In the current section, we shall establish the relationships between ordered regular equivalence relations and weak pseudoorders on an ordered semihypergroup, and discuss homomorphism theorems of ordered semihypergroups by weak pseudoorders.

In order to establish the relationships between ordered regular equivalence relations and weak pseudoorders on an ordered semihypergroup, the following lemma is essential.

#### Lemma 4.1

*Let* (*S*, ∘, ≤) *be an ordered semihypergroup and σ a relation on S*. *Then the following statements are equivalent*:

*σ is a weak pseudoorder on S*.

*There exist an ordered semihypergroup* (*T*, ⋄, ⪯) *and a homomorphism φ* : *S* → *T such that*
$$\begin{array}{}\overrightarrow{ker}\phi :=\{(a,b)\in S\times S\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\phi (a)\u2aaf\phi (b)\}=\sigma .\end{array}$$

#### Proof

(1) ⇒(2). Let *σ* be a weak pseudoorder on *S*. We denote by *σ*^{*} the regular equivalence relation on *S* defined by
$$\begin{array}{}{\sigma}^{\ast}:=\{(a,b)\in S\times S\text{\hspace{0.17em}}|\text{\hspace{0.17em}}(a,b)\in \sigma ,(b,a)\in \sigma \}(=\sigma \cap {\sigma}^{-1}).\end{array}$$

Then, by Theorem 3.8, the set *S*/*σ*^{*} := {(*a*)_{σ*} | *a* ∈ *S*} with the hyperoperation
$\begin{array}{}(a{)}_{{\sigma}^{\ast}}\otimes (b{)}_{{\sigma}^{\ast}}=\bigcup _{c\in a\circ b}(c{)}_{{\sigma}^{\ast}},\end{array}$ for all *a*, *b* ∈ *S* and the order
$$\begin{array}{}{\u2aaf}_{\sigma}:=\{((x{)}_{{\sigma}^{\ast}},(y{)}_{{\sigma}^{\ast}})\in S/{\sigma}^{\ast}\times S/{\sigma}^{\ast}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}(x,y)\in \sigma \}\end{array}$$

is an ordered semihypergroup. Let *T* = (*S*/*σ*^{*}, ⊗, ⪯_{σ}) and *φ* be the mapping of *S* onto *S*/*σ*^{*} defined by *φ* :*S* → *S*/*σ*^{*} | *a* ↦ (*a*)_{σ*}. Then, by Proposition 3.9, *φ* is a homomorphism from *S* onto *S*/*σ*^{*} and clearly,
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ* = *σ*.

(2) ⇒(1). Let (*S*, ∘, ≤) be an ordered semihypergroup. If there exist an ordered semihypergroup (*T*, ⋄, ⪯) and a homomorphism *φ* : *S* → *T* such that
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ* = *σ*, then *σ* is a weak pseudoorder on *S*. Indeed, let (*a*, *b*) ∈ ≤. Then, by hypothesis, *φ*(*a*) ⪯ *φ*(*b*). Thus (*a*, *b*) ∈
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ* = *σ*, and we have ≤⊆ *σ*. Moreover, let (*a*, *b*) ∈ *σ* and (*b*, *c*) ∈ *σ*. Then *φ*(*a*) ⪯ *φ*(*b*) ⪯ *φ*(*c*). Hence *φ*(*a*) ⪯ *φ*(*c*), i.e., (*a*, *c*) ∈
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ* = *σ*. Also, if (*a*, *b*) ∈ *σ*, then *φ*(*a*) ⪯ *φ*(*b*). Since (*T*, ⋄, ⪯) is an ordered semihypergroup, for any *c* ∈ *S* we have *φ*(*a*) ⋄ *φ*(*c*) ⪯ *φ*(*b*)⋄ *φ*(*c*). Since *φ* is a homomorphism from *S* to *T*, we have
$$\begin{array}{}{\displaystyle \bigcup _{x\in a\circ c}\phi (x)=\phi (a)\diamond \phi (c)\u2aaf\phi (b)\diamond \phi (c)=\bigcup _{y\in b\circ c}\phi (y).}\end{array}$$

Thus, for every *x* ∈ *a* ∘ *c* there exists *y* ∈ *b* ∘ *c* such that *φ*(*x*) ⪯ *φ*(*y*), and we have (*x*, *y*) ∈
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ* = *σ*, which implies that *a* ∘ *c* *σ⃗* *b* ∘ *c*. In the same way, it can be shown that *c* ∘ *a* *σ⃗* *c* ∘ *b*. Furthermore, let (*a*, *b*) ∈ *σ*, (*b*, *a*) ∈ *σ* and *c* ∈ *S*. Then *φ*(*a*) ⪯ *φ*(*b*) and *φ*(*b*) ⪯ *φ*(*a*). Thus *φ*(*a*) = *φ*(*b*), which implies that *φ*(*a*) ⋄ *φ*(*c*) = *φ*(*b*)⋄ *φ*(*c*), i.e.,
$\begin{array}{}\bigcup _{x\in a\circ c}\phi (x)=\bigcup _{y\in b\circ c}\phi (y).\end{array}$ Hence, for any *x* ∈ *a* ∘ *c*, there exists *y* ∈ *b* ∘ *c* such that *φ*(*x*) = *φ*(*y*), and we have *φ*(*x*) ⪯*φ*(*y*) and *φ*(*y*) ⪯*φ*(*x*). It thus follows that *x**σ* *y* and *y**σ x*. On the other hand, for any *y*′ ∈ *b* ∘ *c*, there exists *x*′ ∈ *a* ∘ *c* such that *φ*(*x*′) = *φ*(*y*′). Hence we have *φ*(*x*′) ⪯*φ*(*y*′) and *φ*(*y*′) ⪯*φ*(*x*′), which imply that *x*′*σ* *y*′ and *y*′*σ* *x*′. Thus, *a* ∘ *c* *σ͠* *b* ∘ *c*, exactly as required. Similarly, it can be obtained that *c* ∘ *a* *σ͠* *c* ∘ *b*. □

In the following, we give a characterization of ordered regular equivalence relations in terms of weak pseudoorders.

#### Theorem 4.2

*Let* (*S*, ∘, ≤) *be an ordered semihypergroup and ρ an equivalence relation on S*. *Then the following statements are equivalent*:

*ρ is an ordered regular equivalence relation on S*.

*There exists a weak pseudoorder σ on S such that ρ* = *σ* ∩ *σ*^{−1}.

*There exist an ordered semihypergroup T and a homomorphism φ*: *S* → *T such that ρ* = *ker*(*φ*), *where ker**φ* = {(*a*, *b*) ∈ *S* × *S* | *φ* (*a*) = *φ* (*b*)} *is the kernel of φ*.

#### Proof

(1) ⇒(2). Let *ρ* be an ordered regular equivalence relation on *S*. Then there exists an order relation “ ⪯” on the quotient semihypergroup (*S*/*ρ*, ⊗) such that (*S*/*ρ*, ⊗, ⪯) is an ordered semihypergroup, and *φ*: *S* → *S*/*ρ* is a homomorphism. Let *σ* =
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ*. By Lemma 4.1, *σ* is a weak pseudoorder on *S* and it is easy to check that *ρ* = *σ* ∩ *σ*^{−1}.

(2) ⇒(3). For a weak pseudoorder *σ* on *S*, by Lemma 4.1, there exist an ordered semihypergroup *T* and a homomorphism *φ*: *S* → *T* such that *σ* =
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ*. Then we have
$$\begin{array}{}ker\phi =\overrightarrow{ker}\phi \cap (\overrightarrow{ker}\phi {)}^{-1}=\sigma \cap {\sigma}^{-1}=\rho .\end{array}$$

(3) ⇒(1). By hypothesis and Lemma 4.1,
$\begin{array}{}\overrightarrow{ker}\end{array}$*φ* is a weak pseudoorder on *S*. Then, by Theorem 3.8, *ρ* = $\begin{array}{}\overrightarrow{ker}\end{array}$*φ* ∩ ($\begin{array}{}\overrightarrow{ker}\end{array}$*φ*)^{−1} is a regular equivalence relation on *S*. Thus, by the proof of Lemma 4.1, *ρ* is an ordered regular equivalence relation on *S*. □

#### Example 4.3

*We consider a set S*:= {*a*, *b*, *c*, *d*, *e*} *with the following hyperoperation* “∘” *and the order* “≤_{S}”:
$$\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\overline{)\begin{array}{cccccc}\circ & a& b& c& d& e\\ a\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{a\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{c\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{d\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\\ b\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{a\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{b,e\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{c\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{d\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\\ c\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{a\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{c\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{d\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\\ d\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{a\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{c\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{d\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\\ e\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{a\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{c\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{d\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{e\}\end{array}}\\ {\le}_{S}:=\{(a,a),(a,d),(a,e),(b,b),(c,c),(c,e),(d,d),(e,e)\}.\end{array}$$

*We give the covering relation* “≺_{S}” *and the figure of S as follows*:
$$\begin{array}{}{\prec}_{S}=\{(a,d),(a,e),(c,e)\}.\end{array}$$

*Then* (*S*, ∘, ≤_{S}) *is an ordered semihypergroup (see [28])*. *Let ρ be an equivalence relation on S defined as follows*:
$$\begin{array}{}\rho =\{(a,a),(b,b),(c,c),(c,d),(d,c),(d,d),(e,e)\}.\end{array}$$

*Then S*/*ρ* = {{*a*}, {*b*}, {*c*, *d*}, {*e*}}, *and ρ is regular*. *Moreover, we have*

*ρ is an ordered regular equivalence relation on S*. *In fact*, *let S*/*ρ* = {*m*, *n*, *p*, *q*}, *where m* = {*a*}, *n* = {*b*}, *p* = {*c*, *d*}, *q* = {*e*}. *The hyperoperation* “ ⊗” *and the order* “ ⪯” *on S*/*ρ are the follows*:
$$\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\overline{)\begin{array}{ccccc}\otimes & m& n& p& q\\ m\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{m\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{q\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{p\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{q\}\\ n\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{m\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{n,q\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{p\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{q\}\\ p\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{m\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{q\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{p\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{q\}\\ q\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{m\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{q\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{p\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{q\}\end{array}}\\ \u2aaf:=\{(m,m),(n,n),(p,p),(q,q),(m,p),(m,q),(p,q)\}.\end{array}$$

*We give the covering relation* “≺” *and the figure of S*/*ρ*_{2} *as follows*:
$$\begin{array}{}\prec =\{(m,p),(p,q)\}.\end{array}$$

*Then* (*S*/*ρ*, ⊗, ⪯) *is an ordered semihypergroup and the mapping ψ* :*S* → *S*/ *ρ*, *x* ↦ (*x*)_{ρ} *is isotone*. *Hence ρ is an ordered regular equivalence relation on S*.

*Let σ be a relation on S defined as follows*:
$$\begin{array}{}\sigma =\{(a,a),(a,c),(a,d),(a,e),(b,b),(c,c),(c,d),(c,e),(d,c),(d,d),(e,e)\}.\end{array}$$

*With a small amount of effort one can verify that σ is a weak pseudoorder on S*, *and ρ* = *σ* ∩ *σ*^{−1}.

*Let T*:= {*x*, *y*, *z*, *u*, *v*} *with the operation* “∙” *and the order relation* “≤_{T}” *below*:
$$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\overline{)\begin{array}{cccccc}\circ & x& y& z& u& v\\ x\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{x\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{y\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{z\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\\ y\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{x\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{y\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{z\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\\ z\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{x\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{y\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{z\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\\ u\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{x\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{y\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{z\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\\ v\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{x\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{y\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{z\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u\}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\{u,v\}\end{array}}\\ {\le}_{T}:=\{(x,x),(x,z),(x,u),(x,v),(y,y),(y,z),(y,u),(y,v),(z,z),(z,u),(z,v),(u,u),(v,v)\}.\end{array}$$

*We give the covering relation* “≺_{T}” *and the figure of S as follows*:
$$\begin{array}{}{\prec}_{T}=\{(x,z),(y,z),(z,u),(z,v)\}.\end{array}$$

*It is easy to check that* (*T*, ∙, ≤_{T}) *is also an ordered semihypergroup. We define a mapping φ from S to T such that φ*(*a*) = *x*, *φ*(*b*) = *v*, *φ*(*c*) = *φ*(*d*) = *z*, *φ*(*e*) = *u*. *It is a routine matter to verify that φ*: *S* → *T is a homomorphism and ρ* = *ker*(*φ*).

Let *σ* be a weak pseudoorder on an ordered semihypergroup (*S*, ∘, ≤). Then, by Theorem 4.2, *ρ* = *σ* ∩ *σ*^{−1} is an ordered regular equivalence relation on *S*. We denote by *ρ*^{♯} the mapping from *S* onto *S*/*ρ*, i.e., *ρ*^{♯} : *S* → *S*/*ρ* | *x* ↦ (*x*)_{ρ}, which is a homomorphism. In the following, we give out a homomorphism theorem of ordered semihypergroups by weak pseudoorders, which is a generalization of Theorem 1 in [17].

#### Theorem 4.5

*Let* (*S*, ∘, ≤) *and* (*T*, ⋄, ⪯) *be two ordered semihypergroups*, *φ* : *S* → *T* *a homomorphism*. *Then*, *if* *σ* *is a weak pseudoorder on* *S* *such that* *σ* ⊆ $\begin{array}{}\overrightarrow{ker}\phi \end{array}$, *then there exists the unique homomorphism* *f* : *S*/*ρ* → *T* | (*a*)_{ρ} ↦ *φ*(*a*) *such that the diagram*

*commutes*, *where* *ρ* = *σ* ∩ *σ*^{−1}. *Moreover*, *Im*(*φ*) = *Im*(*f*). *Conversely*, *if* *σ* *is a weak pseudoorder on* *S* *for which there exists a homomorphism* *f* : (*S*/*ρ*, ⊗, ⪯_{σ}) → (*T*, ⋄, ⪯) (*ρ* = *σ* ∩ *σ*^{−1}) *such that the above diagram commutes*, *then* *σ* ⊆ $\begin{array}{}\overrightarrow{ker}\phi \end{array}$.

#### Proof

Let *σ* be a weak pseudoorder on *S* such that *σ* ⊆ $\begin{array}{}\overrightarrow{ker}\phi \end{array}$, *f* : *S*/*ρ* → *T* | (*a*)_{ρ} ↦ *φ*(*a*). Then

*f* is well defined. Indeed, if (*a*)_{ρ} = (*b*)_{ρ}, then (*a*, *b*) ∈ *ρ* ⊆ *σ*. Since *σ* ⊆ $\begin{array}{}\overrightarrow{ker}\phi \end{array}$, we have (*φ*(*a*), *φ*(*b*)) ∈⪯. Furthermore, since (*b*, *a*) ∈ *σ* ⊆ $\begin{array}{}\overrightarrow{ker}\phi \end{array}$, we have (*φ*(*b*), *φ*(*a*)) ∈ ⪯. Therefore, *φ*(*a*) = *φ*(*b*).

*f* is a homomorphism and *φ* = *f* ∘ *ρ*^{♯}. In fact, by Lemma 4.1, there exists an order relation “⪯_{σ}” on the quotient semihypergroup (*S*/*ρ*, ⊗) such that (*S*/*ρ*, ⊗, ⪯_{σ}) is an ordered semihypergroup and the mapping *ρ*^{♯} is a homomorphism. Moreover, we have

$$\begin{array}{}{\displaystyle (a{)}_{\rho}{\u2aaf}_{\sigma}(b{)}_{\rho}\Rightarrow (a,b)\in \sigma \subseteq \overrightarrow{ker}\phi}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Rightarrow \phi (a)\u2aaf\phi (b)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Rightarrow f((a{)}_{\rho})\u2aaff((b{)}_{\rho}).\end{array}$$

Also, let (*a*)_{ρ}, (*b*)_{ρ} ∈ *S*/*ρ*. Since *φ* is a homomorphism from *S* to *T*, we have

$$\begin{array}{}{\displaystyle f((a{)}_{\rho})\diamond f((b{)}_{\rho})=\phi (a)\diamond \phi (b)=\bigcup _{c\in a\circ b}\phi (c)=\bigcup _{(c{)}_{\rho}\in (a{)}_{\rho}\otimes (b{)}_{\rho}}f((c{)}_{\rho}).}\end{array}$$

Furthermore, for any *a* ∈ *S*, (*f* ∘ *ρ*^{♯})(*a*) = *f*((*a*)_{ρ}) = *φ*(*a*), and thus *φ* = *f* ∘ *ρ*^{♯}.

We claim that *f* is the unique homomorphism from *S*/*ρ* to *T*. To prove our claim, let *g* is a homomorphism from *S*/*ρ* to *T* such that *φ* = *g* ∘ *ρ*^{♯}. Then

$$\begin{array}{}{\displaystyle f((a{)}_{\rho})=\phi (a)=(g\circ {\rho}^{\mathrm{\u266f}})(a)=g((a{)}_{\rho}).}\end{array}$$

Moreover, *Im*(*f*) = {*f*((*a*)_{ρ}) | *a* ∈ *S*} = {*φ*(*a*) | *a* ∈ *S*} = *Im*(*φ*).

Conversely, let *σ* be a weak pseudoorder on *S*, *f* : *S*/*ρ* → *T* is a homomorphism and *φ* = *f* ∘ *ρ*^{♯}. Then *σ* ⊆ $\begin{array}{}\overrightarrow{ker}\phi \end{array}$. Indeed, by hypothesis, we have

$$\begin{array}{}{\displaystyle (a,b)\in \sigma \iff (a{)}_{\rho}{\u2aaf}_{\sigma}(b{)}_{\rho}\Rightarrow f((a{)}_{\rho})\u2aaff((b{)}_{\rho})}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Rightarrow (f\circ {\rho}^{\mathrm{\u266f}})(a)\u2aaf(f\circ {\rho}^{\mathrm{\u266f}})(b)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Rightarrow \phi (a)\u2aaf\phi (b)\Rightarrow (a,b)\in \overrightarrow{ker}\phi ,\end{array}$$

where the order ⪯_{σ} on *S*/*ρ* is defined as in the proof of Lemma 4.1, that is

$$\begin{array}{}{\displaystyle {\u2aaf}_{\sigma}:=\{((x{)}_{\rho},(y{)}_{\rho})\in S/\rho \times S/\rho \text{\hspace{0.17em}}|\text{\hspace{0.17em}}(x,y)\in \sigma \}.}\end{array}$$ □

#### Corollary 4.6

*Let* (*S*, ∘, ≤) *and* (*T*, ⋄, ⪯) *be two ordered semihypergroups and* *φ*: *S* → *T* *a homomorphism*. *Then* *S*/*ker**φ* ≅ *Im*(*φ*), *where* *ker**φ* *is the kernel of* *φ*.

#### Proof

Let *σ* = $\begin{array}{}\overrightarrow{ker}\phi \end{array}$ and *ρ* = $\begin{array}{}\overrightarrow{ker}\phi \end{array}$ ∩ ($\begin{array}{}\overrightarrow{ker}\phi \end{array}$)^{−1}. Then, by Theorems 4.2 and 4.5, *ρ* is an ordered regular equivalence relation on *S* and *f* : *S*/*ρ* → *T* | (*a*)_{ρ} ↦ *φ*(*a*) is a homomorphism. Moreover, *f* is inverse isotone. In fact, let (*a*)_{ρ}, (*b*)_{ρ} be two elements of *S*/*ρ* such that *f*((*a*)_{ρ}) ⪯ *f*((*b*)_{ρ}). Then *φ*(*a*) ⪯ *φ*(*b*), and we have (*a*, *b*)∈ $\begin{array}{}\overrightarrow{ker}\phi \end{array}$. Thus, by Lemma 4.1, ((*a*)_{ρ}, (*b*)_{ρ}) ∈⪯_{σ}, i.e., (*a*)_{ρ} ⪯_{σ}(*b*)_*ρ*. Clearly, *ρ* = *kerφ*. By Remark 2.4, *S*/*kerφ* ≅ *Im*(*f*). Also, by Theorem 4.5, *Im*(*f*) = *Im*(*φ*). Therefore, *S*/*kerφ* ≅ *Im*(*φ*). □

Let (*S*, ∘, ≤) be an ordered semihypergroup, *ρ*, *θ* be weak pseudoorders on *S* such that *ρ* ⊆*θ*. We define a relation *θ*/*ρ* on (*S*/*ρ*^{*}, ⊗, ⪯_{ρ}) as follows:

$$\begin{array}{}{\displaystyle \theta /\rho :=\{((a{)}_{{\rho}^{\ast}},(b{)}_{{\rho}^{\ast}})\in S/{\rho}^{\ast}\times S/{\rho}^{\ast}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}(a,b)\in \theta \},}\end{array}$$

where ⪯_{ρ} := {((*a*)_{ρ*}, (*b*)_{ρ*}) | (*a*, *b*) ∈ *ρ*}, *ρ*^{*} = *ρ* ∩ *ρ*^{−1}. By Corollary 3.12, *θ*/*ρ* is a weak pseudoorder on *S*/*ρ*^{*}. Moreover, we have the following theorem.

#### Theorem 4.8

*Let* (*S*, ∘, ≤) *be an ordered semihypergroup*, *ρ*, *θ* *be weak pseudoorders on* *S* *such that* *ρ* ⊆*θ*. *Then* (*S*/*ρ*^{*})/(*θ*/*ρ*)^{*} ≅ *S*/*θ*^{*}.

#### Proof

We claim that the mapping *φ* : *S*/*ρ*^{*} → *S*/*θ*^{*} | (*a*)_{ρ*} ↦ (*a*)_{θ*} is a homomorphism. In fact:

*φ* is well-defined. Indeed, let (*a*)_{ρ*} = (*b*)_{ρ*}. Then (*a*, *b*) ∈ *ρ*^{*}. Thus, by the definition of *ρ*^{*}, (*a*, *b*) ∈ *ρ* ⊆*θ* and (*b*, *a*) ∈ *ρ* ⊆*θ*. This implies that (*a*, *b*) ∈ *θ*^{*}, and thus (*a*)_{θ*} = (*b*)_{θ*}.

*φ* is a homomorphism. In fact, let (*a*)_{ρ*}, (*b*)_{ρ*} ∈ *S*/*ρ*^{*}. Then, since *ρ*^{*},*θ*^{*} are both ordered regular equivalence relations on *S*, we have

$$\begin{array}{}{\displaystyle (a{)}_{{\rho}^{\ast}}{\otimes}_{\rho}(b{)}_{{\rho}^{\ast}}=\bigcup _{x\in a\circ b}(x{)}_{{\rho}^{\ast}},(a{)}_{{\theta}^{\ast}}{\otimes}_{\theta}(b{)}_{{\theta}^{\ast}}=\bigcup _{x\in a\circ b}(x{)}_{{\theta}^{\ast}},}\end{array}$$

where “ ⊗_{ρ} ” and “ ⊗_{θ} ” are the hyperactions on *S*/*ρ*^{*} and *S*/*θ*^{*}, respectively. Thus

$$\begin{array}{}{\displaystyle \phi ((a{)}_{{\rho}^{\ast}}){\otimes}_{\theta}\phi ((b{)}_{{\rho}^{\ast}})=(a{)}_{{\theta}^{\ast}}{\otimes}_{\theta}(a{)}_{{\theta}^{\ast}}=\bigcup _{x\in a\circ b}(x{)}_{{\theta}^{\ast}}=\bigcup _{(x{)}_{{\rho}^{\ast}}\in (a{)}_{{\rho}^{\ast}}{\otimes}_{\rho}(b{)}_{{\rho}^{\ast}}}\phi ((x{)}_{{\rho}^{\ast}}).}\end{array}$$

Also, if (*a*)_{ρ*} ⪯_{ρ} (*b*)_{ρ*}, then (*a*, *b*) ∈ *ρ* ⊆*θ*. It implies that (*a*)_{θ*} ⪯_*θ* (*b*)_{θ*}, and thus *φ* is isotone. On the other hand, it is easy to see that *φ* is onto, since

$$\begin{array}{}{\displaystyle Im(\phi )=\{\phi ((a{)}_{{\rho}^{\ast}})\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}a\in S\}=\{(a{)}_{{\theta}^{\ast}}\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}a\in S\}=S/{\theta}^{\ast}.}\end{array}$$

It thus follows from Corollary 4.6 that (*S*/*ρ*^{*})/*kerφ* ≅ *Im*(*φ*) = *S*/*θ*^{*}.

Furthermore, let $\begin{array}{}\overrightarrow{ker}\phi \end{array}$ := {((*a*)_{ρ*}, (*b*)_{ρ*}) ∈ *S*/*ρ*^{*} × *S*/*ρ*^{*} | *φ* ((*a*)_{ρ*}) ⪯_{θ} *φ* ((*b*)_{ρ*})}. Then

$$\begin{array}{}{\displaystyle ((a{)}_{{\rho}^{\ast}},(b{)}_{{\rho}^{\ast}})\in \overrightarrow{ker}\phi \u27fa(a{)}_{{\theta}^{\ast}}{\u2aaf}_{\theta}(b{)}_{{\theta}^{\ast}}\u27fa(a,b)\in \theta \u27fa((a{)}_{{\rho}^{\ast}},(b{)}_{{\rho}^{\ast}})\in \theta /\rho .}\end{array}$$

Therefore, *kerφ* = $\begin{array}{}\overrightarrow{ker}\phi \end{array}$ ∩ ($\begin{array}{}\overrightarrow{ker}\phi \end{array}$)^{−1} = (*θ*/*ρ*) ∩ (*θ*/*ρ*)^{−1} = (*θ*/*ρ*)^{*}. We have thus shown that (*S*/*ρ*^{*})/(*θ*/*ρ*)^{*} ≅ *S*/*θ*^{*}. □

#### Definition 4.9

*Let* (*S*, ∘, ≤) *and* (*T*, ⋄, ⪯) *be two ordered semihypergroups*, *ρ*, *θ* *be two weak pseudoorders on* *S* *and* *T*, *respectively*, *and the mapping* *f* : *S* → *T* *a homomorphism*. *Then*, *f* *is called a* (*ρ*, *θ*)-*homomorphism if* (*x*, *y*) ∈ *ρ* *implies* (*f*(*x*), *f*(*y*)) ∈ *θ*, *for all* *x*, *y* ∈ *S*.

#### Example 4.10

*Consider the ordered semihypergroups* (*S*, ∘, ≤_{S}) *and* (*T*, ∙, ≤_{T}) *given in* *Example 4*.*3*, *and define the relations* *ρ*, *θ* *on* *S* *and* *T*, *respectively*, *as follows*:

$$\begin{array}{}{\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\rho :=\{(a,a),(a,c),(a,d),(a,e),(b,b),(c,c),(c,d),(c,e),(d,c),(d,d),(e,e)\},}\\ \theta :=\{(x,x),(x,z),(x,u),(x,v),(y,y),(y,z),(y,u),(y,v),(z,z),(z,u),(z,v),(u,u),(v,v)\}.\end{array}$$

*It is not difficult to verify that* *ρ*, *θ* *is pseudoorders on* *S* *and* *T*, *respectively*. *Furthermore*, *we define a mapping* *f* *from* *S* *to* *T* *such that* *f*(*a*) = *x*, *f*(*b*) = *v*, *f*(*c*) = *f*(*d*) = *z*, *f*(*e*) = *u*. *By* *Example 4*.*3*, *f* : *S* → *T* *is a homomorphism*. *Hence* *f* *is a* (*ρ*, *θ*)-*homomorphism*.

#### Lemma 4.11

*Let* (*S*, ∘, ≤) *and* (*T*, ⋄, ⪯) *be two ordered semihypergroups*, *ρ*, *θ* *be two weak pseudoorders on* *S* *and* *T*, *respectively*, *and the mapping* *f* : *S* → *T* *a* (*ρ*, *θ*)-*homomorphism*. *Then, the mapping* *f* : (*S*/*ρ*^{*}, ⊗_{ρ}, ⪯_{ρ}) → (*T*/*θ*^{*}, ⊗_{θ}, ⪯_{θ}) *defined by*

$$\begin{array}{}{\displaystyle (\mathrm{\forall}x\in S)\phantom{\rule{thinmathspace}{0ex}}\overline{f}((x{)}_{{\rho}^{\ast}}):=(f(x){)}_{{\theta}^{\ast}}}\end{array}$$

*is a homomorphism of ordered semihypergroups*, *where the orders* ⪯_{ρ}, ⪯_{θ} *on* *S*/*ρ*^{*} *and* *T*/*θ*^{*}, *respectively*, *are both defined as in the proof of Lemma 4.1*.

#### Proof

Let *f* : *S* → *T* be a (*ρ*, *θ*)-homomorphism and *f*: *S*/*ρ*^{*} → *T*/*θ*^{*} | (*x*)_{ρ*} ↦ (*f*(*x*))_{θ*}. Then

*f* is well defined. In fact, let (*x*)_{ρ*}, (*y*)_{ρ*} ∈ *S*/*ρ*^{*} be such that (*x*)_{ρ*} = (*y*)_{ρ*}. Then (*x*, *y*) ∈ *ρ*^{*} ⊆ *ρ*. Since *f* is a (*ρ*, *θ*)-homomorphism, we have (*f*(*x*),*f*(*y*)) ∈ *θ*. It implies that ((*f*(*x*))_{θ*},(*f*(*y*))_{θ*}) ∈⪯_{θ}. Similarly, since (*y*, *x*) ∈ *ρ*, we have ((*f*(*y*))_{θ*},(*f*(*x*))_{θ*}) ∈⪯_{θ}. Therefore, ((*f*(*x*))_{θ*} = (*f*(*y*))_{θ*}, i.e., *f*((*x*)_{ρ*})=*f*((*y*)_{ρ*}).

*f* is a homomorphism. Indeed, let (*x*)_{ρ*},(*y*)_{ρ*} ∈ *S*/*ρ*^{*}. By Theorem 4.2, *ρ*^{*},*θ*^{*} are ordered regular equivalence relations on *S* and *T*, respectively. Since *f* is a homomorphism, by Lemma 2.2 we have

$$\begin{array}{}{\displaystyle \overline{f}((x{)}_{{\rho}^{\ast}}){\otimes}_{\theta}\overline{f}((y{)}_{{\rho}^{\ast}})=(f(x){)}_{{\theta}^{\ast}}{\otimes}_{\theta}(f(y){)}_{{\theta}^{\ast}}=\bigcup _{a\in x\circ y}(f(a){)}_{{\theta}^{\ast}}=\bigcup _{(a{)}_{{\rho}^{\ast}}\in (x{)}_{{\rho}^{\ast}}{\otimes}_{\rho}(y{)}_{{\rho}^{\ast}}}\overline{f}((a{)}_{{\rho}^{\ast}}).}\end{array}$$

Also, since *f* is a (*ρ*, *θ*)-homomorphism, we have

$$\begin{array}{}{\displaystyle (x{)}_{{\rho}^{\ast}}{\u2aaf}_{\rho}(y{)}_{{\rho}^{\ast}}\Rightarrow (x,y)\in \rho \Rightarrow (f(x),f(y))\in \theta}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Rightarrow (f(x){)}_{{\theta}^{\ast}}{\u2aaf}_{\theta}(f(y){)}_{{\theta}^{\ast}}\Rightarrow \overline{f}((x{)}_{{\rho}^{\ast}}){\u2aaf}_{\theta}\overline{f}((y{)}_{{\rho}^{\ast}}).\end{array}$$

Hence *f* is isotone. We have thus shown that *f* is a homomorphism, as required. □

#### Lemma 4.12

*Let* (*S*, ∘, ≤) *and* (*T*, ⋄, ⪯) *be two ordered semihypergroups*, *ρ*, *θ* *be two weak pseudoorders on* *S* *and* *T*, *respectively*, *and the mapping* *f* : *S* → *T* *a* (*ρ*, *θ*)-*homomorphism*. *Then*, *We define a relation on* *S*/*ρ*^{*} *denoted by* *ρ*_{f} *as follows*:

$$\begin{array}{}{\displaystyle {\rho}_{f}:=\{((x{)}_{{\rho}^{\ast}},(y{)}_{{\rho}^{\ast}})\in S/{\rho}^{\ast}\times S/{\rho}^{\ast}\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}(f(x){)}_{{\theta}^{\ast}}{\u2aaf}_{\theta}(f(y){)}_{{\theta}^{\ast}}\}.}\end{array}$$

*Then* *ρ*_{f} *is a weak pseudoorder on* *S*/*ρ*^{*}.

#### Proof

Assume that ((*x*)_{ρ*},(*y*)_{ρ*}) ∈⪯_{ρ}. By Lemma 4.11, *f* is a homomorphism. Then *f*((*x*)_{ρ*}) ⪯_{θ} *f*((*y*)_{ρ*}), i.e., (*f*(*x*))_{θ*}⪯_{θ} (*f*(*y*))_{θ*}. It implies that ((*x*)_{ρ*},(*y*)_{ρ*})∈ *ρ*_{f}, and thus ⪯_{ρ} ⊆*ρ*_{f}. Now, let ((*x*)_{ρ*},(*y*)_{ρ*}) ∈ *ρ*_{f} and ((*y*)_{ρ*},(*z*)_{ρ*}) ∈ *ρ*_{f}. Then (*f*(*x*))_{θ*}⪯_{θ} (*f*(*y*))_{θ*} and (*f*(*y*))_{θ*}⪯_{θ} (*f*(*z*))_{θ*}. Thus, by the transitivity of ⪯_{θ}, (*f*(*x*))_{θ*}⪯_{θ} (*f*(*z*))_{θ*}. This implies that ((*x*)_{ρ*},(*z*)_{ρ*})∈ *ρ*_{f}. Also, let ((*x*)_{ρ*},(*y*)_{ρ*}) ∈ *ρ*_{f} and (*z*)_{ρ*} ∈ *S*/*ρ*^{*}. Then (*f*(*x*))_{θ*}⪯_{θ} (*f*(*y*))_{θ*}. Since (*T*/*θ*^{*}, ⊗_{θ}, ⪯_{θ}) is an ordered semihypergroup, it can be obtained that (*f*(*x*))_{θ*}⊗_{θ} (*f*(*z*))_{θ*}⪯_{θ} (*f*(*y*))_{θ*}⊗_{θ} (*f*(*z*))_{θ*}, that is, *f*((*x*)_{ρ*})⊗_{θ} *f*((*z*)_{ρ*}) ⪯_{θ} *f*((*y*)_{ρ*})⊗_{θ} *f*((*z*)_{ρ*}). Then, since *f* is a homomorphism, we have

$$\begin{array}{}{\displaystyle \bigcup _{(a{)}_{{\rho}^{\ast}}\in (x{)}_{{\rho}^{\ast}}{\otimes}_{\rho}(z{)}_{{\rho}^{\ast}}}\overline{f}((a{)}_{{\rho}^{\ast}}){\u2aaf}_{\theta}\bigcup _{(b{)}_{{\rho}^{\ast}}\in (y{)}_{{\rho}^{\ast}}{\otimes}_{\rho}(z{)}_{{\rho}^{\ast}}}\overline{f}((b{)}_{{\rho}^{\ast}}),}\end{array}$$

which means that $\begin{array}{}\bigcup _{a\in x\circ z}\end{array}$ (*f*(*a*))_{θ*}⪯_{θ} $\begin{array}{}\bigcup _{b\in y\circ z}\end{array}$(*f*(*b*))_{θ*}. Thus, for any *a* ∈ *x* ∘ *z*, there exists *b* ∈ *y* ∘ *z* such that (*f*(*a*))_{θ*}⪯_{θ} (*f*(*b*))_{θ*}. Hence ((*a*)_{ρ*}, (*b*)_{ρ*}) ∈ *ρ*_{f}. It thus implies that (*x*)_{ρ*}⊗_{ρ} (*z*)_{ρ*} *ρ⃗*_{f} (*y*)_{ρ*}⊗_{ρ} (*z*)_{ρ*}. Similarly, it can be shown that (*z*)_{ρ*}⊗_{ρ} (*x*)_{ρ*} *ρ⃗*_{f} (*z*)_{ρ*}⊗_{ρ} (*y*)_{ρ*}. Furthermore, let (*x*)_{ρ*},(*y*)_{ρ*},(*z*)_{ρ*} ∈ *S*/*ρ*^{*} be such that ((*x*)_{ρ*},(*y*)_{ρ*}) ∈ *ρ*_{f} and ((*y*)_{ρ*},(*x*)_{ρ*}) ∈ *ρ*_{f}. Then, similarly as in the above proof, it can be verified that (*x*)_{ρ*}⊗_{ρ} (*z*)_{ρ*} *ρ̃*_{f} (*y*)_{ρ*}⊗_{ρ} (*z*)_{ρ*} and (*z*)_{ρ*}⊗_{ρ} (*x*)_{ρ*} *ρ̃*_{f} (*z*)_{ρ*}⊗_{ρ} (*y*)_{ρ*}. Therefore, *ρ*_{f} is a weak pseudoorder on *S*/*ρ*^{*}. □

By Lemmas 4.11 and 4.12, we immediately obtain the following corollary:

#### Corollary 4.13

*Let* (*S*, ∘, ≤) *and* (*T*, ⋄, ⪯) *be two ordered semihypergroups*, *ρ*, *θ* *be two weak pseudoorders on* *S* *and* *T*, *respectively*, *and the mapping* *f* : *S* → *T* *be a* (*ρ*, *θ*)-*homomorphism*. *Then*, *the following diagram*

*commutates*.

#### Theorem 4.14

*Let* (*S*, ∘, ≤) *and* (*T*, ⋄, ⪯) *be two ordered semihypergroups*, *ρ*, *θ* *be two weak pseudoorders on* *S* *and* *T*, *respectively*, *and the mapping* *f* : *S* → *T* *a* (*ρ*, *θ*)-*homomorphism*. *If* *σ* *is a weak pseudoorder on* *S*/*ρ*^{*} *such that* *σ* ⊆ *ρ*_{f}, *then there exists the unique homomorphism* *φ* : (*S*/*ρ*^{*})/*σ*^{*} → *T*/*θ*^{*} | ((*a*)_{ρ*})_{σ*} ↦ *f*((*a*)_{ρ*}) such that the diagram

*commutes*.

*Conversely*, *if* *σ* *is a weak pseudoorder on* *S*/*ρ*^{*} *for which there exists a homomorphism* *φ* : (*S*/*ρ*^{*})/*σ*^{*} → *T*/*θ*^{*} *such that the above diagram commutes*, *then* *σ* ⊆ *ρ*_{f}.

Let (*S*, ∘, ≤_{S}) and (*T*, ⋄, ≤_{T}) be two ordered semihypergroups. In [25], it is shown that (*S* × *T*, ⋆, ≤_{S×T}) is also an ordered semihypergroup with the hyperoperation “ ⋆ ” and the order relation “ ≤_{S×T} ” below: for any (*s*_{1}, *t*_{1}), (*s*_{2}, *t*_{2}) ∈ *S* × *T*,

$$\begin{array}{}{\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}({s}_{1},{t}_{1})\star ({s}_{2},{t}_{2})=({s}_{1}\circ {s}_{2})\times ({t}_{1}\diamond {t}_{2});}\\ ({s}_{1},{t}_{1}){\le}_{S\times T}({s}_{2},{t}_{2})\iff {s}_{1}{\le}_{S}{s}_{2}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{t}_{1}{\le}_{T}{t}_{2}.\end{array}$$

#### Lemma 4.15

*Let* (*S*, ∘, ≤_{S}) *and* (*T*, ⋄, ≤_{T}) *be ordered semihypergroups*, *ρ*_{1}, *ρ*_{2} *be two weak pseudoorders on* *S* *and* *T*, *respectively*. *We define a relation* *ρ* *on* *S* × *T* *as follows*:

$$\begin{array}{}{\displaystyle (\mathrm{\forall}\phantom{\rule{thinmathspace}{0ex}}({s}_{1},{t}_{1}),({s}_{2},{t}_{2})\in S\times T)\phantom{\rule{thinmathspace}{0ex}}({s}_{1},{t}_{1})\rho ({s}_{2},{t}_{2})\iff {s}_{1}{\rho}_{1}{s}_{2}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}{t}_{1}{\rho}_{2}{t}_{2}.}\end{array}$$

*Then* *ρ* *is a weak pseudoorder on* *S* × *T*.

#### Proof

The verifications of the conditions (1), (2) and (3) of Definition 3.4 are straightforward. We only need show that the condition (4) of Definition 3.4 is satisfied.

Let (*s*_{1}, *t*_{1}), (*s*_{2}, *t*_{2}), (*s*, *t*) ∈ *S* × *T* be such that (*s*_{1}, *t*_{1})*ρ* (*s*_{2}, *t*_{2}) and (*s*_{2}, *t*_{2})*ρ* (*s*_{1}, *t*_{1}). Then *s*_{1}*ρ*_{1} *s*_{2}, *t*_{1}*ρ*_{2} *t*_{2}, *s*_{2}*ρ*_{1} *s*_{1}, *t*_{2}*ρ*_{2} *t*_{1} and *s* ∈ *S*, *t* ∈ *T*. Since *ρ*_{1}, *ρ*_{2} are weak pseudoorders on *S* and *T*, respectively, we have

$$\begin{array}{}{\displaystyle {s}_{1}\circ s\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{\rho}}_{1}\phantom{\rule{thinmathspace}{0ex}}{s}_{2}\circ s\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{t}_{1}\circ t\phantom{\rule{thinmathspace}{0ex}}{\stackrel{~}{\rho}}_{1}\phantom{\rule{thinmathspace}{0ex}}{t}_{2}\circ t.}\end{array}$$

Thus

$$\begin{array}{}{\displaystyle (\mathrm{\forall}x\in {s}_{1}\circ s)\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\exists}u\in {s}_{2}\circ s)\phantom{\rule{thinmathspace}{0ex}}x{\rho}_{1}u\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}u{\rho}_{1}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{\&}\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\forall}{u}^{\prime}\in {s}_{2}\circ s)\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\exists}{x}^{\prime}\in {s}_{1}\circ s)\phantom{\rule{thinmathspace}{0ex}}{x}^{\prime}{\rho}_{1}{u}^{\prime}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{u}^{\prime}{\rho}_{1}{x}^{\prime},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\forall}y\in {t}_{1}\diamond t)\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\exists}v\in {t}_{2}\diamond t)\phantom{\rule{thinmathspace}{0ex}}y{\rho}_{2}v\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}v{\rho}_{2}y\phantom{\rule{thinmathspace}{0ex}}\mathrm{\&}\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\forall}{v}^{\prime}\in {t}_{2}\diamond t)\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\exists}{y}^{\prime}\in {t}_{1}\diamond t)\phantom{\rule{thinmathspace}{0ex}}{y}^{\prime}{\rho}_{2}{v}^{\prime}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{v}^{\prime}{\rho}_{2}{y}^{\prime}.\end{array}$$

Hence we have

$$\begin{array}{}{\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\forall}(x,y)\in ({s}_{1}\circ s)\times ({t}_{1}\diamond t))\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\exists}(u,v)\in ({s}_{2}\circ s)\times ({t}_{2}\diamond t))\phantom{\rule{thinmathspace}{0ex}}(x,y)\phantom{\rule{thinmathspace}{0ex}}\rho \phantom{\rule{thinmathspace}{0ex}}(u,v)\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}(u,v)\phantom{\rule{thinmathspace}{0ex}}\rho \phantom{\rule{thinmathspace}{0ex}}(x,y),}\\ (\mathrm{\forall}({u}^{\prime},{v}^{\prime})\in ({s}_{2}\circ s)\times ({t}_{2}\diamond t))\phantom{\rule{thinmathspace}{0ex}}(\mathrm{\exists}({x}^{\prime},{y}^{\prime})\in ({s}_{1}\circ s)\times ({t}_{1}\diamond t))\phantom{\rule{thinmathspace}{0ex}}({x}^{\prime},{y}^{\prime})\phantom{\rule{thinmathspace}{0ex}}\rho \phantom{\rule{thinmathspace}{0ex}}({u}^{\prime},{v}^{\prime})\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}({u}^{\prime},{v}^{\prime})\phantom{\rule{thinmathspace}{0ex}}\rho \phantom{\rule{thinmathspace}{0ex}}({x}^{\prime},{y}^{\prime}).\end{array}$$

It thus follows that (*s*_{1}∘ *s*) × (*t*_{1}⋄ *t*) *ρ̃* (*s*_{2}∘ *s*) × (*t*_{2}⋄ *t*), i.e., (*s*_{1}, *t*_{1}) ⋆ (*s*, *t*) *ρ̃* (*s*_{2}, *t*_{2}) ⋆ (*s*, *t*). Similarly, it can be verified that (*s*, *t*) ⋆ (*s*_{1}, *t*_{1}) *ρ̃* (*s*, *t*)⋆ (*s*_{2}, *t*_{2}). □

By Theorem 3.8 and Lemma 4.15, ((*S* × *T*)/*ρ*^{*}, ⊗_{ρ}, ⪯_{ρ}) and (*S*/$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$ × *T*/$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$, ⋆′, ⪯_{S×T}) are both ordered semihypergroups, where the hyperoperation ⋆′ and the order relation ⪯_{S×T} on *S*/$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$ × *S*/$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$ are similar to the hyperoperation ⋆ and the order relation ≤_{S×T} on *S* × *T*, respectively. Furthermore, we have the following theorem:

#### Theorem 4.16

*Let* (*S*, ∘, ≤_{S}) *and* (*T*, ⋄, ≤_{T}) *be ordered semihypergroups*, *ρ*_{1}, *ρ*_{2} *be two weak pseudoorders on* *S* *and* *T*, *respectively*. *Then* (*S* × *T*)/*ρ*^{*} ≅ *S*/$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$ × *T*/$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$.

#### Proof

We claim that the mapping *φ* : (*S* × *T*)/*ρ*^{*} → *S*/$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$ × *T*/$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$ | ((*s*, *t*))_{ρ*} ↦ ((*s*)_{$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$}, (*t*)_{$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$}) is an isomorphism. In fact:

*φ* is well-defined. Indeed, let ((*s*_{1}, *t*_{1}))_{ρ*} = ((*s*_{2}, *t*_{2}))_{ρ*}. Then (*s*_{1}, *t*_{1})*ρ*^{*}(*s*_{2}, *t*_{2}). Hence, by the definition of *ρ*^{*}, (*s*_{1}, *t*_{1})*ρ*(*s*_{2}, *t*_{2}) and (*s*_{2}, *t*_{2})*ρ*(*s*_{1}, *t*_{1}). It thus follows that *s*_{1}*ρ*_{1} *s*_{2}, *t*_{1}*ρ*_{2} *t*_{2}, *s*_{2}*ρ*_{1} *s*_{1} and *t*_{2}*ρ*_{2} *t*_{1}, which imply *s*_{1}$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$ *s*_{2} and *t*_{1} $\begin{array}{}{\rho}_{2}^{\ast}\end{array}$ *t*_{2}. Thus, ((*s*_{1})_{$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$},(*t*_{1})_{$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$}) = ((*s*_{2})_{$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$}, (*t*_{2})_{$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$}).

*φ* is an isomorphism. In fact, for any ((*s*_{1}, *t*_{1}))_{ρ*},((*s*_{2}, *t*_{2}))_{ρ*}∈ (*S* × *T*)/*ρ*^{*}, we have

$$\begin{array}{}{\displaystyle \phi ((({s}_{1},{t}_{1}){)}_{{\rho}^{\ast}}){\star}^{\prime}\phi ((({s}_{2},{t}_{2}){)}_{{\rho}^{\ast}})=(({s}_{1}{)}_{{\rho}_{1}^{\ast}},({t}_{1}{)}_{{\rho}_{2}^{\ast}}){\star}^{\prime}(({s}_{2}{)}_{{\rho}_{1}^{\ast}},({t}_{2}{)}_{{\rho}_{2}^{\ast}})}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=(({s}_{1}{)}_{{\rho}_{1}^{\ast}}{\otimes}_{{\rho}_{1}}({s}_{2}{)}_{{\rho}_{1}^{\ast}},({t}_{1}{)}_{{\rho}_{2}^{\ast}}{\otimes}_{{\rho}_{2}}({t}_{2}{)}_{{\rho}_{2}^{\ast}})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=(\bigcup _{s\in {s}_{1}\circ {s}_{2}}(s{)}_{{\rho}_{1}^{\ast}},\bigcup _{t\in {t}_{1}\diamond {t}_{2}}(t{)}_{{\rho}_{2}^{\ast}})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\bigcup _{\genfrac{}{}{0ex}{}{s\in {s}_{1}\circ {s}_{2}}{t\in {t}_{1}\diamond {t}_{2}}}((s{)}_{{\rho}_{1}^{\ast}},(t{)}_{{\rho}_{2}^{\ast}})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\bigcup _{(s,t)\in ({s}_{1}\circ {s}_{2})\times ({t}_{1}\circ {t}_{2})}\phi (((s,t){)}_{{\rho}^{\ast}})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\bigcup _{(s,t)\in ({s}_{1},{t}_{1})\star ({s}_{2},{t}_{2})}\phi (((s,t){)}_{{\rho}^{\ast}})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\bigcup _{((s,t){)}_{{\rho}^{\ast}}\in (({s}_{1},{t}_{1}){)}_{{\rho}^{\ast}}{\otimes}_{\rho}(({s}_{2},{t}_{2}){)}_{{\rho}^{\ast}}}\phi (((s,t){)}_{{\rho}^{\ast}}).\end{array}$$

Moreover, it can be easily seen that *φ* is onto. Also, *φ* is isotone and reverse isotone. Indeed, let ((*s*_{1}, *t*_{1}))_{ρ*},((*s*_{2}, *t*_{2}))_{ρ*} ∈ (*S* × *T*)/*ρ*^{*}. Then we have

$$\begin{array}{}{\displaystyle (({s}_{1},{t}_{1}){)}_{{\rho}^{\ast}}{\u2aaf}_{\rho}(({s}_{2},{t}_{2}){)}_{{\rho}^{\ast}}\u27fa({s}_{1},{t}_{1})\rho ({s}_{2},{t}_{2})\u27fa{s}_{1}{\rho}_{1}{s}_{2}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{t}_{1}{\rho}_{2}{t}_{2}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\u27fa({s}_{1}{)}_{{\rho}_{1}^{\ast}}{\u2aaf}_{{\rho}_{1}}({s}_{2}{)}_{{\rho}_{1}^{\ast}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}({t}_{1}{)}_{{\rho}_{2}^{\ast}}{\u2aaf}_{{\rho}_{2}}({t}_{2}{)}_{{\rho}_{2}^{\ast}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\u27fa(({s}_{1}{)}_{{\rho}_{1}^{\ast}},({t}_{1}{)}_{{\rho}_{2}^{\ast}}){\u2aaf}_{S\times T}(({s}_{2}{)}_{{\rho}_{1}^{\ast}},({t}_{2}{)}_{{\rho}_{2}^{\ast}})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\u27fa\phi ((({s}_{1},{t}_{1}){)}_{{\rho}^{\ast}}){\u2aaf}_{S\times T}\phi ((({s}_{2},{t}_{2}){)}_{{\rho}^{\ast}}).\end{array}$$

Therefore, *φ* is indeed an isomorphism, which means that (*S* × *T*)/*ρ*^{*} ≅ *S*/$\begin{array}{}{\rho}_{1}^{\ast}\end{array}$ × *T*/$\begin{array}{}{\rho}_{2}^{\ast}\end{array}$. □

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