Generated Sets of the Complete Semigroup Binary Relations Defined by Semilattices of the Class Σ8 (X, n + k + 1)

Giuli  Tavdgiridze; Nino  Tsinaridze; Omari  Givradze; Yasha  Diasamidze

doi:10.4236/am.2018.94028

Author(s) Yasha Diasamidze, Omari Givradze, Nino Tsinaridze, Giuli Tavdgiridze

Affiliation(s)
Faculty of Mathematics, Physics and Computer Sciences, Shota Rustaveli State University, Batumi, Georgia.

ABSTRACT

In this article, we study generated sets of the complete semigroups of binary relations defined by X-semilattices unions of the class Σ₈ (X, n + k +1) , and find uniquely irreducible generating set for the given semigroups.

KEYWORDS
Semigroup, Semilattice, Binary Relation

1. Introduction

Let X be an arbitrary nonempty set, 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 $α_{f}$ on the set X that satisfies the condition $α_{f} = \underset{x \in X}{\cup} ({x} \times f (x))$ . The set of all such $α_{f}$ ( $f : X \to D$ ) is denoted by $B_{X} (D)$ . It is easy to prove that $B_{X} (D)$ 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 Æ an empty binary relation or an empty subset of the set X. The condition $(x, y) \in α$ will be written in the form $x α y$ . Further, let $x, y \in X$ , $Y \subseteq X$ , $α \in B_{X} (D)$ , $\overset{⌣}{D} = \underset{Y \in D}{\cup} Y$ and $T \in D$ . We denote by the symbols $y α$ , $Y α$ , $V (D, α)$ , $X^{*}$ and $V (X^{*}, α)$ the following sets:

$\begin{array}{l} y α = {x \in X | y α x}, Y α = \underset{y \in Y}{\cup} y α, V (D, α) = {Y α | Y \in D}, \\ X^{*} = {Y | \emptyset \neq Y \subseteq X}, V (X^{*}, α) = {Y α | \emptyset \neq Y \subseteq X}, \\ D_{T} = {Z \in D | T \subseteq Z}, Y_{T}^{α} = {y \in X | y α = T} . \end{array}$

It is well known the following statements:

Theorem 1.1. Let $D = {\overset{⌣}{D}, Z_{1}, Z_{2}, \dots, Z_{m - 1}}$ be some finite X-semilattice of unions and $C (D) = {P_{0}, P_{1}, P_{2}, \dots, P_{m - 1}}$ be the family of sets of pairwise nonintersecting subsets of the set X (the set Æ can be repeated several times). If j is a mapping of the semilattice D on the family of sets $C (D)$ which satisfies the conditions

$φ = (\begin{array}{l} \overset{⌣}{D} Z_{1} Z_{2} \dots Z_{m - 1} \\ P_{0} P_{1} P_{2} \dots P_{m - 1} \end{array})$

and ${\hat{D}}_{Z} = D \ D_{Z}$ , then the following equalities are valid:

$\overset{⌣}{D} = P_{0} \cup P_{1} \cup P_{2} \cup \dots \cup P_{m - 1}, Z_{i} = P_{0} \cup \underset{T \in {\hat{D}}_{Z_{i}}}{\cup} φ (T) .$ (1.1)

In the sequel these equalities will be called formal.

It is proved that if the elements of the semilattice D are represented in the form (1.1), then among the parameters $P_{i}$ $(0 < i \leq m - 1)$ there exist such parameters that cannot be empty sets for D. Such sets $P_{i}$ are called bases sources, where sets $P_{j}$ $(0 \leq j \leq m - 1)$ , which can be empty sets too are called completeness sources.

It is proved that under the mapping j the number of covering elements of the pre-image of a

bases source is always equal to one, while under the mapping j 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] [2] chapter 11).

Definition 1.1. We say that an element a of the semigroup $B_{X} (D)$ is external if $α \neq δ \circ β$ for all $δ, β \in B_{X} (D) \ {α}$ (see [1] [2] Definition 1.15.1).

It is well known, that if B is all external elements of the semigroup $B_{X} (D)$ and $B^{'}$ is any generated set for the $B_{X} (D)$ , then $B \subseteq B^{'}$ (see [1] [2] Lemma 1.15.1).

Definition 1.2. The representation $α = \underset{T \in D}{\cup} (Y_{T}^{α} \times T)$ of binary relation a is called quasinormal, if $\underset{T \in D}{\cup} Y_{T}^{α} = X$ and $Y_{T}^{α} \cap Y_{T^{'}}^{α} = \emptyset$ for any $T, T^{'} \in D$ ,

$T \neq T^{'}$ (see [1] [2] chapter 1.11).

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

2. Result

Let $Σ_{8} (X, n + k + 1)$ $(3 \leq k \leq n)$ be a class of all X-semilattices of unions whose every element is isomorphic to an X-semilattice of unions $D = {Z_{1}, Z_{2}, \dots, Z_{n + k}, \overset{⌣}{D}}$ , which satisfies the condition:

$\begin{array}{l} Z_{n + i} \subset Z_{i} \subset \overset{⌣}{D}, (i = 1, 2, \dots, k); Z_{j} \subset \overset{⌣}{D}, (j = 1, 2, \dots, n + k); \\ Z_{p} \ Z_{q} \neq \emptyset and Z_{q} \ Z_{p} \neq \emptyset (1 \leq p \neq q \leq n + k) . \end{array}$ (see Figure 1).

Figure 1. Diagram of the semilattice D.

It is easy to see that $\tilde{D} = {Z_{1}, Z_{2}, \dots, Z_{n + k}}$ is irreducible generating set of the semilattice D.

Let $C (D) = {P_{0}, P_{1}, P_{2}, \dots, P_{n + k}}$ be a family of sets, where $P_{0}, P_{1}, P_{2}, \dots, P_{n + k}$ are pairwise disjoint subsets of the set X and $φ = (\begin{array}{l} \overset{⌣}{D} Z_{1} Z_{2} \dots Z_{n + k} \\ P_{0} P_{1} P_{2} \dots P_{n + k} \end{array})$ is a map

ping of the semilattice D onto the family of sets $C (D)$ . Then the formal equalities of the semilattice D have a form:

$\overset{⌣}{D} = \cup_{i = 0}^{n + k} P_{i}; Z_{j} = \cup_{\begin{array}{l} i = 0, \\ i \neq j \end{array}}^{n + k} P_{i}, j = 1, 2, \dots, n; Z_{n + q} = \cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i}, q = 1, 2, \dots, k .$ (2.0)

Here the elements $P_{i} (i = 1, 2, \dots, n + k)$ are bases sources, the element $P_{0}$ are sources of completeness of the semilattice D. Therefore $| X | \geq n + k$ (by symbol $| X |$ we denoted the power of a set X), since $| P_{i} | \geq 1 (i = 1, 2, \dots, n + k)$ (see [1] [2] chapter 11).

In this paper we are learning irreducible generating sets of the semigroup $B_{X} (D)$ defined by semilattices of the class $Σ_{8} (X, n + k + 1)$ .

Note, that it is well known, when $k = 2$ , then generated sets of the complete semigroup of binary relations defined by semilattices of the class $Σ_{8} (X, 2 + 2 + 1) = Σ_{8} (X, 5)$ .

In this paper we suppose, that $3 \leq k \leq n$ .

Remark, that in this case (i.e. $k \geq 3$ ), from the formal equalities of a semilattice D follows, that the intersections of any two elements of a semilattice D is not empty.

Lemma 2.0 If $D \in Σ_{8} (X, n + k + 1)$ , then the following statements are true:

a) $\cap_{i = 1}^{n + k} Z_{i} = P_{0};$

b) $Z_{j + 1} \ Z_{j} = P_{j}, j = 1, 2, \dots, n - 1;$

c) $Z_{q} \ Z_{n + q} = P_{n + q}, q = 1, \dots, k .$

Proof. From the formal equalities of the semilattise D immediately follows the following statements:

$\begin{array}{l} \cap_{i = 1}^{n + k} Z_{i} = P_{0}, Z_{j + 1} \ Z_{j} = (\cup_{\begin{array}{l} i = 0, \\ i \neq j + 1 \end{array}}^{n + k} P_{i}) \ (\cup_{\begin{array}{l} i = 0, \\ i \neq j \end{array}}^{n + k} P_{i}) = P_{j}, j = 1, 2, \dots, n - 1; \\ Z_{q} \ Z_{n + q} = (\cup_{\begin{array}{l} i = 0, \\ i \neq q \end{array}}^{n + k} P_{i}) \ (\cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i}) = P_{n + q}, q = 1, \dots, k . \end{array}$

The statements a), b) and c) of the lemma 2.0 are proved.

Lemma 2.0 is proved.

We denoted the following sets by symbols $D_{1}$ , $D_{2}$ and $D_{3}$ :

$D_{1} = {Z_{1}, Z_{2}, \dots, Z_{k}}, D_{2} = {Z_{k + 1}, Z_{k + 2}, \dots, Z_{n}}, D_{3} = {Z_{n + 1}, Z_{n + 2}, \dots, Z_{n + k}} .$

Lemma 2.1. Let $D \in Σ_{8.0} (X, n + k + 1)$ and $α \in B_{X} (D)$ . Then the following statements are true:

1) Let $T, T^{'} \in D_{2} \cup D_{3}, T \neq T^{'}$ . If $T, T^{'} \in V (D, α)$ , then a is external element of the semigroup $B_{X} (D)$ ;

2) Let $T \in D_{1}$ , $T^{'} \in D_{2} \cup D_{3}$ . If $T^{'} ⊄ T$ and $T, T^{'} \in V (D, α)$ , then a is external element of the semigroup $B_{X} (D)$ .

3) Let $T, T^{'} \in D_{1}$ and $T \neq T^{'}$ . If $T, T^{'} \in V (D, α)$ and $k \geq 3$ , then a is external element of the semigroup $B_{X} (D)$ ;

Proof. Let $Z_{0} = \overset{⌣}{D}$ and $α = δ \circ β$ for some $δ, β \in B_{X} (D) \ {α}$ . If quasinormal representation of binary relation d has a form

$δ = \underset{T \in V (D, δ)}{\cup} (Y_{T}^{δ} \times T)$ ,

then

$α = δ \circ β = \underset{T \in V (D, δ)}{\cup} (Y_{T}^{δ} \times T β)$ . (2.1)

From the formal equalities (2.0) of the semilattice D we obtain that:

$\begin{array}{l} Z_{0} β = \cup_{i = 0}^{n + k} P_{i} β; Z_{j} β = \cup_{\begin{array}{l} i = 0, \\ i \neq j \end{array}}^{n + k} P_{i} β, j = 1, 2, \dots, n; \\ Z_{n + q} β = \cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i} β, q = 1, 2, \dots, k . \end{array}$ (2.2)

where $P_{i} β \neq \emptyset$ for any $P_{i} \neq \emptyset$ $(i = 0, 1, 2, \dots, n + k)$ and $β \in B_{X} (D)$ by definition of a semilattice D from the class $Σ_{8.0} (X, n + k + 1)$ .

Now, let $Z_{m} β = T$ and $Z_{j} β = T^{'}$ for some $T \neq T^{'}$ , $T, T^{'} \in D_{2} \cup D_{3}$ , then from the equalities (2.3) follows that $T = P_{0} β = T^{'}$ since T and $T^{'}$ are minimal elements of the semilattice D and $P_{0} \neq \emptyset$ by preposition. The equality $T = T^{'}$ contradicts the inequality $T \neq T^{'}$ .

The statement a) of the Lemma 2.1 is proved.

Now, let $Z_{m} β = T$ and $Z_{j} β = T^{'}$ , for some $T \in D_{1}$ , $T^{'} \in D_{2} \cup D_{3}$ and $T^{'} ⊄ T$ , then from the equalities 2.3 follows, that

$T^{'} = Z_{j} β = Z_{0} β = \cup_{i = 0}^{n + k} P_{i} β$ , if $j = 0$ , or $T^{'} = Z_{j} β = \cup_{\begin{array}{l} i = 0, \\ i \neq j \end{array}}^{n + k} P_{i} β$ , $1 \leq j \leq n$ , or $T^{'} = Z_{n + q} β = \cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i} β$

where $j = n + q$ . For the $Z_{j} β = T^{'}$ we consider the following cases:

1) If $T^{'} = Z_{0} β = \cup_{i = 0}^{n + k} P_{i} β$ , then we have

$P_{0} β = P_{1} β = \dots = P_{n + k} β = T^{'}$ ,

since $T^{'}$ is a minimal element of a semilattice D. On the other hand,

$T = Z_{m} β = {\begin{cases} \cup_{\begin{array}{l} i = 0, \\ i \neq m \end{array}}^{n + k} P_{i} β = \cup_{\begin{array}{l} i = 0, \\ i \neq m \end{array}}^{n + k} T^{'} = T^{'}, if 1 \leq m \leq n; \\ \cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i} β = \cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} T^{'} = T^{'}, if m = n + q . \end{cases}$

But the equality $T = T^{'}$ contradicts the inequality $T \neq T^{'}$ . Thus we have, that $j \neq 0$ .

2) Let $1 \leq j \leq n$ , i.e. $T^{'} = Z_{j} β = \cup_{\begin{array}{l} i = 0, \\ i \neq j \end{array}}^{n + k} P_{i} β$ , then we have, that

$P_{0} β = P_{1} β = \dots = P_{j - 1} β = P_{j + 1} β = \dots = P_{n + k} β = T^{'}$ ,

since $T^{'}$ is a minimal element of a semilattice D. On the other hand:

$T = Z_{m} β = {\begin{cases} (\cup_{i = 0}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq j \end{array}}^{n + k} P_{i} β) \cup P_{j} β = T^{'} \cup P_{j} β, if m = 0; \\ (\cup_{\begin{array}{l} i = 0, \\ i \neq m \end{array}}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq m, j \end{array}}^{n + k} P_{i} β) \cup P_{j} β = T^{'} \cup P_{j} β, if 1 \leq m \leq n, m \neq j; \\ (\cup_{\begin{array}{l} i = 0, \\ i \neq j, n + j \end{array}}^{n + k} P_{i} β) = T^{'}, if m = n + j; \\ (\cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q, j \end{array}}^{n + k} P_{i} β) \cup P_{j} β = T^{'} \cup P_{j} β, if m = n + q, q \neq j . \end{cases}$

The equality $T = T^{'}$ contradicts the inequality $T \neq T^{'}$ . Also, the equality $T = T^{'} \cup P_{j} β$ $(P_{j} β \in D)$ contradicts the inequality $T \neq T^{'} \cup Z$ for any $Z \in D$ and $T^{'} ⊄ T$ ( $T^{'} ⊄ T$ , by preposition) by definition of a semilattice D.

3) If $j = n + q$ $(1 \leq q \leq k)$ , i.e. $T^{'} = Z_{n + q} β = \cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i} β$ , then we have, that

$P_{0} β = P_{1} β = \dots = P_{q - 1} β = P_{q + 1} β = \dots = P_{n + q - 1} β = P_{n + q + 1} β = \dots = P_{n + k} β = T^{'}$ ,

since $T^{'}$ is a minimal element of a semilattice D. On the other hand:

$T = Z_{m} β = {\begin{cases} (\cup_{i = 0,}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i} β) \cup P_{q} β \cup P_{n + q} β = T^{'} \cup P_{q} β \cup P_{n + q} β, if m = 0; \\ (\cup_{\begin{array}{l} i = 0, \\ i \neq q \end{array}}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq q, n + q \end{array}}^{n + k} P_{i} β) \cup P_{n + q} β = T^{'} \cup P_{n + q} β, if 1 \leq m = q \leq n; \\ (\cup_{\begin{array}{l} i = 0, \\ i \neq m \end{array}}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq m, q, n + q \end{array}}^{n + k} P_{i} β) \cup P_{q} β \cup P_{n + q} β = T^{'} \cup P_{q} β \cup P_{n + q} β, if 1 \leq m \neq q \leq n; \\ (\cup_{\begin{array}{l} i = 0, \\ i \neq m, \end{array}}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq q^{'}, n + q^{'} \end{array}}^{n + k} P_{i} β) = (\cup_{\begin{array}{l} i = 0, \\ i \neq, q^{'}, n + q^{'}, q, n + q \end{array}}^{n + k} P_{i} β) \cup P_{q} β \cup P_{n + q} = T \cup P_{q} β \cup P_{n + q}, \\ if n + 1 \leq m = n + q^{'} \leq n + k, q \neq q^{'} since j \neq m . \end{cases}$

The equality $T = T^{'}$ contradicts the inequality $T \neq T^{'}$ . Also, the equality $T = T^{'} \cup P_{q} β \cup P_{n + q} β$ , or $T = T^{'} \cup P_{n + q} β$ $(P_{q} β, P_{n + q} β \in D)$ contradicts the inequality $T \neq T^{'} \cup Z$ for any $Z \in D$ and $T^{'} ⊄ T$ by definition of a semilattice D.

The statement 2) of the Lemma 2.1 is proved.

Let $T, T^{'} \in D_{1}$ and $T \neq T^{'}$ . If $k \geq 3$ and $Z_{j} β = T^{'}$ , $Z_{m} β = T$ , then from the formal equalities (2.0) of a semilattice D there exists such an element, that $P_{q} \subseteq Z_{j}$ and $P_{q} \subseteq Z_{m}$ , where $0 \leq q \leq m + k$ . So, from the equalities (2.3) follows that $P_{q} β \subseteq Z_{j} β = T^{'}$ and $P_{q} β \subseteq Z_{m} β = T$ . Of from this and from the equalities (2.3) we obtain that there exists such an element $Z \in D$ , for which the equalities $T^{'} = Z \cup Z^{'}$ and $T = Z \cup Z^{″}$ , where $Z^{'}, Z^{″} \in D$ . But such elements by definition of a semilattice D do not exist.

The statement c) of the Lemma 2.1 is proved.

Lemma 2.1 is proved.

Lemma 2.2. Let $D \in Σ_{8} (X, n + k + 1)$ and $α \in B_{X} (D)$ . Then the following statements are true:

1) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} = \emptyset, V (D, α) \cap D_{3} = \emptyset$ . If $| V (D, α) \cap D_{1} | \geq 2$ , then a is external element of the semigroup $B_{X} (D)$ ;

2) Let $V (D, α) \cap D_{1} = \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} = \emptyset$ . If $| V (D, α) \cap D_{2} | \geq 2$ , then a is external element of the semigroup $B_{X} (D)$ ;

3) Let $V (D, α) \cap D_{1} = \emptyset, V (D, α) \cap D_{2} = \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ . If $| V (D, α) \cap D_{3} | \geq 2$ , then a is external element of the semigroup $B_{X} (D)$ ;

4) Let $V (D, α) \cap D_{1} = \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ , then a is external element of the semigroup $B_{X} (D)$ ;

5) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} = \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ . If $| V (D, α) \cap D_{1} | \geq 2$ , $| V (D, α) \cap D_{3} | = 1$ , or $| V (D, α) \cap D_{1} | = 1$ , $| V (D, α) \cap D_{3} | \geq 2$ then a is external element of the semigroup $B_{X} (D)$ ;

6) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} = \emptyset$ , then a is external element of the semigroup $B_{X} (D)$ ;

7) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ , then a is external element of the semigroup $B_{X} (D)$ .

Proof. Let a be any element of the semigroup $B_{X} (D)$ . It is easy that $V (D, α) \in D$ . We consider the following cases:

Let $V (D, α) \cap D_{1} = \emptyset, V (D, α) \cap D_{2} = \emptyset, V (D, α) \cap D_{3} = \emptyset$ , then $V (D, α) \in {\overset{⌣}{D}}$ since $V (D, α)$ is subsemilattice of the semilattice D.

1) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} = \emptyset, V (D, α) \cap D_{3} = \emptyset$ .

If $| V (D, α) \cap D_{1} | = 1$ , then $V (D, α) \in {Z_{j}}$ , or $V (D, α) \in {Z_{j}, \overset{⌣}{D}}$ , where $j = 1, 2, \dots, k$ , since $V (D, α)$ is subsemilattice of the semilattice D.

If $| V (D, α) \cap D_{1} | \geq 2$ , then by statement c) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ .

2) Let $V (D, α) \cap D_{1} = \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} = \emptyset$ .

If $| V (D, α) \cap D_{2} | = 1$ , then $V (D, α) \in {Z_{j}}$ , or $V (D, α) \in {Z_{j}, \overset{⌣}{D}}$ , where $j = k + 1, k + 2, \dots, n$ , since $V (D, α)$ is a subsemilattice of the semilattice D.

If $| V (D, α) \cap D_{2} | \geq 2$ , then by statement a) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ .

3) Let $V (D, α) \cap D_{1} = \emptyset, V (D, α) \cap D_{2} = \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ .

If $| V (D, α) \cap D_{3} | = 1$ , then $V (D, α) \in {Z_{j}}$ , or $V (D, α) \in {Z_{j}, \overset{⌣}{D}}$ , $j = n + 1, n + 2, \dots, n + k$ , since $V (D, α)$ is subsemilattice of the semilattice D.

If $| V (D, α) \cap D_{3} | \geq 2$ , then by statement a) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ .

4) Let $V (D, α) \cap D_{1} = \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ , then by the statement a) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ .

5) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} = \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ .

If $| V (D, α) \cap D_{1} | = 1, | V (D, α) \cap D_{3} | = 1$ , then $V (D, α) = {Z_{n + q}, Z_{q}}$ , or $V (D, α) = {Z_{n + q}, Z_{q}, \overset{⌣}{D}}$ , or $V (D, α) = {Z_{n + q}, Z_{j}, \overset{⌣}{D}}$ where $Z_{1} \leq Z_{j} \leq Z_{k}$ and $q = 1, 2, \dots, k$ .

If $V (D, α) = {Z_{n + q}, Z_{j}, \overset{⌣}{D}}$ where $j \neq q, q = 1, 2, \dots, k$ , then by the statement 2) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ ;

If $| V (D, α) \cap D_{1} | = 1, | V (D, α) \cap D_{3} | \geq 2$ , or $| V (D, α) \cap D_{1} | \geq 2, | V (D, α) \cap D_{3} | = 1$ , then from the statement 1) and 3) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ respectively.

6) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} = \emptyset$ . Then from the statement b) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ .

7) Let $V (D, α) \cap D_{1} \neq \emptyset, V (D, α) \cap D_{2} \neq \emptyset, V (D, α) \cap D_{3} \neq \emptyset$ , then by the statement a) of the Lemma 2.1 follows that a is external element of the semigroup $B_{X} (D)$ .

Lemma 2.2 is proved.

Now we learn the following subsemilattices of the semilattice D:

$\begin{array}{l} A_{1} = {{Z_{n + j}, Z_{j}, \overset{⌣}{D}}}, where j = 1, 2, \dots, k; \\ A_{2} = {{Z_{j}, \overset{⌣}{D}}}, where j = 1, 2, \dots, n + k; \\ A_{3} = {{Z_{n + j}, Z_{j}}}, where j = 1, 2, \dots, k; \\ A_{4} = {{Z_{j}}, {\overset{⌣}{D}}}, where j = 1, 2, \dots, n + k . \end{array}$

We denoted the following sets by symbols $A_{0}$ and $B (A_{0})$ :

$\begin{array}{l} A_{0} = {V (D, α) \subseteq D | V (D, α) \notin A_{1} \cup A_{2} \cup A_{3} \cup A_{4}}, \\ B (A_{0}) = {α \in B_{X} (D) | V (D, α) \in A_{0}} . \end{array}$

By definition of a set $B (A_{0})$ follows that any element of the set is external element of the semigroup $B_{X} (D)$ .

Lemma 2.3. Let $D \in Σ_{8} (X, n + k + 1)$ . If quasinormal representation of a binary relation a has a form

$α = (Y_{n + j}^{α} \times Z_{n + j}) \cup (Y_{j}^{α} \times Z_{j}) \cup (Y_{0}^{α} \times \overset{⌣}{D}),$

where $Y_{n + j}^{α}, Y_{j}^{α}, Y_{0}^{α} \notin {\emptyset}$ and $j = 1, 2, \dots, k$ , then a is generated by elements of the elements of set $B (A_{0})$ .

Proof. 1). Let quasinormal representation of binary relations d and b have a form

$\begin{array}{l} δ = (Y_{n + j}^{δ} \times Z_{n + j}) \cup (Y_{j}^{δ} \times Z_{j}) \cup (Y_{q}^{δ} \times Z_{q}) \cup (Y_{0}^{δ} \times \overset{⌣}{D}), \\ β = (Z_{n + j} \times Z_{n + j}) \cup ((Z_{j} \ Z_{n + j}) \times Z_{j}) \cup ((\overset{⌣}{D} \ Z_{j}) \times Z_{q}) \cup ((X \ \overset{⌣}{D}) \times \overset{⌣}{D}), \end{array}$

where $Y_{n + j}^{α}, Y_{j}^{α}, Y_{q}^{α} \notin {\emptyset}, Z_{1} \leq Z_{q} \leq Z_{k}, q \neq j, j = 1, \dots, k$ .

$\begin{array}{l} Z_{n + j} \cup (Z_{j} \ Z_{n + j}) \cup (\overset{⌣}{D} \ Z_{j}) \cup (X \ \overset{⌣}{D}) \\ = (P_{0} \cup \cup_{\begin{array}{l} i = 1, \\ i \neq j, n + j \end{array}}^{n + k} P_{i}) \cup P_{n + j} \cup P_{j} \cup (X \ \overset{⌣}{D}) = \overset{⌣}{D} \cup (X \ \overset{⌣}{D}) = X \end{array}$ ,

since the representation of a binary relation b is quasinormal and by statement 3) of the Lemma 2.1 binary relations d and b are external elements of the semigroup $B_{X} (D)$ . It is easy to see, that:

$Z_{n + j} β = Z_{n + j},$

$\begin{matrix} Z_{j} β = (P_{0} \cup \cup_{\begin{array}{l} i = 1, \\ i \neq j \end{array}}^{n + k} P_{i}) β = ((P_{0} \cup \cup_{\begin{array}{l} i = 1, \\ i \neq j, n + j \end{array}}^{n + k} P_{i}) \cup P_{n + j}) β \\ = Z_{n + j} β \cup P_{n + j} β = Z_{n + j} \cup Z_{j} = Z_{j}, \end{matrix}$

$\begin{array}{l} Z_{q} β = (P_{0} \cup \cup_{\begin{array}{l} i = 1, \\ i \neq q \end{array}}^{n + k} P_{i}) β = Z_{n + j} \cup Z_{j} \cup Z_{q} = \overset{⌣}{D}, \\ \overset{⌣}{D} β = \cup_{i = 0}^{n + k} P_{i} β = Z_{n + j} \cup Z_{j} \cup Z_{q} = \overset{⌣}{D} \end{array}$

since $Z_{q} \cap Z_{n + j} \neq \emptyset, Z_{q} \cap (Z_{j} \ Z_{n + j}) = P_{n + j} \neq \emptyset, Z_{q} \cap (D \ Z_{j}) = P_{j} \neq \emptyset$ (see equality (2.0))

$\begin{matrix} α = δ \circ β = (Y_{n + j}^{δ} \times Z_{n + j} β) \cup (Y_{j}^{δ} \times Z_{j} β) \cup (Y_{q}^{δ} \times Z_{q} β) \cup (Y_{0}^{δ} \times \overset{⌣}{D} β) \\ = (Y_{n + j}^{δ} \times Z_{n + j}) \cup (Y_{j}^{δ} \times Z_{j}) \cup (Y_{q}^{δ} \times \overset{⌣}{D}) \cup (Y_{0}^{δ} \times \overset{⌣}{D}) \\ = (Y_{n + j}^{δ} \times Z_{n + j}) \cup (Y_{j}^{δ} \times Z_{j}) \cup ((Y_{q}^{δ} \cup Y_{0}^{δ}) \times \overset{⌣}{D}) = α, \end{matrix}$

if $Y_{n + j}^{δ} = Y_{n + j}^{α}$ , $Y_{j}^{δ} = Y_{j}^{α}$ and $Y_{q}^{δ} \cup Y_{0}^{δ} = Y_{0}^{α}$ . Last equalities are possible since $| Y_{q}^{δ} \cup Y_{0}^{δ} | \geq 1$ ( $| Y_{0}^{δ} | \geq 0$ , by preposition).

Lemma 2.3 is proved.

Lemma 2.4. Let $D \in Σ_{8} (X, n + k + 1)$ . If quasinormal representation of a binary relation a has a form $α = (Y_{j}^{α} \times Z_{j}) \cup (Y_{0}^{α} \times \overset{⌣}{D})$ , where $Y_{j}^{α}, Y_{0}^{α} \notin {\emptyset}$ , $j = 1, 2, \dots, n + k$ , then binary relation a is generated by elements of the elements of set $B (A_{0})$ .

Proof. Let quasinormal representation of the binary relations d and b have a form:

$\begin{array}{l} δ = (Y_{j}^{δ} \times Z_{j}) \cup (Y_{q}^{δ} \times Z_{q}) \cup (Y_{0}^{δ} \times \overset{⌣}{D}), \\ β = (Z_{j} \times Z_{j}) \cup ((\overset{⌣}{D} \ Z_{j}) \times Z_{q}) \cup ((X \ \overset{⌣}{D}) \times \overset{⌣}{D}), \end{array}$

where $Y_{j}^{δ}, Y_{q}^{δ} \notin {\emptyset}$ and $Z_{1} \leq Z_{j} \neq Z_{q} \leq Z_{n + k}$ . Then from the statements a), b) and c) of the Lemma 2.1 follows, that d and b are generated by elements of the set $B (A_{0})$ and

$Z_{j} β = Z_{j},$

$Z_{q} β = \overset{⌣}{D}$ , since $Z_{q} \cap (D \ Z_{j}) = P_{j} \neq \emptyset,$

$\overset{⌣}{D} β = \overset{⌣}{D};$

$\begin{matrix} δ \circ β = (Y_{j}^{δ} \times Z_{j} β) \cup (Y_{q}^{δ} \times Z_{q} β) \cup (Y_{0}^{δ} \times \overset{⌣}{D} β) \\ = (Y_{j}^{δ} \times Z_{j}) \cup (Y_{q}^{δ} \times \overset{⌣}{D}) \cup (Y_{0}^{δ} \times \overset{⌣}{D}) \\ = (Y_{j}^{δ} \times Z_{j}) \cup ((Y_{q}^{δ} \cup Y_{0}^{δ}) \times \overset{⌣}{D}) = α, \end{matrix}$

if $Y_{j}^{δ} = Y_{j}^{α}$ , $Y_{q}^{δ} \cup Y_{0}^{δ} = Y_{0}^{α}$ and $q = 1, 2, \dots, n + k$ . Last equalities are possible since $| Y_{q}^{δ} \cup Y_{0}^{δ} | \geq 1$ ( $| Y_{0}^{δ} | \geq 0$ by preposition).

Lemma 2.4 is proved.

Lemma 2.5. Let $D \in Σ_{8} (X, n + k + 1)$ . If quasinormal representation of a binary relation a has a form $α = (Y_{n + j}^{α} \times Z_{n + j}) \cup (Y_{j}^{α} \times Z_{j})$ , where $Y_{n + j}^{α}, Y_{j}^{α} \notin {\emptyset}$ , $j = 1, 2, \dots, k$ , then binary relation a is generated by elements of the elements of set $B (A_{0})$ .

Proof. Let quasinormal representation of a binary relations d, b have a form

$\begin{array}{l} δ = (Y_{n + j}^{δ} \times Z_{n + j}) \cup (Y_{q}^{δ} \times Z_{q}) \cup (Y_{0}^{δ} \times \overset{⌣}{D}), \\ β = (Z_{n + j} \times Z_{n + j}) \cup ((\overset{⌣}{D} \ Z_{n + j}) \times Z_{j}) \cup ((X \ \overset{⌣}{D}) \times \overset{⌣}{D}), \end{array}$

where $Y_{n + j}^{δ}, Y_{q}^{δ} \notin {\emptyset}$ , $j \neq q$ and $j = 1, 2, \dots, k$ . Then from the Lemma 2.2 follows that b is generated by elements of the set $B (A_{0})$ , $δ \in B (A_{0})$ and

$Z_{n + j} β = Z_{n + j},$

$Z_{q} β = Z_{n + j} \cup Z_{j} = Z_{j}$ , since $Z_{q} \cap Z_{n + j} \neq \emptyset$ , $Z_{q} \cap (\overset{⌣}{D} \ Z_{n + j}) = P_{n + j} \neq \emptyset, j \neq q$ (see equality(2.0))

$\overset{⌣}{D} β = Z_{j}$ since $\overset{⌣}{D} \cap (X \ \overset{⌣}{D}) = \emptyset,$

$\begin{matrix} δ \circ β = (Y_{n + j}^{δ} \times Z_{n + j} β) \cup (Y_{q}^{δ} \times Z_{q} β) \cup (Y_{0}^{δ} \times \overset{⌣}{D}) \\ = (Y_{n + j}^{δ} \times Z_{n + j}) \cup (Y_{q}^{δ} \times Z_{j}) \cup (Y_{0}^{δ} \times Z_{j}) \\ = (Y_{n + j}^{δ} \times Z_{n + j}) \cup ((Y_{q}^{δ} \cup Y_{0}^{δ}) \times Z_{j}) = α, \end{matrix}$

if $Y_{n + j}^{δ} = Y_{n + j}^{α}$ and $Y_{q}^{δ} \cup Y_{0}^{δ} = Y_{j}^{α}$ . Last equalities are possible since $| Y_{q}^{δ} \cup Y_{0}^{δ} | \geq 1$ ( $| Y_{0}^{δ} | \geq 0$ by preposition).

Lemma 2.5 is proved.

Lemma 2.6. Let $D \in Σ_{8} (X, n + k + 1)$ . Then the following statements are true:

1) If quasinormal representation of a binary relation a has a form $α = X \times Z_{j}$ $(j = 1, 2, \dots, k)$ , then binary relation a is generated by elements of the set $B (A_{0})$ .

2) If quasinormal representation of a binary relation a has a form $α = X \times \overset{⌣}{D}$ , then binary relation a is generated by elements of the set $B (A_{0})$ .

Proof. 1) Let $T \in D \ (D_{2} \cup D_{3})$ . If quasinormal representation of a binary relations d, b have a form

$\begin{array}{l} δ = (Y_{j}^{δ} \times Z_{j}) \cup (Y_{0}^{δ} \times \overset{⌣}{D}), \\ β = (Z_{n + j} \times Z_{n + j}) \cup ((Z_{j} \ Z_{n + j}) \times Z_{j}) \cup ((X \ \overset{⌣}{D}) \times \overset{⌣}{D}), \end{array}$

where $Y_{j}^{δ}, Y_{0}^{δ} \in {\emptyset}$ , $j = 1, 2, \dots, k$

$\begin{array}{l} Z_{n + j} \cup (Z_{j} \ Z_{n + j}) \cup (X \ \overset{⌣}{D}) \\ = (\cup_{\begin{array}{l} i = 0, \\ i \neq j, n + j \end{array}}^{n + k} P_{i}) \cup (P_{j} \cup P_{n + j}) \cup (X \ \overset{⌣}{D}) = \overset{⌣}{D} \cup (X \ \overset{⌣}{D}) = X \end{array}$

(see equalities (2.0) and (2.1)), then from the Lemma 2.4 follows that d is generated by elements of the set $B (A_{0})$ and from the Lemma 2.3 element b is generated by elements of the set $B (A_{0})$ and

$Z_{j} β = Z_{n + j} \cup Z_{j} = Z_{j},$

$\overset{⌣}{D} β = Z_{j}$ , since $\overset{⌣}{D} \cap (X \ \overset{⌣}{D}) = \emptyset,$

$δ \circ β = (Y_{j}^{δ} \times Z_{j} β) \cup (Y_{0}^{δ} \times \overset{⌣}{D} β) = (Y_{j}^{δ} \times Z_{j}) \cup (Y_{0}^{δ} \times Z_{j}) = X \times Z_{j} = α,$

since representation of a binary relation d is quasinormal.

The statement a) of the lemma 2.6 is proved.

2) Let quasinormal representation of a binary relation d have a form

$δ = (Z_{n + j} \times Z_{q}) \cup ((X \ Z_{n + j}) \times \overset{⌣}{D}),$

where $j \neq q$ , then from the Lemma 2.4 follows that d is generated by elements of the set $B (A_{0})$ and

$Z_{q} δ = (\cup_{\begin{array}{l} i = 0, \\ i \neq q \end{array}}^{n + k} P_{i}) δ = (\cup_{\begin{array}{l} i = 0, \\ i \neq q \end{array}}^{n + k} P_{i} δ) = Z_{q} \cup \overset{⌣}{D} = \overset{⌣}{D}, \overset{⌣}{D} δ = \overset{⌣}{D}$ , since $j \neq q, Z_{q} δ \cap Z_{n + 1} \neq \emptyset$ and $Z_{q} δ \cap (X \ Z_{n + 1}) \neq \emptyset;$

$\begin{matrix} δ \circ δ = (Z_{n + j} \times Z_{q} δ) \cup ((X \ Z_{n + j}) \times \overset{⌣}{D} δ) \\ = (Z_{n + j} \times \overset{⌣}{D}) \cup ((X \ Z_{n + j}) \times \overset{⌣}{D}) = X \times \overset{⌣}{D} = α \end{matrix}$ ,

since representation of a binary relation d is quasinormal.

The statement b) of the lemma 2.6 is proved.

Lemma 2.6 is proved.

Lemma 2.7. Let $D \in Σ_{8} (X, n + k + 1)$ . Then the following statements are true:

a) If $| X \ \overset{⌣}{D} | \geq 1$ and $T \in D_{2} \cup D_{3}$ , then binary relation $α = X \times T$ is generated by elements of the elements of set $B (A_{0})$ ;

b) If $X = \overset{⌣}{D}$ and $T \in D_{2} \cup D_{3}$ , then binary relation $α = X \times T$ is external element for the semigroup $B_{X} (D)$ .

Proof. 1) If quasinormal representation of a binary relation d has a form

$δ = (Y_{0}^{δ} \times \overset{⌣}{D}) \cup \cup_{j = k + 1}^{n + k} (Y_{j}^{δ} \times Z_{j})$ ,

where $Y_{j}^{δ} \neq \emptyset$ for all $j = k + 1, k + 2, \dots, n + k$ , then $δ \in B (A_{0}) \ {α}$ . Let quasinormal representation of a binary relations b have a form

$β = (\overset{⌣}{D} \times T) \cup \underset{t^{'} \in X \ \overset{⌣}{D}}{\cup} ({t^{'}} \times f (t^{'}))$ , where f is any mapping of the set $X \ \overset{⌣}{D}$ in the set $(D_{2} \cup D_{3}) \ {T}$ . It is easy to see, that $β \neq α$ and two elements of the set $D_{2} \cup D_{3}$ belong to the semilattice $V (D, β)$ , i.e. $δ \in B (A_{0}) \ {α}$ . In this case we have that $Z_{j} β = T$ for all $j = k + 1, k + 2, \dots, n + k$ .

$\begin{matrix} δ \circ β = δ = (Y_{0}^{δ} \times \overset{⌣}{D} β) \cup \cup_{j = k + 1}^{n + k} (Y_{j}^{δ} \times Z_{j} β) \\ = (Y_{0}^{δ} \times T) \cup \cup_{j = k + 1}^{n + k} (Y_{j}^{δ} \times T) \\ = ((Y_{0}^{δ} \cup \cup_{j = k + 1}^{n + k} Y_{j}^{δ}) \times T) = X \times T = α, \end{matrix}$

since the representation of a binary relation d is quasinormal. Thus, the element a is generated by elements of the set $B (A_{0})$ .

The statement a) of the lemma 2.7 is proved.

2) Let $X = \overset{⌣}{D}$ , $α = X \times T$ , for some $T \in D_{2} \cup D_{3}$ and $α = δ \circ β$ for some $δ, β \in B_{X} (D) \ {α}$ . Then we obtain that $Z_{j} β = T$ since T is a minimal element of the semilattice D.

Now, let subquasinormal representations $\bar{β}$ of a binary relation b have a form

$\bar{β} = ((\cup_{i = 0}^{n + k} P_{i}) \times T) \cup \underset{t^{'} \in X \ \overset{⌣}{D}}{\cup} ({t^{'}} \times {\bar{β}}_{2} (t^{'}))$ ,

where ${\bar{β}}_{1} = (\begin{array}{l} P_{0} P_{1} P_{2} \dots P_{n + k} \\ T T T \dots T \end{array})$ is normal mapping. But complement mapping ${\bar{β}}_{2}$ is empty, since $X \ \overset{⌣}{D} = \emptyset$ , i.e. in the given case, subquasinormal representation $\bar{β}$ of a binary relation b is defined uniquely. So, we have that $β = \bar{β} = X \times T = α$ (see property 2) in the case 1.1), which contradict the condition, that $β \notin B_{X} (D) \ {α}$ .

Therefore, if $X = \overset{⌣}{D}$ and $α = X \times T$ , for some $T \in D_{2} \cup D_{3}$ , then a is external element of the semigroup $B_{X} (D)$ .

The statement 2) of the Lemma 2.7 is proved.

Lemma 2.7 is proved.

Theorem 2.1. Let $D \in Σ_{8} (X, n + k + 1)$ , $k \geq 3$ , and

$D_{1} = {Z_{1}, Z_{2}, \dots, Z_{k}}, D_{2} = {Z_{k + 1}, Z_{k + 2}, \dots, Z_{n}}, D_{3} = {Z_{n + 1}, Z_{n + 2}, \dots, Z_{n + k}};$

$A_{1} = {{Z_{n + q}, Z_{q}, \overset{⌣}{D}}}, where q = 1, 2, \dots, k;$

$A_{2} = {{Z_{j}, \overset{⌣}{D}}}, where j = 1, 2, \dots, n + k;$

$A_{3} = {{Z_{n + j}, Z_{j}}}, where j = 1, 2, \dots, k;$

$A_{4} = {{Z_{j}}, {\overset{⌣}{D}}}, where j = 1, 2, \dots, n + k;$

$A_{0} = {V (D, α) \subset D | V (D, α) \notin A_{1} \cup A_{2} \cup A_{3} \cup A_{4}},$

$B (A_{0}) = {α \in B_{X} (D) | V (D, α) \in A_{0}},$

$B_{0} = {X \times T | T \notin D_{2} \cup D_{3}}$

Then the following statements are true:

1) If $| X \ \overset{⌣}{D} | \geq 1$ , then the $S_{0} = B (A_{0})$ is irreducible generating set for the semigroup $B_{X} (D)$ ;

2) If $X = \overset{⌣}{D}$ , then the $S_{1} = B_{0} \cup B (A_{0})$ is irreducible generating set for the semigroup $B_{X} (D)$ .

Proof. Let $D \in Σ_{8} (X, n + k + 1)$ , $k \geq 3$ and $| X \ \overset{⌣}{D} | \geq 1$ . First, we proved that every element of the semigroup $B_{X} (D)$ is generated by elements of the set $S_{0}$ . Indeed, let a be an arbitrary element of the semigroup $B_{X} (D)$ . Then quasinormal representation of a binary relation a has a form

$α = (Y_{0}^{α} \times \overset{⌣}{D}) \cup \cup_{i = 1}^{n + k} (Y_{i}^{α} \times Z_{i})$ ,

where $\cup_{i = 0}^{n + k} Y_{i}^{α} = X$ and $Y_{i}^{α} \cap Y_{j}^{α} = \emptyset$ $(0 \leq i \neq j \leq n + k)$ . For the $V (X^{*}, α)$ we consider the following cases:

1) If $V (X^{*}, α) \notin A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$ , then $α \in B (A_{0}) \subseteq S_{0}$ by definition of a set $S_{0}$ .

Now, let $V (X^{*}, α) \in A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$ .

2) If $V (X^{*}, α) \in A_{1}$ , then quasinormal representation of a binary relation a has a form $α = (Y_{n + j}^{α} \times Z_{n + j}) \cup (Y_{j}^{α} \times Z_{j}) \cup (Y_{0}^{α} \times \overset{⌣}{D})$ , where $Y_{n + j}^{α}, Y_{j}^{α}, Y_{0}^{α} \notin {\emptyset}$ $(j = 1, 2, \dots, k)$ and from the Lemma 2.3 follows that a is generated by elements of the elements of set $B (A_{0}) \subseteq S_{0}$ by definition of a set $S_{0}$ .

3) If $V (X^{*}, α) \in A_{2}$ , then quasinormal representation of a binary relation a has a form $α = (Y_{j}^{α} \times Z_{j}) \cup (Y_{0}^{α} \times \overset{⌣}{D})$ , where $Y_{j}^{α}, Y_{0}^{α} \notin {\emptyset}$ , $j = 1, 2, \dots, n + k$ and from the Lemma 2.4 follows that a is generated by elements of the elements of set $B (A_{0}) \subseteq S_{0}$ by definition of a set $S_{0}$ .

4) If $V (X^{*}, α) \in A_{3}$ , then quasinormal representation of a binary relation a has a form $α = (Y_{n + j}^{α} \times Z_{n + j}) \cup (Y_{j}^{α} \times Z_{j})$ , where $Y_{n + j}^{α}, Y_{j}^{α} \notin {\emptyset}$ , $j = 1, 2, \dots, k$ and from the Lemma 2.5 follows that a is generated by elements of the elements of set $B (A_{0}) \subseteq S_{0}$ by definition of a set $S_{0}$ .

Now, let $V (X^{*}, α) \in A_{4}$ , then quasinormal representation of a binary relation a has a form $α = X \times \overset{⌣}{D}$ , or $α = X \times Z_{j}$ , where $j = 1, 2, \dots, n + k$ .

5) If $α = X \times \overset{⌣}{D}$ , then from the statement b) of the Lemma 2.6 follows that binary relation a is generated by elements of the set $B (A_{0})$ .

6) If $α = X \times Z_{j}$ , where $j = 1, 2, \dots, n + k$ , then from the statement a) of the Lemma 2.6 and 2.7 follows that binary relation a is generated by elements of the set $B (A_{0})$ .

Thus, we have that $S_{0}$ is a generating set for the semigroup $B_{X} (D)$ .

If $| X \ \overset{⌣}{D} | \geq 1$ , then the set $S_{0}$ is an irreducible generating set for the semigroup $B_{X} (D)$ since, $S_{0}$ is a set external elements of the semigroup $B_{X} (D)$ .

The statement a) of the Theorem 2.1 is proved.

Now, let $X = \overset{⌣}{D}$ . First, we proved that every element of the semigroup $B_{X} (D)$ is generated by elements of the set $S_{1}$ . The cases 1), 2), 3), 4) and 5) are proved analogously of the cases 1), 2), 3), 4) and 5 given above and consider case, when $V (X^{*}, α) \in A_{1}$ .

If $V (X^{*}, α) = Z_{j}$ , where $j = 1, 2, \dots, k$ , then from the statement a) of the Lemma 2.7 follows that binary relation a is generated by elements of the set $B (A_{0})$ .

If $V (X^{*}, α) = Z_{j}$ , where $Z_{j} \in D_{2} \cup D_{3}$ , then from the statement b) of the Lemma 2.6 follows that binary relation $α = X \times T$ is external element for the semigroup $B_{X} (D)$ .

Thus, we have that $S_{1}$ is a generating set for the semigroup $B_{X} (D)$ .

If $X = \overset{⌣}{D}$ , then the set $S_{1}$ is an irreducible generating set for the semigroup $B_{X} (D)$ since $S_{1}$ is a set external elements of the semigroup $B_{X} (D)$ .

The statement b) of the Theorem 2.1 is proved.

Theorem 2.1 is proved.

Corollary 2.1. Let $D \in Σ_{8} (X, n + k + 1)$ $(k \geq 3)$ and

$D_{1} = {Z_{1}, Z_{2}, \dots, Z_{k}}, D_{2} = {Z_{k + 1}, Z_{k + 2}, \dots, Z_{n}}, D_{3} = {Z_{n + 1}, Z_{n + 2}, \dots, Z_{n + k}};$

$A_{1} = {{Z_{n + q}, Z_{q}, \overset{⌣}{D}}}, where q = 1, 2, \dots, k;$

$A_{2} = {{Z_{j}, \overset{⌣}{D}}}, where j = 1, 2, \dots, n + k;$

$A_{3} = {{Z_{n + j}, Z_{j}}}, where j = 1, 2, \dots, k;$

$A_{4} = {{Z_{j}}, {\overset{⌣}{D}}}, where j = 1, 2, \dots, n + k;$

$A_{0} = {V (D, α) \subset D | V (D, α) \notin A_{1} \cup A_{2} \cup A_{3} \cup A_{4}},$

$B (A_{0}) = {α \in B_{X} (D) | V (D, α) \in A_{0}},$

$B_{0} = {X \times T | T \notin D_{2} \cup D_{3}}$

Then the following statements are true:

1) If $| X \ \overset{⌣}{D} | \geq 1$ , then $S_{0} = B (A_{0})$ is the uniquely defined generating set for the semigroup $B_{X} (D)$ ;

2) If $X = \overset{⌣}{D}$ , then $S_{1} = B_{0} \cup B (A_{0})$ is the uniquely defined generating set for the semigroup $B_{X} (D)$ .

Proof. It is well known, that if B is all external elements of the semigroup $B_{X} (D)$ and $B^{'}$ is any generated set for the $B_{X} (D)$ , then $B \subseteq B^{'}$ (see [1] [2] Lemma 1.15.1). From this follows that the sets $S_{0} = B (A_{0})$ and $S_{1} = B_{0} \cup B (A_{0})$ are defined uniquely, since they are sets external elements of the semigroup $B_{X} (D)$ .

Corollary 2.1 is proved.

It is well-known, that if B is all external elements of the semigroup $B_{X} (D)$ and $B^{'}$ is any generated set for the $B_{X} (D)$ , then $B \subseteq B^{'}$ (Definition 1.1).

In this article, we find irredusible generating set for the complete semigroups of binary relations defined by X-semilattices of unions of the class $Σ_{8} (X, n + k + 1)$ $(k \geq 3)$ . This generating set is uniquely defined, since they are defined by elements of the external elements of the semigroup $B_{X} (D)$ .

Cite this paper
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. doi: 10.4236/am.2018.94028.

References
[1] Diasamidze, Y. and Makharadze, S. (2013) Complete Semigroups of Binary Relations. Kriter, Turkey, 1-519.

[2] Я. И. Диасамидзе, Ш. И. Махарадзе (2017) Полные полугруппы бинарных отношений. Lambert Academic Publishing, 1-692.