r,s\in S$ such that $(xz)(r)>0$ and $(yz)(s)>0$, we have $r{\zeta }_{m,n}s$. Thus $(xz)\overline{\overline{{\zeta }_n^{\ast }}}(yz)$. In the simillar way, we can show that $(zx)\overline{\overline{{\zeta }_n^{\ast }}}(zy)$.

Now, if $x{\zeta }_n^{\ast }y$, then there exists $k\in \mathbb{N}$ and $u_0=x,u_1,\dots ,u_k=y\in S$ such that $u_0=x{\zeta }_n{u}_1{\zeta }_n{u}_2{\zeta }_n\dots {\zeta }_n{u}_k=y.$ By the above result we have $(u_{0}z)=(xz)\overline{\overline{{\zeta }_n^{\ast }}}(u_{1}z)\overline{\overline{{\zeta }_n^{\ast }}}(u_{2}z)\overline{\overline{{\zeta }_n^{\ast }}}\dots \overline{\overline{{\zeta }_n^{\ast }}}(u_{k}z)=(yz),$ and so $(xz)\overline{\overline{{\zeta }_n^{\ast }}}(yz)$. By the simillar way, $(zx)\overline{\overline{{\zeta }_n^{\ast }}}(zy)$.

(ii) By the same manipulation we can prove $(x+z)\overline{\overline{{\zeta }_n^{\ast }}}(y+z)$ and $(z+x)\overline{\overline{{\zeta }_n^{\ast }}}(z+y).$

Corollary 3.6

For any $n\in \mathbb{N}$, the relation ${\zeta }_n^{\ast }$ is a FSR.

Proof.

Let $a,b,a^{\mathrm{\prime }},b^{\mathrm{\prime }}\in S$ and $a{\zeta }_n^{\ast }{a}^{\mathrm{\prime }}$, $b{\zeta }_n^{\ast }{b}^{\mathrm{\prime }}$. Then by Theorem 3.5, we have $(ab)\overline{\overline{{\zeta }_n^{\ast }}}(a^{\mathrm{\prime }}b)$ and $(a^{\mathrm{\prime }}b)\overline{\overline{{\zeta }_n^{\ast }}}(a^{\mathrm{\prime }}{b}^{\mathrm{\prime }})$. Then $(ab)\overline{\overline{{\zeta }_n^{\ast }}}(a^{\mathrm{\prime }}{b}^{\mathrm{\prime }})$. Similarly, we have $(a+b)\overline{\overline{{\zeta }_n^{\ast }}}(a^{\mathrm{\prime }}+b^{\mathrm{\prime }})$.

Proposition 3.7

For any $n\in \mathbb{N}$, we have ${\zeta }_{n+1}^{\ast }\subseteq {\zeta }_n^{\ast }$.

Proof.

Let $x{\zeta }_{n+1}^{\ast }y$, then there exist $k\in \mathbb{N}$, $u_0=x,u_1,\dots ,u_k=y\in S$ such that $(u_0=x){\zeta }_{n+1}{u}_1{\zeta }_{n+1}{u}_2{\zeta }_{n+1}\dots {\zeta }_{n+1}(u_k=y).$ We show that for any $z,t\in S$, (I) $\text{if}z{\zeta }_{n+1}t,\text{ then}z{\zeta }_{n}t.$(I) Since $z{\zeta }_{n+1}t$, there exists $m\in \mathbb{N}$ such that $z{\zeta }_{m,n+1}t$ and so $$\begin{gathered} \mathrm{\exists }(k_1,\dots ,k_m)\in {\mathbb{N}}^m,\mathrm{\exists }\varrho \in {\mathbb{S}}_m, \\ \mathrm{\exists }(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},\mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m);(\varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j\text{ if }{x}_{ij}\notin L_{n+1}(S)) \end{gathered}$$. Such that $\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(z)>0,\text{and}\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(t)>0.$ Now, let ${\varrho }^{\mathrm{\prime }}=\varrho$ and ${x^{\mathrm{\prime }}}_{ij}=x_{ij}$, ${\varrho }_i^{\mathrm{\prime }}={\varrho }_i$ and ${x^{\mathrm{\prime }}}_{ij}\notin L_n(S)$. Since $L_{n+1}(S)\subseteq L_n(S)$, we have ${x^{\mathrm{\prime }}}_{ij}\notin L_{n+1}(S)$. Hence, for $(k_1,\dots ,k_m)\in {\mathbb{N}}^m,{\varrho }^{\mathrm{\prime }}\in {\mathbb{S}}_m,({x^{\mathrm{\prime }}}_{i1},{x^{\mathrm{\prime }}}_{i2},\dots ,{x^{\mathrm{\prime }}}_{i{k}_i})\in S^{k_i},{{\varrho }^{\mathrm{\prime }}}_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m);({\varrho }^{\mathrm{\prime }}(i)=i\text{and}{{\varrho }^{\mathrm{\prime }}}_{{\varrho }^{\mathrm{\prime }}(i)}(j)=j\text{ if }{x^{\mathrm{\prime }}}_{ij}\notin L_n(S)),$ such that $\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x^{\mathrm{\prime }}}_{ij}\right)(z)>0,\text{and}\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{{\varrho }^{\mathrm{\prime }}(i)}}{x^{\mathrm{\prime }}}_{{\varrho }^{\mathrm{\prime }}(i){\varrho }_{{\varrho }^{\mathrm{\prime }}(i)}^{\mathrm{\prime }}(j)}\right)(t)>0.$ Therefore, $z{\zeta }_{n}t$.

Consequently, by (I) we have $(u_0=x){\zeta }_n{u}_1{\zeta }_n{u}_2{\zeta }_n\dots {\zeta }_n(u_k=y)$. Then $x{\zeta }_n^{\ast }y$.

Corollary 3.8

If S is a commutative FHR, then ${ϵ}^{\ast }={\zeta }_n^{\ast }$.

Proof.

It is clear that, ${\zeta }_n^{\ast }\subseteq {ϵ}^{\ast }$. It is enough to show that if S is a commutative, then $ϵ\subseteq {\zeta }_n$. For this, let $aϵb$. Then for some $m\in \mathbb{N}$, $\mathrm{\exists }(k_1,\dots ,k_m)\in {\mathbb{N}}^m,\mathrm{\exists }\varrho \in {\mathbb{S}}_m,\mathrm{\exists }(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},\mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m),$ such that $a\in A_1$ and $b\in A_2$.

Since S is commutative, for any $i,j\in \mathbb{N}$, each element $x_{\varrho (i){\varrho }_{\varrho (i)}(j)}$ can commute with others, and so we can consider $\varrho ,{\varrho }_i$ such that $\varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j\text{ if }{x}_{ij}\notin L_n(S),$ then $b\in A_2$ which implies that $a{\zeta }_{n}b$ and so $ϵ\subseteq {\zeta }_n$.

Example 3.9

Let S be FHR as Example 3.3. Then, by Corollary 3.8, we have ${ϵ}^{\ast }={\zeta }_n^{\ast }$ and so $S/{\zeta }_n^{\ast }=S/{ϵ}^{\ast }\cong S$.

Definition 3.10

Let $R_l$ be a Lie ring. We define $M_0(R_l)=R_l$ and $M_{k+1}(R_l)=\{[x,y]\mid x\in M_k(R_l),y\in R_l\}.$

Remark 3.11

Let φ be a FSR on S. Then, by Theorem 2.6, without loss of generality, one can assume that $(S/\phi ,⊛,⊚)$ is a Lie ring.

Theorem 3.12

$S/{\zeta }_n^{\ast }$ is a nilpotent Lie ring of class at most n + 1.

Proof.

We show that for a FSR φ on S and any $k\in \mathbb{N}$, (I) $M_{k+1}(S/\phi )=\{[\frac{t}{\phi },\frac{s}{\phi }]|t\in L_k(S),s\in S\}.$(I) The proof is based on induction on k. Put $R=S/\phi$ and $\overline{x}=\frac{x}{\phi }$, for any $x\in S$. If k = 0, then $M_1(R)=\{[\overline{t},\overline{s}]\mid t\in L_0(S),s\in S\}$. Now, put $\overline{a}=[\overline{t},\overline{s}]$ where $t\in L_{k+1}(S)$ and $s\in S$, so there exists $x\in L_k(S)$, $y\in S$ and $r\in S$ such that $(xy)(r)>0$ and $(t\oplus (yx))(r)>0$. Then by Theorm 2.6, $\overline{x}⊚\overline{y}=\overline{r}$ and $\overline{t}⊛(\overline{y}⊚\overline{x})=\overline{r}$. Therefore $\overline{t}=[\overline{x},\overline{y}]$. By induction hypotheses we have $\overline{t}\in M_{k+1}(R)$. Hence, $\overline{a}=[\overline{t},\overline{s}]\in M_{k+2}(R)$.

Conversely, let $\overline{a}\in M_{k+2}(R)$. Then by Definition 3.10, $\overline{a}=[\overline{x},\overline{y}]$, where $\overline{x}\in M_{k+1}(R)$ and $\overline{y}\in R$. So induction hypotheses implies that $\overline{x}=[\overline{u},\overline{v}]$, where $u\in L_k(S)$ and $v\in S$. Since $uv,vu\in {\mathcal{F}}^{\ast }(S)$, we have $(uv)(c)>0$ and $(vu)(b)>0$, for some $c,b\in S$. By Definition 2.2, for $b,c$, we have ${\chi }_S(c)=(S+b)(c)$, then there exists $t\in S$ such that $(t+b)(c)>0$. Thus $(t\oplus (vu))(c)=\underset{b\in S}{\bigvee }(t+b)(c)\wedge (vu)(b)>0$, and so $t\in L_{k+1}(S)$. Hence, by $(uv)(c)>0$, $(t\oplus (vu))(c)>0$ and Theorem 2.6, we have $\overline{c}=\overline{t}⊛(\overline{v}⊚\overline{u})=\overline{u}⊚\overline{v}$. Thus, $\overline{t}=[\overline{u},\overline{v}]=\overline{x}$, where $t\in L_{k+1}(S)$ and $y\in S$. Hence, $\overline{a}=[\overline{x},\overline{y}]=[\overline{t},\overline{y}]\in \{[\overline{t},\overline{s}]|\in L_{k+1}(S),s\in S\}.$ Therefore $M_{k+1}(S/\phi ))=\{[\overline{t},\overline{s}]|t\in L_k(S),s\in S\}$.

Now, let $R=S/{\zeta }_n^{\ast }$ and $\overline{x}=\frac{x}{\phi }$, for any $x\in S$. By induction on i, we show that $M_{n+1-i}(R)\subseteq U_i(R)$. Let i = 0, and $\overline{t},\overline{s}\in S/{\zeta }_{n,}^{\ast }$, $t\in L_n(S)$ and $s\in S$. For any $r,s\in S$ such that $(ts)(r)>0,(st)(p)>0$. Assume that $m=1,i=1,k_1=2,\varrho \in {\mathbb{S}}_1,x_{11}=t,x_{12}=s,{\varrho }_1\in {\mathbb{S}}_2,{\varrho }_1(1)=2,{\varrho }_1(2)=1.$ Then $0<(ts)(r)=(x_{11}{x}_{12})(r)$, and $0<(st)(p)=(x_{12}{x}_{11})(p)=x_{\varrho (1){\varrho }_{\varrho (1)}(1)}{x}_{\varrho (1){\varrho }_{\varrho (1)}(2)}$. Then $r{\zeta }_{n,}^{\ast }p$. Therefore, $\overline{r}=\overline{p}$, and so by Theorem 2.6, $\overline{t}⊚\overline{s}=\overline{r}=\overline{p}=\overline{s}⊚\overline{t}$. Then $(\overline{t}⊚\overline{s})-(\overline{s}⊚\overline{t})=\overline{0}$. Thus, $[\overline{t},\overline{s}]=(\overline{t}⊚\overline{s})-(\overline{s}⊚\overline{t})=\overline{0}$. Hence, $M_{n+1}(R)=\{[\overline{t},\overline{s}]|t\in L_k(S),s\in S\}=\{0\}$. Therefore, $\{0\}=M_{n+1}(R)\subseteq U_0(R)=\{0\}$. Now let $M_{n+1-i}(R)\subseteq U_i(R)$ and $a\in M_{n-i}(R)$. Then for any $y\in R$ we have $[a,y]\in M_{n+1-i}(R)\subseteq U_i(R)$. Hence for any $y\in R$, $[[a,y]{,}_{i}y]=0$, and so by definition of $U_{i+1}(R),$ we get $a\in U_{i+1}(R)$. Therefore, $M_{n+1-i}(R)\subseteq U_i(R)$.

Now, let i = n + 1. Then $M_{n+1-(n+1)}(R)\subseteq U_{n+1}(R)$, that is $R=M_0(R)=U_{n+1}(R)$. Hence R is nilpotent of class at most n + 1.

The relation ${\zeta }_n^{\ast }$ which is defined in Definition 3.1, can also be introduced for rings.

Note that every ring is a FHR.

Here, we introduce the ${\zeta }^{\ast }$ relation, which is defined on a finite FHR. This relation connects the FHRs with nilpotent Lie rings.

Definition 3.13

Let S be a finite FHR. Then we define the relations ζ and ${\zeta }^{\ast }$ on S by $\zeta =\bigcap _{n\ge 1}{\zeta }_n\text{and}{\zeta }^{\ast }=\bigcap _{n\ge 1}{\zeta }_n^{\ast }.$

Theorem 3.14

Consider S be a finite FHR. Then the relation ${\zeta }^{\ast }$ is a nilpotent fundamental relation.

Proof.

We show that ${\zeta }^{\ast }$ is a FSR on S such that $S/{\zeta }^{\ast }$ is a nilpotent Lie ring, and if φ is a FSR on S such that $S/\phi$ is a nilpotent Lie ring (of class n + 1), then ${\zeta }^{\ast }\subseteq \phi$.

Since ${\zeta }^{\ast }=\bigcap _{n\ge 1}{\zeta }_n^{\ast }$, by Corollary 3.6, it is easy to see that ${\zeta }^{\ast }$ is a FSR on S. Since S is finite, by Proposition 3.7, there exists $k\in \mathbb{N}$ such that ${\zeta }_{k+1}^{\ast }={\zeta }_k^{\ast }$. Thus $\zeta \ast ={\zeta }_m^{\ast }$, for some $m\in \mathbb{N}$ and so by Theorem 3.12, $S/{\zeta }^{\ast }$ is a nilpotent Lie ring.

Suppose φ is a FSR on S such that $K=S/\phi$ is a nilpotent Lie ring of class n + 1. We show that for any $n\in \mathbb{N},{\zeta }_n\subseteq \phi .$ Let $x{\zeta }_{n}y$. Then there exists $m\in \mathbb{N}$ such that $x{\zeta }_{m,n}y$. Then $\mathrm{\exists }{I}_n$ such that $\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(x)>0\text{ and}\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(y)>0.$ Hence, by Theorem 2.6, for $x,y\in S$, $\frac{x}{\phi }=\underset{i=1}{\overset{m}{⨁}}\underset{j=1}{\overset{k_i}{⨀}}\frac{x_{ij}}{\phi }\text{and}\frac{y}{\phi }=\underset{i=1}{\overset{m}{⨁}}\underset{j=1}{\overset{k_{\varrho (i)}}{⨀}}\frac{x_{\varrho (i){\varrho }_{\varrho (i)}(j)}}{\phi }.$ Since $S/\phi$ is a nilpotent Lie ring of class n + 1, by the proof of Theorem 3.12, we have $\frac{0}{\phi }=M_{n+1}(S/\phi )=\{[\frac{t}{\phi },\frac{s}{\phi }]|t\in L_n(S),s\in S\}$, and so $\frac{0}{\phi }=[\frac{t}{\phi },\frac{s}{\phi }]=(\frac{t}{\phi }⊚\frac{s}{\phi })\ominus (\frac{s}{\phi }⊚\frac{t}{\phi })$. Then $\frac{t}{\phi }⊚\frac{s}{\phi }=\frac{s}{\phi }⊚\frac{t}{\phi }$. Thus, for $x_{ij}\in L_n(S)$ and $v\in S$, we get $\frac{x_{ij}}{\phi }⊚\frac{v}{\phi }=\frac{v}{\phi }⊚\frac{x_{ij}}{\phi }$. Moreover, if $x_{ij}\in L_n(S)$, then $\frac{x_{ij}}{\phi }$ commutes with others. Also, if $x_{ij}\notin L_n(S)$ then $\varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j$, and so $\frac{x_{ij}}{\phi }=\frac{x_{\varrho (i){\varrho }_{\varrho (i)}(j)}}{\phi }$. Consequently, $\frac{x}{\phi }=\frac{y}{\phi }$, which implies that $x\phi y$. Therefore, ${\zeta }_n\subseteq \phi$.

Now, we can prove that ${\zeta }^{\ast }\subseteq \phi$. For this, let $z,t\in S$ and $z{\zeta }^{\ast }t$. Then for any $n\in \mathbb{N}$, $z{\zeta }_n^{\ast }t$ and so there exist $z_0,z_1,\dots ,z_k\in S$ ($k\in \mathbb{N}$) such that $(z=z_0){\zeta }_n{z}_1{\zeta }_n\cdots {\zeta }_n(z_k=t)$. Then we have $(z=z_0)\phi z_1\phi \cdots \phi (z_k=t)$. Hence, ${\zeta }^{\ast }\subseteq \phi$.

4. ζ-Role of a Fuzzy Hyperring

In this section, we introduce the concept of ζ-role and we determine necessary and sufficient conditions such that the relation ζ be transitive.

$\mathbf{N}\mathbf{o}\mathbf{t}\mathbf{e}$. For simplify we use $\mathrm{\forall }{I}_{r\ge 1}$ instead of the following assumption: $\mathrm{\forall }m\in \mathbb{N},k_i\in \mathbb{N}(i=1,2,\dots ,m),(x_{i1},\dots ,x_{i{k}_i})\in S^{k_i}\text{ and}\mathrm{\forall }\varrho \in {\mathbb{S}}_m$, ${\varrho }_i\in {\mathbb{S}}_{k_i};\varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j,\text{ if}{x}_{ij}\notin \bigcup _{r\ge 1}{L}_r(S).$

Definition 4.1

Let $\varphi \ne X\subseteq S$. Then we say that X is an ζ-role of S if the following implication holds: $\mathrm{\forall }{I}_{r\ge 1}\text{ if there exists }x\in X\text{ such that }x\in A_1,\text{ then for any }y\in S\setminus X\text{we have}y\notin A_2,$ where $A_1=\{z\in S|\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(z)>0\}\text{and}{A}_2=\{z\in S|\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(z)>0\}.$

Note that by Definition 4.1, there exists $x\in X$ such that $(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij})(x)>0,$ and for any $y\notin X$, we have $(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)})(y)=0$.

Remark 4.2

Let φ be a FSR on S, $z\in \phi (z)$ be the class of z module φ. Then $\phi (z)$ is an ζ-role of S.

Proof.

Let $\mathrm{\forall }{I}_{r\ge 1}$, there exists $x\in \phi (z)$ such that $x\in A_1$. If for $y\in S\mathrm{\setminus }\phi (z)$ we have $y\in A_2$, then φ is a FSR implies that $x\phi y$. Thus, $y\in \phi (z)$, is a contradiction.

Theorem 4.3

Let $\varphi \ne X\subseteq S$ and $x\in X$. Then the following conditions are equivalent:

(i)	X is an ζ-role of S,
(ii)	if $x\zeta y$, then $y\in X$,
(iii)	if $x{\zeta }^{\ast }y$, then $y\in X$.

Proof.

$(i\Rightarrow ii)$ If $(x,y)$ is a pair of $S^2$, such that $x\in X$ and $x\zeta y$, then for any $n\in \mathbb{N}$, $\mathrm{\exists }{I}_n$ such that $x\in A_1$ and $y\in A_2$. If $y\notin X$, then $y\in S\setminus X$ and so by Definition 4.1, $y\notin A_2$, which is a contradiction. Thus $y\in X$.

$(ii\Rightarrow iii)$ Suppose that $(x,y)$ is a pair of $S^2$ such that $x\in X$ and $x{\zeta }^{\ast }y$. Then there exists $(z_1,\dots ,z_k)\in S^k$($k\in \mathbb{N}$) such that $(x=z_0)\zeta z_1\zeta \cdots \zeta (z_k=y)$. By (ii) (k-times) we obtain $y\in X$.

$(iii\Rightarrow i)$ Let $x\in X$ and $k_i\in \mathbb{N}(i=1,2,\dots ,m)$, $(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i}$ such that $x\in A_1$. Then for any $\varrho \in {\mathbb{S}}_m$ and any ${\varrho }_i\in {\mathbb{S}}_{k_i}\text{such that }(\varrho (i)=i,\text{and}{\varrho }_{\varrho (i)}(j)=j,\text{if}{x}_{ij}\notin \bigcup _{r\ge 1}{L}_r(S))$ and any $y\in S\setminus X$ such that $y\in A_2$, we have $x\zeta y$. Thus $x{\zeta }^{\ast }y$ and by (iii), we obtain $y\in X$, which is a contradiction. Therefore, X is an ζ-role of S.

Theorem 4.4

For any $a\in S$, $\zeta (a)$ is an ζ-role of S if and only if ζ is transitive.

Proof.

$(\Rightarrow )$ Suppose $x{\zeta }^{\ast }y$. Then there exists $(z_1,\dots ,z_k)\in S^k$($k\in \mathbb{N}$) such that $(x=z_0)\zeta z_1\zeta \cdots \zeta (z_k=y)$. Since $\zeta (z_i)$, for any $0\le i\le k$, is an ζ-role, by Theorem 4.3, we have $z_i\in \zeta (z_{i-1})$. Thus $y\in \zeta (x)$ and so $x\zeta y$. Hence ${\zeta }^{\ast }=\zeta$. Therefore, ζ is transitive.

$(\Leftarrow )$ Suppose $x,y\in S$, $z\in \zeta (x)$ and $z\zeta y$. Since ζ is transitive, we conclude that $y\in \zeta (x)$, and so by Theorem 4.3, $\zeta (x)$ is a ζ-role of S.

Remark 4.5

Let θ be a FSR on S and $z\in S$. Then $\theta (z)$ is an ζ-role of S.

Definition 4.6

Consider A be a non-empty subset of S. The intersection of any ζ-role of S, which contains A is called ζ-closure of A in S. We denote the ζ-closure of A by $K(A)$.

$\mathbf{N}\mathbf{o}\mathbf{t}\mathbf{e}$. We use $\mathrm{\exists }{I}_{r\ge 1}$ instead of the following assumption:

$\mathrm{\exists }(k_1,\dots ,k_m)\in {\mathbb{N}}^m,\mathrm{\exists }(x_{i_1},\dots ,x_{i{k}_i})\in S^{k_i}(i=1,\dots ,m),\text{and}\mathrm{\exists }\varrho \in {\mathbb{S}}_m,{\varrho }_i\in {\mathbb{S}}_{k_i};$ $\varrho (i)=i,{\varrho }_{\varrho (i)}(j)=j\text{ if}{x}_{ij}\notin \bigcup _{r\ge 1}{L}_r(S).$

Theorem 4.7

Let $m\in \mathbb{N}$, $\varphi \ne A\subseteq S$ and

(i)	$E_1(A)=A;$
(ii)	$E_{n+1}(A)=\{x\in S|\mathrm{\exists }{I}_{r\ge 1}\text{ such that}x\in A_1\text{ and for some }a\in E_n(A)\text{we have}a\in A_2\}$;
(iii)	$E(A)=\bigcup _{n\ge 1}{E}_n(A)$. Then $E(A)=K(A),$

Proof.

By Definition 4.6, it is enough to prove $E(A)$ is an ζ-role of S. If $A\subseteq B$ and B is an ζ-role, then $E(A)\subseteq B$. Suppose there exists $a\in E(A)$ such that $a\in A_1$, and $\mathrm{\exists }\varrho \in {\mathbb{S}}_m,{\varrho }_i\in {\mathbb{S}}_{k_i}$ such that $\varrho (i)=i,{\varrho }_{\varrho (i)}(j)=j$, if $x_{ij}\notin \bigcup _{r\ge 1}{L}_r(S)$. Since $E=\bigcup _{n\ge 1}{E}_n$, there exists $n\in \mathbb{N}$ such that $a\in E_n(A)$ and $a\in A_1$. Now, if there exists $t\in S\setminus E(A)$ such that $t\in A_2$, then $t\in E_{n+1}(A)$. Thus, $t\in \bigcup _{n\ge 1}{E}_n(A)=E(A)$ which is a contradiction. Thus for any $t\in S\setminus E(A)$, $t\notin A_2$ and so $E(A)$ is an ζ-role of S. We prove the second role by induction on n. We have $E_1(A)=A\subseteq B$. Suppose $E_n(A)\subseteq B$. We show that $E_{n+1}(A)\subseteq B$. If $z\in E_{n+1}(A)$, then $\mathrm{\exists }{I}_{r\ge 1}$ such that $z\in A_1$ and $t\in A_2$, for some $t\in E_n(A)$. Since $E_n(A)\subseteq B$ we have $t\in B$ and $t\in A_2$. Moreover, B is an ζ-role of S, and $z\in A_1$, then $z\in B$.

Theorem 4.8

If $\varphi \ne A\subseteq S$, then $K(A)=\bigcup _{a\in A}K(a)$.

Proof.

It is easy to see that for any $a\in A$, $K(a)\subseteq K(A)$. By Theorem 4.7, we get $K(A)=\bigcup _{n\ge 1}{E}_n(A)$ and $E_1(A)=A=\bigcup _{a\in A}\{a\}$. Suppose $E_n(A)=\bigcup _{a\in A}{E}_n(a)$ holds for n and $z\in E_{n+1}(A)$, then $\mathrm{\exists }{I}_{r\ge 1}$ such that $z\in A_1$ and $a\in A_2$, for some $a\in E_n(A)$. By the hypotheses of induction, $E_n(A)=\bigcup _{b\in A}{E}_n(b)$ and so for $a\in E_n(A)$ we have $a\in \bigcup _{b\in A}{E}_n(b)$, then $a\in E_n(b)$ for some $b\in A$. Thus, $a\in A_2$ for some $a\in E_n(b)$. Hence, $z\in E_{n+1}(b)$, and so $E_{n+1}(A)\subseteq \bigcup _{b\in A}{E}_{n+1}(b)$. Therefore, $K(A)=\bigcup _{a\in A}K(a)$.

Lemma 4.9

For any $n\ge 2$ and $x,y,z\in S$ we have

(i)	$E_n(E_2(z))=E_{n+1}(z)$,
(ii)	$x\in E_n(y)$ if and only if $y\in E_n(x)$.

Proof.

(i) Let $z\in S$. Then by Theorem 4.7, we have $E_2(E_2(z))=\{x\in S|\mathrm{\exists }{I}_{r\ge 1}\text{such that}x\in A_1\text{and for some }a\in E_2(z),a\in A_2\}=E_3(z).$ Now, we prove by induction on n. Suppose $E_n(E_2(z))=E_{n+1}(z)$. Then $\begin{array}{rl}{E}_{n+1}(E_2(z))& =\{x\in S|\mathrm{\exists }{I}_{r\ge 1}\text{such that}x\in A_1\text{ and for some }a\in E_n(E_2(z))a\in A_2\}\\ & =E_{n+2}(z).\end{array}$ (ii) We prove by induction on n. It is easy to see that $x\in E_2(y)$ if and only if $y\in E_2(x)$. Suppose $x\in E_{n+1}(y)$. Then $\mathrm{\exists }{I}_{r\ge 1}$ such that $x\in A_1\text{ and for some }a\in E_n(y),\text{we have}a\in A_2.$ Clearly, $x\in E_1(x)$. Also, $x\in A_1$ and $a\in A_2,$ then $a\in E_2(x)$. By hypotheses of induction, since $a\in E_n(y)$ we have $y\in E_n(a)$ and so by (i), $y\in E_n(E_2(x))=E_{n+1}(x)$. Therefore, $x\in E_n(y)$ if and only if $y\in E_n(x)$.

Theorem 4.10

Let $x,y\in S$.Then $xEy$ if and only if $x\in E(\{y\})$ is an equivalence relation on S.

Proof.

By Theorem 4.7, $W\{x\}=K\{x\}$ and by Definition 4.6, we have $x\in E\{x\}$, thus $xEx$. Then E is reflexive. For transitivity, let $x,y,z\in S$ such that $xEy$ and yEz. Then by Theorem 4.7, $x\in K(y)$ and $y\in K(z)$. Thus, for any ζ-role P which contains z i.e. $z\in P$, we have $K(z)\subseteq P$ and since $y\in K(z)$, we have $y\in P$. Then $K(y)\subseteq P$ and so $x\in P$. Thus for any ζ-role of S, we have $x\in P$, i.e. $x\in K(z)$. Hence, by Theorem 4.7, xEz. Then E is transitive. The symmetrically of E follows directly from Lemma 4.9.

Theorem 4.11

For all $a,b\in S$, aEb if and only if $a{\zeta }^{\ast }b$.

Proof.

$(\Leftarrow )$ Let $a{\zeta }^{\ast }b$. Then for any $n\in \mathbb{N}$, there exists $m\in \mathbb{N}$, such that $a{\zeta }_{m,n}b$. Then $\mathrm{\exists }{I}_n$ such that $a\in A_1,b\in A_2,$ so $a\in E_2(b)$. Thus, by Lemma 4.9, $b\in E_2(a)$. Hence $aEb$ and so ${\zeta }^{\ast }\subset E$.

$(\Rightarrow )$ If xEy, then $x\in E_n(y)$ for some $n\in \mathbb{N}$. Then $\mathrm{\exists }{I}_{r\ge 1}$ such that $x\in A_1$ and for some $x_1\in E_{n-1}(y)$ we have $x_1\in A_2$. Thus, $x{\zeta }_n{x}_1$. Hence continuing this method $\mathrm{\exists }{x}_2,\dots ,x_{n-1}\in S$ such that $x_i\in E_{n-i}(y)$ and $x_{i-1}{\zeta }_n{x}_i$. Then $(x=x_0){\zeta }_n{x}_1{\zeta }_n\dots {\zeta }_n(x_{n-1}=y)$. Therefore, $E\subseteq {\zeta }^{\ast }.$

Remark 4.12

By Theorem 2.6, if $(S,+,\cdot )$ is a FHR, then $(S/{\zeta }^{\ast },⊛,⊚)$ is a ring. We define ${\omega }_S$ by ${\omega }_S=\{x\in S\mid \psi (x)=\frac{0}{{\zeta }^{\ast }}\}={\psi }^{-1}(0_{S/{\zeta }^{\ast }})$, where $\psi :S\to S/{\zeta }^{\ast }$ is the canonical projection.

Lemma 4.13

If M is an ζ-role of S, then ${\psi }^{-1}(\psi (M))=M$.

Proof.

We know $M\subseteq {\psi }^{-1}(\psi (M))$. If $x\in {\psi }^{-1}(\psi (M))$, then there exists $b\in M$ such that $\psi (x)=\psi (b)$. For any $y\in S$, $\psi (y)=\frac{y}{{\zeta }^{\ast }}$ and so $\frac{x}{{\zeta }^{\ast }}=\frac{b}{{\zeta }^{\ast }}$. Thus, $x{\zeta }^{\ast }b$. Since M is an ζ role of S, $x{\zeta }^{\ast }b$ and $b\in M$, then by Theorem 4.3, we have $x\in M$.

Theorem 4.14

${\omega }_S$ is a fuzzy subhyperring of S, which is also an ζ-role of S.

Proof.

It is clear that, ${\omega }_S\subseteq S$. Let $x,y\in {\omega }_S$ and for any $r\in S$, $\overline{r}=\frac{r}{{\zeta }^{\ast }}$ and $(x+y)(r)>0$ ($(x\cdot y)(r)>0$). Then by Theorem 2.6, $\overline{x}⊛\overline{y}=\overline{r}$ ($\overline{x}⊚\overline{y}=\overline{r}$). Then $\overline{r}=\overline{0}$, and so $r\in \omega$. We show that for any $y\in {\omega }_S$, ${\omega }_S+y={\chi }_{{\omega }_S}$. Let $x,y\in {\omega }_S$. Then, by $S+y={\chi }_S$, there exists $u\in S$ such that $(u+y)(x)>0$. Therefore, by Theorem 2.6, $\frac{u}{{\zeta }^{\ast }}⊛\frac{y}{{\zeta }^{\ast }}=\frac{x}{{\zeta }^{\ast }}$. Then $\frac{u}{{\zeta }^{\ast }}=\frac{0}{{\zeta }^{\ast }}$. Thus $u\in {\omega }_S$. Consequently, (I) ${\omega }_S+y={\chi }_{{\omega }_S}.$(I) Hence, ${\omega }_S$ is a fuzzy subhyperring of S. Now we prove that (II) $x\in {\psi }^{-1}(\psi (\{y\}))\iff ({\omega }_S+y)(x)>0.$(II) For this let $x\in S$, $t\in {\omega }_S$ and $(t+y)(x)>0$. Then by Theorem 2.6, $\psi (x)=\psi (t)⊛\psi (y)=0_{S/{\zeta }^{\ast }}⊛\psi (y)=\psi (y)$ and so $x\in {\psi }^{-1}(\psi (\{y\}))$. Conversely, let $x\in {\psi }^{-1}(\psi (\{y\}))$. Then $\psi (x)=\psi (y)$. Since $S+y={\chi }_S$, there is $a\in S$ such that $(a+y)(x)>0$ and so $\psi (y)=\psi (x)=\psi (a)⊛\psi (y)$ (since by Remark 2.6, $S/{\zeta }^{\ast }$ is a ring and $\psi (y)\in S/{\zeta }^{\ast }$). Then $\psi (a)=0_{S/{\zeta }^{\ast }}$ and so $a\in {\psi }^{-1}(0_{S/{\zeta }^{\ast }})={\omega }_S$. Thus by $(a+y)x>0$, we have $({\omega }_S+y)(x)>0$. Also, since $\begin{array}{rl}z\in {\psi }^{-1}(\psi (\{y\}))& ⟺\psi (z)=\psi (y)\\ & ⟺\frac{z}{{\zeta }^{\ast }}=\frac{y}{{\zeta }^{\ast }}\\ & ⟺z{\zeta }^{\ast }y\\ & ⟺z\in \frac{y}{{\zeta }^{\ast }}=E(\{y\})=K(y).(\text{by Theorems }4.10,4.11,4.7,)\end{array}$ we have $K(y)={\psi }^{-1}(\psi (\{y\}))$ which by (I), (II) implies that $K(y)={\omega }_S$, and so ${\omega }_S$ is a ζ-role of S.

By Lemma 4.13 and Theorem 4.14, we have the following corollary.

Corollary 4.15

${\psi }^{-1}(\psi ({\omega }_S))={\omega }_S$.

5. Engel Lie Rings Derived from Fuzzy Hyperrings

In this section, continuing our previous work, we define a FSR, on a FHR S such that the quotient is an Engel Lie ring. Let s be a fixed element of S, unless we notify.

Definition 5.1

We define $L_{(0,s)}(S)=S$, and for $k\ge 0$, $L_{(k+1,s)}(S)=\{t\in S\mid \mathrm{\exists }r\in Ss.t(x\cdot s)(r)>0and(t\oplus (s\cdot x))(r)>0,\mathrm{forsome}x\in L_{(k,s)}(S)\}.$

Now, let $n\in \mathbb{N}$ and ${\nu }_{(1,n,s)}$ be the diagonal relation on S. For every integer $m>1,$ define ${\nu }_{(m,n,s)}$ as follows: $\begin{array}{rl}& a{\nu }_{(m,n,s)}b\iff \mathrm{\exists }{k}_1,\dots ,k_m\in \mathbb{N}(m\in \mathbb{N})\mathrm{and}\mathrm{\exists }\varrho \in {\mathbb{S}}_m,(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},\mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i};\\ & (\varrho (i)=i,{\varrho }_{\varrho (i)}(j)=j\text{ if }{x}_{ij}\notin L_{(n,s)}(S))\text{ and }a\in A_1,b\in A_2\end{array}$ i.e.

$\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}(x_{ij})(a)>0\mathrm{and}(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}(x_{\varrho (i){\varrho }_{\varrho (i)}(j)})(b)>0$ Consider ${\nu }_{(n,s)}=\bigcup _{m\ge 1}{\nu }_{(m,n,s)}$. Then ${\nu }_{(n,s)}$ is symmetric (Similar to Definition ${\zeta }_n$). Define for any $a\in S$, $a(a)=({\chi }_a)(a)=1$, thus ${\nu }_{(n,s)}$ is reflexive. Then ${\nu }_{(n,s)}^{\ast }$, the transitive closure of ${\nu }_{(n,s)}$, is an equivalence relation on S.

In the same way of the Corollary 3.6, we have the following theorem.

Theorem 5.2

For any $n\in \mathbb{N}$ the relation ${\nu }_{(n,s)}^{\ast }$ is a FSR.

Corollary 5.3

For a fixed element $s\in S$ we have ${\nu }_{(n,s)}^{\ast }\subseteq {{\zeta }_n}^{\ast }\subseteq {ϵ}^{\ast }$.

Proof.

Let $z{\nu }_{(n,s)}t$. Then there exists $m\in \mathbb{N}$ such that $z{\nu }_{(m,n,s)}t$ and so $\mathrm{\exists }(k_1,\dots ,k_m)\in {\mathbb{N}}^m,\mathrm{\exists }\varrho \in {\mathbb{S}}_m,\mathrm{\exists }(x_{i1},x_{i2},\dots ,x_{i{k}_i})\in S^{k_i},\mathrm{\exists }{\varrho }_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m);(\varrho (i)=i\text{and}{\varrho }_{\varrho (i)}(j)=j,\text{ if }{x}_{ij}\notin L_{(n,s)}(S)).$ Then $\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_i}{x}_{ij}\right)(z)>0,\mathrm{and}\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{\varrho (i)}}{x}_{\varrho (i){\varrho }_{\varrho (i)}(j)}\right)(t)>0.$ Now, let ${\varrho }^{\mathrm{\prime }}=\varrho$ and ${x^{\mathrm{\prime }}}_{ij}=x_{ij}$, ${\varrho }_i^{\mathrm{\prime }}={\varrho }_i$ and ${x^{\mathrm{\prime }}}_{ij}\notin L_n(S)$. By $L_{(n,s)}(S)\subseteq L_n(S)$, we have ${x^{\mathrm{\prime }}}_{ij}\notin L_{(n,s)}(S)$. Thus, for $(k_1,\dots ,k_m)\in {\mathbb{N}}^m,{\varrho }^{\mathrm{\prime }}\in {\mathbb{S}}_m,({x^{\mathrm{\prime }}}_{i1},{x^{\mathrm{\prime }}}_{i2},\dots ,{x^{\mathrm{\prime }}}_{i{k}_i})\in S^{k_i},{{\varrho }^{\mathrm{\prime }}}_i\in {\mathbb{S}}_{k_i},(i=1,2,\dots ,m),\text{such that}({\varrho }^{\mathrm{\prime }}(i)=i\text{and}{{\varrho }^{\mathrm{\prime }}}_{{\varrho }^{\mathrm{\prime }}(i)}(j)=j\text{ if }{x^{\mathrm{\prime }}}_{ij}\notin L_n(S)),$ such that $\left({⨄}_{i=1}^m\prod _{j=1}^{k_i}{x^{\mathrm{\prime }}}_{ij}\right)(z)>0,\mathrm{and}\left(\underset{i=1}{\overset{m}{⨄}}\prod _{j=1}^{k_{{\varrho }^{\mathrm{\prime }}(i)}}{x^{\mathrm{\prime }}}_{{\varrho }^{\mathrm{\prime }}(i){\varrho }_{{\varrho }^{\mathrm{\prime }}(i)}^{\mathrm{\prime }}(j)}\right)(t)>0.$ Hence, $z{\zeta }_{n}t$ and so ${\nu }_{(n,s)}^{\ast }\subseteq {{\zeta }_n}^{\ast }$. By the similar way, ${{\zeta }_n}^{\ast }\subseteq {ϵ}^{\ast }$.

Similar to the Corollary 3.8, we have:

Corollary 5.4

Let S be a commutative FHR. Then ${\nu }_{(n,s)}^{\ast }={{\zeta }_n}^{\ast }={ϵ}^{\ast }$.

Proposition 5.5

For any $n\in \mathbb{N}$ we have ${\nu }_{(n+1,s)}^{\ast }\subseteq {\nu }_{(n,s)}^{\ast }$.

Proof.

The proof is similar to Proposition 3.7.

Definition 5.6

Let $R_l$ be a Lie ring and $n\in \mathbb{N}$. Then, $R_l$ is an n-Engel if for any $x,y\in R_l,[x{,}_{n}y]=0$. Also, for a fixed element $s\in R_l$ we define the n-th Lie centre of $R_l$ as follows: $U_{(0,s)}(R_l)=\{0\}\text{and}{U}_{(n,s)}(R_l)=\{r\in R_l;[r,\underset{n-\text{times}}{\underbrace{s,\dots ,s}}]=0\}.$

We can conclude that, $R_l$ is an n-Engel if and only if for any $s\in S,U_{(n,s)}(R_l)=R_l$.

Also, let $M_{(0,s)}(R_l)=R_l$ and $M_{(k+1,s)}(R_l)=\{[x,s]\mid x\in M_{(k,s)}(R_l)\}$. Now by Definition 5.6, we have the following theorem.

Theorem 5.7

Let $n\in \mathbb{N}$ and $l_{(n,y)}(R_l)$ be the Lie ideal generated by $M_{(n,y)}(R_l)$. Then Lie ring $R_l$ is an n-Engel if and only if for any $y\in R_l$, $l_{(n,y)}(R_l)=\{0\}$.

Proof.

(⇒) Let y be a fixed element of n-Engel Lie ring $R_l$ and $z\in l_{(n,y)}(R_l)$. Then $z\in [x_1,y]$ for some $x_1\in l_{(n-1,y)}(R_l)$, and so $x_1\in [x_2,y]$ for some $x_2\in l_{(n-2,y)}(R_l)$. Then, $z\in [x_1,y]\subseteq [x_2,y,y]$. Continuing this we have $z\in [x_n{,}_{n}y]$, for some $x_n\in R_l$. By hypotheses for any $x,y\in R_l$, we have $[x_n{,}_{n}y]=0$. Hence, z = 0 and so $l_{(n,y)}(R_l)=\{0\}$.

$(\Leftarrow )$ Let for any $y\in R_l$, $l_{(n,y)}(R_l)=\{0\}$ and x be arbitary elements of $R_l$. By $[x,y]\in l_{(1,y)}(R_l)$, we have $[[x,y],y]\in l_{(2,y)}(R_l)$. Continuing this we have $[x{,}_{n}y]\in l_{(n,y)}(R_l)=0$, and so $[x{,}_{n}y]=0$, which implies that $R_l$ is an n-Engel Lie ring.

Remark 5.8

Let φ be a FSR on S. Then by Theorem 2.6, without loss of generality, we can assume that $(S/\phi ,⊛,⊚)$ is a Lie ring.

Theorem 5.9

If φ is a FSR on S, then for any $k\in \mathbb{N}$, $M_{k+1,\frac{s}{\phi }}\left(\frac{S}{\phi }\right)=\{[\frac{t}{\phi },\frac{s}{\phi }]|t\in L_{k,s}(S)\}.$

Proof.

Put $R=S/\phi$ and $\overline{x}=\frac{x}{\phi }$, for all $x\in S$. If k = 0, then $M_{1,\overline{s}}(R)=\{[\overline{t},\overline{s}]\mid t\in L_{(0,s)}(S)\}$. Now, we show that $M_{k+2,\overline{s}}(R)\supseteq \{[\frac{t}{\phi },\frac{s}{\phi }]|t\in L_{k+1,s}(S)\}.$ Put $\overline{a}=[\overline{t},\overline{s}]$, where $t\in L_{(k+1,s)}(S)$, so there exist $x\in L_{(k,s)}(S)$ and $r\in S$ such that $(xs)(r)>0$ and $(t\oplus (sx))(r)>0$. Then by Theorem 2.6, $\overline{x}⊚\overline{s}=\overline{r}$ and $\overline{t}⊛(\overline{s}⊚\overline{x})=\overline{r}$. Thus $\overline{t}=[\overline{x},\overline{s}]$. By induction hypotheses we have $\overline{t}\in M_{(k+1,\overline{s})}(R)$. Hence $\overline{a}=[\overline{t},\overline{s}]\in M_{(k+2,\overline{s})}(R)$.

Conversely, let $\overline{a}\in M_{k+2,\overline{s}}(R)$. Then $\overline{a}=[\overline{x},\overline{s}]$, where $\overline{x}\in M_{(k+1,\overline{s})}(R)$. So hypotheses of induction implies that $\overline{x}=[\overline{u},\overline{s}]$, where $u\in L_{(k,s)}(S)$. Let $b,c\in S$, $(us)(c)>0$ and $(su)(b)>0$. For $b,c$ we have, $1={\chi }_S(c)=(S+b)(c)$. Then there exists $t\in S$ such that $(t+b)(c)>0$. Thus $(t\oplus su)(c)=\underset{b\in S}{\bigvee }(t+b)(c)\wedge (su)(b)>0,$ and so $t\in L_{(k+1,s)}(S)$. Hence, by Theorem 2.6, $\overline{c}=\overline{t}⊛(\overline{s}⊚\overline{u})=\overline{u}⊚\overline{s}$. Thus, $\overline{t}=[\overline{u},\overline{s}]=\overline{x}.$ Then, $\overline{a}=[\overline{x},\overline{s}]=[\overline{t},\overline{s}]\in \{[\overline{t},\overline{s}]|t\in L_{(k+1,s)}(S),\}.$ Therefore $M_{(k+2,\overline{s})}(S/\phi )=\{[\overline{t},\overline{s}]|t\in L_{(k+1,s)}(S)\}.$

Theorem 5.10

$S/{\nu }_{(n,s)}^{\ast }$ is an $(n+1)$-Engel Lie ring.

Proof.

For any $x\in S$ let $R=S/{\nu }_{(n,s)}^{\ast }$ and $\overline{x}=\frac{x}{{\nu }_{(n,s)}^{\ast }}$. By induction on i, we show that $M_{(n+1-i,\overline{s})}(R)\subseteq U_{(i,\overline{s})}(R)$. Let i = 0, then $\{0\}=M_{(n+1,\overline{s})}(R)\subseteq U_{(0,\overline{s})}(R)=\{0\}$. Let $\overline{a}\in M_{(n-i,\overline{s})}(R)$. Then $[\overline{a},\overline{s}]\in M_{(n+1-i,\overline{s})}(R)$. By hypotheses, we have $M_{(n+1-i,\overline{s})}(R)\subseteq U_{i,\overline{s}}(R)$. Hence $[[\overline{a},\overline{s}]{,}_i\overline{s}]=0$, and so $\overline{a}\in U_{(i+1,\overline{s})}(R)$. Thus, $M_{(n-i,\overline{s})}(R)\subseteq U_{(i+1,\overline{s})}(R)$. If i = n + 1, then $M_{(n+1-(n+1),\overline{s})}(R)\subseteq U_{(n+1,\overline{s})}(R)$ and so $R=M_{(0,\overline{s})}(R)=U_{(n+1,\overline{s})}(R)$. Therefore, R is an $(n+1)$-Engel Lie ring.

6. Conclusion

The fundamental relations on hyperrings were studied by Vougiouklis. Then, commutative fundamental relations on fuzzy hyperrings were introduced by Nozari and Fahimi. Now in this paper, first the smallest equivalence relation ${\zeta }^{\ast }$ on a FHR S, such that the set of equivalence classes was a nilpotent Lie ring, was introduced. Then, the relation ${\nu }_{(n,s)}^{\ast }$ was defined, as the FSR, so that the quotient would be an Engel Lie ring. Finally, theses two different relations were compared.

List of Abbreviations

$\mathcal{F}(H)$, the set of all fuzzy subsets of H,

$\mathcal{F}\mathcal{H}\mathcal{R}$, a fuzzy hyperoperations,

$\mathcal{F}\mathcal{R}$, a fuzzy regular,

$\mathcal{F}\mathcal{S}\mathcal{R}$, a fuzzy strongly regular,

$\mathcal{F}\mathcal{(}\mathcal{S}\mathcal{)}\mathcal{R}$, a fuzzy (strongly) regular.

Disclosure Statement

No potential conflict of interest was reported by the author(s).

Additional information

Notes on contributors

E. Mohammadzadeh

E. Mohammadzadeh is assistant professor at the Payame Noor University, Iran. She has published more than 20 papers in the international journals. Her main scientific interests are algebraic logic, algebraic hyperstructures and fuzzy algebras.

R. A. Borzooei

R. A. Borzooei is full professor at the Shahid Beheshti University, Tehran, Iran. He is currently, Editor In-Chief and founder of “Iranian Journal of Fuzzy Systems” and “Journal of Algebraic Hyperstructures and Logical”, editorial board of six international journals. He has published more than 330 papers in the international journals. His main scientific interests are algebraic logics, ordered algebraic structures, algebraic hyperstructures, fuzzy algebras and fuzzy graphs.

F. Mohammadzadeh

F. Mohammadzadeh is assistant professor at the Payame Noor University, Tehran, Iran. Her main scientific interests are algebraic hyperstructures and fuzzy algebras.

S. S. Ahn

S. S. Ahn is full professor at the Dongguk University, Korea. She works in the Department of Mathematics Education from 1993 to present. She has published more than 150 papers in the international journals. Her main scientific interests are logical algebras, ordered algebras, algebraic hyperstructures, fuzzy algebras and fuzzy graphs.