Generating Sets of the Complete Semigroups of Binary Relations Defined by Semilattices of the Class Σ₂ (X,4)

Bariş Albayrak¹, Omar Givradze², Guladi Partenadze²

¹Department of Banking and Finance, School of Applied Sciences, Çanakkale Onsekiz Mart University, Çanakkale, Turkey.
²Shota Rustaveli State University, Batumi, Georgia.

DOI: 10.4236/am.2018.91002                

Abstract

In this paper, we have studied generating sets of the complete semigroups defined by X-semilattices of the class Σ₂(X, 4).

Keywords

Semigroup, Semilattice, Binary Relations, Idempotent Elements

1. Introduction

Let X be an arbitrary nonempty set and D be a nonempty set of subsets of the set X. If D is closed under the union, then D is called a complete X-semilattice of unions. The union of all elements of the set D is denoted by the symbol $\stackrel{⌣}{D}$ .

Let $B_X$ be the set of all binary relations on X. It is well known that $B_X$ is a semigroup.

Let f be an arbitrary mapping from X into D. Then we denote a binary relation

${\alpha }_f={\displaystyle \underset{x\in X}{\cup }}\left(\left\{x\right\}\times f\left(x\right)\right)$ for each f. The set of all such binary relations is denoted

by $B_X\left(D\right)$ . It is easy to prove that $B_X\left(D\right)$ is a semigroup with respect to the product operation of binary relations. This semigroup $B_X\left(D\right)$ is called a complete semigroup of binary relations defined by an X-semilattice of unions D. This structure was comprehensively investigated in Diasamidze [1] and [2] . We assume that $t\mathrm{,}y\in X$ , $Y\subseteq X$ , $\alpha \in B_X$ , $T\in D$ and $\varnothing \ne D^{\prime }\subseteq D$ . Then we denote following sets

$y\alpha =\left\{x\in X|y\alpha x\right\},Y\alpha ={\displaystyle \underset{y\in Y}{\cup }}y\alpha ,$

$\begin{array}{l}V\left(D,\alpha \right)=\left\{Y\alpha |Y\in D\right\},X^{\ast }=\left\{Y|\varnothing \ne Y\subseteq X\right\}\\ Y_T^{\alpha }=\left\{y\in X|y\alpha =T\right\},V\left(X^{\ast },\alpha \right)=\left\{Y\alpha |\varnothing \ne Y\subseteq X\right\}\\ D_t=\left\{Z^{\prime }\in D|t\in Z^{\prime }\right\},B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}\end{array}$

Let $D=\left\{\stackrel{⌣}{D},Z_1,Z_2,\cdots ,Z_{m-1}\right\}$ be finite X-semilattice of unions and $C\left(D\right)=\left\{P_0,P_1,P_2,\cdots ,P_{m-1}\right\}$ be the family of pairwise nonintersecting subsets of X. If $\phi =\left(\begin{array}{cccc}\stackrel{⌣}{D}& Z_1& \cdots & Z_{m-1}\\ P_0& P_1& \cdots & P_{m-1}\end{array}\right)$ is a mapping from D on $C\left(D\right)$ , then the equalities $\stackrel{⌣}{D}=P_0\cup P_1\cup P_2\cup \cdots \cup P_{m-1}$ and $Z_i=P_0\cup {\displaystyle \underset{T\in D\backslash D_Z}{\cup }}\phi \left(T\right)$ are valid. These equalities are called formal.

Let D be a complete X-semilattice of unions $\alpha \in B_X$ . Then a representation

of a binary relation $\alpha$ of the form $\alpha ={\displaystyle \underset{T\in V\left(X^{\ast },\alpha \right)}{\cup }}\left(Y_T^{\alpha }\times T\right)$ is called quasinormal.

Let $P_0\mathrm{,}{P}_1\mathrm{,}{P}_2,\cdots \mathrm{,}{P}_{m-1}$ be parameters in the formal equalities, $\beta \in B_X\left(D\right)$ , ${\bar{\beta }}_2$

be mapping from $X\backslash \stackrel{⌣}{D}$ to $D$ . Then $\bar{\beta }={\displaystyle \underset{i=0}{\overset{m-1}{\cup }}}\left(P_i\times {\displaystyle \underset{t\in P_i}{\cup }}t\beta \right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

is called subquasinormal represantation of $\beta$ . It can be easily seen that the following statements are true.

a) $\bar{\beta }\in B_X\left(D\right).$

b) ${\displaystyle \underset{i=0}{\overset{m-1}{\cup }}}\left(P_i\times {\displaystyle \underset{t\in P_i}{\cup }}t\beta \right)\subseteq \beta$ and $\beta =\bar{\beta }$ for some ${\bar{\beta }}_2$ .

c) Subquasinormal represantation of $\beta$ is quasinormal.

d) ${\bar{\beta }}_1=\left(\begin{array}{cccc}{P}_0& P_1& \cdots & P_{m-1}\\ P_0\bar{\beta }& P_1\bar{\beta }& \cdots & P_{m-1}\bar{\beta }\end{array}\right)$ is mapping from $C\left(D\right)$ on $D\cup \{\varnothing \}$ .

${\bar{\beta }}_1$ and ${\bar{\beta }}_2$ are respectively called normal and complement mappings for $\beta$ .

Let $\alpha \in B_X\left(D\right)$ . If $\alpha \ne \delta \circ \beta$ for all $\delta \mathrm{,}\beta \in B_X\left(D\right)\backslash \left\{\alpha \right\}$ then $\alpha$ is called external element. Every element of the set $B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}$ is an external element of $B_X\left(D\right)$ .

Theorem 1. [1] Let $X$ be a finite set and $\alpha \mathrm{,}\beta \in B_X\left(D\right)$ . If $\bar{\beta }$ is sub- quasinormal representation of $\beta$ then $\alpha \circ \beta =\alpha \circ \bar{\beta }$ .

Corollary 1. [1] Let ${\tilde{B}}^{\prime }\subseteq \tilde{B}\subseteq B_X\left(D\right)$ . If $\alpha \ne \delta \circ \bar{\beta }$ for $\alpha \in {\tilde{B}}^{\prime }$ , $\delta \in \tilde{B}\backslash \left\{\alpha \right\}$ , $\bar{\beta }\in \tilde{B}\backslash \left\{\alpha \right\}$ and subquasinormal representation of $\beta \in \tilde{B}\backslash \left\{\alpha \right\}$ then $\alpha \ne \delta \circ \beta$ .

It is known that the set of all external elements is subset of any generating set of $B_X\left(D\right)$ in [3] .

2. Results

In this work by symbol ${\Sigma }_{2.2}\left(X\mathrm{,4}\right)$ we denote all semilattices $D=\left\{Z_3,Z_2,Z_1,\stackrel{⌣}{D}\right\}$ of the class ${\Sigma }_2\left(X\mathrm{,4}\right)$ which the intersection of minimal elements $Z_3\cap Z_2=\varnothing$ . This semilattices graphic is given in Figure 1. By using formal equalities, we have $Z_3\cap Z_2=P_0=\varnothing$ . So, the formal equalities of the semilattice D has a form

Figure 1. Graphic of semilattice $D=\left\{Z_3,Z_2,Z_1,D\right\}$ which the intersection of minimal elements $Z_3\cap Z_2=\varnothing$ .

$\begin{array}{l}\stackrel{⌣}{D}=P_1\cup P_2\cup P_3\\ Z_1=P_2\cup P_3\\ Z_2=P_1\cup P_3\\ Z_3=P_2\end{array}$ (1)

Let $\delta \mathrm{,}\bar{\beta }\in B_X\left(D\right)$ . If quasinormal representation of binary relation $\delta$ has a form $\delta =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)$ then

$\delta \circ \bar{\beta }=\left(Y_3^{\delta }\times Z_3\bar{\beta }\right)\cup \left(Y_2^{\delta }\times Z_2\bar{\beta }\right)\cup \left(Y_1^{\delta }\times Z_1\bar{\beta }\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\bar{\beta }\right)$

We denote the set

$\begin{array}{l}{B}_{32}=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=\left\{Z_3,Z_2,\stackrel{⌣}{D}\right\}\right\}\\ B_{21}=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=\left\{Z_2,Z_1,\stackrel{⌣}{D}\right\}\right\}\\ B_{31}=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=\left\{Z_3,Z_1\right\}\right\}\\ {\tilde{B}}_{32}=\left\{\alpha \in B_{32}|\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right),Y_3^{\alpha }\cup Y_2^{\alpha }=X,Y_3^{\alpha }\cap Y_2^{\alpha }=\varnothing \right\}\\ {\tilde{B}}_{21}=\left\{\alpha \in B_{21}|\alpha =\left(Y_2^{\alpha }\times Z_2\right)\cup \left(Y_1^{\alpha }\times Z_1\right),Y_2^{\alpha }\cup Y_1^{\alpha }=X,Y_2^{\alpha }\cap Y_1^{\alpha }=\varnothing \right\}\end{array}$

It is easy to see that

$B_0\cap B_{32}=B_0\cap B_{21}=B_0\cap B_{31}=B_{21}\cap B_{32}=B_{31}\cap B_{32}=B_{21}\cap B_{31}=\varnothing$ .

Lemma 2. Let $D=\left\{Z_3,Z_2,Z_1,\stackrel{⌣}{D}\right\}\in {\Sigma }_{2.2}\left(X,4\right)$ . Then following statements are true for the sets $B_0\mathrm{,}{B}_{32}\mathrm{,}{\tilde{B}}_{32}$ .

a) If $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_1^{\alpha }\times Z_1\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_3^{\alpha }\mathrm{,}{Y}_1^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \varnothing$ , then $\alpha$ is product of some elements of the set $B_0$ .

b) If ${\beta }_0=\left(Z_3\times Z_3\right)\cup \left(\left(X\backslash Z_3\right)\times Z_2\right)$ , then $\left(B_0\circ {\beta }_0\right)\cup {\tilde{B}}_{32}=B_{32}$ .

c) If ${\sigma }_1=\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)$ , then $\left(B_0\circ {\sigma }_1\right)\cup {\tilde{B}}_{21}=B_{21}$ .

d) If ${\sigma }_1=\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)$ , then $B_{32}\circ {\sigma }_1=B_{21}$ .

e) If ${\sigma }_0=\left(Z_3\times Z_3\right)\cup \left(\left(X\backslash Z_3\right)\times Z_1\right)$ , then $B_{32}\circ {\sigma }_0=B_{31}$ .

f) Every element of the set $B_{32}$ is product of elements of the set $B_0\cup {\tilde{B}}_{32}$ .

g) Every element of the set $B_{21}$ is product of elements of the set $B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_1\right\}$ .

Proof. It will be enough to show only a, b and g. The rest can be similarly seen.

a. Let $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_1^{\alpha }\times Z_1\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_3^{\alpha }\mathrm{,}{Y}_1^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \{\varnothing \}$ , $\delta \mathrm{,}\bar{\beta }\in B_0$ . Then quasinormal representation of $\delta$ has a form

$\delta =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)$

where $Y_3^{\delta }\mathrm{,}{Y}_1^{\delta }\mathrm{,}{Y}_0^{\delta }\notin \{\varnothing \}$ . We suppose that

$\bar{\beta }=\left(P_2\times Z_3\right)\cup \left(P_1\times Z_2\right)\cup \left(P_3\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

where ${\bar{\beta }}_1=\left(\begin{array}{cccc}\varnothing & P_1& P_2& P_3\\ \varnothing & Z_2& Z_3& Z_1\end{array}\right)$ is normal mapping for $\bar{\beta }$ and ${\bar{\beta }}_2$ is com-

plement mapping of the set $X\backslash \stackrel{⌣}{D}$ on the set $\tilde{D}$ . So, $\bar{\beta }\in B_0$ since $V\left(X^{\ast },\bar{\beta }\right)=D$ . From the equalities (2.1) and definition of $\bar{\beta }$

$\begin{array}{l}{Z}_3\bar{\beta }=P_2\bar{\beta }=Z_3\\ Z_2\bar{\beta }=\left(P_1\cup P_3\right)\bar{\beta }=P_1\bar{\beta }\cup P_3\bar{\beta }=Z_2\cup Z_1=\stackrel{⌣}{D}\\ Z_1\bar{\beta }=\left(P_2\cup P_3\right)\bar{\beta }=P_2\bar{\beta }\cup P_3\bar{\beta }=Z_3\cup Z_1=Z_1\\ \stackrel{⌣}{D}\bar{\beta }=\left(P_1\cup P_2\cup P_3\right)\bar{\beta }=P_1\bar{\beta }\cup P_2\bar{\beta }\cup P_3\bar{\beta }=Z_2\cup \stackrel{⌣}{D}\cup Z_1=\stackrel{⌣}{D}\end{array}$

$\begin{array}{c}\delta \circ \bar{\beta }=\left(Y_3^{\delta }\times Z_3\bar{\beta }\right)\cup \left(Y_2^{\delta }\times Z_2\bar{\beta }\right)\cup \left(Y_1^{\delta }\times Z_1\bar{\beta }\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\bar{\beta }\right)\\ =\left(Y_3^{\delta }\times \stackrel{⌣}{D}\right)\cup \left(Y_2^{\delta }\times \stackrel{⌣}{D}\right)\cup \left(Y_1^{\delta }\times \stackrel{⌣}{D}\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)\\ =\left(Y_3^{\delta }\times \stackrel{⌣}{D}\right)\cup \left(Y_1^{\delta }\times \stackrel{⌣}{D}\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times \stackrel{⌣}{D}\right)=\alpha .\end{array}$

b. Let $\alpha \in B_0\circ {\beta }_0\cup {\tilde{B}}_{32}$ . Then $\alpha \in B_0\circ {\beta }_0$ or $\alpha \in {\tilde{B}}_{32}$ . If $\alpha \in B_0\circ {\beta }_0$ then $\alpha =\delta \circ {\beta }_0$ for some $\delta \in B_0$ . In this case we have

$\delta =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)$

where $Y_3^{\delta }\mathrm{,}{Y}_1^{\delta }\mathrm{,}{Y}_0^{\delta }\notin \{\varnothing \}$ . Also

$\begin{array}{c}\alpha =\delta \circ {\beta }_0=\left(Y_3^{\delta }\times Z_3{\beta }_0\right)\cup \left(Y_2^{\delta }\times Z_2{\beta }_0\right)\cup \left(Y_1^{\delta }\times Z_1{\beta }_0\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}{\beta }_0\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times \stackrel{⌣}{D}\right)\in B_{32}\backslash {\tilde{B}}_{32}\end{array}$

is satisfied. So, we have $\left(B_0\circ {\beta }_0\right)\cup {\tilde{B}}_{32}\subseteq B_{32}$ . On the other hand, if $\alpha \in {\tilde{B}}_{32}\subseteq B_{32}$ then $\left(B_0\circ {\beta }_0\right)\cup {\tilde{B}}_{32}\subseteq B_{32}$ is satisfied. Conversely, if $\alpha \in B_{32}$ then quasinormal representation of $\alpha$ has a form

$\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$

where $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \{\varnothing \}$ or $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\notin \{\varnothing \}$ and $Y_0^{\alpha }=\varnothing$ . We suppose that $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\notin \{\varnothing \}$ . In this case, we have

$\begin{array}{c}\delta \circ {\beta }_0=\left(Y_3^{\delta }\times Z_3{\beta }_0\right)\cup \left(Y_2^{\delta }\times Z_2{\beta }_0\right)\cup \left(Y_0^{\delta }\times Z_1{\beta }_0\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)=\alpha \end{array}$

for $\delta =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right)\cup \left(Y_0^{\alpha }\times Z_1\right)\in B_0$ . So, we have $B_{32}\subseteq \left(B_0\circ {\beta }_0\right)\cup {\tilde{B}}_{32}$ . Now suppose that $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\notin \{\varnothing \}$ and $Y_0^{\alpha }=\varnothing$ . In this case, we have $\alpha \in {\tilde{B}}_{32}\subseteq \left(B_0\circ {\beta }_0\right)\cup {\tilde{B}}_{32}$ . So, $\left(B_0\circ {\beta }_0\right)\cup {\tilde{B}}_{32}=B_{32}$ .

g. From the statement c, we have that $\left(B_0\circ {\beta }_0\right)\cup {\tilde{B}}_{32}=B_{32}$ where ${\beta }_0\in {\tilde{B}}_{32}$ by definition of ${\beta }_0$ . Thus, every element of the set $B_{32}$ is product of elements of the set $B_0\cup {\tilde{B}}_{32}$ .

Lemma 3. Let $D=\left\{Z_3,Z_2,Z_1,\stackrel{⌣}{D}\right\}\in {\Sigma }_{2.2}\left(X,4\right)$ . If $\left|X\backslash \stackrel{⌣}{D}\right|\ge 1$ then the following statements are true.

a) If $\alpha =X\times \stackrel{⌣}{D}$ then $\alpha$ is product of elements of the set $B_0$ .

b) If $\alpha =X\times Z_1$ then $\alpha$ is product of elements of the set $B_0$ .

c) If $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_1^{\alpha }\times Z_1\right)$ for some $Y_3^{\alpha }\mathrm{,}{Y}_1^{\alpha }\notin \varnothing$ , then $\alpha$ is product of elements of the $B_0$ .

d) If $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_3^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \varnothing$ , then $\alpha$ is product of elements of the $B_0$ .

e) If $\alpha =\left(Y_2^{\alpha }\times Z_2\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_2^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \varnothing$ , then $\alpha$ is product of elements of the $B_0$ .

f) If $\alpha =\left(Y_1^{\alpha }\times Z_1\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_1^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \varnothing$ , then $\alpha$ is product of elements of the $B_0$ .

Proof. c. Let quasinormal representation of $\alpha$ has a form $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_1^{\alpha }\times Z_1\right)$ where $Y_3^{\delta }\mathrm{,}{Y}_1^{\delta }\notin \{\varnothing \}$ . By definition of the semilattice D, $\left|X\right|\ge 3$ . We suppose that $\left|Y_3^{\alpha }\right|\ge 1$ and $\left|Y_1^{\alpha }\right|\ge 2$ . In this case, we suppose that

$\bar{\beta }=\left(P_2\times Z_3\right)\cup \left(\left(P_1\cup P_3\right)\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

where ${\bar{\beta }}_1=\left(\begin{array}{cccc}\varnothing & P_1& P_2& P_3\\ \varnothing & Z_1& Z_3& Z_1\end{array}\right)$ is normal mapping for $\bar{\beta }$ and ${\bar{\beta }}_2$ is comple-

ment mapping of the set $X\times \stackrel{⌣}{D}$ on the set $\tilde{D}\backslash \left\{Z_3,Z_1\right\}=\left\{Z_2\right\}$ (by suppose $\left|X\backslash \stackrel{⌣}{D}\right|\ge 1$ ). So, $\bar{\beta }\in B_0$ since $V\left(X^{\ast },\bar{\beta }\right)=D$ . Also, $Y_3^{\delta }=Y_3^{\alpha }$ and $Y_2^{\delta }\cup Y_1^{\delta }\cup Y_0^{\delta }=Y_1^{\delta }$ since $\left|Y_3^{\delta }\right|\ge 1,\left|Y_2^{\delta }\right|\ge 1,\left|Y_1^{\delta }\right|\ge 1,\left|Y_0^{\delta }\right|\ge 0$ . From the equalities (2.1) and definition of $\bar{\beta }$ we obtain that

$\begin{array}{l}{Z}_3\bar{\beta }=P_2\bar{\beta }=Z_3\\ Z_2\bar{\beta }=\left(P_1\cup P_3\right)\bar{\beta }=P_1\bar{\beta }\cup P_3\bar{\beta }=Z_1\cup Z_1=Z_1\\ Z_1\bar{\beta }=\left(P_2\cup P_3\right)\bar{\beta }=P_2\bar{\beta }\cup P_3\bar{\beta }=Z_3\cup Z_1=Z_1\\ \stackrel{⌣}{D}\bar{\beta }=\left(P_1\cup P_2\cup P_3\right)\bar{\beta }=P_1\bar{\beta }\cup P_2\bar{\beta }\cup P_3\bar{\beta }=Z_1\cup Z_3\cup Z_1=Z_1\end{array}$

$\begin{array}{c}\delta \circ \bar{\beta }=\left(Y_3^{\delta }\times Z_3\bar{\beta }\right)\cup \left(Y_2^{\delta }\times Z_2\bar{\beta }\right)\cup \left(Y_1^{\delta }\times Z_1\bar{\beta }\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\bar{\beta }\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_1\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times Z_1\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(\left(Y_2^{\delta }\cup Y_1^{\delta }\cup Y_0^{\delta }\right)\times Z_1\right)=\alpha \end{array}$

Now, we suppose that $\left|Y_3^{\alpha }\right|\ge 2$ and $\left|Y_1^{\alpha }\right|\ge 1$ . In this case, we suppose that

$\bar{\beta }=\left(\left(P_2\cup P_3\right)\times Z_3\right)\cup \left(P_1\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

where ${\bar{\beta }}_1=\left(\begin{array}{cccc}\varnothing & P_1& P_2& P_3\\ \varnothing & Z_1& Z_3& Z_3\end{array}\right)$ is normal mapping for $\bar{\beta }$ and ${\bar{\beta }}_2$ is com-

plement mapping of the set $X\times \stackrel{⌣}{D}$ on the set $\tilde{D}\backslash \left\{Z_3,Z_1\right\}=\left\{Z_2\right\}$ (by suppose $\left|X\backslash \stackrel{⌣}{D}\right|\ge 1$ ). So, $\bar{\beta }\in B_0$ since $V\left(X^{\ast },\bar{\beta }\right)=D$ . Also, $Y_3^{\delta }\cup Y_1^{\delta }=Y_3^{\alpha }$ and $Y_2^{\delta }\cup Y_0^{\delta }=Y_1^{\alpha }$ since $\left|Y_3^{\delta }\right|\ge \mathrm{1,}\left|Y_2^{\delta }\right|\ge \mathrm{1,}\left|Y_1^{\delta }\right|\ge \mathrm{1,}\left|Y_0^{\delta }\right|\ge 0$ . From the equalities (2.1) and definition of $\bar{\beta }$ we obtain that

$\begin{array}{l}{Z}_3\bar{\beta }=P_2\bar{\beta }=Z_3\\ Z_2\bar{\beta }=\left(P_1\cup P_3\right)\bar{\beta }=P_1\bar{\beta }\cup P_3\bar{\beta }=Z_1\cup Z_3=Z_1\\ Z_1\bar{\beta }=\left(P_2\cup P_3\right)\bar{\beta }=P_2\bar{\beta }\cup P_3\bar{\beta }=Z_3\cup Z_3=Z_3\\ \stackrel{⌣}{D}\bar{\beta }=\left(P_1\cup P_2\cup P_3\right)\bar{\beta }=P_1\bar{\beta }\cup P_2\bar{\beta }\cup P_3\bar{\beta }=Z_1\cup Z_3\cup Z_3=Z_1\end{array}$

$\begin{array}{c}\delta \circ \bar{\beta }=\left(Y_3^{\delta }\times Z_3\bar{\beta }\right)\cup \left(Y_2^{\delta }\times Z_2\bar{\beta }\right)\cup \left(Y_1^{\delta }\times Z_1\bar{\beta }\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\bar{\beta }\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_1\right)\cup \left(Y_1^{\delta }\times Z_3\right)\cup \left(Y_0^{\delta }\times Z_1\right)\\ =\left(\left(Y_3^{\delta }\cup Y_1^{\delta }\right)\times Z_3\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times Z_1\right)=\alpha \end{array}$

Lemma 4. Let $D=\left\{Z_3,Z_2,Z_1,\stackrel{⌣}{D}\right\}\in {\Sigma }_{2.2}\left(X,4\right)$ , ${\sigma }_0=\left(Z_3\times Z_3\right)\cup \left(\left(X\backslash Z_3\right)\times Z_1\right)$ and ${\sigma }_1=\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)$ . If $X=\stackrel{⌣}{D}$ then the following statements are true

a) If $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_3^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is product of elements of the $B_0\cup B_{32}$ .

b) If $\alpha =\left(Y_2^{\alpha }\times Z_2\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_2^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is product of elements of the $B_{32}\cup \left\{{\sigma }_1\right\}$ .

c) If $\alpha =\left(Y_1^{\alpha }\times Z_1\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_1^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is product of elements of the $B_{32}\cup \left\{{\sigma }_0\mathrm{,}{\sigma }_1\right\}$ .

Proof. First, remark that $Z_3{\sigma }_0=Z_3$ , $Z_2{\sigma }_0=\stackrel{⌣}{D}{\sigma }_0=Z_1$ , $Z_3{\sigma }_1=Z_1$ , $Z_2{\sigma }_1=Z_2$ , $\stackrel{⌣}{D}{\sigma }_1=\stackrel{⌣}{D}$ .

a. Let $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$ for some $Y_3^{\alpha }\mathrm{,}{Y}_0^{\alpha }\notin \varnothing$ . In this case, we suppose that

$\delta =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)$

and

${\beta }_1=\left(Z_3\times Z_3\right)\cup \left(\left(Z_1\backslash Z_3\right)\times Z_1\right)\cup \left(\left(X\backslash Z_1\right)\times \stackrel{⌣}{D}\right)$

where $Y_3^{\delta }\mathrm{,}{Y}_2^{\delta }\notin \{\varnothing \}$ . It is easy to see that $\delta \in B_{32}$ and ${\beta }_1$ is generating by elements of the $B_0$ by statement b of Lemma 2. Also, $Y_3^{\delta }=Y_3^{\alpha }$ and $Y_2^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ since $Z_3\bar{\beta }=Z_3$ , $Z_2\bar{\beta }=\stackrel{⌣}{D}\bar{\beta }=\stackrel{⌣}{D}$ and $\left|Y_3^{\delta }\right|\ge \mathrm{1,}\left|Y_2^{\delta }\right|\ge \mathrm{1,}\left|Y_0^{\delta }\right|\ge 0$ . So, $\alpha$ is product of elements of the $B_0\cup B_{32}$ . □

Lemma 5. Let

$D=\left\{Z_3,Z_2,Z_1,\stackrel{⌣}{D}\right\}\in {\Sigma }_{2.2}\left(X,4\right)$

and

${\sigma }_1=\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)$ .

If $\left|X\backslash \stackrel{⌣}{D}\right|\ge 1$ then $S_1=B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_1\right\}$ is an irreducible generating set for the semigroup $B_X\left(D\right)$ .

Proof. First, we must prove that every element of $B_X\left(D\right)$ is product of elements of $S_1$ . Let $\alpha \in B_X\left(D\right)$ and

$\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right)\cup \left(Y_1^{\alpha }\times Z_1\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$

where $Y_3^{\alpha }\cup Y_2^{\alpha }\cup Y_1^{\alpha }\cup Y_0^{\alpha }=X$ and $Y_3^{\alpha }\cap Y_2^{\alpha }=\varnothing$ , $\left(0\le i\ne j\le 3\right)$ . We suppose

that $\left|V\left(X^{\ast },\alpha \right)\right|=1$ . Then we have $V\left(X^{\ast }\mathrm{,}\alpha \right)\in \left\{\left\{Z_3\right\}\mathrm{,}\left\{Z_2\right\}\mathrm{,}\left\{Z_1\right\}\mathrm{,}\left\{\stackrel{⌣}{D}\right\}\right\}$ . If

$V\left(X^{\ast }\mathrm{,}\alpha \right)\in \left\{\left\{Z_3\right\}\mathrm{,}\left\{Z_2\right\}\mathrm{,}\left\{Z_1\right\}\right\}$ then $\alpha =X\times Z_3$ or $\alpha =X\times Z_2$ or $\alpha =X\times Z_1$ . Quasinormal representations of $\delta \mathrm{,}{\beta }_1\mathrm{,}{\beta }_2$ and ${\beta }_3$ has form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)\\ {\beta }_1=\left(\stackrel{⌣}{D}\times Z_3\right)\cup \left(\left(X\backslash \stackrel{⌣}{D}\right)\times Z_2\right)\\ {\beta }_2=\left(\stackrel{⌣}{D}\times Z_2\right)\cup \left(\left(X\backslash \stackrel{⌣}{D}\right)\times Z_1\right)\\ {\beta }_3=\left(\stackrel{⌣}{D}\times Z_1\right)\cup \left(\left(X\backslash \stackrel{⌣}{D}\right)\times Z_2\right)\end{array}$

where $Y_3^{\delta }\mathrm{,}{Y}_2^{\delta }\mathrm{,}{Y}_1^{\delta }\notin \{\varnothing \}$ . So, $\delta \in B_0$ , ${\beta }_1\in {\tilde{B}}_{32}$ and ${\beta }_2\mathrm{,}{\beta }_3\in B_{21}$ since $\left|X\backslash \stackrel{⌣}{D}\right|\ge 1$ . From the definition of $\delta \mathrm{,}{\beta }_1\mathrm{,}{\beta }_2$ and ${\beta }_3$ we obtain that

$\begin{array}{c}\delta \circ {\beta }_1=\left(Y_3^{\delta }\times Z_3{\beta }_1\right)\cup \left(Y_2^{\delta }\times Z_2{\beta }_1\right)\cup \left(Y_1^{\delta }\times Z_1{\beta }_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}{\beta }_1\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_3\right)\cup \left(Y_1^{\delta }\times Z_3\right)\cup \left(Y_0^{\delta }\times Z_3\right)\\ =\left(Y_3^{\delta }\cup Y_2^{\delta }\cup Y_1^{\delta }\cup Y_0^{\delta }\right)\times Z_3=X\times Z_3\end{array}$

$\begin{array}{c}\delta \circ {\beta }_2=\left(Y_3^{\delta }\times Z_3{\beta }_2\right)\cup \left(Y_2^{\delta }\times Z_2{\beta }_2\right)\cup \left(Y_1^{\delta }\times Z_1{\beta }_2\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}{\beta }_2\right)\\ =\left(Y_3^{\delta }\times Z_2\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_2\right)\cup \left(Y_0^{\delta }\times Z_2\right)\\ =\left(Y_3^{\delta }\cup Y_2^{\delta }\cup Y_1^{\delta }\cup Y_0^{\delta }\right)\times Z_2=X\times Z_2\end{array}$

$\begin{array}{c}\delta \circ {\beta }_3=\left(Y_3^{\delta }\times Z_3{\beta }_3\right)\cup \left(Y_2^{\delta }\times Z_2{\beta }_3\right)\cup \left(Y_1^{\delta }\times Z_1{\beta }_3\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}{\beta }_3\right)\\ =\left(Y_3^{\delta }\times Z_1\right)\cup \left(Y_2^{\delta }\times Z_1\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times Z_1\right)\\ =\left(Y_3^{\delta }\cup Y_2^{\delta }\cup Y_1^{\delta }\cup Y_0^{\delta }\right)\times Z_1=X\times Z_1\end{array}$

That means, $X\times Z_1\mathrm{,}X\times Z_2$ and $X\times Z_3$ are generated by $B_0\cup {\tilde{B}}_{32}\mathrm{,}$ $B_0\cup B_{21}$ and $B_0\cup B_{21}$ respectively. By using statement g and h of Lemma 3, we have $X\times Z_1\mathrm{,}X\times Z_2$ and $X\times Z_3$ are generated by $B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_1\right\}$ . On the other hand, if $V\left(X^{\ast },\alpha \right)=\left\{\stackrel{⌣}{D}\right\}$ then $\alpha =X\times \stackrel{⌣}{D}$ By using statement a of Lemma 3, we have $\alpha$ is product of some elemets of $B_0$ .

So, $S_1$ is generating set for the semigroup $B_X\left(D\right)$ . Now, we must prove that $S_1=B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_1\right\}$ is irreducible. Let $\alpha \in S_1$ .

If $\alpha \in B_0$ then $\alpha \ne \sigma \circ \tau$ for all $\sigma \mathrm{,}\tau \in B_X\left(D\right)\backslash \left\{\alpha \right\}$ from Lemma 2. So, $\alpha \ne \sigma \circ \tau$ for all $\sigma \mathrm{,}\tau \in S_1\backslash \left\{\alpha \right\}$ . That means, $\alpha \notin B_0$ .

If $\alpha \in {\tilde{B}}_{32}$ then the quasinormal representation of $\alpha$ has form $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right)$ for some $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\notin \varnothing$ . Let $\alpha =\delta \circ \beta$ for some $\delta \mathrm{,}\beta \in S_1\backslash \left\{\alpha \right\}$ .

We suppose that $\delta \in B_0\backslash \left\{\alpha \right\}$ and $\beta \in S_1\backslash \left\{\alpha \right\}$ . By definition of $B_0$ , quasinormal representation of $\delta$ has form

$\delta =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)$

where $Y_3^{\delta }\mathrm{,}{Y}_2^{\delta }\mathrm{,}{Y}_1^{\delta }\notin \{\varnothing \}$ . By using $Z_3\subset Z_1\subset \stackrel{⌣}{D}$ and $Z_2\subset \stackrel{⌣}{D}$ we have $Z_3\beta$ and $Z_2\beta$ are minimal elements of the semilattice $\left\{Z_3\beta \mathrm{,}{Z}_2\beta \mathrm{,}{Z}_1\beta \mathrm{,}\stackrel{⌣}{D}\beta \right\}$ . Also, we have

$\begin{array}{l}\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right)=\alpha =\delta \circ \beta \\ =\left(Y_3^{\delta }\times Z_3\beta \right)\cup \left(Y_2^{\delta }\times Z_2\beta \right)\cup \left(Y_1^{\delta }\times Z_1\beta \right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\beta \right)\end{array}$

Since $Z_3$ and $Z_2$ are minimal elements of the semilattice $\left\{Z_3\mathrm{,}{Z}_2\mathrm{,}\stackrel{⌣}{D}\right\}$ , this equality is possible only if $Z_3=Z_3\beta$ , $Z_2=Z_2\beta$ or $Z_3=Z_2\beta$ , $Z_2=Z_3\beta$ . By using formal equalities and $P_3\beta \mathrm{,}{P}_2\beta \mathrm{,}{P}_1\beta \in D$ , we obtain

$\begin{array}{lll}{Z}_3=Z_3\beta =P_2\beta \hfill & \text{and}\hfill & Z_2=Z_2\beta =P_1\beta =P_3\beta \hfill \\ Z_2=Z_3\beta =P_2\beta \hfill & \text{and}\hfill & Z_3=Z_2\beta =P_1\beta =P_3\beta \hfill \end{array}$

respectively. Let $Z_3=P_2\beta$ and $Z_2=P_1\beta =P_3\beta$ . If $\bar{\beta }$ is sub-quasinormal representation of $\beta$ then $\delta \circ \beta =\delta \circ \bar{\beta }$ and

$\bar{\beta }=\left(\left(P_1\cup P_3\right)\times Z_2\right)\cup \left(P_2\times Z_3\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

where ${\bar{\beta }}_1=\left(\begin{array}{cccc}\varnothing & P_1& P_2& P_3\\ \varnothing & Z_2& Z_3& Z_2\end{array}\right)$ is normal mapping for $\bar{\beta }$ and ${\bar{\beta }}_2$ is com-

plement mapping of the set $X\times \stackrel{⌣}{D}$ on the set $\tilde{D}=\left\{Z_3,Z_2,Z_1\right\}$ . From formal equalities, we obtain

$\bar{\beta }=\left(Z_2\times Z_2\right)\cup \left(Z_3\times Z_3\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)\in S_1\backslash \left\{\alpha \right\}$

and by using $Z_1\cap Z_2\ne \varnothing ,Z_3\cup Z_2=D$ and $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ , we have

$\begin{array}{c}\delta \circ \bar{\beta }=\left(Y_3^{\delta }\times Z_3\bar{\beta }\right)\cup \left(Y_2^{\delta }\times Z_2\bar{\beta }\right)\cup \left(Y_1^{\delta }\times Z_1\bar{\beta }\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\bar{\beta }\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times \stackrel{⌣}{D}\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times \stackrel{⌣}{D}\right)\ne \alpha \end{array}$

This contradicts with $\alpha =\delta \circ \beta$ . So, $\delta \notin B_0\backslash \left\{\alpha \right\}$ .

Now, we suppose that $\delta \in {\tilde{B}}_{32}\backslash \left\{\alpha \right\}$ and $\beta \in S_1\backslash \left\{\alpha \right\}$ . Similar operations are applied as above, we obtain $\delta \notin {\tilde{B}}_{32}\backslash \left\{\alpha \right\}$ .

Now, we suppose that $\delta ={\sigma }_1$ and $\beta \in S_1\backslash \left\{\alpha \right\}$ . Similar operations are applied as above, we obtain $\delta \ne {\sigma }_1$ .

That means $\alpha \ne \delta \circ \beta$ for any $\alpha \in {\tilde{B}}_{32}$ and $\delta \mathrm{,}\beta \in S_1\backslash \left\{\alpha \right\}$ .

If $\alpha ={\sigma }_1$ , then by the definition of ${\sigma }_1$ , quasinormal representation of $\alpha$ has a form $\alpha =\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)$ . Let $\alpha =\delta \circ \beta$ for some $\delta \mathrm{,}\beta \in S_1\backslash \left\{{\sigma }_1\right\}$ .

We suppose that $\delta \in B_0\backslash \left\{{\sigma }_1\right\}$ and $\beta \in S_1\backslash \left\{{\sigma }_1\right\}$ . By definition of $B_0$ , quasinormal representation of $\delta$ has form

$\delta =\left(Y_3^{\delta }\times Z_3\right)\cup \left(Y_2^{\delta }\times Z_2\right)\cup \left(Y_1^{\delta }\times Z_1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)$

where $Y_3^{\delta }\mathrm{,}{Y}_2^{\delta }\mathrm{,}{Y}_1^{\delta }\notin \{\varnothing \}$ . By using $Z_3\subset Z_1\subset \stackrel{⌣}{D}$ and $Z_2\subset \stackrel{⌣}{D}$ we have $Z_3\beta$ and $Z_2\beta$ are minimal elements of the semilattice $\left\{Z_3\beta \mathrm{,}{Z}_2\beta \mathrm{,}{Z}_1\beta \mathrm{,}\stackrel{⌣}{D}\beta \right\}$ . Also, we have

$\begin{array}{l}\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)=\alpha =\delta \circ \beta \\ =\left(Y_3^{\delta }\times Z_3\beta \right)\cup \left(Y_2^{\delta }\times Z_2\beta \right)\cup \left(Y_1^{\delta }\times Z_1\beta \right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\beta \right)\end{array}$

From $Z_2$ and $Z_1$ are minimal elements of the semilattice $\left\{Z_2\mathrm{,}{Z}_1\mathrm{,}\stackrel{⌣}{D}\right\}$ , this equality is possible only if $Z_2=Z_3\beta$ , $Z_1=Z_2\beta$ or $Z_2=Z_2\beta$ , $Z_1=Z_3\beta$ . By using formal equalities, we obtain

$\begin{array}{lll}{Z}_2=Z_3\beta =P_2\beta \hfill & \text{and}\hfill & Z_1=Z_2\beta =P_1\beta \cup P_3\beta \hfill \\ Z_1=Z_3\beta =P_2\beta \hfill & \text{and}\hfill & Z_2=Z_2\beta =P_1\beta =P_3\beta \hfill \end{array}$

respectively. Let $Z_2=P_2\beta$ and $Z_1=P_1\beta \cup P_3\beta$ where $P_1\beta \mathrm{,}{P}_3\beta \in \left\{Z_3\mathrm{,}{Z}_1\right\}$ . Then subquasinormal representation of $\beta$ has one of the form

${\bar{\beta }}^1=\left(P_1\times Z_3\right)\cup \left(P_2\times Z_2\right)\cup \left(P_3\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

${\bar{\beta }}^2=\left(P_3\times Z_3\right)\cup \left(P_2\times Z_2\right)\cup \left(P_1\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

${\bar{\beta }}^3=\left(P_2\times Z_2\right)\cup \left(\left(P_1\cup P_3\right)\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

where

${\bar{\beta }}_1^1=\left(\begin{array}{cccc}\varnothing & P_1& P_2& P_3\\ \varnothing & Z_3& Z_2& Z_1\end{array}\right)$ , ${\bar{\beta }}_1^2=\left(\begin{array}{cccc}\varnothing & P_1& P_2& P_3\\ \varnothing & Z_1& Z_2& Z_3\end{array}\right)$ , ${\bar{\beta }}_1^3=\left(\begin{array}{cccc}\varnothing & P_1& P_2& P_3\\ \varnothing & Z_1& Z_2& Z_1\end{array}\right)$

are normal mapping for $\bar{\beta }$ , ${\bar{\beta }}_2$ is complement mapping of the set $X\times \stackrel{⌣}{D}$ on the set $\tilde{D}=\left\{Z_3,Z_2,Z_1\right\}$ and $\delta \circ \beta =\delta \circ {\bar{\beta }}_i$ . From formal equalities, we obtain

${\bar{\beta }}^1=\left(\left(Z_2\backslash Z_1\right)\times Z_3\right)\cup \left(\left(Z_1\backslash Z_2\right)\times Z_2\right)\cup \left(\left(Z_2\backslash Z_1\right)\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

${\bar{\beta }}^2=\left(\left(Z_2\cap Z_1\right)\times Z_3\right)\cup \left(\left(Z_1\backslash Z_2\right)\times Z_2\right)\cup \left(\left(Z_2\backslash Z_1\right)\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

${\bar{\beta }}^3=\left(\left(Z_1\backslash Z_2\right)\times Z_2\right)\cup \left(Z_2\times Z_1\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash \stackrel{⌣}{D}}{\cup }}\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)$

and by using $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ , we have

$\begin{array}{c}\delta \circ {\bar{\beta }}^1=\delta \circ {\bar{\beta }}^2=\delta \circ {\bar{\beta }}^3\\ =\left(Y_3^{\delta }\times Z_3{\bar{\beta }}^1\right)\cup \left(Y_2^{\delta }\times Z_2{\bar{\beta }}^1\right)\cup \left(Y_1^{\delta }\times Z_1{\bar{\beta }}^1\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}{\bar{\beta }}^1\right)\\ =\left(Y_3^{\delta }\times Z_2\right)\cup \left(Y_2^{\delta }\times Z_1\right)\cup \left(Y_1^{\delta }\times \stackrel{⌣}{D}\right)\cup \left(Y_0^{\delta }\times \stackrel{⌣}{D}\right)\\ =\left(Y_3^{\delta }\times Z_2\right)\cup \left(Y_2^{\delta }\times Z_1\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times \stackrel{⌣}{D}\right)\ne \alpha \end{array}$

This contradicts with $\alpha =\delta \circ \beta$ . So, $\delta \notin B_0\backslash \left\{{\sigma }_1\right\}$ .

Now, we suppose that $\delta \in {\tilde{B}}_{32}\backslash \left\{{\sigma }_1\right\}$ and $\beta \in S_1\backslash \left\{{\sigma }_1\right\}$ . Similar operations are applied as above, we obtain $\delta \notin {\tilde{B}}_{32}\backslash \left\{{\sigma }_1\right\}$ .

That means $\alpha \ne \delta \circ \beta$ for any $\alpha \in {\tilde{B}}_{32}$ and $\delta \mathrm{,}\beta \in S_1\backslash \left\{\alpha \right\}$ . □

Lemma 6. Let $D=\left\{Z_3,Z_2,Z_1,\stackrel{⌣}{D}\right\}\in {\Sigma }_{2.2}\left(X,4\right)$ , ${\sigma }_0=\left(Z_3\times Z_3\right)\cup \left(\left(X\backslash Z_3\right)\times Z_1\right)$ and ${\sigma }_1=\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)$ . If $X=\stackrel{⌣}{D}$ then $S_2=B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_0,{\sigma }_1\right\}$ is irreducible generating set for the semigroup $B_X\left(D\right)$ .

Theorem 7. Let $D=\left\{Z_3,Z_2,Z_1,\stackrel{⌣}{D}\right\}\in {\Sigma }_{2.2}\left(X,4\right)$ , ${\sigma }_0=\left(Z_3\times Z_3\right)\cup \left(\left(X\backslash Z_3\right)\times Z_1\right)$ and ${\sigma }_1=\left(Z_2\times Z_2\right)\cup \left(\left(X\backslash Z_2\right)\times Z_1\right)$ . If $X$ is a finite set and $\left|X\right|=n$ then the following statements are true

a) If $\left|X\backslash \stackrel{⌣}{D}\right|\ge 1$ then $\left|B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_1\right\}\right|=4^n-3^{n+1}+2^{n+2}-2$

b) If $X=\stackrel{⌣}{D}$ then $\left|B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_0,{\sigma }_1\right\}\right|=4^n-3^{n+1}+2^{n+2}-1$

Proof. Let

$S_n=\left\{{\phi }_i|{\phi }_i\mathrm{<mo>\; :\; </mo>}M=\left\{\mathrm{1,2,}\cdots \mathrm{,}n\right\}\to M=\left\{\mathrm{1,2,}\cdots \mathrm{,}n\right\}\mathrm{,}\text{onetoonemapping}\right\}$

be a group, ${\phi }_{i_1}\mathrm{,}{\phi }_{i_2}\mathrm{,}\cdots \mathrm{,}{\phi }_{i_m}\in S_n$ $\left(m\le n\right)$ and $Y_{{\phi }_1}\mathrm{,}{Y}_{{\phi }_2}\mathrm{,}\cdots \mathrm{,}{Y}_{{\phi }_m}$ be partitioning of

X. It is well known that $k_n^m=\left|\left\{Y_{{\phi }_1},Y_{{\phi }_2},\cdots ,Y_{{\phi }_m}\right\}\right|={\displaystyle \underset{i=1}{\overset{m}{\sum }}}\frac{{\left(-1\right)}^{m+i}}{\left(i-1\right)!\left(m-i\right)!}$ . If $m=2,3,4$

then we have

$\begin{array}{l}{k}_n^2=2^{n-1}-1\\ k_n^3=\frac{1}{2}\cdot 3^{n-1}-2^{n-1}+\frac{1}{2}\\ k_n^4=\frac{1}{6}\cdot 4^{n-1}-\frac{1}{2}\cdot 3^{n-1}+\frac{1}{2}\cdot 2^{n-1}-\frac{1}{6}\end{array}$

If $Y_{{\phi }_1}\mathrm{,}{Y}_{{\phi }_2}$ are any two elements of partitioning of X and $\bar{\beta }=\left(Y_{{\phi }_1}\times T_1\right)\cup \left(Y_{{\phi }_2}\times T_2\right)$ where $T_1\mathrm{,}{T}_2\in D$ and $T_1\ne T_2$ , then the number of different binary relations $\bar{\beta }$ of semigroup $B_X\left(D\right)$ is equal to

$2\cdot k_n^2=2^n-2$ (2)

If $Y_{{\phi }_1}\mathrm{,}{Y}_{{\phi }_2}\mathrm{,}{Y}_{{\phi }_3}$ are any three elements of partitioning of X and $\bar{\beta }=\left(Y_{{\phi }_1}\times T_1\right)\cup \left(Y_{{\phi }_2}\times T_2\right)\cup \left(Y_{{\phi }_3}\times T_3\right)$ where $T_1\mathrm{,}{T}_2\mathrm{,}{T}_3$ are pairwise different elements of D, then the number of different binary relations $\bar{\beta }$ of semigroup $B_X\left(D\right)$ is equal to

$6\cdot k_n^3=3^n-3\cdot 2^n+3$ (3)

If $Y_{{\phi }_1}\mathrm{,}{Y}_{{\phi }_2}\mathrm{,}{Y}_{{\phi }_3}\mathrm{,}{Y}_{{\phi }_4}$ are any four elements of partitioning of X and $\bar{\beta }=\left(Y_{{\phi }_1}\times T_1\right)\cup \left(Y_{{\phi }_2}\times T_2\right)\cup \left(Y_{{\phi }_3}\times T_3\right)\cup \left(Y_{{\phi }_4}\times T_4\right)$ where $T_1\mathrm{,}{T}_2\mathrm{,}{T}_3\mathrm{,}{T}_4$ are pairwise different elements of D, then the number of different binary relations $\bar{\beta }$ of semigroup $B_X\left(D\right)$ is equal to

$24\cdot k_n^4=4^n-4\cdot 3^n+3\cdot 2^n-4$ (4)

Let $\alpha \in B_0$ . Quasinormal represantation of $\alpha$ has form

$\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right)\cup \left(Y_1^{\alpha }\times Z_1\right)\cup \left(Y_0^{\alpha }\times \stackrel{⌣}{D}\right)$

where $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\mathrm{,}{Y}_1^{\alpha }\notin \{\varnothing \}$ . Also, $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\mathrm{,}{Y}_1^{\alpha }$ or $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\mathrm{,}{Y}_1^{\alpha }\mathrm{,}{Y}_0^{\alpha }$ are partitioning of X for $\left|X\right|\ge 4$ . By using Equations (2.3) and (2.4) we obtain

$\left|B_0\right|=4^n-3^{n+1}+3\cdot 2^n-1$

Let $\alpha \in {\tilde{B}}_{32}$ . Quasinormal represantation of $\alpha$ has form $\alpha =\left(Y_3^{\alpha }\times Z_3\right)\cup \left(Y_2^{\alpha }\times Z_2\right)$ where $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }\notin \{\varnothing \}$ . Also, $Y_3^{\alpha }\mathrm{,}{Y}_2^{\alpha }$ are partitioning of X. By using (2.2) we obtain

$\left|{\tilde{B}}_{32}\right|=2^n-2$

So, we have

$\begin{array}{l}\left|B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_1\right\}\right|=4^n-3^{n+1}+2^{n+2}-2\\ \left|B_0\cup {\tilde{B}}_{32}\cup \left\{{\sigma }_0,{\sigma }_1\right\}\right|=4^n-3^{n+1}+2^{n+2}-1\end{array}$

since $B_0\cap {\tilde{B}}_{32}=B_0\cap \left\{{\sigma }_0,{\sigma }_1\right\}={\tilde{B}}_{32}\cap \left\{{\sigma }_0,{\sigma }_1\right\}=\varnothing$ . □

Acknowledgements

Sincere thanks to Prof. Dr. Neşet AYDIN for his valuable suggestions.

Conflicts of Interest

The authors declare no conflicts of interest.

References

[1]	Diasamidze, Y. and Makharadze, S. (2013) Complete Semigroups of Binary Relations. Kriter Yay Nevi, Istanbul.
[2]	Diasamidze, Y., Ayd, N.N. and Erdogan, A. (2016) Generating Set of the Complete Semigroups of Binary Relations. Applied Mathematics, 7, 98-107. https://doi.org/10.4236/am.2016.71009
[3]	Bakuridze, A., Diasamidze, Y. and Givradze, O. Generated Sets of the Complete Semigroups of Binary Relations Defined by Semilattices of the Class Σ1(X, 2)(In Press)