Regular Elements Of The Semigroup Defined By Semilattices Of The Class When

(1)

Regular Elements of the Semigroup

BX( )D

defined

by Semilattices of the Class

Σ3(X,8)

when

Z7 = ∅

Giuli Tavdgiridze , Yasha Diasamidze

In this paper we give a full description of regular elements of the semigroup , which are defined by semilattices of the class . For the case where X-is a finite set we derive formulas by means of which we can calculate the numbers of regular elements of the respective semigroups. In this subsection it is assume that Z₇ = ∅

2010 Mathematical Subject Classification. 20M05

Key words and phrases: semilattice, semigroup, binary relation, regulat element.

Definition 1.1. An element α taken from the semigroup BX

( )

D is colled a regular element

of B_X

( )

D , if in B_X

( )

D there exists an element β such that α β α α  = .

Definition 1.2. A one-to-one mapping ϕ between the complete X − semilattices of unions

D′ and D′′ is called a complete isomorphism if the condition

( )

1 1

T D

D T

ϕ ϕ

′∈ ′

=

_

is fulfilled for each nonempty subset D₁ of the semilattice D′ (see [1],[2] Definition 6.3.2).

Definition 1.3. Let α be some binary relation of the semigroup B_X

( )

D . We say that a

complete isomorphism ϕ between the complete semilatices of unions Q and D′ is a complete

α−isomorphism if

)

a Q=V D

(

,α

)

;

)

b ϕ

( )

∅ = ∅ for ∅ ∈V D

(

,α

)

and ϕ

( )

T α =T for all T∈V D

(

,α

)

(see [1],[2] Definition

6.3.3).

Theorem 1.1. Let D be a finite X −semilattice of unions and α∈B_X

( )

D ; D

( )

α be the set of

those elements T of the semilattice Q V D=

(

,α

) { }

\ ∅ which are nonlimiting elements of the

set Q_T. Then a binary relation α having a quasinormal representation of the form

(

)

( , )

T T V D

Yα T

α α

∈

=

_

× is a regular element of the semigroup B_X

( )

D

iff V D

(

,α

)

is a XI−semilattice of unions and for

α

−isomorphism ϕ of the semilattice

( ) X

B D

( )

Σ₃ X,8

(2)

(

,

)

V D α on some X −subsemilattice D′ of the semilattice D the following conditions are satisfied:

)

a

( )_T T

( )

T D

Yα T

α

ϕ ′ ′∈

⊇





for all T∈D

( )

α ;

)

b Y_Tα ∩ϕ

( )

T ≠ ∅ for all nonlimiting element Tof the set D

( )

α _T (see [1],[2] Theorem 6.3.3).

Theorem 1.2. Let R be the set of all regular elements of the semigroup B_X

( )

D . Then the

following statements are true:

a) R D

( )

′ ∩R D

( )

′′ = ∅ for any D D′ ′′∈Σ, _XI

( )

D and D′≠D′′;

b)

( )

XI

D D

R R D

′∈Σ

′

=

_

;

c) if X is a finite set, then

( )

XI

D D

R R D

′∈Σ

′

=

∑

(see [1],[2] Theorem 6.3.6).

Theorem 1.3. Let D=

{

Z Z Z Z Z Z Z D₇, ₆, ₅, ₄, ₃, ₂, ₁, 

}

∈Σ₃

(

X,8

)

and Z₇ = ∅. Then a binary relation

αof the semigroup B_X( )D that has a quasinormal representation of the form to be given below is a regular element of this semigroup iff there exist a complete α−isomorphism ϕof the semilattice V D( ,α) on some subsemilattice D′ of the semilattice Dthat satisfies at least one of the following conditions:

)

1 α= ∅;

)

2 α=

(

Y₇α×∅ ∪

) (

Y_Tα′×T′

)

, where ∅ ≠ ∈T′ D, YT

{ }

α

′ ∉ ∅ , and satisfies the conditions: Y7

α _{⊇ ∅}

,

( )

T

Yα_′∩ϕ T′ ≠ ∅;

)

3 α=

(

Y₇α×∅ ∪

) (

Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

)

, where ∅ ≠T′⊂T′′∈D



, Y_Tα′,Y_Tα′′∉ ∅

{ }

, and satisfies the

conditions: Y7

α _{⊇ ∅}

, Y7 YT

( )

T α α _ϕ

′ ′

∪ ⊇ , YT

( )

T α _ϕ

′ ∩ ′ ≠ ∅, YT

( )

T α _ϕ

′′∩ ′′ ≠ ∅;

)

4 α=

(

Y₇α×∅ ∪

) (

Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

) (

∪ Y_Tα′′′×T′′′

)

, where ∅ ≠ ⊂T′ T′′⊂T′′′∈D, YT ,YT ,YT

{ }

α α α

′ ′′ ′′′∉ ∅ , and

satisfies the conditions: Y7

α _{⊇ ∅}

, Y7 YT

( )

T α α _ϕ

′ ′

∪ ⊇ , Y7 YT YT

( )

T α α α _ϕ

′ ′′ ′′

∪ ∪ ⊇ , YT

( )

T α _ϕ

′ ∩ ′ ≠ ∅,

( )

T

Yα′′∩ϕ T′′ ≠ ∅, YT

( )

T α _ϕ

′′′∩ ′′′ ≠ ∅;

)

5 α=

(

Y₇α×∅ ∪

) (

Y_Tα× ∪T

) (

Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

)

∪

(

Y₀α×D

)



, where Z7 ≠ ⊂T T′⊂T′′⊂D



, YT ,YT ,YT ,Y0

{ }

α α α α

′ ′′ ∉ ∅ ,

and satisfies the conditions: Y7

α _{⊇ ∅}

, Y7 YT

( )

T α_∪ α _⊇_ϕ

, Y7 YT YT

( )

T α α α _ϕ

′ ′

∪ ∪ ⊇ ,

( )

7 T T T

Yα ∪Yα ∪Yα′∪Yα′′⊇ϕ T′′ , YT

( )

T α _∩_ϕ _{≠ ∅}

, YT

( )

T α _ϕ

′∩ ′ ≠ ∅, YT

( )

T α _ϕ

′′∩ ′′ ≠ ∅, Y0 D

α_{∩ ≠ ∅}

;

)

6 α=

(

Y₇α×∅ ∪

) (

Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

)

∪

(

Y_Tα′∪_T′′×

(

T′∪T′′

)

, where T′\T′′ ≠ ∅, T′′\T′ ≠ ∅, Y YT, T

{ }

α α

′ ′′∉ ∅ and

satisfies the conditions: Y7 YT

( )

T α α _ϕ

′ ′

∪ ⊇ , Y7 YT

( )

T α α _ϕ

′′ ′′

∪ ⊇ , YT

( )

T α _ϕ

′∩ ′ ≠ ∅, YT

( )

T α _ϕ

′′∩ ′′ ≠ ∅;

(3)

)

7 α=

(

Y₇α×∅ ∪

) (

Y_Tα′× ∪T′

) (

Y_Tα′′×T′′

) (

∪Y_Tα′′′×T′′′

)

∪

(

Y_Tα′′ ′′′∪_T ×

(

T′′∪T′′′

)

, where, ∅ ≠T′⊂T′′, ∅ ≠T′⊂T′′′, T′′\T′′′ ≠ ∅,

\

T′′′ T′′ ≠ ∅ Y Y Y_Tα′, _Tα′′, _Tα′′′∉ ∅

{ }

α_{⊇ ∅}

T T

Yα∪ ⊇Yα′ T′, YT YT YT T

α α α ′ ′′ ′′

∪ ∪ ⊇ ,

T T T

Yα∪Yα′∪Yα′′′⊇T′′′ YT T

α

′ ∩ ≠ ∅′ , YT T α

′′∩ ′′≠ ∅, YT T α

′′′∩ ′′′≠ ∅.

)

8 α =

(

Y₇α×∅ ∪

) (

Y_Tα× ∪T

) (

Y₄α×Z₄

) (

∪ Y₂α×Z₂

) (

∪ Y₁α×Z₁

)

∪

(

Y₀α×D

)

, where T∈

{

Z Z6, 5

}

, Y Y Y YT , 4 , 2 , 1

{ }

α α α α_{∉ ∅}

α _{⊇ ∅}

, Y7 YT

( )

T α_∪ α _⊇_ϕ

, Y7 YT Y4

( )

Z4

α_∪ α_∪ α _⊇_ϕ

,

( )

7 T 4 2 2

Yα∪Yα∪Yα∪Yα ⊇ϕ Z , Y7 YT Y4 Y1

( )

Z1

α_{∪ ∪ ∪}α α α_⊇_ϕ

, YT

( )

T α _∩_ϕ _{≠ ∅}

, Y4 Z4

α_{∩ ≠ ∅}

2 2

Yα∩ ≠ ∅Z ,

1 1

Yα∩ ≠ ∅Z

)

9 α=

(

Y₇α×∅ ∪

) (

Y₅α×Z₅

) (

∪ Y₄α×Z₄

) (

∪ Y₃α×Z₃

) (

∪ Y₁α×Z₁

)

∪

(

Y₀α×D

)

, where Z5 ⊂Z3, Z5⊂Z4, 3\ 4

Z Z ≠ ∅, Z4\Z3≠ ∅ Y Y Y Y Y5 , 4 , 3 , 1 , 0

{ }

α α α α α_{∉ ∅}

α _{⊇ ∅}

, Y7 Y5 Z5

α_∪ α_⊇

, Y7 Y5 Y3 Z3

α_{∪ ∪ ⊇}α α

7 5 4 4

Yα∪ ∪ ⊇Yα Yα Z , Y5 Z5

α_{∩ ≠ ∅}

, Y3 Z3

α_∩ _{≠ ∅}

, Y4 Z4

α_∩ _{≠ ∅}

, Y0 D

α_{∩ ≠ ∅}

;

)

10 α=

(

Y₇α×∅ ∪

) (

Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

)

∪

(

Y_Tα′∪_T′′×(T′∪T′′)

) (

∪ Y_Tα′′′×T′′′

)

, where, T′\T′′ ≠ ∅, T′′\T′ ≠ ∅,

T′∪T′′⊂T′′′ Y Y Y_Tα′, _Tα′′, _Tα′′′∉ ∅

{ }

and satisfies the conditions: Y7 YT

( )

T α α _ϕ

′ ′

∪ ⊇ , Y7 YT

( )

T α α _ϕ

′′ ′′ ∪ ⊇ ,

( )

T

Yα′∩ϕ T′ ≠ ∅, YT

( )

T α _ϕ

′′∩ ′′ ≠ ∅, YT

( )

T α _ϕ

′′′∩ ′′′ ≠ ∅;

)

11 α =

(

Y₇α×∅ ∪

) (

Y₆α×Z₆

) (

∪ Y₅α×Z₅

) (

∪ Y₄α×Z₄

) (

∪ Y_Tα× ∪T

)

(

Y₀α×D

)

, where T∈

{

Z Z2, 1

}

, Y Y Y Y6, 5 , T, 0

{ }

α α α α_{∉ ∅}

and satisfies the conditions:, Y7 Y6 Z6

α_∪ α_⊇

, Y7 Y5 Z5

α _∪ α _⊇

( )

7 6 5 4 T

Yα ∪Yα ∪Yα ∪Yα∪Yα ⊇ϕ T , Y6 Z6

α _∩ _{≠ ∅}

,

5 5

Yα∩ ≠ ∅Z YT T α_{∩ ≠ ∅}

, Y0 D

α_{∩ ≠ ∅}

;

)

12 α=

(

Y₇α×∅ ∪

) (

Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

)

∪

(

Y_Tα′∪_T′′×(T′∪T′′)

) (

∪ Y_Tα′′′×T′′′

)

∪

(

Y_Tα′∪ ∪_T′′ _T′′′×(T′∪ ∪T′′ T′′′)

)

, where

\

T′ T′′ ≠ ∅, T′′\T′ ≠ ∅, T′′⊂T′′′,

(

T′∪T′′

)

\T′′′≠ ∅, T′′′\

(

T′∪T′′

)

≠ ∅, , Y Y Y YT, T, T , T T T

{ }

α α α α

′ ′′ ′′′ ′ ′′ ′′′∪ ∪ ∉ ∅ and

satisfies the conditions:, Y7 YT

( )

T α α _ϕ

′ ′

∪ ⊇ , Y7 YT

( )

T α α _ϕ

′′ ′′

∪ ⊇ Y₇α ∪Y_Tα′′∪Y_Tα′′′ ⊇ϕ

( )

T′′′ ,

( )

T

Yα′ ∩ϕ T′ ≠ ∅, YT

( )

T α _ϕ

′′∩ ′′ ≠ ∅ YT

( )

T α _ϕ

′′′∩ ′′′ ≠ ∅;

)

13 α=

(

Y₇α×∅ ∪

) (

Y₅α×Z₅

) (

∪ Y₄α×Z₄

) (

∪ Y₃α×Z₃

) (

∪ Y₂α×Z₂

) (

∪ Y₁α×Z₁

)

∪

(

Y₀α×D

)

, where Z5 ⊂Z3, Z5 ⊂Z4,

3\ 4

Z Z ≠ ∅, Z₄\Z₃≠ ∅, Z₄ ⊂Z₂, Z Z₁\ ₂≠ ∅, Z₂\Z₁≠ ∅, , Y Y Y Y Y Y5, 4, 3 , 2 ,1 , 0

{ }

α α α α α α_{∉ ∅}

and satisfies the conditions: Y7 Z7

α _⊇

, Y7 Y5 Z5

α_∪ α _⊇

7 5 3 3

Yα∪Yα∪Yα ⊇Z , Y7 Y5 Y4 Z4

α_∪ α_∪ α _⊇

, Y7 Y5 Y4 Y1 Z1

α_∪ α_∪ α_∪ α _⊇

,

5 5

Yα∩ ≠ ∅Z , Y3 Z3

α_∩ _{≠ ∅}

, Y4 Z4

α_{∩ ≠ ∅}

1 1

Yα∩ ≠ ∅Z ;

)

14 α=

(

Y₇α×∅ ∪

) (

Y₆α×Z₆

) (

∪ Y₅α×Z₅

) (

∪ Y₄α×Z₄

) (

∪ Y₃α×Z₃

) (

∪ Y₁α×Z₁

)

∪

(

Y₀α×D

)

, where, Z6⊂Z4,

{ }

6, 5 , 4 , 3, 1 , 0

Y Y Y Y Y Yα α α α α α∉ ∅ and satisfies the conditions: Y7 Y5 Z5

α_∪ α_⊇

, Y7 Y6 Z6

α_∪ α_⊇

, Y7 Y5 Y3 Z3

α_∪ α_∪ α _⊇

,

5 5

Yα ∩Z ≠ ∅, Y6 Z6

α _∩ _{≠ ∅}

, Y3 Z3

α_∩ _{≠ ∅}

0

Yα∩ ≠ ∅D ;

)

15 α=

(

Y₇α×∅ ∪

) (

Y₆α×Z₆

) (

∪ Y₅α×Z₅

) (

∪ Y₄α×Z₄

) (

∪ Y₂α×Z₂

) (

∪ Y₁α×Z₁

)

∪

(

Y₀α×D

)

, where Y Y Y Y Y6 , 5, 4, 2, 1

{ }

α α α α α_{∉ ∅}

and satisfies the conditions: Y7 Y6 Z6

α_∪ α _⊇

, Y7 Y5 Z5

α_∪ α _⊇

, Y7 Y6 Y5 Y4 Y2 Z2

α_∪ α_∪ α_∪ α_∪ α _⊇

,

7 6 5 4 1 1

Yα∪Yα∪Yα∪Yα∪Yα ⊇Z Y6 Z6

α _∩ _{≠ ∅}

, Y5 Z5

α _∩ _{≠ ∅}

, Y2 Z2

α _∩ _{≠ ∅}

1 1

Yα∩ ≠ ∅Z ;

)

16 α=

(

Y₇α×∅ ∪ × ∪ × ∪ × ∪ × ∪ × ∪ × ∪ ×

) (

Y₆α Z₆

) (

Y₅α Z

) (

Y₄α Z₄

) (

Y₃α Z₃

) (

Y₂α Z₂

) (

Y₁α Z₁

)

( )

Y₀α D , where, Y Y Y Y Y Y Y6, , , , , ,5 4 3 2 1 0

{ }

α α α α α α α_{∉ ∅}

α_{⊇ ∅}

7 5 5

Yα∪Yα⊇Z , Y7 Y6 Z6

α_∪ α _⊇

, Y7 Y5 Y3 Z3

α_{∪ ∪ ⊇}α α

,

7 5 6 4 2 2

Yα∪ ∪ ∪ ∪ ⊇Yα Yα Yα Yα Z , Y5 Z5

α _∩ _{≠ ∅}

, Y6 Z6

α_∩ _{≠ ∅}

, Y3 Z3

α_∩ _{≠ ∅}

, Y2 Z2

α_∩ _{≠ ∅}

; (see Theorem 1.1 in[3])

Lemma 1.1

(4)

)

a R Q

( )

1 1

∗ ₌ _;

)

b ( )

(

)

\

2 0 2 1 2

T X T

R∗ Q =m ⋅ ′ − ⋅ ′;

)

c ( )

(

) (

\ \

)

\

3 0 2 1 3 2 3

T T T T T X T

R∗ Q =m ⋅ ′ − ⋅ ′′ ′ − ′′ ′ ⋅ ′′;

)

d ( )

(

) (

\ \

) (

\ \

)

\

4 0 2 1 3 2 4 3 4

T T T T T T T T T X T

R∗ Q =m ⋅ ′ − ⋅ ′′ ′ − ′′ ′ ⋅ ′′′ ′′− ′′′ ′′ ⋅ ′′′;

)

e ( )

(

) (

\ \

) (

\ \

)

(

\ \

)

\

5 0 2 1 3 2 4 3 5 4 5

D T D T X D

T T T T T T T T T

R∗ Q =m ⋅ − ⋅ ′ − ′ ⋅ ′′ ′− ′′ ′ ⋅ ′′ − ′′ ⋅

  

;

) ( )

(

\

) (

\

)

\( )

6 2 0 2 1 2 1 4

X T T

T T T T

R∗ Q = ⋅m ⋅ ′ ′′− ⋅ ′′ ′− ⋅ ′∪ ′′

f ;

) ( )

(

)

( )\

(

\ \

) (

\ \

)

\( )

7 2 0 2 1 2 3 2 3 2 5

T T T X T T

T T T T T T T T T

R∗ Q = ⋅m ⋅ ′ − ⋅ ′′∩ ′′′ ′⋅ ′′ ′′′− ′′ ′′′ ⋅ ′′′ ′′ − ′′′ ′′ ⋅ ′′∪′′′

g ;

) ( )

(

) (

4\ 4\

)

( 2 1)\ 4

(

1\ 2 1\ 2

) (

2\ 1 2\ 1

)

\

8 2 0 2 1 3 2 3 4 3 4 3 6 ;

X D Z Z Z

T Z T Z T Z Z Z Z Z Z Z Z

R∗ Q = ⋅m ⋅ − ⋅ − ⋅ ∩ ⋅ − ⋅ − ⋅

h



) ( )

(

5

)

( 3 4)\ 5

(

3\ 4 3\ 4

) (

4\ 3 4\ 3

)

(

\( 3 4) \( 3 4)

)

\

9 2 0 2 1 2 3 2 3 2 6 5 6 ;

D Z Z D Z Z X D

Z Z Z

Z Z Z Z Z Z Z Z Z

R∗ Q = ⋅m ⋅ − ⋅ ∩ ⋅ − ⋅ − ⋅ ∪ − ∪ ⋅

i

  

) ( )

(

\

) (

\

)

(

\( ) \( )

)

\

10 2 0 2 1 2 1 5 4 5

T T T T T T

T T T T X T

R∗ Q = ⋅m ⋅ ′ ′′ − ⋅ ′′ ′ − ⋅ ′′′ ′∪ ′′ − ′′′ ′∪ ′′ ⋅ ′′′

j ;

) ( )

(

\

) (

\

) (

\ 4 \ 4

)

\ \ \

11 2 0 2 1 2 1 5 4 6 5 6

D T D T X D

T T T T T Z T Z

R Q∗ = ⋅m ⋅ ′− ⋅ ′ − ⋅ ′′ − ′′ ⋅_ ′′ − ′′_⋅

 

k

  

;

) ( )

(

\

) (

\

)

(

\( ) \( )

)

\( )

12 0 2 1 2 1 3 2 6

T T T T T T X T T T

T T T T

R∗ Q =m ⋅ ′ ′′′− ⋅ ′′ − ⋅ ′′′ ′∪ ′′ − ′′′ ′∪ ′′ ⋅ ′∪ ∪′′ ′′′

l ;

) ( )

(

5

)

( 3 2)\ 5

(

3\ 4 3\ 4

) (

4\ 2 4\ 2

) (

2\ 1 2\ 1

)

\

13 0

R∗ Q =m ⋅ 2Z − ⋅1 2Z∩Z Z ⋅ 3Z Z −2Z Z ⋅ 3Z Z −2Z Z ⋅ 4Z Z −3Z Z ⋅7X D;

m



) ( )

(

5\ 4

) (

6\ 5

) (

4\ 3 4\ 3

)

(

\ 1 \ 1

)

\

14 0 2 1 2 1 3 2 7 6 7 ;

D Z D Z X D

Z Z Z Z Z Z Z Z

R∗ Q =m ⋅ − ⋅ − ⋅ − ⋅ − ⋅

n

  

) ( )

(

5\ 6

) (

6\ 5

)

( 2 1)\ 4

(

2\ 1 2\ 1

) (

1\ 2 1\ 2

)

\

15 4 0 2 1 2 1 4 5 4 5 4 7

X D Z Z Z

Z Z Z Z Z Z Z Z Z Z Z Z

R∗ Q = ⋅m ⋅ − ⋅ − ⋅ ∩ ⋅ − ⋅ − ⋅

o



) ( )

(

)

(

) (

)

6\ 3 5\ 4 5\ 6 3\ 2 3\ 2 2\ 1 2\ 1 \

16

2R∗ Q = Z Z − ⋅1 2Z Z ⋅ 2Z Z − ⋅1 3Z Z −2Z Z ⋅ 5Z Z −4Z Z ⋅8X D

p



(see Lemma 1.1 in [3])

)

1 Lemma 1.2. Let D=

{

Z Z Z Z Z Z Z D7, 6, 5, 4, 3, 2, ,1

}

∈Σ3

( )

X,8



and Z₇ = ∅. Then R∗( )Q₁ =1.

)

2 Now let binary relation α of the semigroup BX( )D satisfying the condition b) of the

Theorem 1.3 In this case we have Q₂ = ∅{ ,T′}, By definition of the semilattice D follows that

{ }

{

{ } { } { } { } { } { }

}

2 , , , 6 , , 5 , 4 , , 3 , , 2 , , 1

XI

Qϑ = ∅ D ∅ Z ∅ Z ∅ Z ∅ Z ∅ Z ∅ Z

It is easy to see Φ(Q Q2, 2) =1 and Ω

( )

Q2 =7. Assume that

{ }

{ } { } { } { } { } { }

1 , , 2 , 6 , 3 , 5 , 4 , 4 , 5 , 3 , 6 , 2 , 7 , 1

D′= ∅ D D′ = ∅Z D′= ∅ Z D′= ∅ Z D′= ∅ Z D′= ∅ Z D′= ∅ Z

(5)

Lemma 1.3. Let X be a finite set, D=

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z₇ = ∅. Then

( ) \

2 7 2 1 2

D X D

R Q∗ = ⋅_ − ⋅_

 

.

)

c Now let binary relation α of the semigroup B_X

( )

D satisfying the condition c) of the Theorem 1.3 In this case we have Q₂ = ∅{ ,T T′ ′′, }, where T T′ ′′∈, D and ∅ ≠T′⊂T′′. By definition

of the semilattice D follows that

{

} {

}

{ }

{

} {

}

{

{ }

{ } { } { } { } { } { } { }

{ } { }

}

3 1 2 3 4 5 6 6 4

6 2 6 1 5 4 5 3 5 2 5 1 4 2

4 1 3 1

, , , , , , , , , , , , , , , , , , , , ,

, , , , ,

XI

Q Z D Z D Z D Z D Z D Z D Z Z

Z Z Z Z Z Z Z Z Z Z Z Z Z Z

Z Z Z Z

ϑ = ∅ ∅ ∅ ∅ ∅ ∅ ∅

∅ ∅ ∅ ∅ ∅ ∅ ∅

∅ ∅

    

It is easy to see Φ(Q Q₃, ₃) =1 and Ω

( )

Q₃ =16.assume that

{

}

{

}

{

}

{ }

{

}

{

}

{ } { } { } { }

1 1 2 2 3 3 4 4 5 5

6 6 7 6 4 8 6 2 9 6 1 10 5 4

11 5 3 12 5 2 13 5 1 14 4 2

, , , , , , , , , , , , , , ,

, , , , , , , , , , , ,

D Z D D Z D D Z D D Z D D Z D

D Z D D Z Z D Z Z D Z Z D Z Z

D Z Z D Z Z D Z Z D Z Z D

′= ∅ ′= ∅ ′= ∅ ′ = ∅ ′= ∅ ′ = ∅ ′ = ∅ ′= ∅ ′ = ∅ ′ = ∅ ′ = ∅ ′ = ∅ ′ = ∅ ′ = ∅ ′      { }

{ } 15 4 1

16 3 1

, , ,

, ,

Z Z

D Z Z

= ∅ ′ = ∅

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z7 = ∅. Then

( )

6

3 1 5

1

1 3 1 4 2 4

3 5 4 5 4 6

i i

R Q R D R D R D

R D R D R D R D R D R D

∗ = ′ ′ ′ = + ∩ − ′ ′ ′ ′ ′ ′ − ∩ − ∩ − ∩ − ′ ′ ′ ′ ′ ′ − ∩ − ∩ − ∩

∑

Lemma 1.5. Let D=

{

Z Z Z Z Z Z Z D₇, ₆, ₅, ₄, ₃, ₂, ₁,

}

∈ Σ₃(X,8) and Z₇ = ∅. Then

( )

(

)

(

)

(

)

(

)

1 1 1 2 2 2 3 3 3 4 4 4 \ \ \ 3 \ \ \ \ \ \ \ \ \

16 2 1 3 2 3

D Z D Z X D Z

R∗ Q = ⋅ − ⋅ − ⋅ +

_ _ + ⋅ − ⋅_ − _⋅ +     + ⋅ − ⋅_ − _⋅ +     + ⋅ − ⋅ − ⋅ +              

(

)

(

)

(

)

(

)

5 5 5 6 6 6 1 1

1 5 5

1 1

1 3 3

\ \ \ \ \ \ \ \ \ \ \ \ \

16 2 1 3 2 3

16 2 2 1 3 2 3

D Z D Z X D Z

D Z D Z X D Z Z Z

D Z D Z X Z Z Z

  + ⋅ − ⋅_ − _⋅ +     + ⋅ − ⋅ − ⋅ +  _  _ + ⋅ ⋅ − ⋅_ − _⋅ +     − ⋅ ⋅ − ⋅_ − _⋅             

(

)

(

)

(

)

1 1

1 4 4

2 2

2 4 4

3 3

3 5 5

4 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \

16 2 2 1 3 2 3

16 2

D D Z D Z X D Z Z Z

D Z D Z X D Z Z Z

Z Z −   − ⋅ ⋅ − ⋅ − ⋅ − _ _ − ⋅ ⋅ − ⋅_ − _⋅ −     − ⋅ ⋅ − ⋅ − ⋅ −   − ⋅ ⋅          

(

)

(

)

4 4 5 4 4

4 6 6

\ \ \

\

2 1 3 2 3

16 2 2 1 3 2 3

D Z D Z X D Z

D Z D Z X D Z Z Z

(6)

) ′

d Now let binary relation α of the semigroup BX( )D satisfying the condition d) of the

Theorem 1.3 In this case we have Q4= ∅{ ,T Z Z′, , ′}, where T Z Z′, , ′∈D and ∅ ≠T′⊂ ⊂Z Z′. By

definition of the semilattice D follows that

{

}

{

{ }

{

} {

}

{

} {

}

{ }

{

} {

}

{ } { } { }

4 6 4 7 6 2 6 1 5 4 5 3

5 2 5 1 4 2 4 1 3 1

5 4 2 5 4 1 5 3 1 6 4

, , , , , , , , , , , , , , , , , , , ,

, , , , , , , , , , , , , , , , , , ,

, , , , , , , , , , , , , , ,

XI

Q Z Z D Z Z Z D Z Z D Z Z D Z Z D

Z Z D Z Z D Z Z D Z Z D Z Z D

Z Z Z Z Z Z Z Z Z Z Z Z

ϑ = ∅ ∅ ∅ ∅ ∅

∅ ∅ ∅ ∅ ∅

∅ ∅ ∅ ∅

   

{ 2} {, ,∅ Z Z Z6, 4, 1},

( )

Q4 =15. Assume that

{

}

{ }

{

}

{

}

{

}

{

}

{

}

{ }

{

}

{

}

{ } { }

{ }

1 6 4 2 6 2 3 6 1 4 5 4

5 5 3 6 5 2 7 5 1 8 4 2

9 4 1 10 3 1 11 6 4 2 12 6 4 1

13 5 4 2 1

, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

, , , ,

D Z Z D D Z Z D D Z Z D D Z Z D

D Z Z D D Z Z D D Z Z D D Z Z D

D Z Z D D Z Z D D Z Z Z D Z Z Z

D Z Z Z D

′= ∅ ′ = ∅ ′= ∅ ′= ∅

′= ∅ ′= ∅ ′= ∅ ′= ∅

′= ∅ ′ = ∅ ′ = ∅ ′ = ∅

′ = ∅ ′

  

 

{ } { }

4= ∅,Z Z5, 4,Z1 ,D15′ = ∅,Z Z Z5, 3, 1 ,

.

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z7 = ∅. Then

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

10

4 1 2 1 3

1

2 8 3 9 4 6

4 7 6 8 7 9

7 10

i i

R Q R D R D R D R D R D

R D R D R D R D R D R D

R D R D

∗

=

′ ′ ′ ′ ′

= − ∩ − ∩ −

′ ′ ′ ′ ′ ′

− ∩ − ∩ − ∩ −

′ ′ ′ ′ ′ ′

− ∩ − ∩ − ∩ −

′ ′

− ∩

∑

Lemma 1.7 Let D=

{

Z Z Z Z Z Z Z D7, 6, 5, 4, 3, 2, 1,

}

∈ Σ3(X,8) 

and Z7 = ∅. Then

(7)

( )

(

) (

)

(

) (

)

(

) (

)

4 4

6 7 4 6 4 6

2 2

6 7 2 6 2 6

1 1

6 7 1 6 1 6

\ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \

15 2 1 3 2 4 3 4

D Z D Z X D

Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z

R∗ Q = ⋅ − ⋅ − ⋅ − ⋅ +

    + ⋅ − ⋅ − ⋅_ − _⋅ +     + ⋅ − ⋅ − ⋅_ − _⋅ +           

(

) (

)

(

) (

)

(

) (

)

4 4

5 7 4 5 4 5

3 3

5 7 3 5 3 5

2 2

5 7 2 5 2 5

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

15 2 1 3 2 4 3 4

D Z D Z X D

Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z

  + ⋅ − ⋅ − ⋅_ − _⋅ +   + ⋅ − ⋅ − ⋅ − ⋅ + _ _ + ⋅ − ⋅ − ⋅ − ⋅ +           

(

) (

)

(

) (

)

(

) (

)

1 1

5 7 1 5 1 5

2 2

4 7 2 4 2 4

1 1

4 7 1 4 1 4

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

15 2 1 3 2 4 3 4

D Z D Z X D

Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z

  + ⋅ − ⋅ − ⋅_ − _⋅ + _ _ + ⋅ − ⋅ − ⋅_ − _⋅ +     + ⋅ − ⋅ − ⋅_ − _⋅ +           

(

) (

)

(

)

(

)

(

)

(

)

1 1

3 7 1 3 1 3

2 2

6 6 6 7 2 4 4 6 4 6

6 6 6 7 1 4 4 6 4 6

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

15 2 1 3 2 4 3 4

15 3 2 1 3 3 2 4 3 4

15 3 2 1 3 3 2 4

D Z D Z X D

Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z Z Z Z Z

D Z

Z Z Z Z Z Z Z Z Z Z

  + ⋅ − ⋅ − ⋅ − ⋅  _ _ − ⋅ ⋅ − ⋅ ⋅ − ⋅ − ⋅ −   − ⋅ ⋅ − ⋅ ⋅ − ⋅       

(

)

(

)

(

)

(

)

1 1 2 2

4 6 6 7 2 2 2 4 2 4

1 1

4 6 6 7 1 1 1 4 1 4

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 4

15 3 2 1 3 3 2 4 3 4

15 2 2 1 3 3 2 4 3 4

15

D Z X D D Z D Z X D

Z Z Z Z Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z Z Z Z Z

 ₋ _⋅ ₋   _ _ − ⋅ ⋅ − ⋅ ⋅ − ⋅_ − _⋅ −     − ⋅ ⋅ − ⋅ ⋅ − ⋅_ − _⋅ −   −        

(

)

(

)

(

)

(

)

(

)

(

)

2 2

5 5 5 7 2 4 4 5 4 5

1 1

5 5 5 7 1 4 4 5 4 5

5 5 5 7 1 3 3 5 3 5

\ \ \

\ \ \ \ \

\ \ \

\ \ \ \ \

2 2 1 3 3 2 4 3 4

15 2 2 1 3 3 2 4 3 4

15 2 2 1 3 3 2

D Z D Z X D

Z Z Z Z Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z Z Z Z Z

  ⋅ ⋅ − ⋅ ⋅ − ⋅_ − _⋅ −   − ⋅ ⋅ − ⋅ ⋅ − ⋅ − ⋅ −   − ⋅ ⋅ − ⋅ ⋅ − ⋅      

(

)

(

)

(

)

(

)

1 1 2 2

4 5 5 7 2 2 2 4 2 4

1 1

4 5 5 7 1 1 1 4 1 4

\ \ \

\ \ \ \ \

\ \ \

\ \ \ \ \

4 3 4

15 2 2 1 3 3 2 4 3 4

D Z D Z X D D Z D Z X D

Z Z Z Z Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z Z Z Z Z

 ₋ _⋅ ₋       − ⋅ − ⋅ ⋅ − ⋅_ − _⋅ −     − ⋅ − ⋅ ⋅ − ⋅_ − _⋅ −           

(

)

(

)

1 1

3\ 5 5\ 7 1\ 1 1\ 3 1\ 3 \ \ \

15 2 2 1 3 3 2 4 3 4

D Z D Z X D

Z Z Z Z Z Z Z Z Z Z  

− ⋅ ⋅ − ⋅ ⋅ − ⋅_ − _⋅

 

  

) ′

e Now let binary relation α of the semigroup B_X ( )D satisfying the condition e) of the Theorem 1.3 In this case we have Q5= ∅

{

, ,T T T D′ ′′, ,

}

,



where T T T, ′ ′′∈, D and T ⊂T′⊂T′′∈D. By definition of the semilattice D follows that

{

} {

}

{

} {

}

5 6 4 2 6 4 1 5 4 2

5 4 1 5 3 1

, , , , , , , , , , , , , , ,

, , , , , , , , , .

XI

Q Z Z Z D Z Z Z D Z Z Z D

Z Z Z D Z Z Z D

ϑ = ∅ ∅ ∅

∅ ∅

  

 

( )

Q5 =5. Assume that

{

}

{

}

{

}

{

}

{

}

1 6 4 2 2 6 4 1 3 5 4 2

4 5 4 1 5 5 3 1

, , , , , , , , , , , , , , ,

, , , , , , , , ,

D Z Z Z D D Z Z Z D D Z Z Z D

D Z Z Z D D Z Z Z D

′= ∅ ′= ∅ ′= ∅

′ = ∅ ′= ∅

  

 

{

Z Z Z Z Z Z Z D₇, , , , , , ,₆ ₅ ₄ ₃ ₂ ₁ 

}

∈Σ₃

( )

X,8 and Z₇ = ∅. Then

( )5 ( )1 ( )2 ( )3 ( )4 ( )5

R Q∗ = R D′ +R D′ +R D′ +R D′ +R D′

{

Z Z Z Z Z Z Z D₇, ₆, ₅, ₄, ₃, ₂, ₁,

}

∈ Σ₃(X,8) and Z₇ = ∅. Then

(8)

( )

(

) (

)

(

) (

)

(

) (

)

2 2

6 4 6 4 6 2 4 2 4

2 2

6 4 6 4 6 1 4 1 4

2

5 4 5 4 5 2 4 1 4

\ \ \

\ \ \ \

5

\ \ \

\ \ \ \

\

\ \ \ \

5 2 1 3 2 4 3 5 4 5

5 2 1 3 2 4 3 5 4

D Z D Z X D

Z Z Z Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z Z Z Z

D Z

Z Z Z Z Z Z Z Z Z

R Q∗ = ⋅ − ⋅ − ⋅ − ⋅_ − _⋅ +

 

+ ⋅ − ⋅ − ⋅ − ⋅_ − _⋅ +

 

+ ⋅ − ⋅ − ⋅ − ⋅ −

  



(

) (

)

(

) (

)

2

1 1

5 4 5 4 5 1 4 1 4

1 1

5 3 5 3 5 1 3 1 3

\ \

\ \ \

\ \ \ \

\ \ \

\ \ \ \

5

5 2 1 3 2 4 3 5 4 5

D Z X D D Z D Z X D

Z Z Z Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z Z Z Z

 _⋅ ₊

 

 

 

+ ⋅ − ⋅ − ⋅ − ⋅ − ⋅ +

 

 

+ ⋅ − ⋅ − ⋅ − ⋅ − ⋅

 

 

  

) ′

f Now let binary relation α of the semigroup BX( )D satisfying the condition f) of the

Theorem 1.3. In this case we have Q6 ={T T T T, ′ ′′ ′, , ∪T′′}, where T T T, ′ ′′∈, D and T ⊂T′,T ⊂T′′,

\

T′ T′′ ≠ ∅, T′′\T′ ≠ ∅. By definition of the semilattice D follows that

{

}

{

} {

}

{

}

6 XI , 2, 1, , , 6, 5, 4 , , 6, 3, 1 , , 4, 3, 1 , 7, 3, 2,

Qϑ = ∅ Z Z D ∅ Z Z Z ∅ Z Z Z ∅ Z Z Z ∅ Z Z Z D

( )

Q6 =10. Assume that

{

}

{

}

{ } { }

{ } { } { } { }

{

}

{

}

1 2 1 2 1 2 3 6 5 4 4 5 6 4

5 6 3 1 6 3 6 1 7 4 3 1 8 3 4 1

9 3 2 10 2 3

, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

D Z Z D D Z Z D D Z Z Z D Z Z Z

D Z Z Z D Z Z Z D Z Z Z D Z Z Z

D Z Z D D Z Z D

′= ∅ ′ = ∅ ′= ∅ ′ = ∅

′= ∅ ′= ∅ ′ = ∅ ′= ∅

′= ∅ ′ = ∅

 

{

Z Z Z Z Z Z Z D7, 6, 5, 4, 3, 2, ,1

}

∈Σ3

( )

X,8



and Z7 = ∅. If by

( )6

R Q∗ denoted all regular elements of the semigroup BX( )D satisfying the condition f) of the

Theorem 1.3, then

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

10

6 1 10 2 9

1

3 5 4 6 5 7 6 8

7 10 8 9

i i

R Q R D R D R D R D R D

R D R D R D R D R D R D R D R D

R D R D R D R D

∗

=

′ ′ ′ ′ ′

= − ∩ − ∩ −

′ ′ ′ ′ ′ ′ ′ ′

− ∩ − ∩ − ∩ − ∩ −

′ ′ ′ ′

− ∩ − ∩

∑

{

Z Z Z Z Z Z Z D7, 6, 5, 4, 3, 2, 1,

}

∈ Σ3(X,8) 

and Z7 = ∅. Then

( )

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

)

6 3 3 6

2 1 1 2 1

6 5 5 6 4 4 3 3 4 1

2 3 3 2

2 2 1 1 3

\ \ \

6

\ \ \ \ \ \

\

\ \

\ \ \ \

10 2 1 2 1 4 20 2 1 2 1 4

10 2 1 2 1 4

5 2 2 1 2 2

X D Z Z Z Z

Z Z Z Z X Z

Z Z Z Z X Z Z Z Z Z X Z

X D Z Z Z Z

Z D Z Z Z D Z Z

R Q∗ = ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ +

+ ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ +

+ ⋅ − ⋅ − ⋅ − ⋅ ⋅ − ⋅ ⋅



 

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 2

2 1 3 2 2 1

6 1 6 3 3 4 5 6 1 3 4 5 6 6 1 6 3 1

6 3 3 1 3 4 3 1 3 4 6

4 1 1 4 1

\ \ \ \

\ \

\ \ \ \ \ \ \ \ \ \

\ \ \ \ \

\ \ \

1 4 5 2 2 1 2 2 1 4

5 2 2 1 2 2 1 4 5 2 2 1 2 2 1 4

5 2 2 1 2 2 1 4 5 2 2 1 2 2

Z D Z Z Z D X D

X Z Z Z

Z Z Z Z Z Z Z Z X Z Z Z Z Z Z Z Z Z X Z

Z Z Z Z Z Z Z Z Z Z Z

Z Z X Z Z Z

− ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ −

− ⋅ ⋅ − ⋅ ⋅ − ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ −

− ⋅ ⋅ − ⋅ ⋅ − ⋅ − ⋅ ⋅ − ⋅ ⋅

  

(

)

(

)

(

)

(

)

(

)

3 1

3 3

4 3 3 2 3 2 4 3

2 1 2 1

\ \

\ \ \ \

\ \

1 4

5 2 2 1 2 2 1 4 5 2 2 1 2 2 1 4

Z X Z

Z D X D Z D X D

Z Z Z Z Z Z Z Z

Z Z Z Z

− ⋅ − − ⋅ ⋅ − ⋅  ⋅ − ⋅  − ⋅  ⋅ − ⋅ ⋅ − ⋅ 

) ′

g Now let binary relation α of the semigroup BX( )D satisfying the condition g) of the

Theorem 1.3 In this case we have {T T T T T, ,′ ′′ ′′′ ′′, , ∪T′′′}, where T T T T, ,′ ′′ ′′′∈, D, T ⊂T′⊂T′′, T ⊂T′⊂T′′′

, T T\ ′ ≠ ∅ and T′\T ≠ ∅. By definition of the semilattice D follows that

(9)

{

}

{

} {

}

{

}

{

}

7 4 2 1 6 2 1 5 2 1

5 3 2 5 4 3 1

, , , , , , , , , , , , , , ,

, , , , , , , , ,

XI

Z Z Z D Z Z Z Z

ϑ = ∅ ∅ ∅

∅ ∅

  



( )

Q6 =7. assume

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{ } { }

1 4 2 1 2 4 1 2 3 6 2 1 4 6 1 2

5 5 2 1 6 5 1 2 7 5 3 2 8 5 2 3

9 5 4 3 1 10 5 3 4 1

, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

, , , , , , , , ,

D Z Z Z D D Z Z Z D D Z Z Z D D Z Z Z D

D Z Z Z D D Z Z Z D D Z Z Z D D Z Z Z D

D Z Z Z Z D Z Z Z Z

′= ∅ ′= ∅ ′= ∅ ′= ∅

′= ∅ ′= ∅ ′ = ∅ ′= ∅

′ = ∅ ′ = ∅

   

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z7 = ∅. If by R Q( )7

∗

denoted all regular elements of the semigroup BX( )D satisfying the condition g) of the Theorem

1.3 then

( ) ( ) ( ) ( ) ( ) ( )

7

7 1 3 1 5

1

2 4 2 6 2 7

5 8 7 10 8 9

i i

R Q R D R D R D R D R D

R D R D R D R D R D R D

∗ = ′ ′ ′ ′ ′ = − ∩ − ∩ − ′ ′ ′ ′ ′ ′ − ∩ − ∩ − ∩ − ′ ′ ′ ′ ′ ′ − ∩ − ∩ − ∩

∑

{

Z Z Z Z Z Z Z D7, 6, 5, 4, 3, 2, 1,

}

∈ Σ3(X,8) 

and Z7 ≠ ∅. Then

( )

(

)

( )

(

) (

)

(

)

( )

(

) (

)

(

)

( )

(

) (

)

2 1 4

4 2 1 2 1 1 2 1 2

2 1 6

6 2 1 2 1 1 2 1 2

2 1 5

5 2 1 2 1 1 2 1 2

\ \ \ \ \ \ 7 \ \ \ \ \ \ \ \ \ \ \ \

10 2 1 2 3 2 3 2 5

X D Z Z Z

Z Z Z Z Z Z Z Z Z

X D Z Z Z

Z Z Z Z Z Z Z Z Z

X D Z Z Z

Z Z Z Z Z Z Z Z Z

R Q∗ ∩

∩ ∩ = ⋅ − ⋅ ⋅ − ⋅ − ⋅ + + ⋅ − ⋅ ⋅ − ⋅ − ⋅ + + ⋅ − ⋅ ⋅ − ⋅ − ⋅ +   

(

)

( )

(

) (

)

(

)

( )

(

) (

)

(

)

( )

(

)

(

)

2 3 5

5 2 3 2 3 3 2 3 2

4 3 5

5 4 3 4 3 3 4 3 4

2 1

2 1 4

4 6 6 2 1 2 1 1 2 1 2

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

10 2 1 2 3 2 3 2 5

5 2 2 1 2 3 3 2 3 3 2 5

X D Z Z Z

Z Z Z Z Z Z Z Z Z

X D Z Z Z

Z Z Z Z Z Z Z Z Z

Z D Z D X

Z Z Z

Z Z Z Z Z Z Z Z Z Z Z

∩ ∩ ∩ + ⋅ − ⋅ ⋅ − ⋅ − ⋅ + + ⋅ − ⋅ ⋅ − ⋅ − ⋅ + − ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅ − ⋅    

(

)

( )

(

)

(

)

(

)

( )

(

)

(

)

(

)

( )

(

)

2 1

2 1 4

4 5 5 2 1 2 1 1 2 1 2

1 1 2 1 2 2

2 1 4

4 6 6 2 1 2 1

2 2 1 4

4 5 5 2 1 2 1

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

5 2 2 1 2 3 3 2 3 3 2 5

5 2 2 1 2 3 3 2 3

D

Z D Z D X D

Z Z Z

Z Z Z Z Z Z Z Z Z Z Z

Z D Z Z Z Z Z D X D

Z Z Z

Z Z Z Z Z Z Z

Z D Z Z Z

Z Z Z Z Z Z Z

∩ ∩ ∩ − − ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅ − ⋅ − − ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅ − ⋅ − − ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅        

(

)

(

)

( )

(

)

(

)

(

)

( )

(

)

(

)

(

)

( )

1 1 2 1 2

1 2

2 1 4

4 5 5 3 2 3 2 2 3 2 3

2 1

2 1 5

5 5 5 2 1 2 1 3 2 3 2

2 3 5

5 5 5

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

3 2 5

5 2 2 1 2 3 3 2 3 3 2 5

5 2 2 1 2

Z D Z Z Z Z X D

Z D Z D X D

Z Z Z

Z Z Z Z Z Z Z Z Z Z Z

Z D Z D X D

Z Z Z

Z Z Z Z Z Z Z Z Z Z Z

Z Z Z Z Z Z

∩ ∩ ∩ ⋅ − ⋅ − − ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅ − ⋅ − − ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅ − ⋅ − − ⋅ ⋅ − ⋅ ⋅        

(

)

(

)

(

)

( )

(

)

(

)

3 3 2 3 2 2 1 4 3 4 3

3 2 3 5

5 5 5 2 1 4 3 4 3 3 2 3 2

\ \ \ \ \ \ \

\ \

\

\ \ \ \ \ \

3 3 2 3 3 2 5

5 2 2 1 2 3 3 2 3 3 2 5

Z D Z Z Z Z Z Z Z Z Z Z X D

Z D X D

Z Z Z

Z Z Z ∩ Z Z Z Z Z Z Z Z Z Z

⋅ − ⋅ ⋅ − ⋅ −

− ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅ − ⋅

 

) ′

h Now let binary relation α of the semigroup BX( )D satisfying the condition h) of the

Theorem 1.3 In this case we have Q8 =

{

Z T Z Z Z D7, , 4, 2, 1,

}



, where T∈{Z Z₅, ₆}. By definition of the semilattice D follows that

{

} {

}

{

}

8 XI , 6, 4, 2, 1, , , 5, 4, 2, 1,

Qϑ = ∅ Z Z Z Z D ∅ Z Z Z Z D . It is easy to see Φ(Q Q₈, ₈) =2 and Ω( )Q₈ =2. assume

(10)

{

}

{

}

{

}

{

}

1 6 4 2 1 2 6 4 1 2

3 5 4 2 1 4 5 4 1 2

, , , , , , , , , , , ,

, , , , , , , , , , , .

D Z Z Z Z D D Z Z Z Z D

′= ∅ ′ = ∅

 

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z₇ = ∅. If by R Q( )8

∗

denoted all regular elements of the semigroup BX( )D satisfying the condition h) of the Theorem

1.3, then

( ) ( ) ( ) ( ) ( )

(

) (

)

( )

(

)

(

)

(

) (

)

( )

(

) (

)

2 1 4

6 4 6 4 6 1 2 1 2

5 4 5 4 5

2 1 2 1

2 1 4 1 2 1 2 2 1 2 1

8 1 2 3 4

\

\ \ \ \

\ \ \

\ \

\

\ \ \ \ \

4 2 1 3 2 3 4 3 4 3 6 4 2 1 3 2

3 4 3 4 3 6 .

Z Z Z

Z Z Z Z Z Z Z Z Z

X D Z Z Z Z Z

Z Z Z Z

X D Z Z Z Z Z Z Z Z Z Z Z

R Q∗ R D R D R D R D

∩

′ ′ ′ ′

= + + + =

= ⋅ − ⋅ − ⋅ ⋅ − ⋅

⋅ − ⋅ + ⋅ − ⋅ − ⋅

⋅ ⋅ − ⋅ − ⋅



) ′

i Let binary relation α of the semigroup BX( )D satisfying the condition p) of the Theorem

1.3). in this case we have

{

∅,Z Z Z Z D₅, ₄, ₃, ₁, 

}

, By definition of the semilattice D follows that

{

}

9 XI , 5, 4, 3, 1,

Qϑ = ∅ Z Z Z Z D It is easy to see Φ

(

Q Q9, 9

)

=2 and Ω

( )

Q9 =1. If D1′ = ∅

{

,Z Z Z Z D5, 4, 3, 1,

}



,

then R Q( )9 R D( )1

∗ ₌ _′ _, _{( )} _{( )}

9 1

R Q∗ = R D′ and

( )

(

5

)

( 3 4)\ 5

(

3\ 4 3\ 4

) (

4\ 3 4\ 3

)

\

9 2 1 2 3 2 3 2 6

X D Z Z Z

Z Z Z Z Z Z Z Z Z

R Q∗ = − ⋅ ∩ ⋅ − ⋅ − ⋅



) ′

j Now let binary relation α of the semigroup B_X( )D satisfying the condition j) of the Theorem 1.3 In this case we have Q₁₀ ={T T T T, ′ ′′ ′, , ∪T′′,Z}, where T⊂T′, T⊂T′′, T′\T′′ ≠ ∅,

\

T′′ T′ ≠ ∅, T′∪T′′⊂Z. By definition of the semilattice D follows that

{

}

{

} {

}

{ } { }

}

10 6 5 4 6 3 1 4 3 1

6 5 4 2 6 5 4 1

, , , , , , , , , , , , , , ,

, , , , , , , , ,

XI

Z Z Z Z Z Z Z Z

ϑ = ∅ ∅ ∅

∅ ∅

  

.

It is easy to see Φ(Q₁₀,Q₁₀) =2 and Ω

( )

Q₁₀ =6. If

{

}

{

}

{

}

{

}

{

}

{

}

{ } { } { }

{ }

1 6 5 4 2 5 6 4 3 6 3 1

4 3 6 1 5 4 3 1 6 3 4 1

7 6 5 4 2 8 5 6 4 2 9 6 5 4 1

10 5 6 4 1

, , , , , , , , , , , , , , ,

, , , ,

D Z Z Z D D Z Z Z D D Z Z Z D

D Z Z Z Z D Z Z Z Z D Z Z Z Z

D Z Z Z Z

′= ∅ ′= ∅ ′= ∅

′ = ∅

  

Then

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

10 1 2 3 4 5 6

7 8 9 10

R Q R D R D R D R D R D R D

R D R D R D R D

∗ ₌ _′ _∪ _′ _∪ _′ _∪ _′ _∪ _′ _∪ _′ _∪

′ ′ ′ ′

∪ ∪ ∪ ∪ . …(1)

(see Definition 1.9).

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z₇ = ∅. If by R Q( )10

∗

denoted all regular elements of the semigroup B_X( )D satisfying the condition j) of the Theorem 1.3 then

( ) 4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

10 1 3 2 4 3 5 4 6

1

i i

R Q∗ R D R D R D R D R D R D R D R D R D

=

′ ′ ′ ′ ′ ′ ′ ′ ′

=

∑

− ∩ − ∩ − ∩ − ∩

{

Z Z Z Z Z Z Z D7, 6, 5, 4, 3, 2, 1,

}

∈ Σ3(X,8) 

and Z7 = ∅. Then

(11)

( )

(

) (

)

(

)

(

) (

)

(

)

(

) (

)

(

)

(

)

4 4 5 6 6 5

1 1 6 3 3 6

1 1 3 4 4 3

6 1 6 3 3 4 5

\ \ \

\ \

10

\ \ \

\ \

\ \ \

\ \

\ \ \ \

10 2 1 2 1 5 4 5

+10 2 1 2 1 5 4 5

5 2 2 1 2 2

D Z D Z X D

Z Z Z Z

D Z D Z X D

Z Z Z Z

D Z D Z X D

Z Z Z Z

Z Z Z Z Z Z Z Z

R Q∗ = ⋅ − ⋅ − ⋅ − ⋅ +

⋅ − ⋅ − ⋅ − ⋅ +

− ⋅ ⋅ − ⋅ ⋅

  

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1 1 6

1 1 3 4 5 6 6 1 6 3

1 1 6 3 3 1 3 4

4 1

3 1 3 4 4 1 6

\ \ \

\ \ \ \

\ \ \

\

\ \ \ \

1 5 4 5

5 2 2 1 2 2 1 5 4 5

5 2 2 1 2 2

D Z D Z X D

Z Z Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z

Z Z

Z Z Z Z Z Z Z

− ⋅ − ⋅ −

− ⋅ ⋅ − ⋅ ⋅ − ⋅ − ⋅ −

− ⋅ ⋅ − ⋅ ⋅

  

(

3

)

(

\ 2 \ 2

)

\

1 5D Z 4D Z 5X D.

Z _{− ⋅}  ₋  _⋅ 

) ′

k Now let binary relation α of the semigroup BX( )D satisfying the condition k) of the

Theorem 1.3 In this case we have

{

∅,Z Z Z T D6, 5, 4, ,

}



, where T∈{Z Z2, 1}. By definition of the

semilattice D follows that Q11ϑ = ∅XI

{

,Z Z Z Z D6, 5, 4, 2,

} {

, ∅,Z Z Z Z D6, 5, 4, 1,

}

 

.

It is easy to see Φ(Q Q11, 11) =2 and Ω( )Q11 =2. If

{

}

{

}

{

}

{

}

1 6 5 4 2 2 5 6 4 2

3 6 5 4 1 4 5 6 4 1

, , , , , , , , , , , ,

, , , , , , , , , , , .

D Z Z Z Z D D Z Z Z Z D

′= ∅ ′ = ∅

′= ∅ ′= ∅

 

Then

( )11 ( )1 ( )2 ( )3 ( )4 .

R Q∗ =R D′ ∪R D′ ∪R D′ ∪R D′ …(1)

Lemma 1.16 Let X be a finite set, D=

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z₇ ≠ ∅. If by R Q( )11

∗

denoted all regular elements of the semigroup BX( )D satisfying the condition k) of the Theorem

1.3 then

( ) ( ) ( ) ( ) ( )

(

) (

)

(

)

(

) (

)

(

)

2 2

5 6 6 5 2 4 2 4

1 1

5 6 6 5 1 4 1 4

11 1 2 3 4

\ \ \

\ \ \ \

\ \ \

\ \ \ \

+4 2 1 2 1 5 4 6 5 6

+4 2 1 2 1 5 4 6 5 6 .

D Z D Z X D

Z Z Z Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z Z Z Z

R Q∗ = R D′ +R D′ + R D′ + R D′ =

 

⋅ − ⋅ − ⋅ − ⋅_ − _⋅ +

 

 

⋅ − ⋅ − ⋅ − ⋅_ − _⋅

 

  

) ′

l Now let binary relation α of the semigroup BX( )D satisfying the condition l) of the

Theorem 1.3 In this case we have {T T T T, ′ ′′ ′, , ∪T T′′ ′′′ ′, ,T ∪T′′∪T′′′}, where T′\T′′ ≠ ∅, T′′ ′ ≠ ∅\T ,

(T′∪T′′)\T′′′≠ ∅, T′′′\(T′∪T′′)≠ ∅. By definition of the semilattice D follows that

{ }

{

} {

}

{

12 XI , 6, 5, 4, 3, 1 , , 6, 3, 2, 1, , , 4, 3, 2, 1,

Q ϑ = ∅Z Z Z Z Z ∅Z Z Z Z D ∅Z Z Z Z D

It is easy to see Φ(Q12,Q12) =1 and Ω

( )

Q12 =4. If

{

}

{

}

{

}

1 , 6, 5, 4, 3, 1 , 2 , 6, 3, 2, 1, , 3 , 4, 3, 2, 1,

D′= ∅Z Z Z Z Z D′ = ∅ Z Z Z Z D D ′= ∅ Z Z Z Z D

Then

( )12 ( )1 ( )2 ( )3

R Q∗ =R D′ ∪R D′ ∪R D′ …(1)

Lemma 1.17 Let X be a finite set, D=

{

Z Z Z Z Z Z Z D7, , , , , , ,6 5 4 3 2 1

}

∈Σ3

( )

X,8



and Z7 = ∅. If by R Q( )12

∗

denoted all regular elements of the semigroup BX( )D satisfying the condition l) of the Theorem

1.3 then