Generating Sets of the Complete Semigroup of Binary Relations Defined by Semilattices of the Class Σ₈(X, 5)

Nino Tsinaridze

Department of Mathematics, Faculty of Exact Sciences and Education, Batumi Shota Rustaveli State University, Batumi, Georgia.

DOI: 10.4236/am.2024.152010                

Abstract

In this article, we study generating sets of the complete semigroups of binary relations defined by X-semilattices of unions of the class Σ₈(X, 5). Found uniquely irreducible generating set for the given semigroups and when X is finite set formulas for calculating the number of elements in generating sets are derived.

Keywords

Semigroup, Semilattice, Binary Relation

1. Introduction

Let $X\ne \varnothing$ , D is an X-semilattice of unions which is closed with respect to the set-theoretic union of elements from D, f be an arbitrary mapping of the set X in the set D. To each mapping f we put into correspondence a binary relation ${\alpha }_f$

on the set X that satisfies the condition ${\alpha }_f={\displaystyle \underset{x\in X}{\cup }\left(\left\{x\right\}\times f\left(x\right)\right)}$ . The set of all such

${\alpha }_f$ 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 operation of multiplication of binary relations, which is called a complete semigroup of binary relations defined by an X-semilattice of unions D.

We denote by $\varnothing$ an empty subset of the set X or an empty binary relation. The condition $\left(x,y\right)\in \alpha$ will be written in the form $x\alpha y$ .

Let $x,y\in X$ , $Y\subseteq X$ , $\alpha \in B_X\left(D\right)$ , $\stackrel{⌣}{D}={\displaystyle \underset{Y\in D}{\cup }Y}$ and $T\in D$ . We denote by the symbols $y\alpha$ , $Y\alpha$ , $V\left(D,\alpha \right)$ , $X^{\ast }$ and $V\left(X^{\ast },\alpha \right)$ the following sets:

$\begin{array}{l}y\alpha =\left\{x\in X|y\alpha x\right\},Y\alpha ={\displaystyle \underset{y\in Y}{\cup }y\alpha },V\left(D,\alpha \right)=\left\{Y\alpha |Y\in D\right\},\\ X^{\ast }=\left\{Y|\varnothing \ne Y\subseteq X\right\},V\left(X^{\ast },\alpha \right)=\left\{Y\alpha |\varnothing \ne Y\subseteq X\right\},\\ D_T=\left\{Z\in D|T\subseteq Z\right\},Y_T^{\alpha }=\left\{y\in X|y\alpha =T\right\}.\end{array}$

Theorem 1.1. Let $D=\left\{\stackrel{⌣}{D},Z_1,Z_2,\cdots ,Z_{m-1}\right\}$ be some 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 sets of pairwise nonintersecting subsets of the set X (the set $\varnothing$ can be repeated several times). If $\phi$ is a mapping of the semilattice D on the family of sets $C\left(D\right)$ which satisfies the conditions

$\phi =\left(\begin{array}{l}\stackrel{⌣}{D}{Z}_1{Z}_2\cdots Z_{m-1}\\ P_0{P}_1{P}_2\cdots P_{m-1}\end{array}\right)$

and ${\hat{D}}_Z=D\backslash D_Z$ , then the following equalities are valid:

$\begin{array}{l}\stackrel{⌣}{D}=P_0\cup P_1\cup P_2\cup \cdots \cup P_{m-1},\\ Z_i=P_0\cup {\displaystyle \underset{T\in {\hat{D}}_{Z_i}}{\cup }\phi }\left(T\right).\end{array}$

In the sequel these equalities will be called formal. The parameters $P_i$ $\left(0<i\le m-1\right)$ there exist such parameters that cannot be empty sets for D. Such sets $P_i$ are called bases sources, where sets $P_j$ $\left(0\le j\le m-1\right)$, which can be empty sets too are called completeness sources.

It is proved that under the mapping $\phi$ the number of covering elements of the pre-image of a bases source is always equal to one, while under the mapping $\phi$ the number of covering elements of the pre-image of a completeness source either does not exist or is always greater than one (see [1] Theorem 1.1, [2] [3] chapter 11).

Definition 1.1. The representation $\alpha ={\displaystyle \underset{T\in D}{\cup }\left(Y_T^{\alpha }\times T\right)}$ of binary relation $\alpha$ is called quasinormal, if ${\displaystyle \underset{T\in D}{\cup }{Y}_T^{\alpha }}=X$ and $Y_T^{\alpha }\cap Y_{T^{\prime }}^{\alpha }=\varnothing$ for any $T,T^{\prime }\in D$ , $T\ne T^{\prime }$ (see [1] Definition 1.2, [2] , [3] chapter 1.1).

Definition 1.2. Let $\alpha ,\beta \subseteq X\times X$ . Their product $\delta =\alpha \circ \beta$ is defined as follows: $x\delta y$ $\left(x,y\in X\right)$ if there exists an element $z\in X$ such that $x\alpha z\beta y$ (see [1] Definition 1.3, [1] , chapter 1.3).

Definition 1.3. We say that an element $\alpha$ of the semigroup $B_X\left(D\right)$ is external if $\alpha \ne \delta \circ \beta$ for all $\delta ,\beta \in B_X\left(D\right)\backslash \left\{\alpha \right\}$ (see [1] Definition 1.1, [2] [3] Definition 1.15.1).

It is well known, that if B is all external elements of the semigroup $B_X\left(D\right)$ and $B^{\prime }$ is any generated set for the $B_X\left(D\right)$ , then $B\subseteq B^{\prime }$ (see [2] [3] Lemma 1.15.1).

2. Result

Let ${\Sigma }_8\left(X,5\right)$ be a class of all X-semilattices of unions, whose every element is isomorphic to an X-semilattice of unions $D=\left\{T_4,T_3,T_2,T_1,T_0\right\}$ , which satisfies the condition:

$\begin{array}{l}{T}_4\subset T_2\subset T_0,T_3\subset T_1\subset T_0,T_4\backslash T_3\ne \varnothing ,T_3\backslash T_4\ne \varnothing ,\\ T_2\backslash T_1\ne \varnothing ,T_1\backslash T_2\ne \varnothing ,T_2\backslash T_3\ne \varnothing ,T_3\backslash T_2\ne \varnothing ,\\ T_4\backslash T_1\ne \varnothing ,T_1\backslash T_4\ne \varnothing ,T_4\cup T_3=T_4\cup T_1=T_3\cup T_2=T_1\cup T_2=T_0.\end{array}$

(see Figure 1). It is easy to see that $\tilde{D}=\left\{T_4,T_3,T_2,T_1\right\}$ is irreducible generating set of the semilattice D.

Let $C\left(D\right)=\left\{P_0,P_1,P_2,P_3,P_4\right\}$ is a family of sets, where $\phi =\left(\begin{array}{l}{T}_0{T}_1{T}_2{T}_3{T}_4\\ P_0{P}_1{P}_2{P}_3{P}_4\end{array}\right)$ is a mapping of the semilattice D onto the family of sets $C\left(D\right)$ and $P_0,P_1,P_2,P_3,P_4$ are pairwise disjoint subsets of the set X. Then the formal equalities of the semilattice D have a form:

$\begin{array}{l}{T}_0=P_0\cup P_1\cup P_2\cup P_3\cup P_4,\\ T_1=P_0\cup P_2\cup P_3\cup P_4,\\ T_2=P_0\cup P_1\cup P_3\cup P_4,\\ T_3=P_0\cup P_2\cup P_4,\\ T_4=P_0\cup P_1\cup P_3.\end{array}$ (2.1)

Here the element $P_0$ is source of completeness and the elements $P_4,P_3,P_2,P_1$ are basis sources of the semilattice D. Therefore $\left|X\right|\ge 4$ since $\left|P_4\right|\ge 1$ , $\left|P_3\right|\ge 1$ , $\left|P_2\right|\ge 1$ , $\left|P_1\right|\ge 1$ (see Theorem 1.1).

From the formal Equalities (2.1) immediately follows

$\begin{array}{l}{P}_4=T_2\backslash T_4,P_3=\left(T_2\cap T_1\right)\backslash T_3,\\ P_2=T_3\backslash T_2=T_0\backslash T_2,P_1=T_4\backslash T_1,P_0=T_4\cap T_3.\end{array}$ (2.2)

2.1. Generating Sets of the Complete Semigroup of Binary Relations Defined by Semilattices of the Class ${\Sigma }_8\left(X,5\right)$ , When $T_4\cap T_3\ne \varnothing$

In the sequel, we denoted all semilattices $D=\left\{T_4,T_3,T_2,T_1,T_0\right\}$ of the class ${\Sigma }_8\left(X,5\right)$ by symbol ${\Sigma }_{8.0}\left(X,5\right)$ , for which $T_4\cap T_3\ne \varnothing$ . Of the last inequality from the formal Equalities (2.1) of a semilattise D follows that $T_4\cap T_3=P_0\ne \varnothing$ , i.e. $\left|X\right|\ge 5$ .

Figure 1. Diagram of the semilattice D.

We denoted by symbols ${\mathfrak{A}}_4,{\mathfrak{A}}_3,{\mathfrak{A}}_2,{\mathfrak{A}}_1$ the following sets:

$\begin{array}{l}{\mathfrak{A}}_4=\left\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\}\right\},\\ {\mathfrak{A}}_3=\left\{\left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\},\left\{T_4,T_2,T_0\right\},\left\{T_3,T_1,T_0\right\},\left\{T_2,T_1,T_0\right\}\right\},\\ {\mathfrak{A}}_2=\left\{\left\{T_4,T_2\right\},\left\{T_4,T_0\right\},\left\{T_3,T_1\right\},\left\{T_3,T_0\right\},\left\{T_2,T_0\right\}\left\{T_1,T_0\right\}\right\},\\ {\mathfrak{A}}_1=\left\{\left\{T_4\right\},\left\{T_3\right\},\left\{T_2\right\},\left\{T_1\right\},\left\{T_0\right\}\right\}.\end{array}$

Lemma 2.1.1. Let $D\in {\Sigma }_{8.0}\left(X,5\right)$ . Then the following statements are true:

a) Let $T_3,T_4\in V\left(D,\alpha \right)$ , then $\alpha$ is external element of the semigroup $B_X\left(D\right)$ ;

b) Let $Z\in \left\{T_2,T_1\right\}$ , $Z^{\prime }\in \left\{T_4,T_3\right\}$ . If $Z^{\prime }\not\subset Z$ and $Z,Z^{\prime }\in V\left(D,\alpha \right)$ , then $\alpha$ is external element of the semi­group $B_X\left(D\right)$ ;

c) Let $Z,Z^{\prime }\in \left\{T_2,T_1\right\}$ and $Z\ne Z^{\prime }$ . If $V\left(D,\alpha \right)=\left\{T_2,T_1,T_0\right\}$ , then $\alpha$ is external element of the semigroup $B_X\left(D\right)$ .

Proof. Let $\alpha =\delta \circ \beta$ for some $\delta ,\beta \in B_X\left(D\right)\backslash \left\{\alpha \right\}$ . If quasinormal representation of binary relation $\delta$ has a form

$\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),$

then

$\alpha =\delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)$ .(2.1.1)

From the formal Equalities (1) of the semilattice D we obtain that:

$\begin{array}{l}{T}_0\beta =P_0\beta \cup P_1\beta \cup P_2\beta \cup P_3\beta \cup P_4\beta ,\\ T_1\beta =P_0\beta \cup P_2\beta \cup P_3\beta \cup P_4\beta ,\\ T_2\beta =P_0\beta \cup P_1\beta \cup P_3\beta \cup P_4\beta ,\\ T_3\beta =P_0\beta \cup P_2\beta \cup P_4\beta ,\\ T_4\beta =P_0\beta \cup P_1\beta \cup P_3\beta .\end{array}$ (2.1.2)

where $P_k\beta \ne \varnothing$ for any $P_k\ne \varnothing$ $\left(k=0,1,2,3,4\right)$ and $\beta \in B_X\left(D\right)$ . Indeed, by preposition $P_k\ne \varnothing$ for any $k=0,1,2,3,4$ and $\beta \ne \varnothing$ since $\varnothing \notin D$ . Let $y\in P_k$ for some $y\in X$ . Then $y\in T_0$ , $\beta ={\alpha }_f$ for some $f:X\to D$ and ${\alpha }_f={\displaystyle \underset{x\in X}{\cup }\left(\left\{x\right\}\times f\left(x\right)\right)}\supseteq \left\{y\right\}\times f\left(y\right)$ , i.e. there exists an element $t\in f\left(y\right)$ for which $y{\alpha }_{f}t$ and $y\beta t$ . Of this and by definition of a set $P_k\beta$ we obtain that $t\in P_k\beta$ since $y\in P_k$ , $y\beta t$ . Thus, we have that $P_k\beta \ne \varnothing$ , i.e. $P_k\beta \in D$ for any $k=0,1,2,3,4$ .

Now, let $T_i\beta =Z$ and $T_j\beta =Z^{\prime }$ for some $0\le i\ne j\le 4$ and $Z\ne Z^{\prime }$ , $Z,Z^{\prime }\in \left\{T_4,T_3\right\}$ , then from the Equalities (2.2) follows that $Z=P_0\beta =Z^{\prime }$ since Z and $Z^{\prime }$ are minimal elements of the semilattice D. The equality $Z=Z^{\prime }$ contradicts the inequality $Z\ne Z^{\prime }$ .

The statement a) of the Lemma 2.1.1 is proved.

Let $T_i\beta =Z^{\prime }$ , where $Z^{\prime }\in \left\{T_4,T_3\right\}$ and $T_j\beta =Z$ , where $Z\in \left\{T_2,T_1\right\}$ for some $0\le i\ne j\le 4$ . If $0\le i\le 4$ , then from the formal equalities of a semilattice D we obtain that

$\begin{array}{l}{T}_0\beta =P_0\beta \cup P_1\beta \cup P_2\beta \cup P_3\beta \cup P_4\beta =P_0\beta =P_1\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime },\\ T_1\beta =P_0\beta \cup P_2\beta \cup P_3\beta \cup P_4\beta =P_0\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime },\\ T_2\beta =P_0\beta \cup P_1\beta \cup P_3\beta \cup P_4\beta =P_0\beta =P_1\beta =P_3\beta =P_4\beta =Z^{\prime },\\ T_3\beta =P_0\beta \cup P_2\beta \cup P_4\beta =P_0\beta =P_2\beta =P_4\beta =Z^{\prime },\\ T_4\beta =P_0\beta \cup P_1\beta \cup P_3\beta =P_0\beta =P_1\beta =P_3\beta =Z^{\prime }.\end{array}$

since $Z^{\prime }$ is minimal element of the semilattice D. Now, let $i\ne j$ .

1) If $T_0\beta =P_0\beta =P_1\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime }$ and $j=1,2,3,4$ , then we have

$Z=T_1\beta =T_2\beta =T_3\beta =T_4\beta =Z^{\prime }$ ,

which contradicts the inequality $Z\ne Z^{\prime }$ .

2) If $T_1\beta =P_0\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime }$ and $j=0,2,3,4$ , then we have

$Z=T_0\beta =T_2\beta =T_4\beta =Z^{\prime }\cup P_1\beta ,\text{where}{P}_1\beta \in D$

Last equalities are impossible, since $Z\ne Z^{\prime }\cup T$ for any $T\in D$ and $Z\ne Z^{\prime }$ by definition of a semilattice D.

3) If $T_2\beta =P_0\beta =P_1\beta =P_3\beta =P_4\beta =Z^{\prime }$ and $j=0,1,3,4$ , then we have

$Z=T_0\beta =T_2\beta =T_4\beta =Z^{\prime }\cup P_1\beta ,\text{where}{P}_1\beta \in D$

Last equalities are impossible since $Z\ne Z^{\prime }\cup T$ for any $T\in D$ and $Z\ne Z^{\prime }$ by definition of a semilattice D.

4) If $T_3\beta =P_0\beta =P_2\beta =P_4\beta =Z^{\prime }$ and $j=0,1,2,4$ , then we have

$\begin{array}{l}Z=T_0\beta =T_2\beta =T_4\beta =Z^{\prime }\cup P_1\beta \cup P_3\beta ,\\ \text{Z}=T_1\beta =Z^{\prime }\cup P_3\beta ,\text{where}{P}_1\beta ,P_3\beta \in D\end{array}$

Last equalities are impossible since $Z\ne Z^{\prime }\cup T\cup T^{\prime }$ and $Z\ne Z^{\prime }\cup T$ for any $T,T^{\prime }\in D$ , by definition of a semilattice D.

5) If $T_4\beta =P_0\beta =P_1\beta =P_3\beta =Z^{\prime }$ and $j=0,1,2,3$ , then we have

$\begin{array}{l}Z=T_0\beta =T_1\beta =T_3\beta =Z^{\prime }\cup P_2\beta \cup P_4\beta ,\\ Z=T_2\beta =Z^{\prime }\cup P_4\beta ,\text{where}{P}_2\beta ,P_4\beta \in D\end{array}$

Last equalities are impossible since $Z\ne Z^{\prime }\cup T\cup T^{\prime }$ and $Z\ne Z^{\prime }\cup T$ for any $T,T^{\prime }\in D$ , by definition of a semilattice D.

The statement b) of the Lemma 2.1.1 is proved.

Let $Z,Z^{\prime }\in \left\{T_2,T_1\right\}$ , $T_i\beta =Z$ , $T_j\beta =Z^{\prime }$ and $Z\ne Z^{\prime }$ . If $T_i\beta =Z$ where $0\le i\ne j\le 4$ , we consider the following cases:

6) $i=0,j=1,2,3,4$ . Then from the Equality (2.1.2) follows that $Z\subset Z^{\prime }$ , which contradicts the definition of a semilattice D;

7) $i=1,j=0,2,3,4$ .

If $i=1,j=0,3$ . Then from the Equality (2.1.2) follows that $Z^{\prime }\subset Z$ , or $Z\subset Z^{\prime }$ which contradicts the definition of a semilattice D;

If $i=1,j=2,4$ . Then from the Equality (1.4) follows that

$\{\begin{array}{l}{T}_1\beta =\left(P_0\beta \cup P_3\beta \cup P_4\beta \right)\cup P_2\beta ,\\ T_2\beta =\left(P_0\beta \cup P_3\beta \cup P_4\beta \right)\cup P_1\beta ,\end{array}$

where $P_0\beta \cup P_3\beta \cup P_4\beta ,P_2\beta ,P_1\beta \in D$ , i.e. there exists such elements $T,T^{\prime },T^{″}\in D$ , for which $Z=T\cup T^{\prime }$ and $Z^{\prime }=T\cup T^{″}$ . But such element $T\in D$ don’t exist by definition of a semilattice D.

8) $i=2,j=0,1,3,4$ .

If $i=2,j=0,4$ . Then from the Equality (2.1.2) follows that $Z^{\prime }\subset Z$ , or $Z\subset Z^{\prime }$ which contradicts the definition of a semilattice D;

If $i=2,j=1,3$ . In this case analogously for the case 7) we may prove that $Z=T\cup T^{\prime }$ and $Z^{\prime }=T\cup T^{″}$ . But such element $T\in D$ don’t exist by definition of a semilattice D.

9) $i=3,j=0,1,2,4$ .

If $i=3,j=0,1$ . Then from the Equality (2.1.2) follows that $Z^{\prime }\subset Z$ , which contradicts the definition of a semilattice D;

If $i=3,j=2,4$ . Then from the Equality (2.1.2) follows that

$\{\begin{array}{l}{T}_2\beta =P_0\beta \cup \left(P_2\beta \cup P_3\beta \cup P_4\beta \right),\\ T_2\beta =P_0\beta \cup \left(P_1\beta \cup P_3\beta \right),\end{array}$

where $P_0\beta ,P_2\beta \cup P_3\beta \cup P_4\beta ,P_1\beta \cup P_3\beta \in D$ , i.e. there exist such elements $T,T^{\prime },T^{″}\in D$ , for which $Z=T\cup T^{\prime }$ and $Z^{\prime }=T\cup T^{″}$ . But such element $T\in D$ don’t exist by definition of a semilattice D.

10) $i=4,j=0,1,2,3$ .

If $i=4,j=0,2$ . Then from the Equality (2.1.2) follows that $Z\subset Z^{\prime }$ which contradicts the definition of a semilattice D;

If $i=4,j=1,3$ . Then from the Equality (2.1.2) follows that

$\{\begin{array}{l}{T}_1\beta =P_0\beta \cup \left(P_2\beta \cup P_3\beta \cup P_4\beta \right),\\ T_3\beta =P_0\beta \cup \left(P_2\beta \cup P_4\beta \right),\end{array}$

where $P_0\beta ,P_2\beta \cup P_3\beta \cup P_4\beta ,P_2\beta \cup P_4\beta \in D$ , i.e. there exist such elements $T,T^{\prime },T^{″}\in D$ , for which $Z=T\cup T^{\prime }$ and $Z^{\prime }=T\cup T^{″}$ . But such element $T\in D$ do not exist by definition of a semilattice D.

The statement c) of the Lemma 2.1.1 is proved.

Lemma 2.1.1 is proved.

Let $D\in {\Sigma }_{8.0}\left(X,5\right)$ . By symbols ${\mathfrak{A}}_0$ , $B\left({\mathfrak{A}}_0\right)$ and $B_0$ we denoted the following sets:

$\begin{array}{l}{\mathfrak{A}}_0=\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\},\\ \left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\},\left\{T_2,T_1,T_0\right\}\},\\ B\left({\mathfrak{A}}_0\right)=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_0\right\};B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}.\end{array}$

Remark, that the sets $B_0$ and $B\left({\mathfrak{A}}_0\right)$ are external elements for the semigroup $B_X\left(D\right)$ .

Lemma 2.1.2. Let $D\in {\Sigma }_{8.0}\left(X,5\right)$ . Then the following statements are true:

a) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_4^{\alpha },Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

b) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_3^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

Proof. 1). Let quasinormal representation of binary relations $\delta$ and $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_4\times T_4\right)\cup \left(\left(T_2\backslash T_4\right)\times T_2\right)\cup \left(\left(T_0\backslash T_2\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ ,

$\begin{array}{l}{T}_4\cup \left(T_2\backslash T_4\right)\cup \left(T_0\backslash T_2\right)\cup \left(X\backslash T_0\right)\\ =\left(P_0\cup P_1\cup P_3\right)\cup P_4\cup P_2\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X,\end{array}$

(see Equalities (2.1) and (2.2)), then $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_4,T_2\beta =\left(P_0\cup P_1\cup P_3\cup P_4\right)\beta =T_4\cup T_2=T_2,\\ T_1\beta =\left(P_0\cup P_2\cup P_3\cup P_4\right)\beta =T_4\cup T_1=T_0,T_0\beta =T_0.\\ \alpha =\delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_0\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_4^{\alpha }$ , $Y_2^{\delta }=Y_2^{\alpha }$ and $Y_1^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement a) of the lemma 2.1.2 is proved.

2) Let quasinormal representation of binary relations $\delta$ and $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_3\times T_3\right)\cup \left(\left(T_0\backslash T_1\right)\times T_2\right)\cup \left(\left(T_1\backslash T_3\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ ,

$\begin{array}{l}{T}_3\cup \left(T_0\backslash T_1\right)\cup \left(T_1\backslash T_3\right)\cup \left(X\backslash T_0\right)\\ =\left(P_0\cup P_2\cup P_4\right)\cup P_1\cup P_3\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X,\end{array}$

(see Equalities (2.1) and (2.2)), then $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_3,T_2\beta =\left(P_0\cup P_1\cup P_3\cup P_4\right)\beta =T_3\cup T_2\cup T_1=T_0,\\ T_1\beta =\left(P_0\cup P_2\cup P_3\cup P_4\right)\beta =T_3\cup T_1=T_0,T_0\beta =T_0.\\ \alpha =\delta \circ \beta =\left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_0\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_3^{\alpha }$ , $Y_1^{\delta }=Y_1^{\alpha }$ and $Y_2^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_2^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement b) of the lemma 2.1.2 is proved.

Lemma 2.1.2 is proved.

Lemma 2.1.3. Let $D\in {\Sigma }_{8.0}\left(X,5\right)$ . Then the following statements are true:

a) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_2^{\alpha }\times T_2\right),$

where $Y_4^{\alpha },Y_2^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

b) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_4^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

c) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_1^{\alpha }\times T_1\right),$

where $Y_3^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

d) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_3^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

e) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

f) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

g) If quasinormal representation of a binary relation $\alpha$ has a form $\alpha =X\times T_2$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

h) If quasinormal representation of a binary relation $\alpha$ has a form $\alpha =X\times T_1$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

i) If quasinormal representation of a binary relation $\alpha$ has a form $\alpha =X\times T_0$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ .

Proof. 1) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_4\times T_4\right)\cup \left(\left(T_0\backslash T_4\right)\times T_2\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_1^{\delta }\notin \{\varnothing \}$ .

$\begin{array}{l}{T}_4\cup \left(T_0\backslash T_4\right)\cup \left(X\backslash T_0\right)\\ =\left(P_0\cup P_1\cup P_3\right)\cup \left(P_2\cup P_4\right)\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X.\end{array}$

Then from the statement a) of the Lemma 2.1.2 follows that $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ , $\delta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_4,T_1\beta =T_4\cup T_2=T_2,T_0\beta =T_2.\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_2\right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times T_2\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_4^{\alpha }$ , $Y_1^{\delta }\cup Y_0^{\delta }=Y_2^{\alpha }$ . Last equalities are possible since $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement a) of the lemma 2.1.3 is proved.

2) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_4\times T_4\right)\cup \left(\left(T_0\backslash T_4\right)\times T_3\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_1^{\delta }\notin \{\varnothing \}$ .

$\begin{array}{l}{T}_4\cup \left(T_0\backslash T_4\right)\cup \left(X\backslash T_0\right)\\ =\left(P_0\cup P_1\cup P_3\right)\cup \left(P_2\cup P_4\right)\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X.\end{array}$

Then from $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_4,T_1\beta =T_4\cup T_3=T_0,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_0\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_4^{\alpha }$ , $Y_1^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement b) of the lemma 2.1.3 is proved.

3) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_3\times T_3\right)\cup \left(\left(T_0\backslash T_3\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_2^{\delta }\notin \{\varnothing \}$ .

$\begin{array}{l}{T}_3\cup \left(T_0\backslash T_3\right)\cup \left(X\backslash T_0\right)\\ =\left(P_0\cup P_2\cup P_4\right)\cup \left(P_1\cup P_3\right)\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X.\end{array}$

Then from the statement b) of the Lemma 2.1.2 follows that $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ , $\delta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_3\beta =T_3,T_2\beta =T_3\cup T_1=T_1,T_0\beta =T_1.\\ \delta \circ \beta =\left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_1\right)\\ =\left(Y_3^{\delta }\times Z_3\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times T_1\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_3^{\alpha }$ , $Y_2^{\delta }\cup Y_0^{\delta }=Y_1^{\alpha }$ . Last equalities are possible since $\left|Y_2^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement c) of the lemma 2.1.3 is proved.

4) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_3\times T_3\right)\cup \left(\left(T_0\backslash T_3\right)\times T_2\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_3^{\delta },Y_2^{\delta }\notin \{\varnothing \}$ . Then $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_3\beta =T_3,T_2\beta =T_3\cup T_2=T_0,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_0\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_3^{\alpha }$ , $Y_2^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_2^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement d) of the lemma 2.1.3 is proved.

5) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(\left(\left(T_2\cap T_1\right)\backslash T_3\right)\times T_4\right)\cup \left(\left(T_2\backslash T_1\right)\times T_2\right)\cup \left(\left(X\backslash T_4\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$\begin{array}{l}\left(\left(T_2\cap T_1\right)\backslash T_3\right)\cup \left(T_2\backslash T_1\right)\cup \left(X\backslash T_4\right)\\ =\left(P_0\cup P_3\right)\cup P_1\cup \left(X\backslash T_4\right)=T_4\cup \left(X\backslash T_4\right)=X.\end{array}$

(See Equalities (2.1) and (2.2)). Then from the statement b) of the Lemma 2.1.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement a) of the Lemma 2.1.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =\left(P_0\cup P_1\cup P_3\right)\beta =T_4\cup T_2=T_2,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_4^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_2^{\alpha }$ , $Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_4^{\delta }\right|\ge 1$ $\left|Y_0^{\delta }\right|\ge 1$ .

The statement e) of the lemma 2.1.3 is proved.

6) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(\left(\left(T_2\cap T_1\right)\backslash T_4\right)\times T_3\right)\cup \left(\left(T_1\backslash T_2\right)\times T_1\right)\cup \left(\left(X\backslash T_3\right)\times T_0\right),\end{array}$

where $Y_3^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$\begin{array}{l}\left(\left(T_2\cap T_1\right)\backslash T_4\right)\cup \left(T_1\backslash T_2\right)\cup \left(X\backslash T_3\right)\\ =\left(P_0\cup P_4\right)\cup P_2\cup \left(X\backslash T_3\right)=T_3\cup \left(X\backslash T_3\right)=X.\end{array}$

(See Equalities (2.1) and (2.2)). Then from the statement d) of the Lemma 2.1.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement b) of the Lemma 2.1.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_3\beta =\left(P_0\cup P_2\cup P_4\right)\beta =T_3\cup T_1=T_1,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_3^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_1^{\alpha }$ , $Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_4^{\delta }\right|\ge 1$ $\left|Y_0^{\delta }\right|\ge 1$ .

The statement e) of the lemma 2.1.3 is proved.

7) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_1\times T_4\right)\cup \left(\left(T_2\backslash T_1\right)\times T_2\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_2^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$\begin{array}{l}{T}_1\cup \left(T_2\backslash T_1\right)\cup \left(X\backslash T_0\right)\\ =\left(P_0\cup P_2\cup P_3\cup P_4\right)\cup P_1\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X\end{array}$

(see Equalities (2.1) and (2.2)). Then from the statement e) of the Lemma 2.1.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement a) of the Lemma 2.1.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_2\beta =T_4\cup T_2=T_2,T_0\beta =T_2\\ \delta \circ \beta =\left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_2\right)=X\times T_2=\alpha ,\end{array}$

since representation of a binary relation $\delta$ is quasinormal.

The statement g) of the lemma 2.1.3 is proved.

8) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_2\times T_3\right)\cup \left(\left(T_1\backslash T_2\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_1^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$\begin{array}{l}{T}_2\cup \left(T_1\backslash T_2\right)\cup \left(X\backslash T_0\right)\\ =\left(P_0\cup P_1\cup P_3\cup P_4\right)\cup P_2\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X\end{array}$

(see Equalities (2.1) and (2.2)). Then from the statement f) of the Lemma 2.1.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement b) of the Lemma 2.1.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_1\beta =T_3\cup T_1=T_1,T_0\beta =T_1\\ \delta \circ \beta =\left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_1\right)=X\times T_1=\alpha ,\end{array}$

since representation of a binary relation $\delta$ is quasinormal.

The statement h) of the lemma 2.1.3 is proved.

9) Let quasinormal representation of a binary relation $\delta$ has a form

$\delta =\left(T_4\times T_1\right)\cup \left(\left(X\backslash T_4\right)\times T_0\right),$

then

$\begin{array}{l}{T}_1\delta =\left(P_0\cup P_2\cup P_3\cup P_4\right)\delta =T_4\cup T_0=T_0,T_0\delta =T_0\\ \delta \circ \delta =\left(T_4\times T_1\delta \right)\cup \left(\left(X\backslash T_4\right)\times T_0\delta \right)=\left(T_4\times T_0\right)\cup \left(\left(X\backslash T_4\right)\times T_0\right)=X\backslash T_0=\alpha \end{array}$

since representation of a binary relation $\delta$ is quasinormal.

The statement i) of the lemma 2.1.3 is proved.

Lemma 2.1.3 is proved.

Lemma 2..4. Let $D\in {\Sigma }_{8.0}\left(X,5\right)$ . Then the following statements are true:

a) If $\left|X\backslash T_0\right|\ge 1$ and $Z\in \left\{T_4,T_3\right\}$ , then binary relation $\alpha =X\times Z$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

b) If $X=T_0$ and $Z\in \left\{T_4,T_3\right\}$ , then binary relation $\alpha =X\times Z$ is external element for the semigroup $B_X\left(D\right)$ .

Proof. 1) Let quasinormal representation of a binary relation $\delta$ has a form

$\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_3^{\delta }\times T_3\right)\cup \left(Y_0^{\delta }\times T_0\right)$ ,

where $Y_4^{\delta },Y_3^{\delta }\notin \{\varnothing \}$ , then $\delta \in B\left({\mathfrak{A}}_0\right)\backslash \left\{\alpha \right\}$ . If quasinormal representation of a binary relation $\beta$ has a form $\beta =\left(T_0\times Z\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash T_0}{\cup }\left(\left\{t^{\prime }\right\}\times f\left(t^{\prime }\right)\right)}$ , where f is any mapping of the set $X\backslash T_0$ in the set $\left\{T_4,T_3\right\}\backslash \left\{Z\right\}$ . It is easy to see, that $\beta \ne \alpha$ and two elements of the set $\left\{T_4,T_3\right\}$ belong to the semilattice $V\left(D,\beta \right)$ , i.e. $\delta \in B\left({\mathfrak{A}}_0\right)\backslash \left\{\alpha \right\}$ . In this case we have

$\begin{array}{l}{T}_4\beta =T_3\beta =T_0\beta =Z;\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times Z\right)\cup \left(Y_3^{\delta }\times Z\right)\cup \left(Y_0^{\delta }\times Z\right)\\ =\left(\left(Y_4^{\delta }\cup Y_3^{\delta }\cup Y_0^{\delta }\right)\times Z\right)=X\times Z=\alpha ,\end{array}$

since the representation of a binary relation $\delta$ is quasinormal. Thus, element $\alpha$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ .

The statement a) of the lemma 2.1.4 is proved.

2) Let $X=T_0$ , $\alpha =X\times Z$ , for some $Z\in \left\{T_4,T_3\right\}$ and $\alpha =\delta \circ \beta$ for some $\delta ,\beta \in B_X\left(D\right)\backslash \left\{\alpha \right\}$ . Then from the equality (2.1.1) and (2.1.2) we obtain that

$T_4\beta =T_3\beta =T_2\beta =T_1\beta =T_0\beta =Z,P_0\beta =P_1\beta =P_2\beta =P_3\beta =P_4\beta =Z$ ,

since Z is mini­mal element of the semilattice D.

Now, let subquasi­nor­mal representations $\bar{\beta }$ of a binary relation $\beta$ has a form

$\bar{\beta }=\left(\left(P_0\cup P_1\cup P_2\cup P_3\cup P_4\right)\times Z\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash T_0}{\cup }\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)}$ ,

where ${\bar{\beta }}_1=\left(\begin{array}{l}{P}_0{P}_1{P}_2{P}_3{P}_4\\ ZZZZZ\end{array}\right)$ is normal mapping. But complement mapping ${\bar{\beta }}_2$ is empty, since $X\backslash T_0=\varnothing$ , i.e. in the given case, subquasinormal representation $\bar{\beta }$ of a binary relation $\beta$ is defined uniquely. So, we have that $\beta =\bar{\beta }=X\times Z=\alpha$ , which contradicts the condition $\beta \notin B_X\left(D\right)\backslash \left\{\alpha \right\}$ .

Therefore, if $X=T_0$ and $\alpha =X\times Z$ , for some $Z\in \left\{T_4,T_3\right\}$ , then $\alpha$ is external element of the semigroup $B_X\left(D\right)$ .

The statement b) of the lemma 2.1.4 is proved.

Lemma 2.1.4 is proved.

Theorem 2.1.1. Let $D\in {\Sigma }_{8.0}\left(X,5\right)$ and

$\begin{array}{l}{\mathfrak{A}}_0=\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\},\\ \left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\},\left\{T_2,T_1,T_0\right\}\},\\ B\left({\mathfrak{A}}_0\right)=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_0\right\};B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}.\end{array}$

Then the following statements are true:

a) If $\left|X\backslash T_0\right|\ge 1$ , then $S_0=B_0\cup B\left({\mathfrak{A}}_0\right)$ is irreducible generating set for the semigroup $B_X\left(D\right)$ ;

b) If $X=T_0$ , then $S_1=B_0\cup B\left({\mathfrak{A}}_0\right)\cup \left\{X\times T_4,X\times T_3\right\}$ is irreducible genera­ting set for the semigroup $B_X\left(D\right)$ .

Proof. Let $D\in {\Sigma }_{8.0}\left(X,5\right)$ and $\left|X\backslash T_0\right|\ge 1$ . First, we proved that every element of the semigroup $B_X\left(D\right)$ is generating by elements of the set $S_0$ . Indeed, let $\alpha$ be arbitrary element of the semigroup $B_X\left(D\right)$ . Then quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha }\cup Y_3^{\alpha }\cup Y_2^{\alpha }\cup Y_1^{\alpha }\cup Y_0^{\alpha }=X$ and $Y_i^{\alpha }\cap Y_j^{\alpha }=\varnothing$ $\left(0\le i\ne j\le 4\right)$ . For the $\left|V\left(X^{\ast },\alpha \right)\right|$ we consider the following cases:

1) $\left|V\left(X^{\ast },\alpha \right)\right|=5$ . Then $\alpha \in B_0$ and $B_0\subset S_0$ by definition of a set $S_0$ .

2) $\left|V\left(X^{\ast },\alpha \right)\right|=4$ . Then

$V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_4=\left\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\}\right\}\subset {\mathfrak{A}}_0$

i.e. $\alpha \in B\left({\mathfrak{A}}_0\right)$ and $B\left({\mathfrak{A}}_0\right)\subset S_0$ by definition of a set $S_0$ .

3) $\left|V\left(X^{\ast },\alpha \right)\right|=3$ . Then we have

$\begin{array}{l}V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_3=\{\left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\},\left\{T_4,T_2,T_0\right\},\\ \left\{T_3,T_1,T_0\right\},\left\{T_2,T_1,T_0\right\}\}.\end{array}$

By definition of a set ${\mathfrak{A}}_0$ we have $\left\{\left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\},\left\{T_2,T_1,T_0\right\}\right\}\subset {\mathfrak{A}}_0$ , i.e. in this case $\alpha \in B\left({\mathfrak{A}}_0\right)$ and $B\left({\mathfrak{A}}_0\right)\subset S_0$ by definition of a set $S_0$ .

If $V\left(X^{\ast },\alpha \right)\in \left\{\left\{T_4,T_2,T_0\right\},\left\{T_3,T_1,T_0\right\}\right\}$ , then from the statement a) and b) of the Lemma 2.1.2 element $\alpha$ is generating by elements $B\left({\mathfrak{A}}_0\right)$ and $B\left({\mathfrak{A}}_0\right)\subset S_0$ by definition of a set $S_0$ .

4) $\left|V\left(X^{\ast },\alpha \right)\right|=2$ . Then we have

$V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_2=\left\{\left\{T_4,T_2\right\},\left\{T_4,T_0\right\},\left\{T_3,T_1\right\},\left\{T_3,T_0\right\},\left\{T_2,T_0\right\}\left\{T_1,T_0\right\}\right\}$ .

Then from the statement a)-f) of the Lemma 2.1.3 element $\alpha$ is generating by elements $B\left({\mathfrak{A}}_0\right)$ and $B\left({\mathfrak{A}}_0\right)\subset S_0$ by definition of a set $S_0$ .

5) $\left|V\left(X^{\ast },\alpha \right)\right|=1$ . Then we have $V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_1=\left\{\left\{T_4\right\},\left\{T_3\right\},\left\{T_2\right\},\left\{T_1\right\},\left\{T_0\right\}\right\}$ .

If $V\left(X^{\ast },\alpha \right)\in \left\{\left\{T_2\right\},\left\{T_1\right\},\left\{T_0\right\}\right\}$ , then from the statements g), h) and i) of the Lemma 2.1.3 element $\alpha$ is generating by elements $B\left({\mathfrak{A}}_0\right)$ and $B\left({\mathfrak{A}}_0\right)\subset S_0$ by definition of a set $S_0$ .

If $V\left(X^{\ast },\alpha \right)\in \left\{\left\{T_4\right\},\left\{T_3\right\}\right\}$ , then from the statement a) of the Lemma 2.1.4 element $\alpha$ is generating by elements $B\left({\mathfrak{A}}_0\right)$ and $B\left({\mathfrak{A}}_0\right)\subset S_0$ by definition of a set $S_0$ .

Thus, we have that $S_0$ is generating set for the semigroup $B_X\left(D\right)$ .

If $\left|X\backslash T_0\right|\ge 1$ , then the set $S_0$ is irreducible generating set for the semigroup $B_X\left(D\right)$ since $S_0$ is a set external elements of the semigroup $B_X\left(D\right)$ .

The statement a) of the Theorem 2.1.1 is proved.

Now, let $D\in {\Sigma }_{8.0}\left(X,5\right)$ and $X=\stackrel{⌣}{D}$ . First, we proved that every element of the semigroup $B_X\left(D\right)$ is generating by elements of the set $S_1$ . The cases 1), 2), 3) and 4) are proved analogously of the cases 1), 2), 3) and 4) given above and consider case, when

$V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_1=\left\{\left\{T_4\right\},\left\{T_3\right\},\left\{T_2\right\},\left\{T_1\right\},\left\{T_0\right\}\right\}$ .

If $V\left(X^{\ast },\alpha \right)\in \left\{\left\{T_2\right\},\left\{T_1\right\},\left\{T_0\right\}\right\}$ , then from the statements g), h) and i) of the Lemma 2.1.3 element $\alpha$ is generating by elements $B\left({\mathfrak{A}}_0\right)$ and $B\left({\mathfrak{A}}_0\right)\subset S_1$ by definition of a set $S_1$ .

If $V\left(X^{\ast },\alpha \right)\in \left\{\left\{T_4\right\},\left\{T_3\right\}\right\}$ , then $\alpha \in S_1$ by definition of a set $S_1$ .

Thus, we have that $S_1$ is generating set for the semigroup $B_X\left(D\right)$ .

If $X=T_0$ , then the set $S_1$ is irreducible generating set for the semigroup $B_X\left(D\right)$ since $S_1$ is a set external elements of the semigroup $B_X\left(D\right)$ .

The statement b) of the Theorem 2.1.1 is proved.

Theorem 2.1.1 is proved.

Theorem 2.1.2. Let $n\ge 6$ , $D=\left\{T_4,T_3,T_2,T_1,T_0\right\}\in {\Sigma }_{8.0}\left(X,5\right)$ and

$\begin{array}{l}{\mathfrak{A}}_0=\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\},\\ \left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\},\left\{T_2,T_1,T_0\right\}\},\\ B\left({\mathfrak{A}}_0\right)=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_0\right\};B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}.\end{array}$

Then the following statements are true:

a) If $\left|X\backslash T_0\right|\ge 1$ , then the number $\left|S_0\right|$ elements of the set $S_0=B_0\cup B\left({\mathfrak{A}}_0\right)$ is equal to

$\left|S_0\right|=5^n-2\cdot 3^n+1$ .

b) If $X=T_0$ , then the number $\left|S_1\right|$ elements of the set $S_1=B_0\cup B\left({\mathfrak{A}}_0\right)\cup \left\{X\times T_4,X\times T_3\right\}$ is equal to

$\left|S_1\right|=5^n-2\cdot 3^n+3$ .

Proof. Let number of a set X is equal to $n\ge 6$ , i.e. $\left|X\right|=n\ge 6$ . Let $S_n=\left\{{\phi }_1,{\phi }_2,\cdots ,{\phi }_{n!}\right\}$ is a group all one to one mapping of a set $M=\left\{1,2,\cdots ,n\right\}$ on the set M and ${\phi }_{i_1},{\phi }_{i_2},\cdots ,{\phi }_{i_m}$ $\left(m\le n\right)$ are arbitrary elements of the group $S_n$ , $Y_{{\phi }_1},Y_{{\phi }_2},\cdots ,Y_{{\phi }_m}$ are arbitrary partitioning of a set X. By symbol $k_n^m$ we denote the number elements of a set $\left\{Y_{{\phi }_1},Y_{{\phi }_2},\cdots ,Y_{{\phi }_m}\right\}$ . It is well known, that

$k_n^m={\displaystyle \underset{i=1}{\overset{m}{\sum }}\frac{{\left(-1\right)}^{m+i}}{\left(i-1\right)!\cdot \left(m-i\right)!}\cdot i^{n-1}}$ .

If $m=2,3,4,5$ , then we have

$\begin{array}{l}{k}_n^2=2^{n-1}-1,k_n^3=\frac{1}{2}\cdot 3^{n-1}-2^{n-1}+\frac{1}{2},k_n^4=\frac{1}{6}\cdot 4^{n-1}-\frac{1}{2}\cdot 3^{n-1}+\frac{1}{2}\cdot 2^{n-1}-\frac{1}{6},\\ k_n^5=\frac{1}{24}\cdot 5^{n-1}-\frac{1}{6}\cdot 4^{n-1}+\frac{1}{4}\cdot 3^{n-1}-\frac{1}{6}\cdot 2^{n-1}+\frac{1}{24}.\end{array}$

If $Y_{{\phi }_1},Y_{{\phi }_2}$ are any two elements partitioning of a set X and $\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)$ , where $Z_1,Z_2\in D$ and $Z_1\ne Z_2$ . Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$2\cdot k_n^2=2^n-2$ . (2.1.3)

If $Y_{{\phi }_1},Y_{{\phi }_2},Y_{{\phi }_3}$ are any tree elements partitioning of a set X and

$\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)\cup \left(Y_{{\phi }_3}\times Z_3\right)$ ,

where $Z_1,Z_2,Z_3$ are pairwise different elements of a given semilattice D. Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$6\cdot k_n^3=3^n-3\cdot 2^n+3$ . (2.1.4)

If $Y_{{\phi }_1},Y_{{\phi }_2},Y_{{\phi }_3},Y_{{\phi }_4}$ are any four elements partitioning of a set X and

$\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)\cup \left(Y_{{\phi }_3}\times Z_3\right)\cup \left(Y_{{\phi }_4}\times Z_4\right)$ ,

where $Z_1,Z_2,Z_3,Z_4$ are pairwise different elements of a given semilattice D. Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$24\cdot k_n^4=4^n-4\cdot 3^n+3\cdot 2^{n+1}-4$ . (2.1.5)

If $Y_{{\phi }_1},Y_{{\phi }_2},Y_{{\phi }_3},Y_{{\phi }_4},Y_{{\phi }_5}$ are any four elements partitioning of a set X and

$\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)\cup \left(Y_{{\phi }_3}\times Z_3\right)\cup \left(Y_{{\phi }_4}\times Z_4\right)\cup \left(Y_{{\phi }_5}\times Z_5\right)$ ,

where $Z_1,Z_2,Z_3,Z_4,Z_5$ are pairwise different elements of a given semilattice D. Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$120\cdot k_n^5=5^n-5\cdot 4^n+10\cdot 3^n-10\cdot 2^n+5.$ (2.1.6)

If $\alpha \in B_0$ , then quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or a system $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }$ are partitioning of the set X.

If the system $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }$ , or a system $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }$ are partitioning of the set X. Of this and from the equalities (2.1.4), (2.1.5) and (2.1.6) follows that

$\begin{array}{c}\left|B_0\right|=\left(5^n-5\cdot 4^n+10\cdot 3^n-10\cdot 2^n+5\right)+\left(4^n-4\cdot 3^n+6\cdot 2^n-4\right)\\ =5^n-4\cdot 4^n+6\cdot 3^n-4\cdot 2^n+1.\end{array}$

If $\alpha \in B\left({\mathfrak{A}}_0\right)$ , then by definition of a set $B\left({\mathfrak{A}}_0\right)$ the quasinormal representation of a binary relation $\alpha$ has a form:

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_3^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_3^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_3^{\alpha },Y_2^{\alpha }\notin \{\varnothing \}$ , or $Y_3^{\alpha },Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_2^{\alpha },Y_1^{\alpha }\in \{\varnothing \}$ , or $Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\in \{\varnothing \}$ are partitioning of the set X respectively.

Of this and from the equality (2.1.3), (2.1.4) and (2.1.5) follows that

$\begin{array}{c}\left|B\left({\mathfrak{A}}_0\right)\right|=4\cdot \left(2^n-2\right)+8\cdot \left(3^n-3\cdot 2^n+3\right)+4\cdot \left(4^n-4\cdot 3^n+6\cdot 2^n-4\right)\\ =4\cdot 4^n-8\cdot 3^n+4\cdot 2^n.\end{array}$

So, we have

$\begin{array}{l}\left|S_0\right|=\left|B_0\cup B\left({\mathfrak{A}}_0\right)\right|=\left(5^n-4\cdot 4^n+6\cdot 3^n-4\cdot 2^n+1\right)+\left(4\cdot 4^n-8\cdot 3^n+4\cdot 2^n\right)\\ =5^n-2\cdot 3^n+1,\\ \left|S_1\right|=\left|B_0\cup B\left({\mathfrak{A}}_0\right)\cup \left\{X\times T_4,X\times T_3\right\}\right|=5^n-2\cdot 3^n+3\end{array}$

Since

$B_0\cap B\left({\mathfrak{A}}_0\right)=B_0\cap \left\{X\times T_4,X\times T_3,X\times T_2\right\}=B\left({\mathfrak{A}}_0\right)\cap \left\{X\times T_4,X\times T_3,X\times T_2\right\}=\varnothing$ .

Theorem 2.1.2 is proved.

2.2. Generating Sets of the Complete Semigroup of Binary Relations Defined by Semilattices of the Class ${\Sigma }_8\left(X,5\right)$ , When $T_4\cap T_3=\varnothing$

In the sequel, we denoted all semilattices $D=\left\{T_4,T_3,T_2,T_1,T_0\right\}$ of the class ${\Sigma }_8\left(X,5\right)$ by symbol ${\Sigma }_{8.1}\left(X,5\right)$ for which $T_4\cap T_3=\varnothing$ . Of the last equality from the formal equalities of a semilattise D follows that $T_4\cap T_3=P_0=\varnothing$ , i.e. $\left|X\right|\ge 4$ since $P_4\ne \varnothing$ , $P_3\ne \varnothing$ , $P_2\ne \varnothing$ , $P_1\ne \varnothing$ .

In this case, the formal equalities of the semilattice D have a form:

$\begin{array}{l}{T}_0=P_1\cup P_2\cup P_3\cup P_4,\\ T_1=P_2\cup P_3\cup P_4,\\ T_2=P_1\cup P_3\cup P_4,\\ T_3=P_2\cup P_4,\\ T_4=P_1\cup P_3.\end{array}$ (2.2.1)

From the formal equalities of the semilattise D immediately follows, that:

$P_4=T_2\backslash T_4,P_3=T_1\backslash T_3,P_2=T_1\backslash T_2,P_1=T_2\backslash T_1$ . (2.2.2)

In this case we suppose that $D\in {\Sigma }_{8.1}\left(X,5\right)$ .

By symbols ${\mathfrak{A}}_4,{\mathfrak{A}}_3,{\mathfrak{A}}_2$ and ${\mathfrak{A}}_1$ we denoted the follo­wing sets:

$\begin{array}{l}{\mathfrak{A}}_4=\left\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\}\right\},\\ {\mathfrak{A}}_3=\left\{\left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\},\left\{T_4,T_2,T_0\right\},\left\{T_3,T_1,T_0\right\},\left\{T_2,T_1,T_0\right\}\right\},\\ {\mathfrak{A}}_2=\left\{\left\{T_4,T_2\right\},\left\{T_4,T_0\right\},\left\{T_3,T_1\right\},\left\{T_3,T_0\right\},\left\{T_2,T_0\right\}\left\{T_1,T_0\right\}\right\},\\ {\mathfrak{A}}_1=\left\{\left\{T_4\right\},\left\{T_3\right\},\left\{T_2\right\},\left\{T_1\right\},\left\{T_0\right\}\right\}.\end{array}$

Lemma 2.2.1. Let $D\in {\Sigma }_{8.1}\left(X,5\right)$ . Then the following statements are true:

a) Let $Z,Z^{\prime }\in \left\{T_4,T_3,T_2\right\}$ , $Z\ne Z^{\prime }$ . If $Z,Z^{\prime }\in V\left(D,\alpha \right)$ , then $\alpha$ is external element of the semi­group $B_X\left(D\right)$ ;

b) Let $Z\in \left\{T_2,T_1\right\}$ , $Z^{\prime }\in \left\{T_4,T_3\right\}$ . If $Z\not\subset Z^{\prime }$ and $Z,Z^{\prime }\in V\left(D,\alpha \right)$ , then $\alpha$ is external element of the semi­group $B_X\left(D\right)$ .

Proof. Let $\alpha =\delta \circ \beta$ for some $\delta ,\beta \in B_X\left(D\right)\backslash \left\{\alpha \right\}$ . If quasinormal representation of binary relation $\delta$ has a form

$\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),$

then

$\alpha =\delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)$ .(2.2.3)

From the formal equalities (2.2.1) of the semilattice D we obtain that:

$\begin{array}{l}{T}_0\beta =P_1\beta \cup P_2\beta \cup P_3\beta \cup P_4\beta ,\\ T_1\beta =P_2\beta \cup P_3\beta \cup P_4\beta ,\\ T_2\beta =P_1\beta \cup P_3\beta \cup P_4\beta ,\\ T_3\beta =P_2\beta \cup P_4\beta ,\\ T_4\beta =P_1\beta \cup P_3\beta ,\end{array}$ (2.2.4)

where $P_i\beta \ne \varnothing$ for any $P_i\ne \varnothing$ $\left(i=1,2,3,4\right)$ and $\beta \in B_X\left(D\right)$ . Indeed, by preposition $P_i\ne \varnothing$ for any $i=1,2,3,4$ and $\beta \ne \varnothing$ since $\varnothing \notin D$ . Let $y\in P_i$ for some $y\in X$ , then $y\in \stackrel{⌣}{D}$ , $\beta ={\alpha }_f$ for some $f:X\to D$ and ${\alpha }_f={\displaystyle \underset{x\in X}{\cup }\left(\left\{x\right\}\times f\left(x\right)\right)}\supseteq \left\{y\right\}\times f\left(y\right)$ , i.e. there exists an element $z\in f\left(y\right)$ for which $y{\alpha }_{f}z$ and $y\beta z$ . Of this and by definition of a set $P_i\beta$ we obtain that $z\in P_i\beta$ since $y\in P_i$ , $y\beta z$ . Thus, we have $P_i\beta \ne \varnothing$ , i.e. $P_i\beta \in D$ for any $i=1,2,3,4$ .

Now, let $T_i\beta =Z$ and $T_j\beta =Z^{\prime }$ for some $0\le i\ne j\le 4$ and $Z\ne Z^{\prime }$ , $Z,Z^{\prime }\in \left\{T_4,T_3\right\}$ , then from the Equalities (2.2.4) follows that $Z=P_0\beta =Z^{\prime }$ since Z and $Z^{\prime }$ are minimal elements of the semilattice D. The equality $Z=Z^{\prime }$ contradicts the inequality $Z\ne Z^{\prime }$ .

The statement a) of the Lemma 2.2.1 is proved.

Let $T_i\beta =Z^{\prime }$ , where $Z^{\prime }\in \left\{T_4,T_3\right\}$ and $T_j\beta =Z$ , $Z\in \left\{T_2,T_1\right\}$ for some $0\le i\ne j\le 4$ . If $0\le i\le 4$ , then from the formal equalities of a semilattice D we obtain that

$\begin{array}{l}{T}_0\beta =P_1\beta \cup P_2\beta \cup P_3\beta \cup P_4\beta =P_1\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime },\\ T_1\beta =P_2\beta \cup P_3\beta \cup P_4\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime },\\ T_2\beta =P_1\beta \cup P_3\beta \cup P_4\beta =P_1\beta =P_3\beta =P_4\beta =Z^{\prime },\\ T_3\beta =P_2\beta \cup P_4\beta =P_2\beta =P_4\beta =Z^{\prime },\\ T_4\beta =P_1\beta \cup P_3\beta =P_1\beta =P_3\beta =Z^{\prime },\end{array}$

since $Z^{\prime }$ is minimal element of the semilattice D.

Now, let $i\ne j$ .

1) If $T_0\beta =P_1\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime }$ and $j=1,2,3,4$ , then we have

$Z=T_1\beta =T_2\beta =T_3\beta =T_4\beta =Z^{\prime }$ ,

which contradicts the inequality $Z\ne Z^{\prime }$ .

2) If $T_1\beta =P_2\beta =P_3\beta =P_4\beta =Z^{\prime }$ and $j=0,2,3,4$ , then we have

$\begin{array}{l}Z=T_0\beta =T_2\beta =T_4\beta =Z^{\prime }\cup P_1\beta ,\text{where}{P}_1\beta \in D;\\ Z=T_3\beta =Z^{\prime }.\end{array}$

Last equalities are impossible since $Z\ne Z^{\prime }\cup T$ for any $T\in D$ and $Z\ne Z^{\prime }$ by definition of a semilattice D.

3) If $T_2\beta =P_1\beta =P_3\beta =P_4\beta =Z^{\prime }$ and $j=0,1,3,4$ , then we have

$\begin{array}{l}Z=T_0\beta =T_2\beta =T_4\beta =Z^{\prime }\cup P_1\beta ,\text{where}{P}_1\beta \in D;\\ Z=T_3\beta =Z^{\prime }.\end{array}$

Last equalities are impossible since for any $T\in D$ and $Z\ne Z^{\prime }$ by definition of a semilattice D.

4) If $T_3\beta =P_2\beta =P_4\beta =Z^{\prime }$ and $j=0,1,2,4$ , then we have

$\begin{array}{l}Z=T_0\beta =T_2\beta =T_4\beta =Z^{\prime }\cup P_1\beta \cup P_3\beta ,\\ Z=T_1\beta =Z^{\prime }\cup P_3\beta ,\text{where}{P}_1\beta ,P_3\beta \in D.\end{array}$

Last equalities are impossible since $Z\ne Z^{\prime }\cup T\cup T^{\prime }$ and $Z\ne Z^{\prime }\cup T$ for any $T,T^{\prime }\in D$ , by definition of a semilattice D.

5) If $T_4\beta =P_1\beta =P_3\beta =Z^{\prime }$ and $j=0,1,2,3$ , then we have

$\begin{array}{l}Z=T_0\beta =T_1\beta =T_3\beta =Z^{\prime }\cup P_2\beta \cup P_4\beta ,\\ Z=T_2\beta =Z^{\prime }\cup P_4\beta ,\text{where}{P}_2\beta ,P_4\beta \in D.\end{array}$

Last equalities are impossible since $Z\ne Z^{\prime }\cup T\cup T^{\prime }$ and $Z\ne Z^{\prime }\cup T$ for any $T,T^{\prime }\in D$ , by definition of a semilattice D.

The statement b) of the Lemma 2.2.1 is proved.

Lemma 2.2.1 is proved.

Let $D\in {\Sigma }_{8.1}\left(X,5\right)$ . We denoted the following sets by symbols ${\mathfrak{A}}_0$ , $B\left({\mathfrak{A}}_0\right)$ and $B_0$ :

$\begin{array}{l}{\mathfrak{A}}_0=\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\},\\ \left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\}\},\\ B\left({\mathfrak{A}}_0\right)=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_0\right\};B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}.\end{array}$

Remark, that the sets $B_0$ and $B\left({\mathfrak{A}}_0\right)$ are external elements for the semigroup $B_X\left(D\right)$ .

Lemma 2.2.2. Let $D\in {\Sigma }_{8.1}\left(X,5\right)$ . Then the following statements are true:

a) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_4^{\alpha },Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

b) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_3^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

c) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B_0\cup B\left({\mathfrak{A}}_0\right)$ .

Proof. 1). Let quasinormal representation of binary relations $\delta$ and $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_4\times T_4\right)\cup \left(\left(T_2\backslash T_4\right)\times T_2\right)\cup \left(\left(T_0\backslash T_2\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ ,

$\begin{array}{l}{T}_4\cup \left(T_2\backslash T_4\right)\cup \left(T_0\backslash T_2\right)\cup \left(X\backslash T_0\right)\\ =\left(P_1\cup P_3\right)\cup P_4\cup P_2\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X,\end{array}$

(see Equalities (2.2.1) and (2.2.2)), then $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_4,T_2\beta =\left(P_1\cup P_3\cup P_4\right)\beta =T_4\cup T_2=T_2,\\ T_1\beta =\left(P_2\cup P_3\cup P_4\right)\beta =T_4\cup T_1=T_0,T_0\beta =T_0.\\ \alpha =\delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_0\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_4^{\alpha }$ , $Y_2^{\delta }=Y_2^{\alpha }$ and $Y_1^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement a) of the lemma 2.2.2 is proved.

2) Let quasinormal representation of binary relations $\delta$ and $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_3\times T_3\right)\cup \left(\left(T_0\backslash T_1\right)\times T_2\right)\cup \left(\left(T_1\backslash T_3\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ ,

$\begin{array}{l}{T}_3\cup \left(T_0\backslash T_1\right)\cup \left(T_1\backslash T_3\right)\cup \left(X\backslash T_0\right)\\ =\left(P_2\cup P_4\right)\cup P_1\cup P_3\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X,\end{array}$

(see Equalities (2.2.1) and (2.2.2)), then $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_3,T_2\beta =\left(P_1\cup P_3\cup P_4\right)\beta =T_3\cup T_2\cup T_1=T_0,\\ T_1\beta =\left(P_2\cup P_3\cup P_4\right)\beta =T_3\cup T_1=T_0,T_0\beta =T_0.\\ \alpha =\delta \circ \beta =\left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_0\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_3^{\alpha }$ , $Y_1^{\delta }=Y_1^{\alpha }$ and $Y_2^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_2^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement b) of the lemma 2.2.2 is proved.

3) Let quasinormal representation of binary relations $\delta$ and $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_3^{\delta }\times T_3\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(\left(T_2\backslash T_1\right)\times T_4\right)\cup \left(\left(T_1\backslash T_2\right)\times T_3\right)\cup \left(\left(T_1\backslash T_3\right)\times T_2\right)\\ \cup \left(\left(T_2\backslash T_4\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\alpha },Y_3^{\alpha }\notin \{\varnothing \}$ ,

$\begin{array}{l}\left(T_2\backslash T_1\right)\cup \left(T_1\backslash T_2\right)\cup \left(T_1\backslash T_3\right)\cup \left(T_2\backslash T_4\right)\cup \left(X\backslash T_0\right)\\ =P_1\cup P_2\cup P_3\cup P_4\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X,\end{array}$

(see Equalities (2.2.1) and (2.2.2)), then $\delta \in B\left({\mathfrak{A}}_0\right)$ , $\beta \in B_0$ and

$\begin{array}{l}{T}_4\beta =\left(P_1\cup P_3\right)\beta =T_4\cup T_2=T_2,\\ T_3\beta =\left(P_2\cup P_4\right)\beta =T_3\cup T_1=T_1,T_0\beta =T_2\cup T_1=T_0,\\ \alpha =\delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times T_2\right)\cup \left(Y_3^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_2^{\alpha }$ , $Y_3^{\delta }=Y_1^{\alpha }$ and $Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_4^{\delta }\right|\ge 1$ , $\left|Y_3^{\delta }\right|\ge 1$ and $\left|Y_0^{\delta }\right|\ge 0$ .

The statement c) of the lemma 2.2.2 is proved.

Lemma 2.2.2 is proved.

Lemma 2.2.3. Let $D\in {\Sigma }_{8.1}\left(X,5\right)$ . Then the following statements are true:

a) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_2^{\alpha }\times T_2\right),$

where $Y_4^{\alpha },Y_2^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

b) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_4^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

c) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_1^{\alpha }\times T_1\right),$

where $Y_3^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

d) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_0^{\alpha }\times T_0\right),$

where $Y_3^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

e) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

f) If quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

g) If quasinormal representation of a binary relation $\alpha$ has a form $\alpha =X\times T_2$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

h) If quasinormal representation of a binary relation $\alpha$ has a form $\alpha =X\times T_1$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

i) If quasinormal representation of a binary relation $\alpha$ has a form $\alpha =X\times T_0$ , then $\alpha$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ .

Proof. 1) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_4\times T_4\right)\cup \left(\left(T_0\backslash T_4\right)\times T_2\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_1^{\delta }\notin \{\varnothing \}$ .

$\begin{array}{l}{T}_4\cup \left(T_0\backslash T_4\right)\cup \left(X\backslash T_0\right)\\ =\left(P_1\cup P_3\right)\cup \left(P_2\cup P_4\right)\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X.\end{array}$

Then from the statement a) of the Lemma 2.2.2 follows that $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ , $\delta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_4,T_1\beta =T_4\cup T_2=T_2,T_0\beta =T_2.\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_2\right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times T_2\right)=\alpha ,\end{array}$

If $Y_4^{\delta }=Y_4^{\alpha }$ , $Y_1^{\delta }\cup Y_0^{\delta }=Y_2^{\alpha }$ . Last equalities are possible since $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement a) of the lemma 2.2.3 is proved.

2) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_4\times T_4\right)\cup \left(\left(T_0\backslash T_4\right)\times T_3\right)\cup \left(\left(X\backslash T\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_1^{\delta }\notin \{\varnothing \}$ .

$\begin{array}{l}{T}_4\cup \left(T_0\backslash T_4\right)\cup \left(X\backslash T_0\right)\\ =\left(P_1\cup P_3\right)\cup \left(P_2\cup P_4\right)\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X.\end{array}$

Then from $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =T_4,T_1\beta =T_4\cup T_3=T_0,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_1^{\delta }\times T_0\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_4^{\delta }\times T_4\right)\cup \left(\left(Y_1^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_4^{\alpha }$ , $Y_1^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_1^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement b) of the lemma 2.2.3 is proved.

3) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_3\times T_3\right)\cup \left(\left(T_0\backslash T_3\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_2^{\delta }\notin \{\varnothing \}$ .

$\begin{array}{l}{T}_3\cup \left(T_0\backslash T_3\right)\cup \left(X\backslash T_0\right)\\ =\left(P_2\cup P_4\right)\cup \left(P_1\cup P_3\right)\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X.\end{array}$

Then from the statement b) of the Lemma 2.2.2 follows that $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ , $\delta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_3\beta =T_3,T_2\beta =T_3\cup T_1=T_1,T_0\beta =T_1.\\ \delta \circ \beta =\left(Y_2^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_1\right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times T_1\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_3^{\alpha }$ , $Y_2^{\delta }\cup Y_0^{\delta }=Y_1^{\alpha }$ . Last equalities are possible since $\left|Y_2^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement c) of the lemma 2.2.3 is proved.

4) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_3\times T_3\right)\cup \left(\left(T_0\backslash T_3\right)\times T_2\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_3^{\delta },Y_2^{\delta }\notin \{\varnothing \}$ . Then $\delta ,\beta \in B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_3\beta =T_3,T_2\beta =T_3\cup T_2=T_0,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_2^{\delta }\times T_0\right)\cup \left(Y_0^{\delta }\times T_0\right)\\ =\left(Y_3^{\delta }\times T_3\right)\cup \left(\left(Y_2^{\delta }\cup Y_0^{\delta }\right)\times T_0\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_3^{\alpha }$ , $Y_2^{\delta }\cup Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_2^{\delta }\cup Y_0^{\delta }\right|\ge 1$ ( $\left|Y_0^{\delta }\right|\ge 0$ by preposition).

The statement d) of the lemma 2.2.3 is proved.

5) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(\left(\left(T_2\cap T_1\right)\backslash T_3\right)\times T_4\right)\cup \left(\left(T_2\backslash T_1\right)\times T_2\right)\cup \left(\left(X\backslash T_4\right)\times T_0\right),\end{array}$

where $Y_4^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$\left(\left(T_2\cap T_1\right)\backslash T_3\right)\cup \left(T_2\backslash T_1\right)\cup \left(X\backslash T_4\right)=P_3\cup P_1\cup \left(X\backslash T_4\right)=T_4\cup \left(X\backslash T_4\right)=X.$

(See Equalities (2.2.1) and (2.2.2)). Then from the statement b) of the Lemma 2.2.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement a) of the Lemma 2.2.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_4\beta =\left(P_1\cup P_3\right)\beta =T_4\cup T_2=T_2,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_4^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right)=\alpha ,\end{array}$

if $Y_4^{\delta }=Y_2^{\alpha }$ , $Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_4^{\delta }\right|\ge 1$ $\left|Y_0^{\delta }\right|\ge 1$ .

The statement e) of the lemma 2.2.3 is proved.

6) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_3^{\delta }\times T_3\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(\left(\left(T_2\cap T_1\right)\backslash T_4\right)\times T_3\right)\cup \left(\left(T_1\backslash T_2\right)\times T_1\right)\cup \left(\left(X\backslash T_3\right)\times T_0\right),\end{array}$

where $Y_3^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$\left(\left(T_2\cap T_1\right)\backslash T_4\right)\cup \left(T_1\backslash T_2\right)\cup \left(X\backslash T_3\right)=P_4\cup P_2\cup \left(X\backslash T_3\right)=T_3\cup \left(X\backslash T_3\right)=X.$

(see Equalities (2.2.1) and (2.2.2)). Then from the statement d) of the Lemma 2.2.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement b) of the Lemma 2.2.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_3\beta =\left(P_2\cup P_4\right)\beta =T_3\cup T_1=T_1,T_0\beta =T_0.\\ \delta \circ \beta =\left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_3^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right)=\alpha ,\end{array}$

if $Y_3^{\delta }=Y_1^{\alpha }$ , $Y_0^{\delta }=Y_0^{\alpha }$ . Last equalities are possible since $\left|Y_4^{\delta }\right|\ge 1$ $\left|Y_0^{\delta }\right|\ge 1$ .

The statement e) of the lemma 2.2.3 is proved.

7) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_1\times T_4\right)\cup \left(\left(T_2\backslash T_1\right)\times T_2\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_2^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$T_1\cup \left(T_2\backslash T_1\right)\cup \left(X\backslash T_0\right)=\left(P_2\cup P_3\cup P_4\right)\cup P_1\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X$

(see Equalities (2.2.1) and (2.2.2)). Then from the statement e) of the Lemma 2.2.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement a) of the Lemma 2.2.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_2\beta =T_4\cup T_2=T_2,T_0\beta =T_2\\ \delta \circ \beta =\left(Y_2^{\delta }\times T_2\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_2^{\delta }\times T_2\right)\cup \left(Y_0^{\delta }\times T_2\right)=X\times T_2=\alpha ,\end{array}$

since representation of a binary relation $\delta$ is quasinormal.

The statement g) of the lemma 2.2.3 is proved.

8) Let quasinormal representation of a binary relations $\delta$ , $\beta$ have a form

$\begin{array}{l}\delta =\left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_0\right),\\ \beta =\left(T_2\times T_3\right)\cup \left(\left(T_1\backslash T_2\right)\times T_1\right)\cup \left(\left(X\backslash T_0\right)\times T_0\right),\end{array}$

where $Y_1^{\delta },Y_0^{\delta }\notin \{\varnothing \}$ ,

$T_2\cup \left(T_1\backslash T_2\right)\cup \left(X\backslash T_0\right)=\left(P_1\cup P_3\cup P_4\right)\cup P_2\cup \left(X\backslash T_0\right)=T_0\cup \left(X\backslash T_0\right)=X$

(see Equalities (2.2.1) and (2.2.2)). Then from the statement f) of the Lemma 2.2.3 follows that $\delta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and from the statement b) of the Lemma 2.2.2 element $\beta$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ and

$\begin{array}{l}{T}_1\beta =T_3\cup T_1=T_1,T_0\beta =T_1\\ \delta \circ \beta =\left(Y_1^{\delta }\times T_1\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)=\left(Y_1^{\delta }\times T_1\right)\cup \left(Y_0^{\delta }\times T_1\right)=X\times T_1=\alpha ,\end{array}$

since representation of a binary relation $\delta$ is quasinormal.

The statement h) of the lemma 2.2.3 is proved.

9) Let quasinormal representation of a binary relation $\delta$ has a form

$\delta =\left(T_4\times T_1\right)\cup \left(\left(X\backslash T_4\right)\times T_0\right),$

then

$\begin{array}{l}{T}_1\delta =\left(P_2\cup P_3\cup P_4\right)\delta =T_4\cup T_0=T_0,T_0\delta =T_0\\ \delta \circ \delta =\left(T_4\times T_1\delta \right)\cup \left(\left(X\backslash T_4\right)\times T_0\delta \right)=\left(T_4\times T_0\right)\cup \left(\left(X\backslash T_4\right)\times T_0\right)=X\backslash T_0=\alpha \end{array}$

since representation of a binary relation $\delta$ is quasinormal.

The statement i) of the lemma 2.2.3 is proved.

Lemma 2.2.3 is proved.

Lemma 2.2.4. Let $D\in {\Sigma }_{8.1}\left(X,5\right)$ . Then the following statements are true:

a) If $\left|X\backslash T_0\right|\ge 1$ and $Z\in \left\{T_4,T_3\right\}$ , then binary relation $\alpha =X\times Z$ is generating by elements of the elements of set $B\left({\mathfrak{A}}_0\right)$ ;

b) If $X=T_0$ and $Z\in \left\{T_4,T_3\right\}$ , then binary relation $\alpha =X\times Z$ is external element for the semigroup $B_X\left(D\right)$ .

Proof. 1) Let quasinormal representation of a binary relation $\delta$ has a form

$\delta =\left(Y_4^{\delta }\times T_4\right)\cup \left(Y_3^{\delta }\times T_3\right)\cup \left(Y_0^{\delta }\times T_0\right)$ ,

where $Y_4^{\delta },Y_3^{\delta }\notin \{\varnothing \}$ , then $\delta \in B\left({\mathfrak{A}}_0\right)\backslash \left\{\alpha \right\}$ . If quasinormal representation of a binary relation $\beta$ has a form $\beta =\left(T_0\times T\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash T_0}{\cup }\left(\left\{t^{\prime }\right\}\times f\left(t^{\prime }\right)\right)}$ , where f is any mapping of the set $X\backslash T_0$ in the set $\left\{T_4,T_3\right\}\backslash \left\{Z\right\}$ . It is easy to see, that $\beta \ne \alpha$ and two elements of the set $\left\{T_4,T_3\right\}$ belong to the semilattice $V\left(D,\beta \right)$ , i.e. $\delta \in B\left({\mathfrak{A}}_0\right)\backslash \left\{\alpha \right\}$ . In this case we have

$\begin{array}{l}{T}_4\beta =T_3\beta =T_0\beta =Z;\\ \delta \circ \beta =\left(Y_4^{\delta }\times T_4\beta \right)\cup \left(Y_3^{\delta }\times T_3\beta \right)\cup \left(Y_0^{\delta }\times T_0\beta \right)\\ =\left(Y_4^{\delta }\times Z\right)\cup \left(Y_3^{\delta }\times Z\right)\cup \left(Y_0^{\delta }\times Z\right)\\ =\left(\left(Y_4^{\delta }\cup Y_3^{\delta }\cup Y_0^{\delta }\right)\times Z\right)=X\times Z=\alpha ,\end{array}$

since the representation of a binary relation $\delta$ is quasinormal. Thus, element $\alpha$ is generating by elements of the set $B\left({\mathfrak{A}}_0\right)$ .

The statement a) of the lemma 2.2.4 is proved.

2) Let $X=T_0$ , $\alpha =X\times Z$ , for some $Z\in \left\{T_4,T_3\right\}$ and $\alpha =\delta \circ \beta$ for some $\delta ,\beta \in B_X\left(D\right)\backslash \left\{\alpha \right\}$ . Then from the Equalities (2.2.3) and (2.2.4) we obtain that

$T_4\beta =T_3\beta =T_2\beta =T_1\beta =T_0\beta =Z,P_1\beta =P_2\beta =P_3\beta =P_4\beta =Z$ ,

since Z is minimal element of the semilattice D.

Now, let subquasinormal representations $\bar{\beta }$ of a binary relation $\beta$ has a form

$\bar{\beta }=\left(\left(P_1\cup P_2\cup P_3\cup P_4\right)\times Z\right)\cup {\displaystyle \underset{t^{\prime }\in X\backslash T_0}{\cup }\left(\left\{t^{\prime }\right\}\times {\bar{\beta }}_2\left(t^{\prime }\right)\right)}$ ,

where ${\bar{\beta }}_1=\left(\begin{array}{l}{P}_0{P}_1{P}_2{P}_3{P}_4\\ \varnothing ZZZZ\end{array}\right)$ is normal mapping. But complement mapping ${\bar{\beta }}_2$ is empty, since $X\backslash T_0=\varnothing$ , i.e. in the given case, subquasinormal representation $\bar{\beta }$ of a binary relation $\beta$ is defined uniquely. So, we have that $\beta =\bar{\beta }=X\times Z=\alpha$ , which contradicts the condition $\beta \notin B_X\left(D\right)\backslash \left\{\alpha \right\}$ .

Therefore, if $X=T_0$ and $\alpha =X\times Z$ , for some $Z\in \left\{T_4,T_3\right\}$ , then $\alpha$ is external element of the semigroup $B_X\left(D\right)$ .

The statement b) of the lemma 2.2.4 is proved.

lemma 2.2.4 is proved.

Theorem 2.2.1. Let $D\in {\Sigma }_{8.1}\left(X,5\right)$ and

$\begin{array}{l}{\mathfrak{A}}_0=\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\},\\ \left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\}\},\\ B\left({\mathfrak{A}}_0\right)=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_0\right\};B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}.\end{array}$

Then the following statements are true:

a) If $\left|X\backslash T_0\right|\ge 1$ , then $S_0=B_0\cup B\left({\mathfrak{A}}_0\right)$ is irreducible generating set for the semigroup.

b) If $X=T_0$ , then $S_1=B_0\cup B\left({\mathfrak{A}}_0\right)\cup \left\{X\times T_4,X\times T_3\right\}$ is irreducible genera­ting set for the semigroup $B_X\left(D\right)$ .

Proof. The theorem 2.2.1 we may prove analogously of the theorems 2.1.1.

Theorem 2.2.2. Let $n\ge 6$ , $D=\left\{T_4,T_3,T_2,T_1,T_0\right\}\in {\Sigma }_{8.1}\left(X,5\right)$ and

$\begin{array}{l}{\mathfrak{A}}_0=\{\left\{T_4,T_3,T_2,T_0\right\},\left\{T_4,T_3,T_1,T_0\right\},\left\{T_4,T_2,T_1,T_0\right\},\left\{T_3,T_2,T_1,T_0\right\},\\ \left\{T_4,T_3,T_0\right\},\left\{T_4,T_1,T_0\right\},\left\{T_3,T_2,T_0\right\}\},\\ B\left({\mathfrak{A}}_0\right)=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)\in {\mathfrak{A}}_0\right\};B_0=\left\{\alpha \in B_X\left(D\right)|V\left(X^{\ast },\alpha \right)=D\right\}.\end{array}$

Then the following statements are true:

a) If $\left|X\backslash T_0\right|\ge 1$ , then the number $\left|S_0\right|$ elements of the set $S_0=B_0\cup B\left({\mathfrak{A}}_0\right)$ is equal to

$\left|S_0\right|=5^n-3\cdot 3^n+2\cdot 2^n+2$ .

b) If $X=T_0$ , then the number $\left|S_1\right|$ elements of the set $S_1=B_0\cup B\left({\mathfrak{A}}_0\right)\cup \left\{X\times T_4,X\times T_3\right\}$ is equal to

$\left|S_1\right|=5^n-3\cdot 3^n+2\cdot 2^n+4$ .

Proof. Let number of a set X is equal to $n\ge 6$ , i.e. $\left|X\right|=n\ge 6$ . Let $S_n=\left\{{\phi }_1,{\phi }_2,\cdots ,{\phi }_{n!}\right\}$ is a group all one to one mapping of a set $M=\left\{1,2,\cdots ,n\right\}$ on the set M and ${\phi }_{i_1},{\phi }_{i_2},\cdots ,{\phi }_{i_m}$ $\left(m\le n\right)$ are arbitrary elements of the group $S_n$ , $Y_{{\phi }_1},Y_{{\phi }_2},\cdots ,Y_{{\phi }_m}$ are arbitrary partitioning of a set X. By symbol $k_n^m$ we denote the number elements of a set $\left\{Y_{{\phi }_1},Y_{{\phi }_2},\cdots ,Y_{{\phi }_m}\right\}$ . It is well known, that

$k_n^m={\displaystyle \underset{i=1}{\overset{m}{\sum }}\frac{{\left(-1\right)}^{m+i}}{\left(i-1\right)!\cdot \left(m-i\right)!}\cdot i^{n-1}}$ .

If $m=2,3,4,5$ , then we have

$\begin{array}{l}{k}_n^2=2^{n-1}-1,k_n^3=\frac{1}{2}\cdot 3^{n-1}-2^{n-1}+\frac{1}{2},k_n^4=\frac{1}{6}\cdot 4^{n-1}-\frac{1}{2}\cdot 3^{n-1}+\frac{1}{2}\cdot 2^{n-1}-\frac{1}{6},\\ k_n^5=\frac{1}{24}\cdot 5^{n-1}-\frac{1}{6}\cdot 4^{n-1}+\frac{1}{4}\cdot 3^{n-1}-\frac{1}{6}\cdot 2^{n-1}+\frac{1}{24}.\end{array}$

If $Y_{{\phi }_1},Y_{{\phi }_2}$ are any two elements partitioning of a set X and $\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)$ , where $Z_1,Z_2\in D$ and $Z_1\ne Z_2$ . Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$2\cdot k_n^2=2^n-2$ . (2.2.5)

If $Y_{{\phi }_1},Y_{{\phi }_2},Y_{{\phi }_3}$ are any tree elements partitioning of a set X and

$\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)\cup \left(Y_{{\phi }_3}\times Z_3\right)$ ,

where $Z_1,Z_2,Z_3$ are pairwise different elements of a given semilattice D. Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$6\cdot k_n^3=3^n-3\cdot 2^n+3$ . (2.2.6)

If $Y_{{\phi }_1},Y_{{\phi }_2},Y_{{\phi }_3},Y_{{\phi }_4}$ are any four elements partitioning of a set X and

$\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)\cup \left(Y_{{\phi }_3}\times Z_3\right)\cup \left(Y_{{\phi }_4}\times Z_4\right)$ ,

where $Z_1,Z_2,Z_3,Z_4$ are pairwise different elements of a given semilattice D. Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$24\cdot k_n^4=4^n-4\cdot 3^n+3\cdot 2^{n+1}-4$ . (2.2.7)

If $Y_{{\phi }_1},Y_{{\phi }_2},Y_{{\phi }_3},Y_{{\phi }_4},Y_{{\phi }_5}$ are any four elements partitioning of a set X and

$\bar{\beta }=\left(Y_{{\phi }_1}\times Z_1\right)\cup \left(Y_{{\phi }_2}\times Z_2\right)\cup \left(Y_{{\phi }_3}\times Z_3\right)\cup \left(Y_{{\phi }_4}\times Z_4\right)\cup \left(Y_{{\phi }_5}\times Z_5\right)$ ,

where $Z_1,Z_2,Z_3,Z_4,Z_5$ are pairwise different elements of a given semilattice D. Then number of different binary relations $\bar{\beta }$ of a semigroup $B_X\left(D\right)$ is equal to

$120\cdot k_n^5=5^n-5\cdot 4^n+10\cdot 3^n-10\cdot 2^n+5.$ (2.2.8)

If $\alpha \in B_0$ , then quasinormal representation of a binary relation $\alpha$ has a form

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or a system $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }$ are partitioning of the set X.

If the system $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }$ , or a system $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }$ are partitioning of the set X. Of this from the Equalities (2.2.7) and (2.2.8) follows that

$\begin{array}{c}\left|B_0\right|=\left(5^n-5\cdot 4^n+10\cdot 3^n-10\cdot 2^n+5\right)+\left(4^n-4\cdot 3^n+6\cdot 2^n-4\right)\\ =5^n-4\cdot 4^n+6\cdot 3^n-4\cdot 2^n+1.\end{array}$

If $\alpha \in B\left({\mathfrak{A}}_0\right)$ , then by definition of a set $B\left({\mathfrak{A}}_0\right)$ the quasinormal representation of a binary relation $\alpha$ has a form:

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_3^{\alpha },Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_3^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_3^{\alpha },Y_2^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_3^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_3^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_4^{\alpha }\times T_4\right)\cup \left(Y_1^{\alpha }\times T_1\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_4^{\alpha },Y_1^{\alpha }\notin \{\varnothing \}$ , or $Y_4^{\alpha },Y_1^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively;

$\alpha =\left(Y_3^{\alpha }\times T_3\right)\cup \left(Y_2^{\alpha }\times T_2\right)\cup \left(Y_0^{\alpha }\times T_0\right)$ ,

where $Y_3^{\alpha },Y_2^{\alpha }\notin \{\varnothing \}$ , or $Y_3^{\alpha },Y_2^{\alpha },Y_0^{\alpha }\notin \{\varnothing \}$ are partitioning of the set X respectively.

Of this and from the Equality (2.2.5), (2.2.6) and (2.2.7) follows that

$\begin{array}{c}\left|B\left({\mathfrak{A}}_0\right)\right|=3\cdot \left(2^n-2\right)+7\cdot \left(3^n-3\cdot 2^n+3\right)+4\cdot \left(4^n-4\cdot 3^n+6\cdot 2^n-4\right)\\ =4\cdot 4^n-9\cdot 3^n+6\cdot 2^n+1.\end{array}$

So, we have that:

$\begin{array}{l}\left|S_0\right|=\left|B_0\cup B\left({\mathfrak{A}}_0\right)\right|\\ =\left(5^n-4\cdot 4^n+6\cdot 3^n-4\cdot 2^n+1\right)+\left(4\cdot 4^n-9\cdot 3^n+6\cdot 2^n+1\right)\\ =5^n-3\cdot 3^n+2\cdot 2^n+2,\\ \left|S_1\right|=\left|B_0\cup B\left({\mathfrak{A}}_0\right)\cup \left\{X\times T_4,X\times T_3\right\}\right|=5^n-3\cdot 3^n+2\cdot 2^n+4.\end{array}$

Since

$B_0\cap B\left({\mathfrak{A}}_0\right)=B_0\cap \left\{X\times T_4,X\times T_3,X\times T_2\right\}=B\left({\mathfrak{A}}_0\right)\cap \left\{X\times T_4,X\times T_3,X\times T_2\right\}=\varnothing$ .

Theorem 2.2.2 is proved.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

References

[1]	Diasamidze, Y., Givradze, O., Tsinaridze, N. and Tavdgiridze, G. (2018) Generated Sets of the Complete Semigroup Binary Relations Defined by Semilattices of the Class Σ8(X, n+k+1). Applied Mathematics, 9, 369-382. https://doi.org/10.4236/am.2018.94028
[2]	Diasamidze, Y. and Makharadze, S. (2013) Complete Semigroups of Binary Relations. Kriter, Turkey, 1-519.
[3]	Диасамидзе, Я.И. and Махарадзе, Ш.И. (2017) Полные полугруппы бинарных отношений. Lambert Academic Publishing, Saarbrucken, 1-692.