Regular Elements of Bx(D) Defined by the Class ∑1(X,10) I

(1)

http://dx.doi.org/10.4236/am.2016.79078

Regular Elements of

B

_X

( )

D

Defined by the

Class

Σ

₁

(

X

,10

)

−

I

Yasha Diasamidze

1

_{, Nino Tsinaridze}

1

_{, Neşet Aydn}

2

_{, Ali Erdoğan}

3

1_{Shota Rustavelli University, Batumi, Georgia}

2_{Çanakkale Onsekiz Mart University, Çanakkale, Turkey} 3_{Hacettepe University, Ankara, Turkey}

Received 15 January 2016; accepted 24 May 2016; published 27 May 2016

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

Abstract

In order to obtain some results in the theory of semigroups, the concept of regularity, introduced by J. V. Neumann for elements of rings, is useful. In this work, all regular elements of semigroup defined by semilattices of the class Σ₁

(

X,10

)

are studied. When X has finitely many elements, we have given the number of regular elements.

Keywords

Semilattice, Semigroup, Binary Relation

1. Introduction

Let D be a nonempty set of subsets of a given set X, closed under union. Such a set D is called a complete

X-semilattice of unions. For any map f from X to D, we define a binary relation.

{ }

( )

(

)

.

f x X

x f x

α

∈

=

_

×

The set of all α_f, denoted by B_X

( )

D , is a subsemigroup of B_X semigroup of all binary relations on X. (See [1]-[6].)

All notations, symbol and required definitions used in this work can be found in [7]. Recall the following results.

Lemma 1. [1], Corollary 1.18.1. Let Y =

{

y y₁, ₂,,y_k

}

and D_j=

{

T₁,,T_j

}

be two sets where k≥1

(2)

j j

T ∈D′ is given by _{s k j}

(

,

)

₌ _jk ₋

(

_j₋1

)

k_.

Theorem 1. [1], Theorem 1.18.1. Let D_j =

{

T T₁, ₂,,T_j

}

. Let Y⊆X be nonempty sets. Then the number of mappings from X to D such that _j f y

( )

=T_j for some y∈Y is equal to s= jX Y\ ⋅

(

jY − −

(

j 1

)

Y

)

.

2. Results

Let X be a nonempty set, D a X-semilattice of union with the conditions (see Figure 1);

(

)

9 4 1 9 5 1

9 6 1 9 6 2

9 6 3 9 7 3

9 8 3

1 2 1 3 2 3 4 2

4 3 4 7 4 8 5 2

5 3 5 7 5 8 7 1

7 2 8 1 8 2

, ,

;

\ 1 3 or 4 8 ;

i j

Z Z Z D Z Z Z D

Z Z Z D

Z Z i j i j

Z Z Z Z Z Z Z Z

Z Z Z Z Z Z D

⊂ ⊂ ⊂ ⊂ ⊂ ⊂

⊂ ⊂ ⊂

≠ ∅ ≤ ≠ ≤ ≤ ≠ ≤

∪ = ∪ = ∪ = ∪

= ∪ = ∪ = ∪ = ∪

= ∪ = ∪ = ∪ =

 



4 5 4 6 5 6 1

6 7 6 8 7 8 3

; ; .

Z Z Z Z Z Z Z

∪ = ∪ = ∪ =

(1)

The class of X-semilattices where each element is isomorphic to D is denoted by Σ₁

(

X,10

)

.

An element α∈B_X

( )

D is called regular if

α β α α

  = for some β ∈B_X. Our aim in this work is to identify all regular elements of B_X

( )

D where D is given above.

Definition 1. The complete X-semilattice of unions is called an XI-semilattice of unions if Λ

(

D D, _t

)

∈D and Z =



t Z∈Λ

(

D D, t

)

for any nonempty Z in D. Here Λ

(

D D, t

)

is an exact lower bound of Dt in D where

{

|

}

.

t

D = Z′∈D t∈Z′

The following Lemma is well known (see [7], Lemma 3).

Lemma 2. All semilattices in the form of the diagrams in Figure 2 are XI-semilattices.

Figure 1. Diagram of semilattice of unions D.

(3)

Definition 2. Let D′ and D′′ be two X-semilattices of unions. A one to one map from D′ to D′′ is said to be a complete isomorphism if

(

)

( )

1

T D

D T

ϕ ϕ

′∈

′

∪ =



for D₁⊆D′.

Definition 3. [1], Definition 6.3.3. Let α∈B_X

( )

D . We say that a complete isomorphism ϕ:Q→D′ is a complete α-isomorphism if

a) Q=V D

(

,α

)

;

b) ϕ

( )

∅ = ∅ for ∅ ∈V D

(

,α

)

and ϕ

( )

T α=T for any T∈V D

(

,α

)

.

The following subsemilattices are all XI-semilattices of the X-semilattices of unions D. a) Q₁=

{ }

T , where T∈D (see diagram 1 of the Figure 3);

b) Q₂=

{

T T, ′

}

, where T T, ′∈D and T⊂T′ (see diagram 2 of the Figure 3);

c) Q₃ =

{

T T T, ′ ′′,

}

, where T T T, ′ ′′∈, D and T⊂T′⊂T′′ (see diagram 3 of the Figure 3); d) Q₄=

{

Z T T D₉, , ′, 

}

, where T T, ′∈D and Z₉⊂ ⊂T T′⊂D (see diagram 4 of the Figure 3);

e) Q₅=

{

T T T T, ′ ′′ ′, , ∪T′′

}

where T T T, ′ ′′∈, D , T⊂T′ , T⊂T′′ , T′\T′′ ≠ ∅ , T′′\T′ ≠ ∅ , (see diagram 5 of the Figure 3);

f) Q₆=

{

Z T T T₉, , ′, ∪T D′, 

}

, where T T, ′∈D, Z₉ ⊂T, Z₉⊂T′, T T\ ′ ≠ ∅, T′\T≠ ∅, T∪T′⊂D (see diagram 6 of the Figure 3);

g) Q₇ =

{

Z Z T T D₉, ₆, , ′, 

}

, where T T, ′∈D, Z₆ ⊂T, Z₆⊂T′, T T\ ′ ≠ ∅, T′\T≠ ∅, T∪T′=D (see diagram 7 of the Figure 3);

h) Q₈ =

{

Z T T T₉, , ′, ∪T Z D′, , 

}

, where Z₉⊂T, Z₉⊂T′⊂Z, T T\ ′ ≠ ∅, T′\T≠ ∅,

(

T∪T′

)

\Z ≠ ∅,

(

)

\

Z T∪T′ ≠ ∅, T∪ ∪ =T′ Z D (see diagram 8 of the Figure 3);

For each i=1, 2,, 8 we set Q_iϑ_XI =

{

D′⊂D D| ′is isomorphic to Q_i

}

. One can see that

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

{ }

{

}

{

}

{

} {

}

{

} {

}

{

}

{

}

{

}

{

} {

}

{

}

{

}

{

}

{

}

{

} {

}

1 9 8 7 6 5 4 3 2 1

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

9 3 9 2 9 1 8 3 8 7 3

7 6 3 6 2 6 1 6 5 1

5 4 1 4 3 2 1

8 9 8

, , , , , , , , , ;

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

, , , , , , , , , , , ;

, ,

XI

Q Z Z Z Z Z Z Z Z Z D

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

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

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

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

Q Z Z Z

ϑ

= =

=

 



 

    



{

} {

}

{

} {

}

{

} {

}

6 3 2 9 8 6 3 1 9 7 6 3 2

9 7 6 3 1 9 6 5 3 1 9 6 5 2 1

9 6 4 3 1 9 6 4 2 1

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

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

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

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

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

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

  

 

Assume that D′∈Q_iϑ_XI and denote by the symbol R D

( )

′ the set of all regular elements α of the semigroup

(4)

( )

X

B D , for which the semilattices D′ and Q_i are mutually α-isomorphic and V D

(

,α

)

=D′ and

( )

i XI i

D Q

R Q R D

ϑ

∗

′∈

′

=

_

(see [1], Definition 6.3.5).

The following results have the key role in this study.

Theorem 2. Let R_D be the set of all regular elements of the semigroup B_X

( )

D . Then the following state- ments are true:

a) R D

( )

′ ∩R D

( )

′′ = ∅ for any D D′ ′′∈ Σ, _XI

( )

D and D′≠D′′;

b) _{( )}

( )

XI

D D D

R =



_′∈Σ R D′ ;

c) if X is a finite set, then _{( )}

( )

XI

D D D

R =

∑

_′∈Σ R D′ (see [1], Theorem 6.3.6).

Lemma 3. Let

ϕ

be isomorphism between Q_i and D_i′ semilattices, T∈Q_i, T∈D_i′ and ϕ

( )

T =T . If X is a finite set and Q_iϑ_XI =m_i

(

i=1, 2,, 8

)

, then the following equalities are true:

a) R Q

( )

₁ =1;

b) R Q

( )

₂ =m₂⋅

(

2T T′\ − ⋅1 2

)

X T\ ′;

c) R Q

( )

₃ =m₃⋅

(

2T T′\ − ⋅1

) (

3T T′′ ′\ −2T T′′ ′\

)

⋅3X T\ ′′;

d)

( )

(

\ 9

) (

\ \

)

(

\ \

)

\

4 4 2 1 3 2 4 3 4 ;

D T D T X D

T Z T T T T

R Q =m ⋅ − ⋅ ′ − ′ ⋅ ′ − ′ ⋅

  

e) R Q

( )

₅ = ⋅2 m₅⋅

(

2T T′ ′′\ − ⋅1

) (

2T T′′ ′\ − ⋅1 4

)

X T\( ′∪T′′);

f)

( )

6 2 6

(

2 \ 1

) (

2 \ 1

)

(

5 \( ) 4 \( )

)

5 \ ;

D T T D T T X D

T T T T

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

  

g)

( )

(

6\9

)

( )\ 6

(

\ \

) (

\ \

)

\

7 2 7 2 1 2 3 2 3 2 5 ;

T T Z T T T T T T T T X D

Z Z

R Q = ⋅m ⋅ − ⋅ ∩′ ⋅ ′ − ′ ⋅ ′ − ′ ⋅



h) R Q

( )

₈ = ⋅2 m₈⋅

(

2T Z\ − ⋅1

) (

2T T′\ − ⋅1

)

(

3Z\(T∪T′) −2Z\(T∪T′)

)

⋅6X D\ .



Proof. The propositions a), b), c) and d) immediately follow from ([1], Theorem 6.3.5 and Theorem 13.1.2), while the equalities e), f), g) and h) follow from ([1], Theorem 6.3.5, Corolaries 13.3.4-5-6 and 13.7.3). □

3. Regular Elements of the Complete Semigroups of Binary Relations of the Class

(

X

)

1

,10

Σ

, When

∅ ∉

D

and

Z

₉

≠ ∅

Theorem 3. Let D=

{

Z Z Z Z Z Z Z Z Z D₉, ₈, ₇, ₆, ₅, ₄, ₃, ₂, ₁, 

}

∈ Σ₁

(

X,10

)

and Z₉≠ ∅. Then a binary relation α of the semigroup B_X

( )

D whose quasinormal representation has the form ₍ ₎

(

)

, T , T V D Y T

α α

α=

_

_∈ × will be a regular element of this semigroup iff there exist a complete α-isomorphism

ϕ

of the semilattice V D

(

,α

)

on some subsemilattice D′ of the semilattice D which satisfies at least one of the following conditions:

a)

α

= ×X T, for some T∈D;

b) α =

(

Y_Tα×T

) (

∪ Y_Tα′ ×T′

)

, for some T T, ′∈D, T⊂T′ and YT ,YT

{ }

α α

′ ∉ ∅ which satisfies the condi-

tions: Y_Tα ⊇

ϕ

( )

T , Y_Tα′ ∩

ϕ

( )

T′ ≠ ∅;

c) α =

(

Y_Tα×T

) (

∪ Y_Tα′ ×T′

) (

∪ Y_Tα′′×T′′

)

, for some T T T, ′ ′′∈, D , T ⊂T′⊂T′′, and YT ,YT ,YT

{ }

α α α ′ ′′∉ ∅

which satisfies the conditions: Y_Tα ⊇

ϕ

( )

T , Y_Tα ∪Y_Tα′ ⊇

ϕ

( )

T′ , YT

( )

T

α

_ϕ

′ ∩ ′ ≠ ∅, YT

( )

T

α

_ϕ

′′∩ ′′ ≠ ∅;

d) α=

(

Y_Tα×T

) (

∪ Y_Tα_′ ×T′

) (

∪ Y_Tα_′′×T′′

)

∪

(

Y₀α×D

)

, for some T T T, ′ ′′∈, D, T ⊂T′⊂T′′⊂D and

{ }

0

, , ,

T T T

Yα Yα′ Yα′′ Yα∉ ∅ which satisfies the conditions: YT

( )

T

α _⊇

_ϕ

, YT YT

( )

T

α α

_ϕ

′ ′

∪ ⊇ , YT YT YT

( )

T

α α α

_ϕ

′ ′′ ′′

∪ ∪ ⊇ ,

( )

T

Yα′ ∩

ϕ

T′ ≠ ∅, YT

( )

T

α

_ϕ

′′∩ ′′ ≠ ∅, Y0

( )

D

α _∩_ϕ  _{≠ ∅}_;

e) α=

(

Y_Tα×T

) (

∪ Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

)

∪

(

Y_Tα′∪_T′′×

(

T′∪T′′

)

, where T T T, ′ ′′∈, D, T⊂T′, T⊂T′′,

\

T′ T′′ ≠ ∅, T′′\T′ ≠ ∅, YT ,YT ,YT

{ }

α α α

′ ′′∉ ∅ and satisfies the conditions: YT YT

( )

T

α α

_ϕ

′ ′

∪ ⊇ ,

( )

T T

Yα∪Yα′′ ⊇

ϕ

T′′ , YT

( )

T

α

_ϕ

′ ∩ ′ ≠ ∅, YT

( )

T

α

_ϕ

(5)

f) α =

(

Y₉α×Z₉

) (

∪ Y₆α×Z₆

) (

∪ Y_Tα×T

) (

∪ Y_Tα′×T′

)

∪

(

Y₀α×D

)

, where, Y9 ,Y6 ,YT ,YT

{ }

α α α α

′ ∉ ∅ ,

{

3 2 1

}

, , ,

T T′∈ Z Z Z , T≠T′, T∪T′=D and satisfies the conditions: Y₉α ⊇

ϕ

( )

Z₉ , Y₉α∪Y₆α ⊇

ϕ

( )

Z₆ ,

( )

9 6 T

Yα∪Yα∪Yα ⊇

ϕ

T , Y₉α∪Y₆α∪Y_Tα′ ⊇

ϕ

( )

T′ , Y6

( )

Z6

α_∩

_ϕ

_{≠ ∅}_,

( )

T

Yα∩

ϕ

T ≠ ∅, Y_Tα′ ∩

ϕ

( )

T′ ≠ ∅;

g) α=

(

Y₉α×Z₉

) (

∪ Y_Tα×T

) (

∪ Y_Tα_′ ×T′

)

∪

(

Y_Tα_∪_T_′×

(

T∪T′

)

∪

(

Y₀α×D

)

, where YT ,YT ,Y₀

{ }

α α α

′ ∉ ∅ ,

{

8, 7, 6, 5

}

T∈ Z Z Z Z , T′∈

{

Z Z Z Z₇, ₆, ₅, ₄

}

, T ≠T′, and satisfies the conditions: Y₉α∪Y_Tα ⊇

ϕ

( )

T ,

( )

9 T

Yα∪Yα′ ⊇

ϕ

T′ , YT

( )

T

α_∩

_ϕ

_{≠ ∅}_,

( )

T

Yα′ ∩

ϕ

T′ ≠ ∅, Y0

( )

D

α_∩_ϕ  _{≠ ∅}_;

h)

(

) (

)

(

)

) (

)

(

)

6

9 9 T 6 6 T Z 6 T 0

Yα Z Yα T Yα Z Yα T Z Yα T Yα D

α= × ∪ × ∪ × ∪ ∪ × ∪ ∪ ′× ′ ∪ ×



, where

{ }

9 , T , 6 , T

Yα Yα Yα Yα′ ∉ ∅ , T Z\ 6≠ ∅, Z6\T≠ ∅, Z9⊂Z6⊂T′ and satisfies the conditions: Y9

( )

Z9

α _⊇

_ϕ

,

( )

9 6 6

Yα∪Yα ⊇

ϕ

Z , Y₉α ∪Y_Tα ⊇

ϕ

( )

T , Y₉α∪Y₆α∪Y_Tα′ ⊇

ϕ

( )

T′ , YT

( )

Z6

α_∩

_ϕ

_{≠ ∅}_,

( )

T

Yα′ ∩

ϕ

T ≠ ∅,

( )

T

Yα′ ∩

ϕ

T′ ≠ ∅.

Proof. In this case from Lemma 2 it follows that diagrams 1-8 given in Figure 2 exhaust all diagrams of

XI-subsemilattices of the semilattice D. A quasinormal representation of regular elements of the semigroup

( )

X

B D , which are defined by these XI-semilattices, may have one of the form listed above. Then the validity of theorem immediately follows from ([1], Theorem 13.1.1, Theorem 13.3.1 and Theorem 13.7.1). □

Lemma 4. Let D=

{

Z Z Z₉, ₈, ₇,Z Z Z₆, ₅, ₄,Z Z₃, ₂,Z D₁, 

}

∈ Σ₁

(

X,10

)

and Z₉ ≠ ∅. Let R Q∗

( )

₁ be set of all regular elements of B_X

( )

D such that each element satisfies the condition of a) of Theorem 3. Then

( )

1 10

R∗ Q = .

Proof. Let binary relation α of the semigroup B_X

( )

D satisfy the condition a) of Theorem 3. Then quasinormal representation of a binary relation α has a form

α

=X×T for some T∈D. It is easy to see that

α α α = for all T∈D, i.e. binary relation α is a regular element of the semigroup B_X

( )

D . Therefore

( )

1 10.

R∗ Q = □ Now let binary relation α of the semigroup B_X

( )

D satisfy the condition b) of Theorem 3 (see diagram 2 of the Figure 3). In this case we have Q₂=

{

T T, ′

}

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

{

} {

}

{

} {

}

{

}

{

}

{

}

{

}

{

}

{

} {

}

{

}

{

}

{

}

{

}

{

} {

} { }

}

2 9 8 9 7 9 6 9 5 9 4 9 3

9 2 9 1 9 8 3 8 7 3

7 6 3 6 2 6 1 6 5 1

5 4 1 4 3 2 1

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

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

XI

Q Z Z Z Z Z Z Z Z Z Z Z Z

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

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

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

ϑ =

 

    

It is easy to see that there is only one isomorphism from Q₂ to itself. That is Φ

(

Q Q₂, ₂

)

=1 and

( )

Q2 24

Ω = . If

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

1 9 2 5 3 2 4 9 8 5 9 7

6 9 6 7 9 5 8 9 4 9 9 3 10 9 2

11 9 1 12 8 3 13 8 14 7 3 15 7

16 6 3 17 6 2 18 6 1 19 6

, , , , , , , , , ,

, , , , , , , ,

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 Z Z D Z D D Z Z D Z D

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

′= ′= ′= ′= ′=

′= ′= ′= ′= ′ =

′ = ′ = ′ = ′ = ′ =

′ = ′ = ′ = ′ =

  

 



_{

_}

{

}

{

}

{

}

{ }

20 5 1

21 4 1 22 4 23 3 24 1

, ,

, , , , , , , .

D Z Z

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

′ =

′ = ′ =  ′ =  ′ = 

then

( )

24

( )

2 1

.

i i

R∗ Q R D

=

′

=



(2)

Lemma 5. Let X be a finite set,

{

9, 8, 7, 6, 5, 4, 3, 2, 1,

}

1

(

,10

)

D= Z Z Z Z Z Z Z Z Z D ∈ Σ X

(6)

condition b) of Theorem 3. Then

( )

(

\ 9

)

\

2 24 2 1 2 .

D Z X D

R∗ Q = ⋅ − ⋅

 

Proof. Let Z Z, ′∈D, Z ⊂Z′, D′=

{

Z Z, ′

}

and α∈R D

( )

′ . Then quasinormal representation of a binary relation α has a form α =

(

Y_Tα×T

) (

∪ Y_Tα′×T′

)

for some T T, ′∈D, T⊂T′, YT ,YT

{ }

α α

′ ∉ ∅ and by state-

ment b) of theorem 3 satisfies the conditions Y_Tα ⊇Z and Y_Tα′ ∩Z′≠ ∅. By definition of the semilattice D we

have Z⊇Z₉ and D⊇Z′, i.e., Y_Tα ⊇Z₉ and Y_Tα′ ∩D ≠ ∅. It follows that α∈R D

( )

1′ . Therefore we have

( )

2

( )

1 .

R Q∗ =R D′ (3)

From this equality and by statement b) of Lemma 3 it immediately follows that

( )

(

\ 9

)

\

2 24 2 1 2 .

D Z X D

R∗ Q = ⋅ − ⋅

 

□

Let binary relation α of the semigroup B_X

( )

D satisfy the condition c) of Theorem 3 (see diagram 3 of the

Figure 3). In this case we have Q₂=

{

T T T, ′ ′′,

}

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

{

} {

}

{

} {

}

{

} {

}

{

} {

}

{

} {

}

{

}

3 9 8 9 7 9 6 9 5 9 4

9 3 9 2 9 1 9 8 3 9 7 3

9 6 3 9 6 2 9 6 1 9 5 1 9 4 1

8 3 7 3 6 3 6 2 6 1

5 1 4 1

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

, , , ,

XI

Q 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 Z Z Z Z Z Z Z Z Z

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

Z Z D Z Z

ϑ

=     

  

    



_{

_}

,D .

It is easy to see Φ

(

Q Q₃, ₃

)

=1 and Ω

( )

Q₃ =22. If

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

1 9 8 2 9 7 3 9 6 4 9 5

5 9 4 6 9 3 7 9 2 8 9 1 9 9 8 3 10 9 7 3 11 9 6 3 12 9 6 2 13 9 6 1 14 9 5 1 15 9 4 1

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

, , , , , , , ,

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 Z Z Z

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

′= ′= ′= ′=

′= ′ = ′ = ′ =

′ = ′ = ′ =

   

{

}

{

}

{

}

{

}

{

}

{

}

{

}

16 8 3

17 7 3 18 6 3 19 6 2 20 6 1

21 5 1 22 4 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

′ =

′ = ′ = ′ = ′ =

′ = ′ =



   

 

then

( )

22

( )

3 1

.

i i

R∗ Q R D

=

′

=



(4)

{

9, 8, 7, 6, 5, 4, 3, 2, 1,

}

1

(

,10

)

D= Z Z Z Z Z Z Z Z Z D ∈ Σ X

and Z₉ ≠ ∅. Let R Q

( )

3

∗

be the set of all regular elements of B_X

( )

D such that each element satisfies the condition c) of Theorem 3. Then

( )

( ) 5( )3

( )

8

3

1 ,

i j k

i j k M Q

R Q∗ R D R D R D

= ∈

′ ′ ′

=

∑

−

∑

∩

where

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

{

}

5 3 2, 6 , 3, 6 , 3, 7 , 3, 8 , 4, 8 , 5, 8 .

M Q =

(7)

quasinormal representation of a binary relation α has a form α =

(

Y_Tα×T

) (

∪ Y_Tα′×T′

) (

∪ Y_Tα′′×T′′

)

for some

, , ,

T T T′ ′′∈D T ⊂T′⊂T′′, Y_Tα,Y_Tα′,Y_Tα′′∉ ∅

{ }

and by statement c) of Theorem 3 satisfies the conditions

T i

Yα ⊇Y, Y_Tα∪Y_Tα_′ ⊇Y_i′, Y_Tα_′ ∩ ≠ ∅Y_i′ and Y_Tα_′′ ∩ ≠ ∅Y_i′′ . By definition of the semilattice D we have Y_i ⊇Z₉,

i

D⊇Y′′. From this and by the condition Y_Tα ⊇Y_i, Y_Tα∪Y_Tα_′ ⊇Y_i′, Y_Tα_′ ∩ ≠ ∅Y_i′ , Y_Tα_′′ ∩ ≠ ∅Y_i′′ we have

9, , , ,

T T T i T i T

Yα ⊇Z Yα∪Yα′ ⊇Y Y′ α′ ∩ ≠ ∅Y′ Yα′′∩ ≠ ∅D i.e. α∈R D

( )

′_j , where D′j =

{

Z Y D₉, ,i′

}



. It follows that R D

( )

_i′ ⊆R D

( )

′_j , From the last inclusion and by definition of the semilattice D we have R D

( )

_i′ ⊆R D

( )

′_j for all

( )

i j, ∈M₁

( )

Q₃ , where

( ) ( ) (

{

) (

)

(

) (

)

}

1 3 9,1 , 10, 2 , 11, 3 , 12, 3 , 13, 3 , 14, 4 , 15, 5 ,

16, 6 , 17, 6 , 18, 6 , 19, 7 , 20, 8 , 21, 8 , 22, 8 .

M Q =

Therefore the following equality

( )

8

( )

3 1

i i

R∗ Q R D

=

′

=

_

(5)

holds. Now, let D_i′=

{

Z Y D₉, ,_i 

}

,D′_j=

{

Z Y D₉, _j, 

}

∈

{

D D₁′ ′, ₂,,D₈′

}

, D_i′≠D′_j and α∈R D

( )

_i′ ∩R D

( )

′_j . Then for the binary relation α we have

9 9

, , , ,

, , , .

T T T i T i T

T T T j T j T

Y Z Y Y Y Y Y Y D

α α α α α

α α α′ α′ ′′α

′ ′ ′′

⊇ ∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

 

From the last condition it follows that YT Z Y9, T YT Yi Yj

α α α

′

⊇ ∪ ⊇ ∪ .

1) Yi∪Yj =D



. Then we have that

(

Y_Tα∪Y_Tα′

)

∩Y_Tα′′ ⊇

(

Y_i∪Y_j

)

∩Y_Tα′′ =D∩Y_Tα′′ ≠ ∅. But the inequality

(

YT YT

)

YT

α α α

′ ′′

∪ ∩ ≠ ∅ contradicts the condition that representation of binary relation α is quasinormal. So, the equality R D

( )

_i′ ∩R D

( )

′_j = ∅ is true. From the last equality and by definition of the semilattice D we have

( )

i

( )

j

R D′ ∩R D′ = ∅ for all

( )

i j, ∈M₂

( )

Q₃ , where

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

{

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

}

2 3 1, 4 , 1, 5 , 1, 7 , 1, 8 , 2, 4 , 2, 5 , 2, 7 , 2, 8 ,

4, 6 , 4, 7 , 5, 6 , 5, 7 , 6, 7 , 6, 8 , 7, 8 .

M Q =

2) Di′= , ,

{

Z Y D₉ i

}

, ,Dj′=

{

Z Y D₉ j,

}

, ,Dk′=

{

Z Y₉ i∪Y Dj,

}

,∈

{

D D₁′ ′₂, ,D₈′

}

  

 , D_i′≠D′_j, D_i′≠D′_k, D′_j≠D_k′,

( )

i

( )

j R D R D

α∈ ′ ∩ ′ and α∈R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ are true. Then we have 9

9

, , , ,

T T T i T i T

T T T j T j T

α α α α α

′ ′ ′′

⊇ ∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

 

and

(

)

9

, , , ,

, , ,

T T T i T i T

T T T j T j T

T T T i j T i j T

Y Z Y Y Y Y Y Y Y Y D

α α α α α

′ ′ ′′

⊇ ∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

⊇ ∪ ⊇ ∪ ∩ ∪ ≠ ∅ ∩ ≠ ∅

 



respectively, i.e., α∈R D

( )

_i′ ∩R D

( )

′_j or α∈R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ if and only if

9, , , , .

T T T i j T i T j T

Yα ⊇Z Yα∪Yα′ ⊇ ∪Y Y Yα′ ∩ ≠ ∅Y Yα′ ∩Y ≠ ∅ Yα′′∩ ≠ ∅D



Therefore the equality R D

( )

_i′ ∩R D

( )

′_j =R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ is true. From the last equality and by definition of the semilattice D we have R D

( )

_i′ ∩R D

( )

′_j =R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ for all

(

i j k, ,

)

∈M₃

( )

Q₃ , where

( ) (

{

) (

)

}

3 3 1, 2, 6 , 1, 3, 6 , 2, 3, 6 , 3, 4, 8 , 3, 5, 8 , 4, 5, 8 .

M Q =

(8)

p q

D′ ≠D′, p q, ∈

{

i j k t, , ,

}

, p≠q, α∈R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ and

( )

i

( )

j

( )

k

( )

t

R D R D R D R D

α∈ ′ ∩ ′ ∩ ′ ∩ ′ are true. Then we have

9 9 9

, , , ,

, , ,

T T T i T i T

T T T j T j T

T T T k T k T

α α α α α

′ ′ ′′

⊇ ∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

  

and

(

)

9 9

9 '

9

, , , ,

, , ,

T T T i T i T

T T T j T j T

T T T k _T k T

T T T i j k T i j k T

Y Z Y Y Y Y Y Y Y Y Y Y D

α α α α α

′ ′ ′′

′ ′′

′ ′ ′′

⊇ ∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

⊇ ∪ ⊇ ∪ ∪ ∩ ∪ ∪ ≠ ∅ ∩ ≠ ∅

  



respectively, i.e., α∈R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ and α∈R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ ∩R D

( )

_t′ if and only if

9, , , , , .

T T T i j k T i T j T k T

Yα ⊇Z Yα∪Yα′ ⊇ ∪ ∪Y Y Y Yα′ ∩ ≠ ∅Y Yα′ ∩Y ≠ ∅ Yα′ ∩Y ≠ ∅ Yα′′∩ ≠ ∅D



Therefore the equality R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ =R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ ∩R D

( )

_t′ is true. From the last equality and by definition of the semilattice D we have

( )

i

( )

j

( )

k

( )

i

( )

j

( )

k

( )

t

R D′ ∩R D′ ∩R D′ =R D′ ∩R D′ ∩R D′ ∩R D′

for all

(

i j k t, , ,

)

∈M₄

( )

Q₃ , where

( ) (

{

) (

)

}

4 3 1, 2, 3, 6 , 3, 4, 5, 8 .

M Q =

Now, by equality (4) and conditions 1), 2) and 3) it follows that the following equality is true

( )

( ) 5( )3

( )

8

3

1 ,

i j k

i j k M Q

R Q∗ R D R D R D

= ∈

′ ′ ′

=

∑

−

∑

∩

where

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

{

}

5 3 2, 6 , 3, 6 , 3, 7 , 3, 8 , 4, 8 , 5, 8 .

M Q = □

Lemma 7. Let D′ =

{

Z Y D₉, , 

}

, D′′=

{

Z Y D₉, ′, 

}

where Y Y, ′∈D and Y′ ⊇Y . If quasinormal repre- sentation of binary relation α of the semigroup B_X

( )

D has a form α =

(

Y_Tα×T

) (

∪ Y_Tα_′×T′

)

∪

(

Y₀α×D

)

for some T T, ′∈D, T⊂T′⊂D and YT ,YT ,Y₀

{ }

α α α

′ ∉ ∅ , then α∈R D

( )

′ ∩R D

( )

′′ iff

9, , , 0 .

T T T T

Yα ⊇Z Yα∪Yα′ ⊇Y Y′ α′ ∩ ≠ ∅Y Yα∩ ≠ ∅D Proof. If α∈R D

( )

′ ∩R D

( )

′′ , then by statement c) of Theorem 3 we have

9 0

, , , ;

, , , .

T T T T

α α α α α

′ ′

⊇ ∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

′ ′

⊇ ∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅



 (6)

From the last condition we have

9, , , 0 ,

T T T T

Yα ⊇Z Yα∪Yα_′ ⊇Y Y′ α_′ ∩ ≠ ∅Y Yα∩ ≠ ∅D (7) since Y′ ⊇Y by assumption.

On the other hand, if the conditions of (7) holds, then (6) immediately follows, i.e.

α

∈R D

( )

′ ∩R D

( )

′′ . Lemma is proved. □

Lemma 8. Let D′=

{

Z Y D₉, , 

}

,D′′=

{

Z Y D₉, ′, 

}

∈

{

D D₁′ ′, ₂,,D₈′

}

, Y′ ⊇Y and X be a finite set. Then the following equality holds

( )

\

(

\9

)

(

\ \

)

\

22 2Y Y 2Y Z 1 3D Y 2D Y 3X D.

R D′ ∩R D′′ = ⋅ ′ ⋅ − ⋅ ′ − ′ ⋅

  

(9)

( )

R D R D

α

∈ ′ ∩ ′′ and a quasinormal representation of a regular binary relation α has a form

(

YT T

) (

YT T

)

(

Y0 D

)

α α α

α= × ∪ ′ × ′ ∪ ×  for some T T, ′∈D , T⊂T′⊂D



and Y_Tα,Y_Tα′,Y₀α∉ ∅

{ }

. Then ac-

cording to Lemma 7, we have

9, , , 0

T T T T

Yα ⊇Z Yα∪Yα′ ⊇Y Y′ α′ ∩ ≠ ∅Y Yα∩ ≠ ∅D



(8)

Further, let f_α be a mapping of the set X in the semilattice D satisfying the conditions f_α

( )

t =t

α

for all

t∈X. f₀α, f1α, f2α and f3α are the restrictions of the mapping fα on the sets Z9, Y′\Z9, D Y\ ′

 ,

\

X D respectively. It is clear that the intersection of elements of the set

{

Z₉,Y′\Z₉,D Y\ ′,X \D

}

is an empty set, and Z₉∪

(

Y′\Z₉

)

∪

(

D Y\ ′

) (

∪ X \D

)

=X . We are going to find properties of the maps f₀_α, f₁_α,

2

f_α, f₃_α.

1) t∈Z₉. Then by the properties (1) we have Z9 YT

α

⊆ , i.e., t YT

α

∈ and tα=T by definition of the set

T

Yα. Therefore f0α

( )

t =T for all t∈Z9.

2) t∈Y′\Z₉. Then by the properties (1) we have Y \Z₉ Y YT YT

α α

′

′ ⊆ ′⊆ ∪ , i.e., t∈Y_Tα∪Y_Tα′ and

{

,

}

t

α

∈ T T′ by definition of the sets YT

α

and YT

α

′ . Therefore f1α

( ) {

t ∈ T T, ′

}

for all t∈Y′\Z9.

By suppose we have that YT Y

α

′ ∩ ≠ ∅, i.e. t′α=T′ for some t′∈Y. If t′∈Z9, then t Z9 YT

α

′∈ ⊆ .

Therefore t′ =α T. That is contradiction to the equality t′α=T′, while T ≠T′ by definition of the semilattice D.

Therefore f1α

( )

t′ =T′ for some t′∈Y Z\ 9.

3) t∈D Y\ ′. Then by properties (1) we have D Y\ ′ ⊆ ⊆D X =Y_Tα∪Y_Tα′ ∪Y₀α, i.e., t YT YT Y0

α α α

′

∈ ∪ ∪ and

{

, ,

}

tα∈ T T D′  by definition of the sets Y_Tα, Y_Tα′ and Y0

α_{. Therefore}

_{( )}

{

}

3 , ,

fα t ∈ T T D′



for all t∈D Y\ ′. By suppose we have that Y0 D

α_{∩ ≠ ∅}

, i.e. t′′ =α D for some t′′∈D. If t′′∈Y′, then t Y YT YT

α α

′

′′∈ ⊆′ ∪ .

Therefore t′′

α

∈

{

T T, ′

}

by definition of the set Y_Tα and Y_Tα′. We have contradiction to the equality t′′ =α D.

Therefore f₃_α

( )

t′′ =D for some t′′∈D Y\ ′.

4) t∈X\D. Then by definition of a quasinormal representation of a binary relation α and by property (1) we have t∈X D\ ⊆X =Y_Tα∪Y_Tα′ ∪Y₀α, i.e. tα∈

{

T T D, ′, 

}

by definition of the sets YT ,YT

α α

′ and Y0

α

. There- fore f₄_α

( )

t ∈

{

T T D, ′, 

}

for all t∈X \D.

We have seen that for every binary relation

α

∈R D

( )

′ ∩R D

( )

′′ there exists ordered system

(

f0α,f1α,f2α,f3α

)

. It is obvious that for disjoint binary relations there exist disjoint ordered systems.

Further, let

{ }

{

}

0: 9 , :1 \ 9 , ,

f Z → T f Y′ Z → T T′

{

}

{

}

2: \ , , , 3: \ , ,

f D Y ′→ T T D′  f X D→ T T D′ 

be such mappings that satisfy the conditions:

( )

0

f t =T for all t∈Z₉;

( ) {

}

1 ,

f t ∈ T T′ for all t∈Y′\Z₉ and f t1

( )

′ =T′ for some t′∈Y Z\ 9;

( )

{

}

2 , ,

f t ∈ T T D′  for all t∈D Y\ ′ and f₂

( )

t′′ =D for some t′′∈D Y\ ′;

( )

{

}

3 , ,

f t ∈ T T D′  for all t∈X \D.

Now we define a map f from X to the semilattice D, which satisfies the condition:

( )

0 9

1 9

2 3

, if , , if \ ,

, if \ , , if \ .

f t t Z

f t t Y Z

f t

f t t D Y

f t t X D

 ∈

 _∈ _′



=  _∈ _′



 ∈



 

Further, let

(

{ }

( )

)

x X x f x

β =

_

_∈ × , YT

{

t t| T

}

β ₌

_β

₌

, YT

{

t t| T

}

β

_β

′ = = ′ and Y0

{

t t| D

}

β ₌ _β ₌  _{. Then}

binary relation

β

may be represented by

(

YT T

) (

YT T

)

(

Y0 D

)

β β β

β = × ∪ ′ × ′ ∪ ×



(10)

9, , , 0

T T T T

Yβ ⊇Z Yβ ∪Yβ′ ⊇Y Y′ β′ ∩ ≠ ∅Y Yβ∩ ≠ ∅D .

(By suppose f t₁

( )

′ =T′ for some t′∈Y Z\ ₉ and f₂

( )

t′′ =D for some t′′∈D Y\ ′), i.e., by lemma 7 we have that

β

∈R D

( )

′ ∩R D

( )

′′ .

Therefore for every binary relation

α

∈R D

( )

′ ∩R D

( )

′′ and ordered system

(

f₀_α,f₁_α,f₂_α,f₃_α

)

there exists one to one mapping.

By Lemma 1 and by Theorem 1 the number of the mappings f₀α,f₁α,f₂α,f₃α,f₄α are respectively

( \ 9) (\ \ 9)

(

\9

)

\ \ \

1, 2Y Z′ Y Z ⋅ 2Y Z −1 , 3D Y′ −2D Y′, 3X D.

  

Note that the number ₂Y′\(Y∪Z9) ⋅

(

₂Y Z\ 9 − ⋅₁

)

(

₃D Y\ ′ −₂D Y\ ′

)

⋅₃X D\

  

does not depend on choice of chains

T⊂T′⊂T′′

(

T T T, ′ ′′∈, D

)

of the semilattice D. Since the number of such different chains of the semilattice

D is equal to 22, for arbitrary T T T, ′ ′′∈, D where T ⊂T′⊂T′′, the number of regular elements of the set

( )

R D′ ∩R D′′ is equal to

( )

_{22 2}Y Y\

(

₂Y Z\ 9 ₁

)

(

₃D Y\ ₂D Y\

)

₃X D\ _.

R D′ ∩R D′′ = ⋅ ′ ⋅ − ⋅ ′ − ′ ⋅

  

□

Therefore we obtain

( )

(

)

(

)

( )

(

)

(

)

( )

(

)

(

)

( )

(

)

(

)

( )

3 3

3 8 8 9

3 3

3 7 7 9

3 3

3 6 6 9

2 2

2 6 6 9

\ \ \ \ \ 1 6 \ \ \ \ \ 2 6 \ \ \ \ \ 3 6 \ \ \ \ \ 3 7 3

22 2 2 1 3 2 3 ,

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

D Z D Z X D

Z Z Z Z

R D R D

R D ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′            

( )

(

)

(

)

( )

(

)

(

)

( )

(

)

(

)

1 1

1 6 6 9

1 1

1 5 5 9

1 1 4 9 1 4 \ \ \ \ \ 8 \ \ \ \ \ 4 8 \ \ \ \ \ 5 8

22 2 2 1 3 2 3 ,

22 2 2 1 3 2 3 .

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

R D

R D R D

′ ∩ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅          (9)

Lemma 9. Let X be a finite set, D=

{

Z Z Z Z Z Z Z Z Z D₉, ₈, ₇, ₆, ₅, ₄, ₃, ₂, ₁,

}

∈ Σ₁

(

X,10

)

and Z₉ ≠ ∅. Let

( )

3

R Q∗ be set of all regular elements of BX

( )

D such that each element satisfies the condition c) of Theorem

3. Then

( )

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

8 8 7 7

8 9 7 9

6 6 5 5

6 9 5 9

4 4 3 3

4 9 3 9

\ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

22 2 1 3 2 3 22 2 1 3 2 3

22 2 1 3 2 3 22 2 1 3 2 .3

22

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

Z Z Z Z

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

Z Z Z Z

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

Z Z Z Z

R∗ Q = ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅

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

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

2 2 1 1

2 9 1 9

3 3 3 3

3 8 8 9 3 7 7 9

3 3

3 6 6 9 2 6 6 9

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

2 1 3 2 3 22 2 1 3 2 3

22 2 2 1 3 2 3 22 2 2 1 3 2 3

22 2 2 1 3 2 3 22 2 2 1 3

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

Z Z Z Z

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

Z Z Z Z Z Z Z Z

D Z D Z X D D

Z Z Z Z Z Z Z Z

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

(

)

(

)

(

)

(

)

(

)

(

)

(

)

2 2

1 1 1 1

1 6 6 9 1 5 5 9

1 1 4 9 1 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 3

22 2 2 1 3 2 3 22 2 2 1 3 2 3

22 2 2 1 3 2 3 .

Z D Z X D

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

Z Z Z Z Z Z Z Z

D Z D Z X D

(11)

Proof. The given Lemma immediately follows from Lemma 6 and from the Equalities (5).

Now let a binary relation α of the semigroup B_X

( )

D satisfy the condition (d) of Theorem 3 (see diagram 4 of the Figure 3). In this case we have Q₄=

{

Z T T D₉, , ′, 

}

, where T T, ′∈D and Z₉⊂ ⊂T T′⊂D. By definition of the semilattice D it follows that

{

} {

}

{

} {

}

4 9 8 3 9 7 3 9 6 3 9 6 2

9 6 1 9 5 1 9 4 1

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

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

XI

Q 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 Z D

ϑ =    

  

It is easy to see Φ

(

Q Q₄, ₄

)

=1 and Ω

( )

Q₄ =7. If

{

}

{

}

{

}

{

}

{

}

{

}

{

}

1 9 8 3 2 9 7 3 3 9 6 3

4 9 6 2 5 9 6 1 6 9 5 1

7 9 4 1

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

, , , ;

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

D Z Z Z D

′= ′= ′= ′= ′= ′= ′ =        then

( )

4 7

( )

1

.

i i

R∗ Q R D

=

′

=



(10)

{

9, 8, 7, 6, 5, 4, 3, 2, 1,

}

1

(

,10

)

D= Z Z Z Z Z Z Z Z Z D ∈ Σ X

and Z₉ ≠ ∅. Let R Q∗

( )

₄ be set of all regular elements of B_X

( )

D such that each element satisfies the condition d) of Theorem 3. Then

( )

(

) (

)

(

)

(

) (

)

(

)

(

) (

)

(

)

(

) (

)

(

)

(

)

3 3

8 9 3 8 3 8

3 3

7 9 3 7 3 7

3 3

6 9 3 6 3 6

2 2

6 9 2 6 2 6

6 9 1

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

7 2 1 3 2 4 3 4

7 2 1 3

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

D Z D Z X D

Z Z Z Z Z Z

Z Z Z

R∗ Q = ⋅ − ⋅ − ⋅ − ⋅

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

(

)

(

)

(

) (

)

(

)

(

) (

)

(

)

1 1

6 1 6

1 1

5 9 1 5 1 5

1 1

4 9 1 4 1 4

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

2 4 3 4

7 2 1 3 2 4 3 4

7 2 1 3 2 4 3 4 .

D Z D Z X D

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

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

Proof. Let D_i′=

{

Z Y Y D₉, , ,_i _i′ 

}

,D′_j =

{

Z Y Y D₉, _j, _j′, 

}

∈

{

D D₁′ ′, ₂,,D₇′

}

, D_i′≠D′_j and α∈D_i′∩D′_j. Then

(

9 9

) (

)

(

0

)

= Y Z YT T YT T Y D

α α α α

α × ∪ × ∪ ′× ′ ∪ ×



, where Z9⊂ ⊂T T′⊂D



, Y9 ,YT ,YT ,Y0

{ }

α α α α

′ ∉ ∅ and the

following inclusions and inequalities are true

9 9 9 9

0

9 9 9 9

0

, , ,

, , ;

, , ,

, , .

T i T T i

T i T i

T j T T j

T j T j

Y Z Y Y Y Y Y Y Y

Y Y Y Y Y D

Y Z Y Y Y Y Y Y Y

Y Y Y Y Y D

α α α α α α

α α α

α α α α α α

α α α

′ ′ ′ ′ ′ ⊇ ∪ ⊇ ∪ ∪ ⊇ ′ ∩ ≠ ∅ ∩ ≠ ∅ ∩ ≠ ∅ ′ ⊇ ∪ ⊇ ∪ ∪ ⊇ ′ ∩ ≠ ∅ ∩ ≠ ∅ ∩ ≠ ∅  

From this it follows that

9 9, 9 T i j, 9 T T i j.

Yα ⊇Z Yα∪Yα ⊇ ∪Y Y Yα∪Yα∪Yα′ ⊇ ∪Y′ Y′

We consider the following cases.

1) Y_i∪Y_j =D or Y_i′∪Y_j′=D. Then we have

(

Y₉α∪Y_Tα∪Y_Tα′

)

∩Y₀α ⊇D∩Y₀α ≠ ∅ 