Since the conditions in Theorem 3.1 are not immediate, we aim to find some particular hyperrings, where the relation *ε*_{m} is transitive. For doing this, we will first define the concept of *m*-complete part and then we will prove that *ε*_{m} is transitive on *m*-idempotent hyperfields.

The main role of the complete parts of a semihypergroup, introduced by Koskas [3] and very well recalled by Antampoufis et al. in the survey [2], is played in finding the *β*^{*} class of each element. In particular, a nonempty subset *A* of a semihypergroup (*H*, ⋅) is called a *complete part* of *H* if, for any nonzero natural number *n* and any elements *a*_{1}, …, *a*_{n} of *H*, the following implication holds:

$$\begin{array}{}{\displaystyle A\cap \prod _{i=1}^{n}{a}_{i}\ne \mathrm{\varnothing}\Rightarrow \prod _{i=1}^{n}{a}_{i}\subseteq A.}\end{array}$$

In other words, the complete part *A* absorbs every hyperproduct containing at least one element of *A*. In particular, for any element *x* ∈ *A*, the class *β*^{*}(*x*) is a complete part of *H*. Moreover, the intersection of all complete parts of *H* containing *A* is called the *complete closure* of *A* in *H*, denoted by *C*(*A*). Besides, *β*^{*}(*x*) = *C*(*x*), for any *x* ∈ *H*.

As already mentioned before, Vougiouklis [16] defined the relation *γ* on a hyperring *R*, proving that its transitive closure *γ*^{*} is the smallest strongly regular relation defined on *R* such that the quotient *R*/*γ*^{*} is a ring. Later on Mirvakili et al. [19] studied the transitivity property of this relation, introducing the notion of *complete part* on hyperrings as follows: a nonempty subset *M* of a hyperring *R* is a complete part if, for any natural number *n*, any *i* = 1, 2, …, *n*, any natural number *k*_{i} and arbitrary elements *z*_{i1}, …, *z*_{iki} ∈ *R*, we have

$$\begin{array}{}{\displaystyle M\cap \sum _{i=1}^{n}(\prod _{j=1}^{{k}_{i}}{z}_{ij})\ne \mathrm{\varnothing}\Rightarrow \sum _{i=1}^{n}(\prod _{j=1}^{{k}_{i}}{z}_{ij})\subseteq M.}\end{array}$$

Now we will extend these definitions to the case of hyperrings, aiming to prove that the class *ε*^{*}(*x*) of an element *x* in the hyperring *R* is an *m*-complete part of *R*.

#### Definition 4.1

*We say that a nonempty subset* *A* *of a (semi)hyperring* (*R*, +, ⋅) *is an* *m*-*complete part of* *R* *if*
$\begin{array}{}{\displaystyle A\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing}}\end{array}$ *implies that*
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}\subseteq A}\end{array}$, *for all* *n* ∈ ℕ *and* *z*_{1}, …, *z*_{n} ∈ *R*. *The intersection of all* *m*-*complete parts of* *R* *containing a nonempty subset* *A* *of* *R* *is called the* *m*-*complete closure of* *A* *and it is denoted by* 𝓒_{m}(*A*).

#### Example 4.2

*Consider the following hyperoperations on the set* *R* = {*a*, *b*, *c*, *d*}:

$$\begin{array}{}{\displaystyle \begin{array}{ccccc}+& a& b& c& d\\ a& \{b,c\}& \{b,d\}& \{b,d\}& \{b,d\}\\ b& \{b,d\}& \{b,d\}& \{b,d\}& \{b,d\}\\ c& \{b,d\}& \{b,d\}& \{b,d\}& \{b,d\}\\ d& \{b,d\}& \{b,d\}& \{b,d\}& \{b,d\}\end{array}\phantom{\rule{2em}{0ex}}\begin{array}{ccccc}\cdot & a& b& c& d\\ a& \{b,d\}& \{b,d\}& \{b,d\}& \{b,d\}\\ b& \{b,d\}& \{b,d\}& \{b,d\}& \{b,d\}\\ c& \{b,d\}& \{b,d\}& \{b,d\}& \{b,d\}\\ d& \{b,d\}& \{b,d\}& \{b,d\}& \{b,d\}\end{array}}\end{array}$$

*Then* (*R*, +, ⋅) *is a semihyperring. For every m* ≥ 2 *and for all* *z*_{1}, …, *z*_{n} ∈ *R*, *we have*
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}=\{b,d\}.}\end{array}$ *It follows that the subsets* *A*_{1} = {*b*, *d*}, *A*_{2} = {*a*, *b*, *d*} *and* *A*_{3} = {*c*, *b*, *d*} *are all proper* *m*-*complete parts of* *R*, *i.e*. *m*-*complete parts of* *R*, *different by* *R*.

#### Theorem 4.3

*Let* *ρ* *be a strongly regular equivalence relation on* *R*. *Then* *ρ*(*a*) *is an* *m*-*complete part of* *R*, *for all* *a* ∈ *R*.

#### Proof

Since *ρ* is a strongly regular relation on *R*, it follows that the quotient *R*/*ρ* is a ring (with the addition “⊕” and the multiplication “⊙”). Let *a* ∈ *R* and
$\begin{array}{}{\displaystyle \rho (a)\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing},}\end{array}$ for arbitrary elements *z*_{1}, …, *z*_{n} ∈ *R*. Hence, there exists
$\begin{array}{}{\displaystyle y\in \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ such that *ρ*(*y*) = *ρ*(*a*). Consider the strong homomorphism *π* : *R* ⟶ *R*/*ρ* defined by *π*(*x*) = *ρ*(*x*), for all *x* ∈ *R*, where *R*/*ρ* is a ring (a trivial hyperring). Thus,

$$\begin{array}{}{\displaystyle \pi (\sum _{i=1}^{n}{z}_{i}^{m})=\underset{i=1}{\overset{n}{\u2a01}}(\underset{m\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}times}{\underset{\u23df}{\rho ({z}_{i})\odot \dots \odot \rho ({z}_{i})}})=\rho (y)=\rho (a),}\end{array}$$

which implies that
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}\subseteq \rho (a).}\end{array}$ This completes the proof. □

For a nonempty subset *A* of a (semi)hyperring *R*, denote

$$\begin{array}{}{\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{K}_{1}^{m}(A)=A}\\ {\displaystyle {K}_{n+1}^{m}(A)=\{x\in R\mid \mathrm{\exists}({z}_{1},\dots ,{z}_{n})\in {R}^{n}\phantom{\rule{thinmathspace}{0ex}};\phantom{\rule{thinmathspace}{0ex}}x\in \sum _{i=1}^{n}{z}_{i}^{m}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{K}_{n}^{m}(A)\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing}\}}\\ {\displaystyle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{K}^{m}(A)=\bigcup _{n\ge 1}{K}_{n}^{m}(A).}\end{array}$$

Moreover, for any *x* ∈ *R* and any natural number *n*, for simplicity we denote
$\begin{array}{}{\displaystyle {K}_{n}^{m}(\{x\})={K}_{n}^{m}(x).}\end{array}$

#### Lemma 4.4

*For any nonempty subset* *A* *of a hyperring* *R*, *the set* *K*^{m}(*A*) *is an* *m*-*complete part of* *R*.

#### Proof

Let *K*^{m}(*A*) ∩
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ ≠ ∅, for arbitrary elements *z*_{1}, … *z*_{n} ∈ *R*. Then, there exists *t* ∈ ℕ such that
$\begin{array}{}{\displaystyle {K}_{t}^{m}}\end{array}$(*A*) ∩
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ ≠ ∅, which implies that
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}\subseteq {K}_{t+1}^{m}(A)\subseteq {K}^{m}(A).}\end{array}$ Thus, *K*^{m}(*A*) is an *m*-complete part of *R*, containing *A*.□

#### Theorem 4.5

*K*^{m}(*A*) = 𝓒_{m}(*A*), *for any nonempty subset* *A* *of* *R*.

#### Proof

By Lemma 4.4, we have 𝓒_{m}(*A*) ⊆ *K*^{m}(*A*). Now, let *M* be an *m*-complete part of *R* containing *A*. Clearly,
$\begin{array}{}{\displaystyle {K}_{1}^{m}}\end{array}$(*A*) = *A* ⊆ *M*. Suppose that
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*A*) ⊆ *M*. Let *x* ∈
$\begin{array}{}{\displaystyle {K}_{n+1}^{m}}\end{array}$(*A*). Then there exist the elements *z*_{1}, …, *z*_{n} ∈ *R* such that
$\begin{array}{}{\displaystyle x\in \sum _{i=1}^{n}{z}_{i}^{m}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\mathrm{\varnothing}\ne {K}_{n}^{m}(A)\cap \sum _{i=1}^{n}{z}_{i}^{m}\subseteq M\cap \sum _{i=1}^{n}{z}_{i}^{m}.}\end{array}$ Since *M* is an *m*-complete part, it follows that
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ ⊆ *M* and so
$\begin{array}{}{\displaystyle {K}_{n+1}^{m}}\end{array}$(*A*) ⊆ *M*. Hence, *K*^{m}(*A*) ⊆ *M*, and thus *K*^{m}(*A*) ⊆ 𝓒_{m}(*A*).□

#### Example 4.6

*Consider the semihyperring* *R* *in Example 4.2. One obtains that the* *m*-*complete closure of* *A*_{2} *is* *A*_{2} *itself*, *for any natural number m* ≥ 2. *Moreover*, *if we consider the* *γ*-*relation on* *R*, *then we have* ∑∏*z*_{i} = {*b*, *c*} := *P* *or* ∑∏*z*_{i} = {*b*, *d*} := *Q*, *for any finite hypersums of finite hyperproducts of elements* *z*_{i} ∈ *R*. *Since P* ∩ *Q* ≠ ∅ *and P* ⊈ *Q* *and Q* ⊈ *P*, *then* *P* *and* *Q* *are not complete parts of* *R*. *Besides*, *A*_{2} ∩ *P* ≠ ∅, *but P* ⊈ *A*_{2}. *This means that* *A*_{2} *is not a complete part*, *but only an* *m*-*complete part*.

#### Theorem 4.7

*For all nonempty subsets* *A* *of* *R*, *it holds* 𝓒_{m}(*A*) =
$\begin{array}{}{\displaystyle \bigcup _{a\in A}}\end{array}$𝓒_{m}(*a*).

#### Proof

Clearly we have the inclusion 𝓒_{m}(*a*) ⊆ 𝓒_{m}(*A*), for all *a* ∈ *A*. Hence,
$\begin{array}{}{\displaystyle \bigcup _{a\in A}}\end{array}$𝓒_{m}(*a*) ⊆ 𝓒_{m}(*A*).

Conversely, we show that
$\begin{array}{}{\displaystyle {K}_{n}^{m}(A)\subseteq \bigcup _{a\in A}{K}_{n}^{m}(a),}\end{array}$ by induction on ”*n*”. For *n* = 1, we have
$\begin{array}{}{\displaystyle {K}_{1}^{m}}\end{array}$(*A*) = *A* =
$\begin{array}{}{\displaystyle \bigcup _{a\in A}\{a\}=\bigcup _{a\in A}{K}_{1}^{m}(a).}\end{array}$. Now, suppose that
$\begin{array}{}{\displaystyle {K}_{n}^{m}(A)\subseteq \bigcup _{a\in A}{K}_{n}^{m}(a)}\end{array}$ and take an arbitrary *x* ∈
$\begin{array}{}{\displaystyle {K}_{n+1}^{m}}\end{array}$(*A*). Then, *x* ∈
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ and
$\begin{array}{}{\displaystyle \mathrm{\varnothing}\ne {K}_{n}^{m}(A)\cap \sum _{i=1}^{n}{z}_{i}^{m},}\end{array}$ for some elements *z*_{1}, …, *z*_{n} ∈ *R*. Thus, there exists *y* ∈ *R* such that *x*, *y* ∈
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ and
$\begin{array}{}{\displaystyle y\in {K}_{n}^{m}(A)\subseteq \bigcup _{a\in A}{K}_{n}^{m}(a).}\end{array}$ This implies that there exists *a*′ ∈ *A* such that
$\begin{array}{}{\displaystyle y\in {K}_{n}^{m}({a}^{\prime})\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing},}\end{array}$ meaning that *x* ∈
$\begin{array}{}{\displaystyle {K}_{n+1}^{m}}\end{array}$(*a*′), and thus
$\begin{array}{}{\displaystyle {K}_{n+1}^{m}(A)\subseteq \bigcup _{a\in A}{K}_{n+1}^{m}(a).}\end{array}$

If *x* ∈ 𝓒_{m}(*A*), it follows that *x* ∈ *K*^{m}(*A*) =
$\begin{array}{}{\displaystyle \bigcup _{n\ge 1}{K}_{n}^{m}(A)\subseteq \bigcup _{n\ge 1}(\bigcup _{a\in A}{K}_{n}^{m}(a)).}\end{array}$ Then, for some *a*′ ∈ *A* and *n* ≥ 1, we have *x* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*a*′) ⊆ *K*^{m}(*a*′) = 𝓒_{m}(*a*′) ⊆ ⋃_{a∈A} 𝓒_{m}(*a*). This completes the proof. □

In the following we will give an equivalent description of the relation
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$ on hyperrings, using the notion of *m*-complete part. First we will prove some properties of the *m*-complete parts.

#### Lemma 4.8

$\begin{array}{}{\displaystyle {K}_{n}^{m}({K}_{2}^{m}(x))={K}_{n+1}^{m}(x),}\end{array}$ *for all* *x* ∈ *R* *and* *n* ≥ 2.

#### Proof

We prove it by induction on ”*n*”. For *n* = 2, we have

$$\begin{array}{}{\displaystyle {K}_{2}^{m}({K}_{2}^{m}(x))=\{x\in R\mid \mathrm{\exists}({z}_{1},\dots ,{z}_{n})\in {R}^{n}\phantom{\rule{thinmathspace}{0ex}};\phantom{\rule{thinmathspace}{0ex}}x\in \sum _{i=1}^{n}{z}_{i}^{m}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{K}_{1}^{m}({K}_{2}^{m}(x))\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing}\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\{x\in R\mid \mathrm{\exists}({z}_{1},\dots ,{z}_{n})\in {R}^{n}\phantom{\rule{thinmathspace}{0ex}};\phantom{\rule{thinmathspace}{0ex}}x\in \sum _{i=1}^{n}{z}_{i}^{m}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{K}_{2}^{m}(x)\cap \sum _{i=1}^{n}{z}_{i}^{m}\ne \mathrm{\varnothing}\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={K}_{3}^{m}(x).}\end{array}$$

Now, suppose that
$\begin{array}{}{\displaystyle {K}_{n-1}^{m}({K}_{2}^{m}(x))={K}_{n}^{m}(x)}\end{array}$ and take an arbitrary element
$\begin{array}{}{\displaystyle y\in {K}_{n}^{m}({K}_{2}^{m}(x)).}\end{array}$ Then there exist *z*_{1}, …, *z*_{n} ∈ *R* such that
$\begin{array}{}{\displaystyle y\in \sum _{i=1}^{n}{z}_{i}^{m}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\mathrm{\varnothing}\ne {K}_{n-1}^{m}({K}_{2}^{m}(x))\cap \sum _{i=1}^{n}{z}_{i}^{m}.\text{\hspace{0.17em}}\text{Thus},\text{\hspace{0.17em}}\mathrm{\varnothing}\ne {K}_{n}^{m}(x)\cap \sum _{i=1}^{n}{z}_{i}^{m},\text{\hspace{0.17em}}\text{and so}\text{\hspace{0.17em}}y\in {K}_{n+1}^{m}(x).}\end{array}$ Hence,
$\begin{array}{}{\displaystyle {K}_{n}^{m}({K}_{2}^{m}(x))\subseteq {K}_{n+1}^{m}(x).}\end{array}$ Similarly, we have
$\begin{array}{}{\displaystyle {K}_{n+1}^{m}(x)\subseteq {K}_{n}^{m}({K}_{2}^{m}(x)),}\end{array}$ that completes the proof. □

#### Lemma 4.9

*For all* *n* ≥ 2 *and* *x*, *y* ∈ *R*, *x* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*y*) *if and only if* *y* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*x*).

#### Proof

We prove the result by induction on ”*n*”. Let *n* = 2. Then *x* ∈
$\begin{array}{}{\displaystyle {K}_{2}^{m}}\end{array}$(*y*) if and only if, for *z*_{1}, …, *z*_{n} ∈ *R*, we have *x* ∈
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ and ∅ ≠ {*y*} ∩
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$, meaning that, for *z*_{1}, …, *z*_{n} ∈ *R*, *y* ∈
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ and ∅ ≠ {*x*} ∩
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$. This is equivalent with *y* ∈
$\begin{array}{}{\displaystyle {K}_{2}^{m}}\end{array}$(*x*). Suppose now that *x* ∈
$\begin{array}{}{\displaystyle {K}_{n-1}^{m}}\end{array}$(*y*) if and only if *y* ∈
$\begin{array}{}{\displaystyle {K}_{n-1}^{m}}\end{array}$(*x*) and take *x* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*y*); then there exist *z*, *z*_{1}, …, *z*_{n} ∈ *R* such that
$\begin{array}{}{\displaystyle x\in \sum _{i=1}^{n}{z}_{i}^{m},z\in {K}_{n-1}^{m}(y)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}z\in \sum _{i=1}^{n}{z}_{i}^{m}.}\end{array}$ Hence, by induction procedure,
$\begin{array}{}{\displaystyle y\in {K}_{n-1}^{m}(z)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}z\in {K}_{2}^{m}(x),}\end{array}$ which implies that
$\begin{array}{}{\displaystyle y\in {K}_{n-1}^{m}({K}_{2}^{m}(x))={K}_{n}^{m}(x)}\end{array}$ by Lemma 4.8. Similarly, *y* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*x*) implies that *x* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*y*).□

Define on a hyperring *R* the relation *θ* as follows: *xθy* if and only if *x* ∈ *K*^{m}(*y*), for all *x*, *y* ∈ *R*.

#### Corollary 4.10

*The relation* *θ* *is an equivalence on* *R*.

#### Proof

For all *x* ∈ *R*, we have *x* ∈
$\begin{array}{}{\displaystyle {K}_{1}^{m}}\end{array}$(*x*) ⊆ *K*^{m}(*x*). Hence, *θ* is reflexive. Now, let *x* ∈ *K*^{m}(*y*), for *x*, *y* ∈ *R*. By Theorem 4.5, there exists *n* ≥ 1 such that *x* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*y*), which implies that *y* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*x*), by Lemma 4.9. Then, *y* ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*x*) ⊆ *K*^{m}(*x*). Similarly, the converse is valid. Thus, *θ* is symmetric. Moreover, let *xθy* and *yθz* for *x*, *y*, *z* ∈ *R*. Hence, *x* ∈ 𝓒_{m}(*y*) and *y* ∈ 𝓒_{m}(*z*). Let *A* be an *m*-complete part of *R* containing *z*. Since *y* ∈ 𝓒_{m}(*z*) and 𝓒_{m}(*z*) ⊆ *A*, it follows that *y* ∈ *A*. Hence, 𝓒_{m}(*y*) ⊆ *A* and thus *x* ∈ *A*. Therefore, *x* ∈
$\begin{array}{}{\displaystyle \bigcap _{z\in A}}\end{array}$*A* = 𝓒_{m}(*z*) = *K*^{m}(*z*) and so *xθz*. □

#### Theorem 4.11

*For all* *x* ∈ *R*,
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$(*x*) = *θ*(*x*).

#### Proof

If *xε*_{m}y, then *x* ∈
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ and ∅ ≠ {*y*} ∩
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$, for some elements *z*_{1}, …, *z*_{n} ∈ *R*. Hence, *x* ∈
$\begin{array}{}{\displaystyle {K}_{2}^{m}}\end{array}$(*y*) ⊆ *K*^{m}(*y*). Thus, *ε*_{m} ⊆ *θ* and
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$ ⊆ *θ*^{*} = *θ*.

Conversely, let *xθy*, that is, *x* ∈ *K*^{m}(*y*), which implies that *x* ∈
$\begin{array}{}{\displaystyle {K}_{n+1}^{m}}\end{array}$(*y*), for *n* ∈ ℕ. Then, there exist *x*_{1}, *z*_{1}, …, *z*_{n} ∈ *R* such that
$\begin{array}{}{\displaystyle x\in \sum _{i=1}^{n}{z}_{i}^{m},{x}_{1}\in {K}_{n}^{m}(y)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{x}_{1}\in \sum _{i=1}^{n}{z}_{i}^{m}.\text{\hspace{0.17em}}\text{Hence},\text{\hspace{0.17em}}\{x,{x}_{1}\}\subseteq \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ and so *xε*_{m} *x*_{1}. Similarly, since *x*_{1} ∈
$\begin{array}{}{\displaystyle {K}_{n}^{m}}\end{array}$(*y*), there exists *x*_{2} ∈
$\begin{array}{}{\displaystyle {K}_{n-1}^{m}}\end{array}$(*y*) such that *x*_{1}*ε*_{m} *x*_{2}. By continuing this process, we can obtain
$\begin{array}{}{\displaystyle {x}_{3}\in {K}_{n-2}^{m}(y),\dots ,{x}_{n-1}\in {K}_{2}^{m}(y)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{x}_{n}\in {K}_{1}^{m}(y)=\{y\}}\end{array}$ such that *xε*_{m} *x*_{2}*ε*_{m} x_{3}…*ε*_{m}*x*_{n–1}*ε*_{m} x_{n} = *y*. Hence,
$\begin{array}{}{\displaystyle x{\epsilon}_{m}^{\ast}y\text{\hspace{0.17em}}\text{and so}\text{\hspace{0.17em}}\theta \subseteq {\epsilon}_{m}^{\ast}.}\end{array}$ Therefore the proof is completed.□

Now, we recall that a hyperring (*R*, +, ⋅) is said to be a *hyperfield*, if (*R*, ⋅) is a hypergroup. Moreover, a *strong homomorphism* from a hyperring (*R*, +, ⋅) to a hyperring (*S*, ⊕, ⊙) is a map *f* : *R* ⟶ *S* such that *f*(*x* + *y*) = *f*(*x*) ⊕ *f*(*y*) and *f*(*x* ⋅ *y*) = *f*(*x*) ⊙ *f*(*y*), for all *x*, *y* ∈ *R*. Considering the *ε*_{m} relation on *R*, it can be seen that the map *φ*_{m} : *R* ⟶ *R*/
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$ is a strong homomorphism.

In the following we will consider *R* a hyperfield satisfying relation (2) (this is a crucial assumption in the proofs of the next results) such that *R*/
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$ has a unit element denoted by
$\begin{array}{}{\displaystyle {1}_{R/{\epsilon}_{m}^{\ast}}.\text{\hspace{0.17em}}\text{Set}\text{\hspace{0.17em}}{\omega}_{R}^{m}={\phi}_{m}^{-1}({1}_{R/{\epsilon}_{m}^{\ast}})=\{x\in R\mid {\phi}_{m}(x)={1}_{R/{\epsilon}_{m}^{\ast}}\}.}\end{array}$ We will state some properties of the *m*-complete parts of hyperfields satisfying relation (2).

#### Theorem 4.12

*If* (*R*, +, ⋅) *is a hyperfield and* *A* ⊆ *R*, *then*
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*A*)) = *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$.

#### Proof

Let *x* ∈
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*A*)). Then there exists *y* ∈ *A* such that *φ*_{m}(*x*) = *φ*_{m}(*y*). Since (*R*, ⋅) is a hypergroup, it follows that there exists *t* ∈ *R* such that *x* ∈ *y* ⋅ *t*, which implies that *φ*_{m}(*x*) = *φ*_{m}(*y*) ⊙ *φ*_{m}(*t*). Since (*R*/
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$, ⊙) is a group and *φ*_{m}(*x*) = *φ*_{m}(*y*), it results *φ*_{m}(*t*) =
$\begin{array}{}{\displaystyle {1}_{R/{\epsilon}_{m}^{\ast}}}\end{array}$. Thus,
$\begin{array}{}{\displaystyle t\in {\phi}_{m}^{-1}({1}_{R/{\epsilon}_{m}^{\ast}})={\omega}_{R}^{m},}\end{array}$ and so *x* ∈ *y* ⋅ *t* ⊆ *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. Then
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*A*)) ⊆ *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$.

Conversely, let *x* ∈ *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. Then *x* ∈ *a* ⋅ *y*, for some *a* ∈ *A* and *y* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, that is *φ*_{m}(*y*) =
$\begin{array}{}{\displaystyle {1}_{R/{\epsilon}_{m}^{\ast}}}\end{array}$. Hence *φ*_{m}(*x*) = *φ*_{m}(*a* ⋅ *y*) = *φ*_{m}(*a*) ⊙ *φ*_{m}(*y*) = *φ*_{m}(*a*) ⊙
$\begin{array}{}{\displaystyle {1}_{R/{\epsilon}_{m}^{\ast}}}\end{array}$ = *φ*_{m}(*a*) ∈ *φ*_{m}(*A*), and so *x* ∈
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*A*)). This completes the proof. □

#### Theorem 4.13

*If* (*R*, +, ⋅) *is a hyperfield and* *A* ⊆ *R*, *then* 𝓒_{m}(*A*) = *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$.

#### Proof

It is easy to see that
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*A*)) = {*x* ∈ *R* | ∃*a* ∈ *A* : *φ*_{m}(*x*) = *φ*_{m}(*a*)}. Also, we have *φ*_{m}(*x*) = *φ*_{m}(*a*) if and only if
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$(*x*) =
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$(*a*), equivalently with *θ*(*x*) = *θ*(*a*), meaning that *x* ∈ *K*^{m}(*a*) = 𝓒_{m}(*a*), by Theorem 4.11. Hence, *x* ∈ 𝓒_{m}(*A*) if and only if *x* ∈
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*A*)), thus *x* ∈ *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, by Theorem 4.7 and Theorem 4.12. Therefore, 𝓒_{m}(*A*) = *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$.□

#### Corollary 4.14

*Let* *R* *be a hyperfield*. *A* *is an* *m*-*complete part of* *R* *if and only if* *A* = *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$.

#### Proof

If *A* is an *m*-complete part, then 𝓒_{m}(*A*) = *A*. Hence, *A* = 𝓒_{m}(*A*) = *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, by Theorem 4.13. Conversely, if *A* = *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, then *A* = 𝓒_{m}(*A*) by Theorem 4.13, and so *A* is an *m*-complete part of *R*.□

By Theorem 4.12, we have
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}\cdot {\omega}_{R}^{m}={\phi}_{m}^{-1}({\phi}_{m}({\omega}_{R}^{m}))={\omega}_{R}^{m}.\text{\hspace{0.17em}}\text{Hence},\text{\hspace{0.17em}}{\omega}_{R}^{m}}\end{array}$ is an *m*-complete part of the hyperfield (*R*, +, ⋅), by Corollary 4.14.

Moreover, notice that, for two subsets *A* and *B* of the hyperfield *R* such that one of them is an *m*-complete part of *R* (assume that *A* is so), we have (*A* ⋅ *B*) ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ = (*A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$) ⋅ *B* = *A* ⋅ *B*, by Corollary 4.14. Hence, *A* ⋅ *B* is an *m*-complete part of *R*, by Corollary 4.14.

#### Theorem 4.15

*Let* (*R*, +, ⋅) *be a hyperfield. Then every* *m*-*complete part subhypergroup of* (*R*, ⋅) *is invertible. Moreover*, *it is closed*.

#### Proof

Let *A* be an *m*-complete part of *R* such that (*A*, ⋅) is a subhypergroup of (*R*, ⋅). Take *x* ∈ *A* ⋅ *y* for *x*, *y* ∈ *R*. Thus, *x* ∈ *a* ⋅ *y*, for *a* ∈ *A*, which implies that *φ*_{m}(*x*) = *φ*_{m}(*a*) ⊙ *φ*_{m}(*y*). Since *φ*_{m}(*A*) is a subgroup of *R*/
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$, we have *φ*_{m}(*y*) = *φ*_{m}(*a*)^{–1} ⊙ *φ*_{m}(*x*) ∈ *φ*_{m}(*A*)⊙ *φ*_{m}(*x*) = *φ*_{m}(*A* ⋅ *x*). Besides, *A* ⋅ *x* is an *m*-complete part of *R*, hence *y* ∈
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*A* ⋅ *x*)) = 𝓒_{m}(*A* ⋅ *x*) = *A* ⋅ *x*. Then, *A* is invertible on the left. Similarly, we can show that *A* is invertible on the right. Therefore, *A* is invertible, and by consequence it is also closed.□

#### Theorem 4.16

*Let* (*R*, +, ⋅) *be a hyperfield and* 𝓢_{Cm}(*R*) *be the set of all* *m*-*complete parts of* *R* *which are subhypergroups of* (*R*, ⋅). *Then*,
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}=\bigcap _{A\in {\mathcal{S}}_{{C}_{m}}(R)}A.}\end{array}$

#### Proof

We know that
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ is an *m*-complete part of *R*. Let *x* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. For all *t*, *y* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, we have *φ*_{m}(*x*) =
$\begin{array}{}{\displaystyle {1}_{R/{\epsilon}_{m}^{\ast}}}\end{array}$ = *φ*_{m}(*t*) ⊙ *φ*_{m}(*y*) = *φ*_{m}(*t* ⋅ *y*). Hence, *x* ∈
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*t* ⋅ *ω*_{m})) = 𝓒_{m}(*t* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$) = *t* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, and so
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}\subseteq t\cdot {\omega}_{R}^{m}\text{\hspace{0.17em}}\text{for all}\text{\hspace{0.17em}}t\in {\omega}_{R}^{m}.}\end{array}$ Clearly,
$\begin{array}{}{\displaystyle t\cdot {\omega}_{R}^{m}\subseteq {\omega}_{R}^{m}.\text{\hspace{0.17em}}\text{Then}\text{\hspace{0.17em}}{\omega}_{R}^{m}=t\cdot {\omega}_{R}^{m},\text{\hspace{0.17em}}\text{for all}\text{\hspace{0.17em}}t\in {\omega}_{R}^{m},}\end{array}$ which implies that
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ is a subhypergroup of (*R*, ⋅). Thus,
$\begin{array}{}{\displaystyle \bigcap _{A\in {\mathcal{S}}_{{C}_{m}}(R)}A\subseteq {\omega}_{R}^{m}.}\end{array}$ Now, let *A* ∈ 𝓢_{Cm}(*R*). By Corollary 4.14, *A* = *A* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. Hence, for every *x* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, there exist *a*, *b* ∈ *A* such that *a* ∈ *b* ⋅ *x*, and so *a* ∈ *A* ⋅ *x*. By Theorem 4.15, *A* is invertible, and therefore *x* ∈ *A* ⋅ *a* ⊆ *A*. Then
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ ⊆ *A*, and thus
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}\subseteq \bigcap _{A\in {\mathcal{S}}_{{C}_{m}}(R)}A.}\end{array}$ Hence, the proof is complete.□

We recall that a hyperring (*R*, +, ⋅) is said to be *m*-*idempotent* ([12]) if there exists a constant *m*, 2 ≤ *m* ∈ ℕ, such that *x* ∈ *x*^{m}, for all *x* ∈ *R*.

#### Example 4.17

*[12] Consider the Krasner hyperring* *R* = {0, *a*, *b*} *with the hyperaddition and the multiplication defined as follows [20]*:

$$\begin{array}{}{\displaystyle \begin{array}{cccc}+& 0& a& b\\ 0& \{0\}& \{a\}& \{b\}\\ a& \{a\}& \{a,b\}& R\\ b& \{b\}& R& \{a,b\}\end{array}\phantom{\rule{1em}{0ex}}\begin{array}{cccc}\cdot & 0& a& b\\ 0& 0& 0& 0\\ a& 0& b& a\\ b& 0& a& b\end{array}}\end{array}$$

*For every odd number* *m* ∈ ℕ, *we have* 0^{m} = 0, *a*^{m} = *a* *and* *b*^{m} = *b*. *Hence*, *R* *is* *m*-*idempotent*, *for all odd natural numbers* *m*.

*Besides*, *since* *a*^{2} = *a* ⋅ *a* = *b*, *it follows that* *R* *is not an* 2-*idempotent hyperring*. *Similarly*, *one proves that*, *for all even numbers* *m* ∈ ℕ, *the hyperring* *R* *is not* *m*-*idempotent*.

#### Example 4.18

*The hyperring defined in Example 3.2 is an* *m*-*idempotent hyperring (satisfying relation* (2)*)*, *for all* *m*, 2 ≤ *m* ∈ ℕ *[12]*.

#### Example 4.19

*Define on the set* *R* = {0, 1} *two hyperoperations as follows*:

$$\begin{array}{}{\displaystyle \begin{array}{ccc}\u229e& 0& 1\\ 0& \{0\}& R\\ 1& R& \{1\}\end{array}\phantom{\rule{2em}{0ex}}\begin{array}{ccc}\u22a1& 0& 1\\ 0& \{0\}& \{0\}\\ 1& \{0\}& R\end{array}}\end{array}$$

*Then*, (*R*, ⊞, ⊡) *is an* *m*-*idempotent hyperring satisfying relation* (2), *for all* *m*, 2 ≤ *m* ∈ ℕ.

#### Example 4.20

*Similarly*, *take the same support set* *R* = {0, 1} *and define on* *R* *the two hyperoperations as follows*:

$$\begin{array}{}{\displaystyle \begin{array}{ccc}\oplus & 0& 1\\ 0& \{0\}& R\\ 1& R& R\end{array}\phantom{\rule{2em}{0ex}}\begin{array}{ccc}\odot & 0& 1\\ 0& \{0\}& R\\ 1& \{0\}& R\end{array}}\end{array}$$

*The hyperring* (*R*, ⊕, ⊙) *is* *m*-*idempotent*, *for all* *m*, 2 ≤ *m* ∈ ℕ, *and satisfies relation* (2).

Now we give an example of *m*-idempotent hyperfield.

#### Example 4.21

*Define on* *R* = {0, 1} *two hyperoperations as follows*:

$$\begin{array}{}{\displaystyle \begin{array}{ccc}+& 0& 1\\ 0& \{0\}& R\\ 1& R& \{1\}\end{array}\phantom{\rule{1em}{0ex}}\begin{array}{ccc}\cdot & 0& 1\\ 0& \{0\}& R\\ 1& R& \{1\}\end{array}}\end{array}$$

*Then*, (*R*, +, ⋅) *is an* *m*-*idempotent hyperfield satisfying relation* (2).

Now, for all *a* ∈ *R*, put
$\begin{array}{}{\displaystyle \mathcal{X}(a)=\bigcup \{\sum _{i=1}^{n}{A}_{i}^{m}\mid a\in \sum _{i=1}^{n}{A}_{i}^{m}\},}\end{array}$ where *A*_{1}, …, *A*_{n} ⊆ *R*.

#### Theorem 4.22

*Let* *R* *be an* *m*-*idempotent hyperfield*. *Then* 𝓧(*a*) *is an* *m*-*complete part of* *R*, *for all* *a* ∈ *R*.

#### Proof

Suppose that *a* ∈ *R* and
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ ∩ 𝓧(*a*) ≠ ∅, for some *z*_{1}, …, *z*_{n} ∈ *R*. Then there exists *z* ∈ *R* such that *z* ∈
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ and *z* ∈ *A*, where *A* =
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{A}_{i}^{m}}\end{array}$, for *A*_{1}, …, *A*_{n} ⊆ *R*. Since (*R*, ⋅) is a hypergroup, there exist *w*, *b* ∈ *R* such that *z*_{n} ∈ *w* ⋅ *a* and *a* ∈ *z* ⋅ *b*. Also, for all 1 ≤ *i* ≤ *n* – 1 we have *z*_{i} ∈ *z*_{i} ⋅ *R*. Hence,

$$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}\subseteq \sum _{i=1}^{n-1}{z}_{i}^{m}+(w\cdot a{)}^{m}\subseteq ({z}_{1}\cdot R{)}^{m}+\dots +({z}_{n-1}\cdot R{)}^{m}+(w\cdot A\cdot b{)}^{m},}\end{array}$$

since *z* ∈ *A*. Moreover, *R* is *m*-idempotent and we have *b* ∈ *b*^{m}, thus

$$\begin{array}{}{\displaystyle a\in z\cdot b\subseteq (\sum _{i=1}^{n}{z}_{i}^{m})\cdot b\subseteq {z}_{1}^{m}\cdot {b}^{m}+\dots +{z}_{n-1}^{m}\cdot {b}^{m}+(w\cdot a\cdot b{)}^{m}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\subseteq ({z}_{1}\cdot R{)}^{m}+\dots +({z}_{n-1}\cdot R{)}^{m}+(w\cdot A\cdot b{)}^{m}.}\end{array}$$

Therefore, (*z*_{1} ⋅ *R*)^{m} + … + (*z*_{n–1} ⋅ *R*)^{m} + (*w* ⋅ *A* ⋅ *b*)^{m} ⊆ 𝓧(*a*) and so
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{z}_{i}^{m}}\end{array}$ ⊆ 𝓧(*a*). Then, 𝓧(*a*) is an *m*-complete part of *R*.□

#### Theorem 4.23

*Let* *R* *be an* *m*-*idempotent hyperfield*. *Then* 𝓧(*a*) =
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$, *for every* *a* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$.

#### Proof

It is not difficult to see that
$\begin{array}{}{\displaystyle a\cdot {\omega}_{R}^{m}={\omega}_{R}^{m},}\end{array}$ for all *a* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. Hence,
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ = *a* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ = 𝓒_{m}(*a*) ⊆ 𝓧(*a*), by Theorem 4.13 and Theorem 4.22. Now, let *a* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ and *x* ∈ 𝓧(*a*). Then there exists *A* =
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{A}_{i}^{m}}\end{array}$ ⊆ 𝓧(*a*) such that *x* ∈ *A*. Since *a* ∈ *A*, then {*x*, *a*} ⊆ *A*. Since *R* satisfies relation (2), there exist *x*_{i} ∈ *A*_{i}, for 1 ≤ *i* ≤ *n*, such that {*x*, *a*} ⊆
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{x}_{i}^{m}}\end{array}$ and so
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$(*x*) =
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$(*a*). Then *φ*_{m}(*x*) = *φ*_{m}(*a*). Hence, *x* ∈
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*a*)) = *a* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ =
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. Therefore, the proof is complete.□

#### Theorem 4.24

*The relation* *ε*_{m} *is transitive on* *m*-*idempotent hyperfields*.

#### Proof

Let *R* be an *m*-idempotent hyperfield and
$\begin{array}{}{\displaystyle x{\epsilon}_{m}^{\ast}y,}\end{array}$ for *x*, *y* ∈ *R*. Hence, *x* ∈
$\begin{array}{}{\displaystyle {\phi}_{m}^{-1}}\end{array}$(*φ*_{m}(*y*)) = *y* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. Similarly, we have *y* ∈ *x* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ which implies that {*x*, *y*} ⊆ *x* ⋅
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$. Then, there exist *t*, *z* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ such that *x* ∈ *x* ⋅ *t* and *y* ∈ *x* ⋅ *z*. By Theorem 4.23, *z* ∈
$\begin{array}{}{\displaystyle {\omega}_{R}^{m}}\end{array}$ = 𝓧(*t*) and thus {*t*, *z*} ⊆ *A* =
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{A}_{i}^{m}}\end{array}$ for *A* ⊆ 𝓧(*t*). Since *R* is *m*-idempotent, {*x*, *y*} ⊆ *x* ⋅ *A* ⊆
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}}\end{array}$(*x* ⋅ *A*_{i})^{m}. So, there exist *z*_{i} ∈ *x* ⋅ *A*_{i} for every 1 ≤ *i* ≤ *n* such that {*x*, *y*} ⊆
$\begin{array}{}{\displaystyle \sum _{i=1}^{n}{x}_{i}^{m}}\end{array}$. Therefore, *xε*_{m}y and so
$\begin{array}{}{\displaystyle {\epsilon}_{m}^{\ast}}\end{array}$ = *ε*_{m}.□

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