1. Introduction

Applied Mathematics

2152-7385

Scientific Research Publishing

10.4236/am.2024.152010

AM-131636

Articles

Physics&Mathematics

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

Nino

Tsinaridze

₁^*

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

29022024

15021691976, February 202426, February 2024 29, February 2024

2014

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

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.

Semigroup Semilattice Binary Relation

1. Introduction

Let X ≠ ∅ , 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 = ∪ x ∈ X ( { x } × f ( x ) ) . The set of all such

α f 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 ∅ an empty subset of the set X or an empty binary relation. The condition ( x , y ) ∈ α will be written in the form x α y .

Let x , y ∈ X , Y ⊆ X , α ∈ B X ( D ) , D ⌣ = ∪ Y ∈ D Y and T ∈ D . We denote by the symbols y α , Y α , V ( D , α ) , X ∗ and V ( X ∗ , α ) the following sets:

y α = { x ∈ X | y α x } , Y α = ∪ y ∈ Y y α , V ( D , α ) = { Y α | Y ∈ D } , X ∗ = { Y | ∅ ≠ Y ⊆ X } , V ( X ∗ , α ) = { Y α | ∅ ≠ Y ⊆ X } , D T = { Z ∈ D | T ⊆ Z } , Y T α = { y ∈ X | y α = T } .

Theorem 1.1. Let D = { D ⌣ , Z 1 , Z 2 , ⋯ , Z m − 1 } be some finite X-semilattice of unions and C ( D ) = { P 0 , P 1 , P 2 , ⋯ , P m − 1 } be the family of sets of pairwise nonintersecting subsets of the set X (the set ∅ can be repeated several times). If φ is a mapping of the semilattice D on the family of sets C ( D ) which satisfies the conditions

φ = ( D ⌣ Z 1 Z 2 ⋯ Z m − 1 P 0 P 1 P 2 ⋯ P m − 1 )

and D ^ Z = D \ D Z , then the following equalities are valid:

D ⌣ = P 0 ∪ P 1 ∪ P 2 ∪ ⋯ ∪ P m − 1 , Z i = P 0 ∪ ∪ T ∈ D ^ Z i φ ( T ) .

In the sequel these equalities will be called formal. The parameters P i ( 0 < i ≤ 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 ≤ j ≤ m − 1 ) , which can be empty sets too are called completeness sources.

It is proved that under the mapping φ the number of covering elements of the pre-image of a bases source is always equal to one, while under the mapping φ 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 α = ∪ T ∈ D ( Y T α × T ) of binary relation α is called quasinormal, if ∪ T ∈ D Y T α = X and Y T α ∩ Y T ′ α = ∅ for any T , T ′ ∈ D , T ≠ T ′ (see [1] Definition 1.2, [2] , [3] chapter 1.1).

Definition 1.2. Let α , β ⊆ X × X . Their product δ = α ∘ β is defined as follows: x δ y ( x , y ∈ X ) if there exists an element z ∈ X such that x α z β y (see [1] Definition 1.3, [1] , chapter 1.3).

Definition 1.3. We say that an element α of the semigroup B X ( D ) is external if α ≠ δ ∘ β for all δ , β ∈ B X ( D ) \ { α } (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 ( D ) and B ′ is any generated set for the B X ( D ) , then B ⊆ B ′ (see [2] [3] Lemma 1.15.1).

2. Result

Let Σ 8 ( X , 5 ) be a class of all X-semilattices of unions, whose every element is isomorphic to an X-semilattice of unions D = { T 4 , T 3 , T 2 , T 1 , T 0 } , which satisfies the condition:

T 4 ⊂ T 2 ⊂ T 0 , T 3 ⊂ T 1 ⊂ T 0 , T 4 \ T 3 ≠ ∅ , T 3 \ T 4 ≠ ∅ , T 2 \ T 1 ≠ ∅ , T 1 \ T 2 ≠ ∅ , T 2 \ T 3 ≠ ∅ , T 3 \ T 2 ≠ ∅ , T 4 \ T 1 ≠ ∅ , T 1 \ T 4 ≠ ∅ , T 4 ∪ T 3 = T 4 ∪ T 1 = T 3 ∪ T 2 = T 1 ∪ T 2 = T 0 .

(see Figure 1). It is easy to see that D ˜ = { T 4 , T 3 , T 2 , T 1 } is irreducible generating set of the semilattice D.

Let C ( D ) = { P 0 , P 1 , P 2 , P 3 , P 4 } is a family of sets, where φ = ( T 0 T 1 T 2 T 3 T 4 P 0 P 1 P 2 P 3 P 4 ) is a mapping of the semilattice D onto the family of sets C ( D ) 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:

T 0 = P 0 ∪ P 1 ∪ P 2 ∪ P 3 ∪ P 4 , T 1 = P 0 ∪ P 2 ∪ P 3 ∪ P 4 , T 2 = P 0 ∪ P 1 ∪ P 3 ∪ P 4 , T 3 = P 0 ∪ P 2 ∪ P 4 , T 4 = P 0 ∪ P 1 ∪ P 3 . (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 | X | ≥ 4 since | P 4 | ≥ 1 , | P 3 | ≥ 1 , | P 2 | ≥ 1 , | P 1 | ≥ 1 (see Theorem 1.1).

From the formal Equalities (2.1) immediately follows

P 4 = T 2 \ T 4 , P 3 = ( T 2 ∩ T 1 ) \ T 3 , P 2 = T 3 \ T 2 = T 0 \ T 2 , P 1 = T 4 \ T 1 , P 0 = T 4 ∩ T 3 . (2.2)

2.1. Generating Sets of the Complete Semigroup of Binary Relations Defined by Semilattices of the Class Σ 8 ( X , 5 ) , When T 4 ∩ T 3 ≠ ∅

In the sequel, we denoted all semilattices D = { T 4 , T 3 , T 2 , T 1 , T 0 } of the class Σ 8 ( X , 5 ) by symbol Σ 8.0 ( X , 5 ) , for which T 4 ∩ T 3 ≠ ∅ . Of the last inequality from the formal Equalities (2.1) of a semilattise D follows that T 4 ∩ T 3 = P 0 ≠ ∅ , i.e. | X | ≥ 5 .

We denoted by symbols A 4 , A 3 , A 2 , A 1 the following sets:

A 4 = { { T 4 , T 3 , T 2 , T 0 } , { T 4 , T 3 , T 1 , T 0 } , { T 4 , T 2 , T 1 , T 0 } , { T 3 , T 2 , T 1 , T 0 } } , A 3 = { { T 4 , T 3 , T 0 } , { T 4 , T 1 , T 0 } , { T 3 , T 2 , T 0 } , { T 4 , T 2 , T 0 } , { T 3 , T 1 , T 0 } , { T 2 , T 1 , T 0 } } , A 2 = { { T 4 , T 2 } , { T 4 , T 0 } , { T 3 , T 1 } , { T 3 , T 0 } , { T 2 , T 0 } { T 1 , T 0 } } , A 1 = { { T 4 } , { T 3 } , { T 2 } , { T 1 } , { T 0 } } .

Lemma 2.1.1. Let D ∈ Σ 8.0 ( X , 5 ) . Then the following statements are true:

a) Let T 3 , T 4 ∈ V ( D , α ) , then α is external element of the semigroup B X ( D ) ;

b) Let Z ∈ { T 2 , T 1 } , Z ′ ∈ { T 4 , T 3 } . If Z ′ ⊄ Z and Z , Z ′ ∈ V ( D , α ) , then α is external element of the semigroup B X ( D ) ;

c) Let Z , Z ′ ∈ { T 2 , T 1 } and Z ≠ Z ′ . If V ( D , α ) = { T 2 , T 1 , T 0 } , then α is external element of the semigroup B X ( D ) .

Proof. Let α = δ ∘ β for some δ , β ∈ B X ( D ) \ { α } . If quasinormal representation of binary relation δ has a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) ,

then

α = δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 3 δ × T 3 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) .(2.1.1)

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

T 0 β = P 0 β ∪ P 1 β ∪ P 2 β ∪ P 3 β ∪ P 4 β , T 1 β = P 0 β ∪ P 2 β ∪ P 3 β ∪ P 4 β , T 2 β = P 0 β ∪ P 1 β ∪ P 3 β ∪ P 4 β , T 3 β = P 0 β ∪ P 2 β ∪ P 4 β , T 4 β = P 0 β ∪ P 1 β ∪ P 3 β . (2.1.2)

where P k β ≠ ∅ for any P k ≠ ∅ ( k = 0 , 1 , 2 , 3 , 4 ) and β ∈ B X ( D ) . Indeed, by preposition P k ≠ ∅ for any k = 0 , 1 , 2 , 3 , 4 and β ≠ ∅ since ∅ ∉ D . Let y ∈ P k for some y ∈ X . Then y ∈ T 0 , β = α f for some f : X → D and α f = ∪ x ∈ X ( { x } × f ( x ) ) ⊇ { y } × f ( y ) , i.e. there exists an element t ∈ f ( y ) for which y α f t and y β t . Of this and by definition of a set P k β we obtain that t ∈ P k β since y ∈ P k , y β t . Thus, we have that P k β ≠ ∅ , i.e. P k β ∈ D for any k = 0 , 1 , 2 , 3 , 4 .

Now, let T i β = Z and T j β = Z ′ for some 0 ≤ i ≠ j ≤ 4 and Z ≠ Z ′ , Z , Z ′ ∈ { T 4 , T 3 } , then from the Equalities (2.2) follows that Z = P 0 β = Z ′ since Z and Z ′ are minimal elements of the semilattice D. The equality Z = Z ′ contradicts the inequality Z ≠ Z ′ .

The statement a) of the Lemma 2.1.1 is proved.

Let T i β = Z ′ , where Z ′ ∈ { T 4 , T 3 } and T j β = Z , where Z ∈ { T 2 , T 1 } for some 0 ≤ i ≠ j ≤ 4 . If 0 ≤ i ≤ 4 , then from the formal equalities of a semilattice D we obtain that

T 0 β = P 0 β ∪ P 1 β ∪ P 2 β ∪ P 3 β ∪ P 4 β = P 0 β = P 1 β = P 2 β = P 3 β = P 4 β = Z ′ , T 1 β = P 0 β ∪ P 2 β ∪ P 3 β ∪ P 4 β = P 0 β = P 2 β = P 3 β = P 4 β = Z ′ , T 2 β = P 0 β ∪ P 1 β ∪ P 3 β ∪ P 4 β = P 0 β = P 1 β = P 3 β = P 4 β = Z ′ , T 3 β = P 0 β ∪ P 2 β ∪ P 4 β = P 0 β = P 2 β = P 4 β = Z ′ , T 4 β = P 0 β ∪ P 1 β ∪ P 3 β = P 0 β = P 1 β = P 3 β = Z ′ .

since Z ′ is minimal element of the semilattice D. Now, let i ≠ j .

1) If T 0 β = P 0 β = P 1 β = P 2 β = P 3 β = P 4 β = Z ′ and j = 1 , 2 , 3 , 4 , then we have

Z = T 1 β = T 2 β = T 3 β = T 4 β = Z ′ ,

which contradicts the inequality Z ≠ Z ′ .

2) If T 1 β = P 0 β = P 2 β = P 3 β = P 4 β = Z ′ and j = 0 , 2 , 3 , 4 , then we have

Z = T 0 β = T 2 β = T 4 β = Z ′ ∪ P 1 β , where P 1 β ∈ D

Last equalities are impossible, since Z ≠ Z ′ ∪ T for any T ∈ D and Z ≠ Z ′ by definition of a semilattice D.

3) If T 2 β = P 0 β = P 1 β = P 3 β = P 4 β = Z ′ and j = 0 , 1 , 3 , 4 , then we have

Z = T 0 β = T 2 β = T 4 β = Z ′ ∪ P 1 β , where P 1 β ∈ D

Last equalities are impossible since Z ≠ Z ′ ∪ T for any T ∈ D and Z ≠ Z ′ by definition of a semilattice D.

4) If T 3 β = P 0 β = P 2 β = P 4 β = Z ′ and j = 0 , 1 , 2 , 4 , then we have

Z = T 0 β = T 2 β = T 4 β = Z ′ ∪ P 1 β ∪ P 3 β , Z = T 1 β = Z ′ ∪ P 3 β , where P 1 β , P 3 β ∈ D

Last equalities are impossible since Z ≠ Z ′ ∪ T ∪ T ′ and Z ≠ Z ′ ∪ T for any T , T ′ ∈ D , by definition of a semilattice D.

5) If T 4 β = P 0 β = P 1 β = P 3 β = Z ′ and j = 0 , 1 , 2 , 3 , then we have

Z = T 0 β = T 1 β = T 3 β = Z ′ ∪ P 2 β ∪ P 4 β , Z = T 2 β = Z ′ ∪ P 4 β , where P 2 β , P 4 β ∈ D

Last equalities are impossible since Z ≠ Z ′ ∪ T ∪ T ′ and Z ≠ Z ′ ∪ T for any T , T ′ ∈ D , by definition of a semilattice D.

The statement b) of the Lemma 2.1.1 is proved.

Let Z , Z ′ ∈ { T 2 , T 1 } , T i β = Z , T j β = Z ′ and Z ≠ Z ′ . If T i β = Z where 0 ≤ i ≠ j ≤ 4 , we consider the following cases:

6) i = 0 , j = 1 , 2 , 3 , 4 . Then from the Equality (2.1.2) follows that Z ⊂ Z ′ , 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 ′ ⊂ Z , or Z ⊂ Z ′ which contradicts the definition of a semilattice D;

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

{ T 1 β = ( P 0 β ∪ P 3 β ∪ P 4 β ) ∪ P 2 β , T 2 β = ( P 0 β ∪ P 3 β ∪ P 4 β ) ∪ P 1 β ,

where P 0 β ∪ P 3 β ∪ P 4 β , P 2 β , P 1 β ∈ D , i.e. there exists such elements T , T ′ , T ″ ∈ D , for which Z = T ∪ T ′ and Z ′ = T ∪ T ″ . But such element T ∈ 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 ′ ⊂ Z , or Z ⊂ Z ′ 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 ∪ T ′ and Z ′ = T ∪ T ″ . But such element T ∈ 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 ′ ⊂ Z , which contradicts the definition of a semilattice D;

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

{ T 2 β = P 0 β ∪ ( P 2 β ∪ P 3 β ∪ P 4 β ) , T 2 β = P 0 β ∪ ( P 1 β ∪ P 3 β ) ,

where P 0 β , P 2 β ∪ P 3 β ∪ P 4 β , P 1 β ∪ P 3 β ∈ D , i.e. there exist such elements T , T ′ , T ″ ∈ D , for which Z = T ∪ T ′ and Z ′ = T ∪ T ″ . But such element T ∈ 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 ⊂ Z ′ which contradicts the definition of a semilattice D;

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

{ T 1 β = P 0 β ∪ ( P 2 β ∪ P 3 β ∪ P 4 β ) , T 3 β = P 0 β ∪ ( P 2 β ∪ P 4 β ) ,

where P 0 β , P 2 β ∪ P 3 β ∪ P 4 β , P 2 β ∪ P 4 β ∈ D , i.e. there exist such elements T , T ′ , T ″ ∈ D , for which Z = T ∪ T ′ and Z ′ = T ∪ T ″ . But such element T ∈ 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 ∈ Σ 8.0 ( X , 5 ) . By symbols A 0 , B ( A 0 ) and B 0 we denoted the following sets:

A 0 = { { T 4 , T 3 , T 2 , T 0 } , { T 4 , T 3 , T 1 , T 0 } , { T 4 , T 2 , T 1 , T 0 } , { T 3 , T 2 , T 1 , T 0 } , { T 4 , T 3 , T 0 } , { T 4 , T 1 , T 0 } , { T 3 , T 2 , T 0 } , { T 2 , T 1 , T 0 } } , B ( A 0 ) = { α ∈ B X ( D ) | V ( X ∗ , α ) ∈ A 0 } ; B 0 = { α ∈ B X ( D ) | V ( X ∗ , α ) = D } .

Remark, that the sets B 0 and B ( A 0 ) are external elements for the semigroup B X ( D ) .

Lemma 2.1.2. Let D ∈ Σ 8.0 ( X , 5 ) . Then the following statements are true:

a) If quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 2 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

b) If quasinormal representation of a binary relation α has a form

α = ( Y 3 α × T 3 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 3 α , Y 1 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

Proof. 1). Let quasinormal representation of binary relations δ and β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 4 × T 4 ) ∪ ( ( T 2 \ T 4 ) × T 2 ) ∪ ( ( T 0 \ T 2 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 α , Y 2 α , Y 1 α ∉ { ∅ } ,

T 4 ∪ ( T 2 \ T 4 ) ∪ ( T 0 \ T 2 ) ∪ ( X \ T 0 ) = ( P 0 ∪ P 1 ∪ P 3 ) ∪ P 4 ∪ P 2 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X ,

(see Equalities (2.1) and (2.2)), then δ , β ∈ B ( A 0 ) and

T 4 β = T 4 , T 2 β = ( P 0 ∪ P 1 ∪ P 3 ∪ P 4 ) β = T 4 ∪ T 2 = T 2 , T 1 β = ( P 0 ∪ P 2 ∪ P 3 ∪ P 4 ) β = T 4 ∪ T 1 = T 0 , T 0 β = T 0 . α = δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × T 4 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 0 ) ∪ ( Y 0 δ × T 0 ) = ( Y 4 δ × T 4 ) ∪ ( Y 2 δ × T 2 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × T 0 ) = α ,

if Y 4 δ = Y 4 α , Y 2 δ = Y 2 α and Y 1 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 1 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement a) of the lemma 2.1.2 is proved.

2) Let quasinormal representation of binary relations δ and β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 3 × T 3 ) ∪ ( ( T 0 \ T 1 ) × T 2 ) ∪ ( ( T 1 \ T 3 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 3 α , Y 2 α , Y 1 α ∉ { ∅ } ,

T 3 ∪ ( T 0 \ T 1 ) ∪ ( T 1 \ T 3 ) ∪ ( X \ T 0 ) = ( P 0 ∪ P 2 ∪ P 4 ) ∪ P 1 ∪ P 3 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X ,

(see Equalities (2.1) and (2.2)), then δ , β ∈ B ( A 0 ) and

T 4 β = T 3 , T 2 β = ( P 0 ∪ P 1 ∪ P 3 ∪ P 4 ) β = T 3 ∪ T 2 ∪ T 1 = T 0 , T 1 β = ( P 0 ∪ P 2 ∪ P 3 ∪ P 4 ) β = T 3 ∪ T 1 = T 0 , T 0 β = T 0 . α = δ ∘ β = ( Y 3 δ × T 3 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 0 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) = ( Y 3 δ × T 3 ) ∪ ( Y 1 δ × T 1 ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × T 0 ) = α ,

if Y 3 δ = Y 3 α , Y 1 δ = Y 1 α and Y 2 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 2 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 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 ∈ Σ 8.0 ( X , 5 ) . Then the following statements are true:

a) If quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 2 α × T 2 ) ,

where Y 4 α , Y 2 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

b) If quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

c) If quasinormal representation of a binary relation α has a form

α = ( Y 3 α × T 3 ) ∪ ( Y 1 α × T 1 ) ,

where Y 3 α , Y 1 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

d) If quasinormal representation of a binary relation α has a form

α = ( Y 3 α × T 3 ) ∪ ( Y 0 α × T 0 ) ,

where Y 3 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

e) If quasinormal representation of a binary relation α has a form

α = ( Y 2 α × T 2 ) ∪ ( Y 0 α × T 0 ) ,

where Y 2 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

f) If quasinormal representation of a binary relation α has a form

α = ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 1 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

g) If quasinormal representation of a binary relation α has a form α = X × T 2 , then α is generating by elements of the elements of set B ( A 0 ) ;

h) If quasinormal representation of a binary relation α has a form α = X × T 1 , then α is generating by elements of the elements of set B ( A 0 ) ;

i) If quasinormal representation of a binary relation α has a form α = X × T 0 , then α is generating by elements of the elements of set B ( A 0 ) .

Proof. 1) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 4 × T 4 ) ∪ ( ( T 0 \ T 4 ) × T 2 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 δ , Y 1 δ ∉ { ∅ } .

T 4 ∪ ( T 0 \ T 4 ) ∪ ( X \ T 0 ) = ( P 0 ∪ P 1 ∪ P 3 ) ∪ ( P 2 ∪ P 4 ) ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X .

Then from the statement a) of the Lemma 2.1.2 follows that β is generating by elements of the set B ( A 0 ) , δ ∈ B ( A 0 ) and

T 4 β = T 4 , T 1 β = T 4 ∪ T 2 = T 2 , T 0 β = T 2 . δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × T 4 ) ∪ ( Y 1 δ × T 2 ) ∪ ( Y 0 δ × T 2 ) = ( Y 4 δ × T 4 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × T 2 ) = α ,

if Y 4 δ = Y 4 α , Y 1 δ ∪ Y 0 δ = Y 2 α . Last equalities are possible since | Y 1 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement a) of the lemma 2.1.3 is proved.

2) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 4 × T 4 ) ∪ ( ( T 0 \ T 4 ) × T 3 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 δ , Y 1 δ ∉ { ∅ } .

T 4 ∪ ( T 0 \ T 4 ) ∪ ( X \ T 0 ) = ( P 0 ∪ P 1 ∪ P 3 ) ∪ ( P 2 ∪ P 4 ) ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X .

Then from δ , β ∈ B ( A 0 ) and

T 4 β = T 4 , T 1 β = T 4 ∪ T 3 = T 0 , T 0 β = T 0 . δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × T 4 ) ∪ ( Y 1 δ × T 0 ) ∪ ( Y 0 δ × T 0 ) = ( Y 4 δ × T 4 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × T 0 ) = α ,

if Y 4 δ = Y 4 α , Y 1 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 1 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement b) of the lemma 2.1.3 is proved.

3) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 3 × T 3 ) ∪ ( ( T 0 \ T 3 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 δ , Y 2 δ ∉ { ∅ } .

T 3 ∪ ( T 0 \ T 3 ) ∪ ( X \ T 0 ) = ( P 0 ∪ P 2 ∪ P 4 ) ∪ ( P 1 ∪ P 3 ) ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X .

Then from the statement b) of the Lemma 2.1.2 follows that β is generating by elements of the set B ( A 0 ) , δ ∈ B ( A 0 ) and

T 3 β = T 3 , T 2 β = T 3 ∪ T 1 = T 1 , T 0 β = T 1 . δ ∘ β = ( Y 3 δ × T 3 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 1 ) ∪ ( Y 0 δ × T 1 ) = ( Y 3 δ × Z 3 ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × T 1 ) = α ,

if Y 3 δ = Y 3 α , Y 2 δ ∪ Y 0 δ = Y 1 α . Last equalities are possible since | Y 2 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement c) of the lemma 2.1.3 is proved.

4) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 3 × T 3 ) ∪ ( ( T 0 \ T 3 ) × T 2 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 3 δ , Y 2 δ ∉ { ∅ } . Then δ , β ∈ B ( A 0 ) and

T 3 β = T 3 , T 2 β = T 3 ∪ T 2 = T 0 , T 0 β = T 0 . δ ∘ β = ( Y 3 δ × T 3 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 0 ) ∪ ( Y 0 δ × T 0 ) = ( Y 3 δ × T 3 ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × T 0 ) = α ,

if Y 3 δ = Y 3 α , Y 2 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 2 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement d) of the lemma 2.1.3 is proved.

5) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 0 δ × T 0 ) , β = ( ( ( T 2 ∩ T 1 ) \ T 3 ) × T 4 ) ∪ ( ( T 2 \ T 1 ) × T 2 ) ∪ ( ( X \ T 4 ) × T 0 ) ,

where Y 4 δ , Y 0 δ ∉ { ∅ } ,

( ( T 2 ∩ T 1 ) \ T 3 ) ∪ ( T 2 \ T 1 ) ∪ ( X \ T 4 ) = ( P 0 ∪ P 3 ) ∪ P 1 ∪ ( X \ T 4 ) = T 4 ∪ ( X \ T 4 ) = X .

(See Equalities (2.1) and (2.2)). Then from the statement b) of the Lemma 2.1.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement a) of the Lemma 2.1.2 element β is generating by elements of the set B ( A 0 ) and

T 4 β = ( P 0 ∪ P 1 ∪ P 3 ) β = T 4 ∪ T 2 = T 2 , T 0 β = T 0 . δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) = α ,

if Y 4 δ = Y 2 α , Y 0 δ = Y 0 α . Last equalities are possible since | Y 4 δ | ≥ 1 | Y 0 δ | ≥ 1 .

The statement e) of the lemma 2.1.3 is proved.

6) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 0 δ × T 0 ) , β = ( ( ( T 2 ∩ T 1 ) \ T 4 ) × T 3 ) ∪ ( ( T 1 \ T 2 ) × T 1 ) ∪ ( ( X \ T 3 ) × T 0 ) ,

where Y 3 δ , Y 0 δ ∉ { ∅ } ,

( ( T 2 ∩ T 1 ) \ T 4 ) ∪ ( T 1 \ T 2 ) ∪ ( X \ T 3 ) = ( P 0 ∪ P 4 ) ∪ P 2 ∪ ( X \ T 3 ) = T 3 ∪ ( X \ T 3 ) = X .

(See Equalities (2.1) and (2.2)). Then from the statement d) of the Lemma 2.1.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement b) of the Lemma 2.1.2 element β is generating by elements of the set B ( A 0 ) and

T 3 β = ( P 0 ∪ P 2 ∪ P 4 ) β = T 3 ∪ T 1 = T 1 , T 0 β = T 0 . δ ∘ β = ( Y 3 δ × T 3 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 3 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) = α ,

if Y 3 δ = Y 1 α , Y 0 δ = Y 0 α . Last equalities are possible since | Y 4 δ | ≥ 1 | Y 0 δ | ≥ 1 .

The statement e) of the lemma 2.1.3 is proved.

7) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 1 × T 4 ) ∪ ( ( T 2 \ T 1 ) × T 2 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 2 δ , Y 0 δ ∉ { ∅ } ,

T 1 ∪ ( T 2 \ T 1 ) ∪ ( X \ T 0 ) = ( P 0 ∪ P 2 ∪ P 3 ∪ P 4 ) ∪ P 1 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X

(see Equalities (2.1) and (2.2)). Then from the statement e) of the Lemma 2.1.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement a) of the Lemma 2.1.2 element β is generating by elements of the set B ( A 0 ) and

T 2 β = T 4 ∪ T 2 = T 2 , T 0 β = T 2 δ ∘ β = ( Y 2 δ × T 2 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 2 ) = X × T 2 = α ,

since representation of a binary relation δ is quasinormal.

The statement g) of the lemma 2.1.3 is proved.

8) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 2 × T 3 ) ∪ ( ( T 1 \ T 2 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 1 δ , Y 0 δ ∉ { ∅ } ,

T 2 ∪ ( T 1 \ T 2 ) ∪ ( X \ T 0 ) = ( P 0 ∪ P 1 ∪ P 3 ∪ P 4 ) ∪ P 2 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X

(see Equalities (2.1) and (2.2)). Then from the statement f) of the Lemma 2.1.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement b) of the Lemma 2.1.2 element β is generating by elements of the set B ( A 0 ) and

T 1 β = T 3 ∪ T 1 = T 1 , T 0 β = T 1 δ ∘ β = ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 1 ) = X × T 1 = α ,

since representation of a binary relation δ is quasinormal.

The statement h) of the lemma 2.1.3 is proved.

9) Let quasinormal representation of a binary relation δ has a form

δ = ( T 4 × T 1 ) ∪ ( ( X \ T 4 ) × T 0 ) ,

then

T 1 δ = ( P 0 ∪ P 2 ∪ P 3 ∪ P 4 ) δ = T 4 ∪ T 0 = T 0 , T 0 δ = T 0 δ ∘ δ = ( T 4 × T 1 δ ) ∪ ( ( X \ T 4 ) × T 0 δ ) = ( T 4 × T 0 ) ∪ ( ( X \ T 4 ) × T 0 ) = X \ T 0 = α

since representation of a binary relation δ is quasinormal.

The statement i) of the lemma 2.1.3 is proved.

Lemma 2.1.3 is proved.

Lemma 2..4. Let D ∈ Σ 8.0 ( X , 5 ) . Then the following statements are true:

a) If | X \ T 0 | ≥ 1 and Z ∈ { T 4 , T 3 } , then binary relation α = X × Z is generating by elements of the elements of set B ( A 0 ) ;

b) If X = T 0 and Z ∈ { T 4 , T 3 } , then binary relation α = X × Z is external element for the semigroup B X ( D ) .

Proof. 1) Let quasinormal representation of a binary relation δ has a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 3 δ × T 3 ) ∪ ( Y 0 δ × T 0 ) ,

where Y 4 δ , Y 3 δ ∉ { ∅ } , then δ ∈ B ( A 0 ) \ { α } . If quasinormal representation of a binary relation β has a form β = ( T 0 × Z ) ∪ ∪ t ′ ∈ X \ T 0 ( { t ′ } × f ( t ′ ) ) , where f is any mapping of the set X \ T 0 in the set { T 4 , T 3 } \ { Z } . It is easy to see, that β ≠ α and two elements of the set { T 4 , T 3 } belong to the semilattice V ( D , β ) , i.e. δ ∈ B ( A 0 ) \ { α } . In this case we have

T 4 β = T 3 β = T 0 β = Z ; δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 3 δ × T 3 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × Z ) ∪ ( Y 3 δ × Z ) ∪ ( Y 0 δ × Z ) = ( ( Y 4 δ ∪ Y 3 δ ∪ Y 0 δ ) × Z ) = X × Z = α ,

since the representation of a binary relation δ is quasinormal. Thus, element α is generating by elements of the set B ( A 0 ) .

The statement a) of the lemma 2.1.4 is proved.

2) Let X = T 0 , α = X × Z , for some Z ∈ { T 4 , T 3 } and α = δ ∘ β for some δ , β ∈ B X ( D ) \ { α } . Then from the equality (2.1.1) and (2.1.2) we obtain that

T 4 β = T 3 β = T 2 β = T 1 β = T 0 β = Z , P 0 β = P 1 β = P 2 β = P 3 β = P 4 β = Z ,

since Z is minimal element of the semilattice D.

Now, let subquasinormal representations β ¯ of a binary relation β has a form

β ¯ = ( ( P 0 ∪ P 1 ∪ P 2 ∪ P 3 ∪ P 4 ) × Z ) ∪ ∪ t ′ ∈ X \ T 0 ( { t ′ } × β ¯ 2 ( t ′ ) ) ,

where β ¯ 1 = ( P 0 P 1 P 2 P 3 P 4 Z Z Z Z Z ) is normal mapping. But complement mapping β ¯ 2 is empty, since X \ T 0 = ∅ , i.e. in the given case, subquasinormal representation β ¯ of a binary relation β is defined uniquely. So, we have that β = β ¯ = X × Z = α , which contradicts the condition β ∉ B X ( D ) \ { α } .

Therefore, if X = T 0 and α = X × Z , for some Z ∈ { T 4 , T 3 } , then α is external element of the semigroup B X ( D ) .

The statement b) of the lemma 2.1.4 is proved.

Lemma 2.1.4 is proved.

Theorem 2.1.1. Let D ∈ Σ 8.0 ( X , 5 ) and

Then the following statements are true:

a) If | X \ T 0 | ≥ 1 , then S 0 = B 0 ∪ B ( A 0 ) is irreducible generating set for the semigroup B X ( D ) ;

b) If X = T 0 , then S 1 = B 0 ∪ B ( A 0 ) ∪ { X × T 4 , X × T 3 } is irreducible generating set for the semigroup B X ( D ) .

Proof. Let D ∈ Σ 8.0 ( X , 5 ) and | X \ T 0 | ≥ 1 . First, we proved that every element of the semigroup B X ( D ) is generating by elements of the set S 0 . Indeed, let α be arbitrary element of the semigroup B X ( D ) . Then quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 3 α × T 3 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α ∪ Y 3 α ∪ Y 2 α ∪ Y 1 α ∪ Y 0 α = X and Y i α ∩ Y j α = ∅ ( 0 ≤ i ≠ j ≤ 4 ) . For the | V ( X ∗ , α ) | we consider the following cases:

1) | V ( X ∗ , α ) | = 5 . Then α ∈ B 0 and B 0 ⊂ S 0 by definition of a set S 0 .

2) | V ( X ∗ , α ) | = 4 . Then

V ( X ∗ , α ) ∈ A 4 = { { T 4 , T 3 , T 2 , T 0 } , { T 4 , T 3 , T 1 , T 0 } , { T 4 , T 2 , T 1 , T 0 } , { T 3 , T 2 , T 1 , T 0 } } ⊂ A 0

i.e. α ∈ B ( A 0 ) and B ( A 0 ) ⊂ S 0 by definition of a set S 0 .

3) | V ( X ∗ , α ) | = 3 . Then we have

V ( X ∗ , α ) ∈ A 3 = { { T 4 , T 3 , T 0 } , { T 4 , T 1 , T 0 } , { T 3 , T 2 , T 0 } , { T 4 , T 2 , T 0 } , { T 3 , T 1 , T 0 } , { T 2 , T 1 , T 0 } } .

By definition of a set A 0 we have { { T 4 , T 3 , T 0 } , { T 4 , T 1 , T 0 } , { T 3 , T 2 , T 0 } , { T 2 , T 1 , T 0 } } ⊂ A 0 , i.e. in this case α ∈ B ( A 0 ) and B ( A 0 ) ⊂ S 0 by definition of a set S 0 .

If V ( X ∗ , α ) ∈ { { T 4 , T 2 , T 0 } , { T 3 , T 1 , T 0 } } , then from the statement a) and b) of the Lemma 2.1.2 element α is generating by elements B ( A 0 ) and B ( A 0 ) ⊂ S 0 by definition of a set S 0 .

4) | V ( X ∗ , α ) | = 2 . Then we have

V ( X ∗ , α ) ∈ A 2 = { { T 4 , T 2 } , { T 4 , T 0 } , { T 3 , T 1 } , { T 3 , T 0 } , { T 2 , T 0 } { T 1 , T 0 } } .

Then from the statement a)-f) of the Lemma 2.1.3 element α is generating by elements B ( A 0 ) and B ( A 0 ) ⊂ S 0 by definition of a set S 0 .

5) | V ( X ∗ , α ) | = 1 . Then we have V ( X ∗ , α ) ∈ A 1 = { { T 4 } , { T 3 } , { T 2 } , { T 1 } , { T 0 } } .

If V ( X ∗ , α ) ∈ { { T 2 } , { T 1 } , { T 0 } } , then from the statements g), h) and i) of the Lemma 2.1.3 element α is generating by elements B ( A 0 ) and B ( A 0 ) ⊂ S 0 by definition of a set S 0 .

If V ( X ∗ , α ) ∈ { { T 4 } , { T 3 } } , then from the statement a) of the Lemma 2.1.4 element α is generating by elements B ( A 0 ) and B ( A 0 ) ⊂ S 0 by definition of a set S 0 .

Thus, we have that S 0 is generating set for the semigroup B X ( D ) .

If | X \ T 0 | ≥ 1 , then the set S 0 is 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 2.1.1 is proved.

Now, let D ∈ Σ 8.0 ( X , 5 ) and X = D ⌣ . First, we proved that every element of the semigroup B X ( D ) 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 ( X ∗ , α ) ∈ A 1 = { { T 4 } , { T 3 } , { T 2 } , { T 1 } , { T 0 } } .

If V ( X ∗ , α ) ∈ { { T 2 } , { T 1 } , { T 0 } } , then from the statements g), h) and i) of the Lemma 2.1.3 element α is generating by elements B ( A 0 ) and B ( A 0 ) ⊂ S 1 by definition of a set S 1 .

If V ( X ∗ , α ) ∈ { { T 4 } , { T 3 } } , then α ∈ S 1 by definition of a set S 1 .

Thus, we have that S 1 is generating set for the semigroup B X ( D ) .

If X = T 0 , then the set S 1 is 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 2.1.1 is proved.

Theorem 2.1.1 is proved.

Theorem 2.1.2. Let n ≥ 6 , D = { T 4 , T 3 , T 2 , T 1 , T 0 } ∈ Σ 8.0 ( X , 5 ) and

Then the following statements are true:

a) If | X \ T 0 | ≥ 1 , then the number | S 0 | elements of the set S 0 = B 0 ∪ B ( A 0 ) is equal to

| S 0 | = 5 n − 2 ⋅ 3 n + 1 .

b) If X = T 0 , then the number | S 1 | elements of the set S 1 = B 0 ∪ B ( A 0 ) ∪ { X × T 4 , X × T 3 } is equal to

| S 1 | = 5 n − 2 ⋅ 3 n + 3 .

Proof. Let number of a set X is equal to n ≥ 6 , i.e. | X | = n ≥ 6 . Let S n = { φ 1 , φ 2 , ⋯ , φ n ! } is a group all one to one mapping of a set M = { 1 , 2 , ⋯ , n } on the set M and φ i 1 , φ i 2 , ⋯ , φ i m ( m ≤ n ) are arbitrary elements of the group S n , Y φ 1 , Y φ 2 , ⋯ , Y φ m are arbitrary partitioning of a set X. By symbol k n m we denote the number elements of a set { Y φ 1 , Y φ 2 , ⋯ , Y φ m } . It is well known, that

k n m = ∑ i = 1 m ( − 1 ) m + i ( i − 1 ) ! ⋅ ( m − i ) ! ⋅ i n − 1 .

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

k n 2 = 2 n − 1 − 1 , k n 3 = 1 2 ⋅ 3 n − 1 − 2 n − 1 + 1 2 , k n 4 = 1 6 ⋅ 4 n − 1 − 1 2 ⋅ 3 n − 1 + 1 2 ⋅ 2 n − 1 − 1 6 , k n 5 = 1 24 ⋅ 5 n − 1 − 1 6 ⋅ 4 n − 1 + 1 4 ⋅ 3 n − 1 − 1 6 ⋅ 2 n − 1 + 1 24 .

If Y φ 1 , Y φ 2 are any two elements partitioning of a set X and β ¯ = ( Y φ 1 × Z 1 ) ∪ ( Y φ 2 × Z 2 ) , where Z 1 , Z 2 ∈ D and Z 1 ≠ Z 2 . Then number of different binary relations β ¯ of a semigroup B X ( D ) is equal to

2 ⋅ k n 2 = 2 n − 2 . (2.1.3)

If Y φ 1 , Y φ 2 , Y φ 3 are any tree elements partitioning of a set X and

β ¯ = ( Y φ 1 × Z 1 ) ∪ ( Y φ 2 × Z 2 ) ∪ ( Y φ 3 × Z 3 ) ,

where Z 1 , Z 2 , Z 3 are pairwise different elements of a given semilattice D. Then number of different binary relations β ¯ of a semigroup B X ( D ) is equal to

6 ⋅ k n 3 = 3 n − 3 ⋅ 2 n + 3 . (2.1.4)

If Y φ 1 , Y φ 2 , Y φ 3 , Y φ 4 are any four elements partitioning of a set X and

β ¯ = ( Y φ 1 × Z 1 ) ∪ ( Y φ 2 × Z 2 ) ∪ ( Y φ 3 × Z 3 ) ∪ ( Y φ 4 × Z 4 ) ,

where Z 1 , Z 2 , Z 3 , Z 4 are pairwise different elements of a given semilattice D. Then number of different binary relations β ¯ of a semigroup B X ( D ) is equal to

24 ⋅ k n 4 = 4 n − 4 ⋅ 3 n + 3 ⋅ 2 n + 1 − 4 . (2.1.5)

If Y φ 1 , Y φ 2 , Y φ 3 , Y φ 4 , Y φ 5 are any four elements partitioning of a set X and

β ¯ = ( Y φ 1 × Z 1 ) ∪ ( Y φ 2 × Z 2 ) ∪ ( Y φ 3 × Z 3 ) ∪ ( Y φ 4 × Z 4 ) ∪ ( Y φ 5 × Z 5 ) ,

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 β ¯ of a semigroup B X ( D ) is equal to

120 ⋅ k n 5 = 5 n − 5 ⋅ 4 n + 10 ⋅ 3 n − 10 ⋅ 2 n + 5. (2.1.6)

If α ∈ B 0 , then quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 3 α × T 3 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 3 α , Y 2 α , Y 1 α ∉ { ∅ } , or a system Y 4 α , Y 3 α , Y 2 α , Y 1 α , Y 0 α are partitioning of the set X.

If the system Y 4 α , Y 3 α , Y 2 α , Y 1 α , or a system Y 4 α , Y 3 α , Y 2 α , Y 1 α , Y 0 α are partitioning of the set X. Of this and from the equalities (2.1.4), (2.1.5) and (2.1.6) follows that

| B 0 | = ( 5 n − 5 ⋅ 4 n + 10 ⋅ 3 n − 10 ⋅ 2 n + 5 ) + ( 4 n − 4 ⋅ 3 n + 6 ⋅ 2 n − 4 ) = 5 n − 4 ⋅ 4 n + 6 ⋅ 3 n − 4 ⋅ 2 n + 1.

If α ∈ B ( A 0 ) , then by definition of a set B ( A 0 ) the quasinormal representation of a binary relation α has a form:

α = ( Y 4 α × T 4 ) ∪ ( Y 3 α × T 3 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 3 α , Y 2 α ∉ { ∅ } , or Y 4 α , Y 3 α , Y 2 α , Y 0 α ∉ { ∅ } are partitioning of the set X respectively;

α = ( Y 4 α × T 4 ) ∪ ( Y 3 α × T 3 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 3 α , Y 1 α ∉ { ∅ } , or Y 4 α , Y 3 α , Y 1 α , Y 0 α ∉ { ∅ } are partitioning of the set X respectively;

α = ( Y 4 α × T 4 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 2 α , Y 1 α ∉ { ∅ } , or Y 4 α , Y 2 α , Y 1 α , Y 0 α ∉ { ∅ } are partitioning of the set X respectively;

α = ( Y 3 α × T 3 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 3 α , Y 2 α , Y 1 α ∉ { ∅ } , or Y 3 α , Y 2 α , Y 1 α , Y 0 α ∉ { ∅ } are partitioning of the set X respectively;

α = ( Y 4 α × T 4 ) ∪ ( Y 3 α × T 3 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 3 α ∉ { ∅ } , or Y 4 α , Y 3 α , Y 0 α ∉ { ∅ } are partitioning of the set X respectively;

α = ( Y 4 α × T 4 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 1 α ∉ { ∅ } , or Y 4 α , Y 1 α , Y 0 α ∉ { ∅ } are partitioning of the set X respectively;

α = ( Y 3 α × T 3 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 0 α × T 0 ) ,

where Y 3 α , Y 2 α ∉ { ∅ } , or Y 3 α , Y 2 α , Y 0 α ∉ { ∅ } are partitioning of the set X respectively;

α = ( Y 2 α × T 2 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 2 α , Y 1 α ∈ { ∅ } , or Y 2 α , Y 1 α , Y 0 α ∈ { ∅ } 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

| B ( A 0 ) | = 4 ⋅ ( 2 n − 2 ) + 8 ⋅ ( 3 n − 3 ⋅ 2 n + 3 ) + 4 ⋅ ( 4 n − 4 ⋅ 3 n + 6 ⋅ 2 n − 4 ) = 4 ⋅ 4 n − 8 ⋅ 3 n + 4 ⋅ 2 n .

So, we have

| S 0 | = | B 0 ∪ B ( A 0 ) | = ( 5 n − 4 ⋅ 4 n + 6 ⋅ 3 n − 4 ⋅ 2 n + 1 ) + ( 4 ⋅ 4 n − 8 ⋅ 3 n + 4 ⋅ 2 n ) = 5 n − 2 ⋅ 3 n + 1 , | S 1 | = | B 0 ∪ B ( A 0 ) ∪ { X × T 4 , X × T 3 } | = 5 n − 2 ⋅ 3 n + 3

Since

B 0 ∩ B ( A 0 ) = B 0 ∩ { X × T 4 , X × T 3 , X × T 2 } = B ( A 0 ) ∩ { X × T 4 , X × T 3 , X × T 2 } = ∅ .

Theorem 2.1.2 is proved.

2.2. Generating Sets of the Complete Semigroup of Binary Relations Defined by Semilattices of the Class Σ 8 ( X , 5 ) , When T 4 ∩ T 3 = ∅

In the sequel, we denoted all semilattices D = { T 4 , T 3 , T 2 , T 1 , T 0 } of the class Σ 8 ( X , 5 ) by symbol Σ 8.1 ( X , 5 ) for which T 4 ∩ T 3 = ∅ . Of the last equality from the formal equalities of a semilattise D follows that T 4 ∩ T 3 = P 0 = ∅ , i.e. | X | ≥ 4 since P 4 ≠ ∅ , P 3 ≠ ∅ , P 2 ≠ ∅ , P 1 ≠ ∅ .

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

T 0 = P 1 ∪ P 2 ∪ P 3 ∪ P 4 , T 1 = P 2 ∪ P 3 ∪ P 4 , T 2 = P 1 ∪ P 3 ∪ P 4 , T 3 = P 2 ∪ P 4 , T 4 = P 1 ∪ P 3 . (2.2.1)

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

P 4 = T 2 \ T 4 , P 3 = T 1 \ T 3 , P 2 = T 1 \ T 2 , P 1 = T 2 \ T 1 . (2.2.2)

In this case we suppose that D ∈ Σ 8.1 ( X , 5 ) .

By symbols A 4 , A 3 , A 2 and A 1 we denoted the following sets:

Lemma 2.2.1. Let D ∈ Σ 8.1 ( X , 5 ) . Then the following statements are true:

a) Let Z , Z ′ ∈ { T 4 , T 3 , T 2 } , Z ≠ Z ′ . If Z , Z ′ ∈ V ( D , α ) , then α is external element of the semigroup B X ( D ) ;

b) Let Z ∈ { T 2 , T 1 } , Z ′ ∈ { T 4 , T 3 } . If Z ⊄ Z ′ and Z , Z ′ ∈ V ( D , α ) , then α is external element of the semigroup B X ( D ) .

Proof. Let α = δ ∘ β for some δ , β ∈ B X ( D ) \ { α } . If quasinormal representation of binary relation δ has a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) ,

then

α = δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 3 δ × T 3 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) .(2.2.3)

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

T 0 β = P 1 β ∪ P 2 β ∪ P 3 β ∪ P 4 β , T 1 β = P 2 β ∪ P 3 β ∪ P 4 β , T 2 β = P 1 β ∪ P 3 β ∪ P 4 β , T 3 β = P 2 β ∪ P 4 β , T 4 β = P 1 β ∪ P 3 β , (2.2.4)

where P i β ≠ ∅ for any P i ≠ ∅ ( i = 1 , 2 , 3 , 4 ) and β ∈ B X ( D ) . Indeed, by preposition P i ≠ ∅ for any i = 1 , 2 , 3 , 4 and β ≠ ∅ since ∅ ∉ D . Let y ∈ P i for some y ∈ X , then y ∈ D ⌣ , β = α f for some f : X → D and α f = ∪ x ∈ X ( { x } × f ( x ) ) ⊇ { y } × f ( y ) , i.e. there exists an element z ∈ f ( y ) for which y α f z and y β z . Of this and by definition of a set P i β we obtain that z ∈ P i β since y ∈ P i , y β z . Thus, we have P i β ≠ ∅ , i.e. P i β ∈ D for any i = 1 , 2 , 3 , 4 .

Now, let T i β = Z and T j β = Z ′ for some 0 ≤ i ≠ j ≤ 4 and Z ≠ Z ′ , Z , Z ′ ∈ { T 4 , T 3 } , then from the Equalities (2.2.4) follows that Z = P 0 β = Z ′ since Z and Z ′ are minimal elements of the semilattice D. The equality Z = Z ′ contradicts the inequality Z ≠ Z ′ .

The statement a) of the Lemma 2.2.1 is proved.

Let T i β = Z ′ , where Z ′ ∈ { T 4 , T 3 } and T j β = Z , Z ∈ { T 2 , T 1 } for some 0 ≤ i ≠ j ≤ 4 . If 0 ≤ i ≤ 4 , then from the formal equalities of a semilattice D we obtain that

T 0 β = P 1 β ∪ P 2 β ∪ P 3 β ∪ P 4 β = P 1 β = P 2 β = P 3 β = P 4 β = Z ′ , T 1 β = P 2 β ∪ P 3 β ∪ P 4 β = P 2 β = P 3 β = P 4 β = Z ′ , T 2 β = P 1 β ∪ P 3 β ∪ P 4 β = P 1 β = P 3 β = P 4 β = Z ′ , T 3 β = P 2 β ∪ P 4 β = P 2 β = P 4 β = Z ′ , T 4 β = P 1 β ∪ P 3 β = P 1 β = P 3 β = Z ′ ,

since Z ′ is minimal element of the semilattice D.

Now, let i ≠ j .

1) If T 0 β = P 1 β = P 2 β = P 3 β = P 4 β = Z ′ and j = 1 , 2 , 3 , 4 , then we have

Z = T 1 β = T 2 β = T 3 β = T 4 β = Z ′ ,

which contradicts the inequality Z ≠ Z ′ .

2) If T 1 β = P 2 β = P 3 β = P 4 β = Z ′ and j = 0 , 2 , 3 , 4 , then we have

Z = T 0 β = T 2 β = T 4 β = Z ′ ∪ P 1 β , where P 1 β ∈ D ; Z = T 3 β = Z ′ .

Last equalities are impossible since Z ≠ Z ′ ∪ T for any T ∈ D and Z ≠ Z ′ by definition of a semilattice D.

3) If T 2 β = P 1 β = P 3 β = P 4 β = Z ′ and j = 0 , 1 , 3 , 4 , then we have

Z = T 0 β = T 2 β = T 4 β = Z ′ ∪ P 1 β , where P 1 β ∈ D ; Z = T 3 β = Z ′ .

Last equalities are impossible since for any T ∈ D and Z ≠ Z ′ by definition of a semilattice D.

4) If T 3 β = P 2 β = P 4 β = Z ′ and j = 0 , 1 , 2 , 4 , then we have

Z = T 0 β = T 2 β = T 4 β = Z ′ ∪ P 1 β ∪ P 3 β , Z = T 1 β = Z ′ ∪ P 3 β , where P 1 β , P 3 β ∈ D .

Last equalities are impossible since Z ≠ Z ′ ∪ T ∪ T ′ and Z ≠ Z ′ ∪ T for any T , T ′ ∈ D , by definition of a semilattice D.

5) If T 4 β = P 1 β = P 3 β = Z ′ and j = 0 , 1 , 2 , 3 , then we have

Z = T 0 β = T 1 β = T 3 β = Z ′ ∪ P 2 β ∪ P 4 β , Z = T 2 β = Z ′ ∪ P 4 β , where P 2 β , P 4 β ∈ D .

Last equalities are impossible since Z ≠ Z ′ ∪ T ∪ T ′ and Z ≠ Z ′ ∪ T for any T , T ′ ∈ 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 ∈ Σ 8.1 ( X , 5 ) . We denoted the following sets by symbols A 0 , B ( A 0 ) and B 0 :

A 0 = { { T 4 , T 3 , T 2 , T 0 } , { T 4 , T 3 , T 1 , T 0 } , { T 4 , T 2 , T 1 , T 0 } , { T 3 , T 2 , T 1 , T 0 } , { T 4 , T 3 , T 0 } , { T 4 , T 1 , T 0 } , { T 3 , T 2 , T 0 } } , B ( A 0 ) = { α ∈ B X ( D ) | V ( X ∗ , α ) ∈ A 0 } ; B 0 = { α ∈ B X ( D ) | V ( X ∗ , α ) = D } .

Remark, that the sets B 0 and B ( A 0 ) are external elements for the semigroup B X ( D ) .

Lemma 2.2.2. Let D ∈ Σ 8.1 ( X , 5 ) . Then the following statements are true:

a) If quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 2 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

b) If quasinormal representation of a binary relation α has a form

α = ( Y 3 α × T 3 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 3 α , Y 1 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

c) If quasinormal representation of a binary relation α has a form

α = ( Y 2 α × T 2 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 2 α , Y 1 α ∉ { ∅ } , then α is generating by elements of the elements of set B 0 ∪ B ( A 0 ) .

Proof. 1). Let quasinormal representation of binary relations δ and β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 4 × T 4 ) ∪ ( ( T 2 \ T 4 ) × T 2 ) ∪ ( ( T 0 \ T 2 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 α , Y 2 α , Y 1 α ∉ { ∅ } ,

T 4 ∪ ( T 2 \ T 4 ) ∪ ( T 0 \ T 2 ) ∪ ( X \ T 0 ) = ( P 1 ∪ P 3 ) ∪ P 4 ∪ P 2 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X ,

(see Equalities (2.2.1) and (2.2.2)), then δ , β ∈ B ( A 0 ) and

T 4 β = T 4 , T 2 β = ( P 1 ∪ P 3 ∪ P 4 ) β = T 4 ∪ T 2 = T 2 , T 1 β = ( P 2 ∪ P 3 ∪ P 4 ) β = T 4 ∪ T 1 = T 0 , T 0 β = T 0 . α = δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × T 4 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 0 ) ∪ ( Y 0 δ × T 0 ) = ( Y 4 δ × T 4 ) ∪ ( Y 2 δ × T 2 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × T 0 ) = α ,

if Y 4 δ = Y 4 α , Y 2 δ = Y 2 α and Y 1 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 1 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement a) of the lemma 2.2.2 is proved.

2) Let quasinormal representation of binary relations δ and β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 3 × T 3 ) ∪ ( ( T 0 \ T 1 ) × T 2 ) ∪ ( ( T 1 \ T 3 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 3 α , Y 2 α , Y 1 α ∉ { ∅ } ,

T 3 ∪ ( T 0 \ T 1 ) ∪ ( T 1 \ T 3 ) ∪ ( X \ T 0 ) = ( P 2 ∪ P 4 ) ∪ P 1 ∪ P 3 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X ,

(see Equalities (2.2.1) and (2.2.2)), then δ , β ∈ B ( A 0 ) and

T 4 β = T 3 , T 2 β = ( P 1 ∪ P 3 ∪ P 4 ) β = T 3 ∪ T 2 ∪ T 1 = T 0 , T 1 β = ( P 2 ∪ P 3 ∪ P 4 ) β = T 3 ∪ T 1 = T 0 , T 0 β = T 0 . α = δ ∘ β = ( Y 3 δ × T 3 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 0 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) = ( Y 3 δ × T 3 ) ∪ ( Y 1 δ × T 1 ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × T 0 ) = α ,

if Y 3 δ = Y 3 α , Y 1 δ = Y 1 α and Y 2 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 2 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement b) of the lemma 2.2.2 is proved.

3) Let quasinormal representation of binary relations δ and β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 3 δ × T 3 ) ∪ ( Y 0 δ × T 0 ) , β = ( ( T 2 \ T 1 ) × T 4 ) ∪ ( ( T 1 \ T 2 ) × T 3 ) ∪ ( ( T 1 \ T 3 ) × T 2 ) ∪ ( ( T 2 \ T 4 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 α , Y 3 α ∉ { ∅ } ,

( T 2 \ T 1 ) ∪ ( T 1 \ T 2 ) ∪ ( T 1 \ T 3 ) ∪ ( T 2 \ T 4 ) ∪ ( X \ T 0 ) = P 1 ∪ P 2 ∪ P 3 ∪ P 4 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X ,

(see Equalities (2.2.1) and (2.2.2)), then δ ∈ B ( A 0 ) , β ∈ B 0 and

T 4 β = ( P 1 ∪ P 3 ) β = T 4 ∪ T 2 = T 2 , T 3 β = ( P 2 ∪ P 4 ) β = T 3 ∪ T 1 = T 1 , T 0 β = T 2 ∪ T 1 = T 0 , α = δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 3 δ × T 3 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × T 2 ) ∪ ( Y 3 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) = α ,

if Y 4 δ = Y 2 α , Y 3 δ = Y 1 α and Y 0 δ = Y 0 α . Last equalities are possible since | Y 4 δ | ≥ 1 , | Y 3 δ | ≥ 1 and | Y 0 δ | ≥ 0 .

The statement c) of the lemma 2.2.2 is proved.

Lemma 2.2.2 is proved.

Lemma 2.2.3. Let D ∈ Σ 8.1 ( X , 5 ) . Then the following statements are true:

a) If quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 2 α × T 2 ) ,

where Y 4 α , Y 2 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

b) If quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

c) If quasinormal representation of a binary relation α has a form

α = ( Y 3 α × T 3 ) ∪ ( Y 1 α × T 1 ) ,

where Y 3 α , Y 1 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

d) If quasinormal representation of a binary relation α has a form

α = ( Y 3 α × T 3 ) ∪ ( Y 0 α × T 0 ) ,

where Y 3 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

e) If quasinormal representation of a binary relation α has a form

α = ( Y 2 α × T 2 ) ∪ ( Y 0 α × T 0 ) ,

where Y 2 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

f) If quasinormal representation of a binary relation α has a form

α = ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 1 α , Y 0 α ∉ { ∅ } , then α is generating by elements of the elements of set B ( A 0 ) ;

g) If quasinormal representation of a binary relation α has a form α = X × T 2 , then α is generating by elements of the elements of set B ( A 0 ) ;

h) If quasinormal representation of a binary relation α has a form α = X × T 1 , then α is generating by elements of the elements of set B ( A 0 ) ;

i) If quasinormal representation of a binary relation α has a form α = X × T 0 , then α is generating by elements of the elements of set B ( A 0 ) .

Proof. 1) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 4 × T 4 ) ∪ ( ( T 0 \ T 4 ) × T 2 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 δ , Y 1 δ ∉ { ∅ } .

T 4 ∪ ( T 0 \ T 4 ) ∪ ( X \ T 0 ) = ( P 1 ∪ P 3 ) ∪ ( P 2 ∪ P 4 ) ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X .

Then from the statement a) of the Lemma 2.2.2 follows that β is generating by elements of the set B ( A 0 ) , δ ∈ B ( A 0 ) and

If Y 4 δ = Y 4 α , Y 1 δ ∪ Y 0 δ = Y 2 α . Last equalities are possible since | Y 1 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement a) of the lemma 2.2.3 is proved.

2) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 4 × T 4 ) ∪ ( ( T 0 \ T 4 ) × T 3 ) ∪ ( ( X \ T ) × T 0 ) ,

where Y 4 δ , Y 1 δ ∉ { ∅ } .

T 4 ∪ ( T 0 \ T 4 ) ∪ ( X \ T 0 ) = ( P 1 ∪ P 3 ) ∪ ( P 2 ∪ P 4 ) ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X .

Then from δ , β ∈ B ( A 0 ) and

if Y 4 δ = Y 4 α , Y 1 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 1 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement b) of the lemma 2.2.3 is proved.

3) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 3 × T 3 ) ∪ ( ( T 0 \ T 3 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 4 δ , Y 2 δ ∉ { ∅ } .

T 3 ∪ ( T 0 \ T 3 ) ∪ ( X \ T 0 ) = ( P 2 ∪ P 4 ) ∪ ( P 1 ∪ P 3 ) ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X .

Then from the statement b) of the Lemma 2.2.2 follows that β is generating by elements of the set B ( A 0 ) , δ ∈ B ( A 0 ) and

T 3 β = T 3 , T 2 β = T 3 ∪ T 1 = T 1 , T 0 β = T 1 . δ ∘ β = ( Y 2 δ × T 3 β ) ∪ ( Y 2 δ × T 2 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 1 ) ∪ ( Y 0 δ × T 1 ) = ( Y 3 δ × T 3 ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × T 1 ) = α ,

if Y 3 δ = Y 3 α , Y 2 δ ∪ Y 0 δ = Y 1 α . Last equalities are possible since | Y 2 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement c) of the lemma 2.2.3 is proved.

4) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 3 × T 3 ) ∪ ( ( T 0 \ T 3 ) × T 2 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 3 δ , Y 2 δ ∉ { ∅ } . Then δ , β ∈ B ( A 0 ) and

if Y 3 δ = Y 3 α , Y 2 δ ∪ Y 0 δ = Y 0 α . Last equalities are possible since | Y 2 δ ∪ Y 0 δ | ≥ 1 ( | Y 0 δ | ≥ 0 by preposition).

The statement d) of the lemma 2.2.3 is proved.

5) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 0 δ × T 0 ) , β = ( ( ( T 2 ∩ T 1 ) \ T 3 ) × T 4 ) ∪ ( ( T 2 \ T 1 ) × T 2 ) ∪ ( ( X \ T 4 ) × T 0 ) ,

where Y 4 δ , Y 0 δ ∉ { ∅ } ,

( ( T 2 ∩ T 1 ) \ T 3 ) ∪ ( T 2 \ T 1 ) ∪ ( X \ T 4 ) = P 3 ∪ P 1 ∪ ( X \ T 4 ) = T 4 ∪ ( X \ T 4 ) = X .

(See Equalities (2.2.1) and (2.2.2)). Then from the statement b) of the Lemma 2.2.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement a) of the Lemma 2.2.2 element β is generating by elements of the set B ( A 0 ) and

T 4 β = ( P 1 ∪ P 3 ) β = T 4 ∪ T 2 = T 2 , T 0 β = T 0 . δ ∘ β = ( Y 4 δ × T 4 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 4 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) = α ,

if Y 4 δ = Y 2 α , Y 0 δ = Y 0 α . Last equalities are possible since | Y 4 δ | ≥ 1 | Y 0 δ | ≥ 1 .

The statement e) of the lemma 2.2.3 is proved.

6) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 3 δ × T 3 ) ∪ ( Y 0 δ × T 0 ) , β = ( ( ( T 2 ∩ T 1 ) \ T 4 ) × T 3 ) ∪ ( ( T 1 \ T 2 ) × T 1 ) ∪ ( ( X \ T 3 ) × T 0 ) ,

where Y 3 δ , Y 0 δ ∉ { ∅ } ,

( ( T 2 ∩ T 1 ) \ T 4 ) ∪ ( T 1 \ T 2 ) ∪ ( X \ T 3 ) = P 4 ∪ P 2 ∪ ( X \ T 3 ) = T 3 ∪ ( X \ T 3 ) = X .

(see Equalities (2.2.1) and (2.2.2)). Then from the statement d) of the Lemma 2.2.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement b) of the Lemma 2.2.2 element β is generating by elements of the set B ( A 0 ) and

T 3 β = ( P 2 ∪ P 4 ) β = T 3 ∪ T 1 = T 1 , T 0 β = T 0 . δ ∘ β = ( Y 3 δ × T 3 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 3 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) = α ,

if Y 3 δ = Y 1 α , Y 0 δ = Y 0 α . Last equalities are possible since | Y 4 δ | ≥ 1 | Y 0 δ | ≥ 1 .

The statement e) of the lemma 2.2.3 is proved.

7) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 1 × T 4 ) ∪ ( ( T 2 \ T 1 ) × T 2 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 2 δ , Y 0 δ ∉ { ∅ } ,

T 1 ∪ ( T 2 \ T 1 ) ∪ ( X \ T 0 ) = ( P 2 ∪ P 3 ∪ P 4 ) ∪ P 1 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X

(see Equalities (2.2.1) and (2.2.2)). Then from the statement e) of the Lemma 2.2.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement a) of the Lemma 2.2.2 element β is generating by elements of the set B ( A 0 ) and

T 2 β = T 4 ∪ T 2 = T 2 , T 0 β = T 2 δ ∘ β = ( Y 2 δ × T 2 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 2 δ × T 2 ) ∪ ( Y 0 δ × T 2 ) = X × T 2 = α ,

since representation of a binary relation δ is quasinormal.

The statement g) of the lemma 2.2.3 is proved.

8) Let quasinormal representation of a binary relations δ , β have a form

δ = ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 0 ) , β = ( T 2 × T 3 ) ∪ ( ( T 1 \ T 2 ) × T 1 ) ∪ ( ( X \ T 0 ) × T 0 ) ,

where Y 1 δ , Y 0 δ ∉ { ∅ } ,

T 2 ∪ ( T 1 \ T 2 ) ∪ ( X \ T 0 ) = ( P 1 ∪ P 3 ∪ P 4 ) ∪ P 2 ∪ ( X \ T 0 ) = T 0 ∪ ( X \ T 0 ) = X

(see Equalities (2.2.1) and (2.2.2)). Then from the statement f) of the Lemma 2.2.3 follows that δ is generating by elements of the set B ( A 0 ) and from the statement b) of the Lemma 2.2.2 element β is generating by elements of the set B ( A 0 ) and

T 1 β = T 3 ∪ T 1 = T 1 , T 0 β = T 1 δ ∘ β = ( Y 1 δ × T 1 β ) ∪ ( Y 0 δ × T 0 β ) = ( Y 1 δ × T 1 ) ∪ ( Y 0 δ × T 1 ) = X × T 1 = α ,

since representation of a binary relation δ is quasinormal.

The statement h) of the lemma 2.2.3 is proved.

9) Let quasinormal representation of a binary relation δ has a form

δ = ( T 4 × T 1 ) ∪ ( ( X \ T 4 ) × T 0 ) ,

then

T 1 δ = ( P 2 ∪ P 3 ∪ P 4 ) δ = T 4 ∪ T 0 = T 0 , T 0 δ = T 0 δ ∘ δ = ( T 4 × T 1 δ ) ∪ ( ( X \ T 4 ) × T 0 δ ) = ( T 4 × T 0 ) ∪ ( ( X \ T 4 ) × T 0 ) = X \ T 0 = α

since representation of a binary relation δ 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 ∈ Σ 8.1 ( X , 5 ) . Then the following statements are true:

a) If | X \ T 0 | ≥ 1 and Z ∈ { T 4 , T 3 } , then binary relation α = X × Z is generating by elements of the elements of set B ( A 0 ) ;

b) If X = T 0 and Z ∈ { T 4 , T 3 } , then binary relation α = X × Z is external element for the semigroup B X ( D ) .

Proof. 1) Let quasinormal representation of a binary relation δ has a form

δ = ( Y 4 δ × T 4 ) ∪ ( Y 3 δ × T 3 ) ∪ ( Y 0 δ × T 0 ) ,

where Y 4 δ , Y 3 δ ∉ { ∅ } , then δ ∈ B ( A 0 ) \ { α } . If quasinormal representation of a binary relation β has a form β = ( T 0 × T ) ∪ ∪ t ′ ∈ X \ T 0 ( { t ′ } × f ( t ′ ) ) , where f is any mapping of the set X \ T 0 in the set { T 4 , T 3 } \ { Z } . It is easy to see, that β ≠ α and two elements of the set { T 4 , T 3 } belong to the semilattice V ( D , β ) , i.e. δ ∈ B ( A 0 ) \ { α } . In this case we have

since the representation of a binary relation δ is quasinormal. Thus, element α is generating by elements of the set B ( A 0 ) .

The statement a) of the lemma 2.2.4 is proved.

2) Let X = T 0 , α = X × Z , for some Z ∈ { T 4 , T 3 } and α = δ ∘ β for some δ , β ∈ B X ( D ) \ { α } . Then from the Equalities (2.2.3) and (2.2.4) we obtain that

T 4 β = T 3 β = T 2 β = T 1 β = T 0 β = Z , P 1 β = P 2 β = P 3 β = P 4 β = Z ,

since Z is minimal element of the semilattice D.

Now, let subquasinormal representations β ¯ of a binary relation β has a form

β ¯ = ( ( P 1 ∪ P 2 ∪ P 3 ∪ P 4 ) × Z ) ∪ ∪ t ′ ∈ X \ T 0 ( { t ′ } × β ¯ 2 ( t ′ ) ) ,

where β ¯ 1 = ( P 0 P 1 P 2 P 3 P 4 ∅ Z Z Z Z ) is normal mapping. But complement mapping β ¯ 2 is empty, since X \ T 0 = ∅ , i.e. in the given case, subquasinormal representation β ¯ of a binary relation β is defined uniquely. So, we have that β = β ¯ = X × Z = α , which contradicts the condition β ∉ B X ( D ) \ { α } .

Therefore, if X = T 0 and α = X × Z , for some Z ∈ { T 4 , T 3 } , then α is external element of the semigroup B X ( D ) .

The statement b) of the lemma 2.2.4 is proved.

lemma 2.2.4 is proved.

Theorem 2.2.1. Let D ∈ Σ 8.1 ( X , 5 ) and

Then the following statements are true:

a) If | X \ T 0 | ≥ 1 , then S 0 = B 0 ∪ B ( A 0 ) is irreducible generating set for the semigroup.

b) If X = T 0 , then S 1 = B 0 ∪ B ( A 0 ) ∪ { X × T 4 , X × T 3 } is irreducible generating set for the semigroup B X ( D ) .

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

Theorem 2.2.2. Let n ≥ 6 , D = { T 4 , T 3 , T 2 , T 1 , T 0 } ∈ Σ 8.1 ( X , 5 ) and

Then the following statements are true:

a) If | X \ T 0 | ≥ 1 , then the number | S 0 | elements of the set S 0 = B 0 ∪ B ( A 0 ) is equal to

| S 0 | = 5 n − 3 ⋅ 3 n + 2 ⋅ 2 n + 2 .

b) If X = T 0 , then the number | S 1 | elements of the set S 1 = B 0 ∪ B ( A 0 ) ∪ { X × T 4 , X × T 3 } is equal to

| S 1 | = 5 n − 3 ⋅ 3 n + 2 ⋅ 2 n + 4 .

k n m = ∑ i = 1 m ( − 1 ) m + i ( i − 1 ) ! ⋅ ( m − i ) ! ⋅ i n − 1 .

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

2 ⋅ k n 2 = 2 n − 2 . (2.2.5)

If Y φ 1 , Y φ 2 , Y φ 3 are any tree elements partitioning of a set X and

β ¯ = ( Y φ 1 × Z 1 ) ∪ ( Y φ 2 × Z 2 ) ∪ ( Y φ 3 × Z 3 ) ,

where Z 1 , Z 2 , Z 3 are pairwise different elements of a given semilattice D. Then number of different binary relations β ¯ of a semigroup B X ( D ) is equal to

6 ⋅ k n 3 = 3 n − 3 ⋅ 2 n + 3 . (2.2.6)

If Y φ 1 , Y φ 2 , Y φ 3 , Y φ 4 are any four elements partitioning of a set X and

β ¯ = ( Y φ 1 × Z 1 ) ∪ ( Y φ 2 × Z 2 ) ∪ ( Y φ 3 × Z 3 ) ∪ ( Y φ 4 × Z 4 ) ,

where Z 1 , Z 2 , Z 3 , Z 4 are pairwise different elements of a given semilattice D. Then number of different binary relations β ¯ of a semigroup B X ( D ) is equal to

24 ⋅ k n 4 = 4 n − 4 ⋅ 3 n + 3 ⋅ 2 n + 1 − 4 . (2.2.7)

If Y φ 1 , Y φ 2 , Y φ 3 , Y φ 4 , Y φ 5 are any four elements partitioning of a set X and

β ¯ = ( Y φ 1 × Z 1 ) ∪ ( Y φ 2 × Z 2 ) ∪ ( Y φ 3 × Z 3 ) ∪ ( Y φ 4 × Z 4 ) ∪ ( Y φ 5 × Z 5 ) ,

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 β ¯ of a semigroup B X ( D ) is equal to

120 ⋅ k n 5 = 5 n − 5 ⋅ 4 n + 10 ⋅ 3 n − 10 ⋅ 2 n + 5. (2.2.8)

If α ∈ B 0 , then quasinormal representation of a binary relation α has a form

α = ( Y 4 α × T 4 ) ∪ ( Y 3 α × T 3 ) ∪ ( Y 2 α × T 2 ) ∪ ( Y 1 α × T 1 ) ∪ ( Y 0 α × T 0 ) ,

where Y 4 α , Y 3 α , Y 2 α , Y 1 α ∉ { ∅ } , or a system Y 4 α , Y 3 α , Y 2 α , Y 1 α , Y 0 α are partitioning of the set X.

If the system Y 4 α , Y 3 α , Y 2 α , Y 1 α , or a system Y 4 α , Y 3 α , Y 2 α , Y 1 α , Y 0 α are partitioning of the set X. Of this from the Equalities (2.2.7) and (2.2.8) follows that