Regular Elements of BX (D) Defined by the Class ∑1(X,10) Ⅱ

(1)

http://dx.doi.org/10.4236/am.2016.79079

Regular Elements of

B

_X

( )

D

Defined by the

Class

Σ

₁

(

X

,10

)

−

II

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

This part of the paper is the continuotion of paper “Regular Elements of BX

( )

D Defined by the

Class Σ1

(

X,10

)

−I”.

Keywords

Semilattice, Semigroup, Binary Relation

1. Introduction

This work is continuation of the paper “Regular Elements of B_X

( )

D Defined by the Class Σ₁

(

X,10

)

−I” whose sections are labelled 1-2-3. In the introduction and the second section, some definitions and well known results are stated with references. In Section 3 we give a full description of regular elements of the semigroup

( )

X

B D when an empty set is not included in D and Z₉ ≠ ∅.

In the present work our aim is to identify regular elements of thesemigroup B_X

( )

D when ∅ ∈D and 9 .

Z = ∅

The method used in this part does not differ from the method given in [1].

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

(

X

)

1

,10

Σ

, When

∅ ∈

D

and

Z

₉

= ∅

We denoted the following semilattices by symbols:

(2)

2) Q2 = ∅

{

,T

}

, where T∈D (see diagram 2 of the Figure 1);

3) Q3= ∅

{

, ,T T′

}

, where T T, ′∈D and ∅ ⊂ ⊂T T′ (see diagram 3 of the Figure 1); 4) Q₄ = ∅

{

, ,T T D′, 

}

, where T T, ′∈D and ∅ ⊂ ⊂T T′⊂D (see diagram 4 of the Figure 1); 5) Q₅ = ∅

{

, ,T T T′, ∪T′

}

where T T, ′∈D, T T\ ′ ≠ ∅, T′\T ≠ ∅, (see diagram 5 of the Figure 1); 6) Q₆= ∅

{

, ,T T T′, ∪T D′, 

}

, where T T, ′∈D, T T\ ′ ≠ ∅, T′\T ≠ ∅, T∪T′⊂D (see diagram 6 of the Figure 1);

7) Q₇ = ∅

{

,Z T T D₆, , ′, 

}

, where T T, ′∈D, Z₆⊂T, Z₆⊂T′, T T\ ′ ≠ ∅, T′\T ≠ ∅, T∪T′=D (see diagram 7 of the Figure 1);

8) Q₈= ∅

{

, ,T T T′, ∪T Z D′, ,

}

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

(

T∪T′

)

\Z≠ ∅,

(

)

\

[image:2.595.130.498.254.375.2]

Z T∪T′ ≠ ∅, T∪ ∪ =T′ Z D (see diagram 8 of the Figure 1);

Figure 1. Diagram of all XI-subsemilattices of semi lattices of unions D.

Note that the semilattices 1)-8), which are given by diagram 1-8 of the Figure 1 always are XI-semilattices (see [2], Lemma 1.2.3).

Remark that

{ }

{

} {

}

{

} {

}

{

} {

}

{

}

{

} {

}

1

2 8 7 6 5 4 3 2 1

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

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

;

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

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

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

XI

Q

Q D Z Z Z Z Z Z Z Z

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

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

ϑ ϑ

ϑ = ∅

= ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅

…

= ∅ ∅ ∅ ∅

∅ ∅ ∅ ∅



   

  

_{



_}

. D

Lemma 1. Let ϕ be an 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_i

ϑ

_XI =m_i

(

i=1, 2,,8

)

and Z₉ = ∅, then the following equalities are true:

1) R Q

( )

₁ =1;

2) R Q

( )

₂ =m₂⋅

(

2T − ⋅1 2

)

X T\ ;

3) R Q

( )

₃ =m₃⋅

(

2T − ⋅1

) (

3T T′\ −2T T′\

)

⋅3X T\ ′;

4) R Q

( )

₄ =m₄⋅

(

2T − ⋅1

) (

3T T′\ −2T T′\

)

⋅

(

4D T\′ −3D T\′

)

⋅4X D\ ;

  

5) R Q

( )

₅ = ⋅2 m₅⋅

(

2T T\ ′ − ⋅1

) (

2T T′\ − ⋅1 4

)

X T\( ∪T′);

6) R Q

( )

₆ = ⋅2 m₆⋅

(

2T T\ ′ − ⋅1

) (

2T T′\ − ⋅1

)

(

5D T\( ∪T′) −4D T\( ∪T′)

)

⋅5X D\ ;

  

7)

( )

(

6

)

( )\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

R Q = ⋅m − ∩ ′ ′ − ′ ′ − ′

(3)

8) R Q

( )

₈ = ⋅2 m₈⋅

(

2T Z\ − ⋅1

) (

2T T′\ − ⋅1

)

(

3Z T\( ∪T′) −2Z T\( ∪T′)

)

⋅6X D\ ; 

Proof. Let Z₉ = ∅. Then given Lemma immediately follows from ([1], Lemma 3). □

Theorem 1. 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 a 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:

•

α

= ∅;

• α =

(

Y_∅α× ∅ ∪

) (

Y_Tα×T

)

, for some T∈D and Y_Tα ≠ ∅ which satisfies the condition Y_Tα∩

ϕ

( )

T ≠ ∅; • α =

(

Y∅α× ∅ ∪

) (

Y_Tα×T

) (

∪ Y_Tα′×T′

)

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

{ }

α α

′ ∉ ∅ which satisfies the conditions: Y∅α∪Y_Tα ⊇

ϕ

( )

T , YT

( )

T

α_∩

_ϕ

_{≠ ∅}_,

( )

T

Yα′ ∩

ϕ

T′ ≠ ∅; • α=

(

Y∅α× ∅ ∪

) (

Y_Tα×T

) (

∪ Y_Tα′×T′

)

∪

(

Y₀α×D

)



, for some T T, ′∈D, ∅ ⊂ ⊂T T′⊂D and

{ }

0 , ,

T T

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

( )

T

α α

_ϕ

∅ ∪ ⊇ , Y YT YT

( )

T

α α α

_ϕ

′

∅ ∪ ∪ ⊇ ′ ,

( )

T

Yα∩

ϕ

T ≠ ∅, Y_Tα′ ∩

ϕ

( )

T′ ≠ ∅, Y0

( )

D α_∩_ϕ  _{≠ ∅}_;

• α =

(

Y∅α× ∅ ∪

) (

Y_Tα×T

) (

∪ Y_Tα′×T′

)

∪

(

Y_{T T}α∪′×

₍

T∪T′

₎

)

, where T T, ′∈D, T T\ ′ ≠ ∅, T′\T ≠ ∅,

{ }

,

T T

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

( )

T

α α

_ϕ

∅ ∪ ⊇ , Y YT

( )

T

α α

_ϕ

′

∅ ∪ ⊇ ′ , YT

( )

T

α _∩

_ϕ

_{≠ ∅} ,

( )

T

Yα′ ∩

ϕ

T′ ≠ ∅;

• α =

(

Y∅α× ∅ ∪

) (

Y₆α×Z₆

) (

∪ Y_Tα×T

) (

∪ Y_Tα′ ×T′

)

∪

(

Y₀α×D

)



, where, Y₆ ,YT ,YT

{ }

α α α

′∉ ∅ ,

{

3 2 1

}

, , ,

T T′∈ Z Z Z , T ≠T′, T∪T′=D and satisfies the conditions: Y_∅α∪Y₆α ⊇

ϕ

( )

Z₆ ,

( )

6 T

Y∅α∪Yα∪Yα ⊇

ϕ

T , Y Y6 YT

( )

T

α α α

_ϕ

′

∅ ∪ ∪ ⊇ ′ , Y6

( )

Z6

α_∩

_ϕ

_{≠ ∅}_,

( )

T

Yα∩

ϕ

T ≠ ∅, Y_Tα′ ∩

ϕ

( )

T′ ≠ ∅. • α=

(

Y∅α× ∅ ∪

) (

Y_Tα×T

) (

∪ Y_Tα′×T′

)

∪

(

Y_{T T}α∪ ′×

(

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: Y∅α ∪Y_Tα ⊇

ϕ

( )

T ,

( )

T

Y∅α∪Yα′ ⊇

ϕ

T′ , YT

( )

T

α_∩

_ϕ

_{≠ ∅}_,

( )

T

Yα′ ∩

ϕ

T′ ≠ ∅, Y0

( )

D α_∩_ϕ  _{≠ ∅}_;

•

(

) (

)

(

)

) (

)

(

)

6

6 6 6 0

T T Z T

Yα Yα T Yα Z Yα T Z Yα T Yα D

α = ∅ × ∅ ∪ × ∪ × ∪ ∪ × ∪ ∪ ′ × ′ ∪ ×



, where YT ,Y₆ ,YT

{ }

α α α

′∉ ∅ , 6

\

T Z ≠ ∅, Z₆\T≠ ∅, ∅ ⊂Z₆⊂T′ and satisfies the conditions: Y∅α∪Y₆α ⊇

ϕ

( )

Z₆ ,

( )

T

Y_∅α∪Yα ⊇

ϕ

T , Y∅α∪Y₆α∪Y_Tα′ ⊇

ϕ

( )

T′ , YT

( )

Z6 α_∩

_ϕ

_{≠ ∅}

, Y_Tα′ ∩

ϕ

( )

T ≠ ∅, YT

( )

T

α

_ϕ

′ ∩ ′ ≠ ∅.

Proof. Let Z₉ = ∅. Then given Theorem immediately follows from ([1], Theorem 2). □

Lemma 2. 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 = ∅. Let R Q∗

( )

₁ be set of all regular elements of the semigroup BX

( )

D such that each element satisfies the condition a) of Theorem 1.

Then R∗

( )

Q₁ =1.

Now let a binary relation α of the semigroup B_X

( )

D satisfy the condition b) of Theorem 1 (see diagram 2 of the Figure 1). In this case we have Q₂= ∅

{

,T

}

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

{

} {

}

{

} {

}

{ }

}

2 XI , 8 , , 7 , , 6 , , 5 , , 4 , , 3 , , 2 , , 1 , , .

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

It is easy to see Φ

(

Q Q₂, ₂

)

=1 and Ω

( )

Q₂ =9. If

{ }

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

1 2 8 3 7 4 6

5 5 6 4 7 3 8 2 9 1

, , , , , , , ,

, , , , , , , , , ,

D D D Z D Z D Z

D Z D Z D Z D Z D Z

′= ∅ ′= ∅ ′= ∅ ′= ∅

′= ∅ ′= ∅ ′= ∅ ′= ∅ ′= ∅



(4)

( )

9

( )

2 1

i i

R∗ Q R D =

′

=

_

(1) (see remark page 5 in [1]).

Lemma 3. 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

( )

2

R Q∗ be set of all regular elements of the semigroup BX

( )

D such that each element satisfies the condition

b) of Theorem 1. Then

( )

\

2 9 2 1 2 .

D X D

R∗ Q = ⋅ − ⋅

 

Proof. Let Z∈D, D′ = ∅

{

,Z

}

and

α

∈R D

( )

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

(

Y_∅α× ∅ ∪

) (

Y_Tα×T

)

for some T∈D, YT

α _{≠ ∅}

and by statement b) of Theorem 1 satisfies the conditions Y∅α ⊇ ∅ and YT Z

α_{∩ ≠ ∅}

. By definition of the semilattice D we have D ⊇Z, i.e., Y∅α ⊇ ∅ and Y_Tα∩ ≠ ∅D . It follows that

α

∈R D

( )

₁′ . Therefore the inclusion R D

( )

′ ⊆R D

( )

₁′ holds. By the Equality (1) we have

( )

2

( )

1 .

R Q∗ =R D′ (2) From this equality and by statement b) of Lemma 1 it immediately follows that

( )

\

2 9 2 1 2 .

D X D

R∗ Q = ⋅ − ⋅

 

□

Let binary relation α of the semigroup B_X

( )

D satisfy the condition c) of Theorem 1 (see diagram 3 of the Figure 1). In this case we have Q₂=

{

∅, ,T T′

}

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

{

} {

}

{

} {

}

{

} {

}

{

} {

}

3 8 7 6 5 4 3

2 1 8 3 7 3 6 3 6 2

6 1 5 1 4 1

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

, , , , , , , , .

XI

Q Z D Z D 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

ϑ

= ∅ ∅ ∅ ∅ ∅ ∅

∅ ∅ ∅ ∅ ∅ ∅

∅ ∅ ∅

     

 

(

Q Q₃, ₃

)

=1 and Ω

( )

Q₃ =15. If-1

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

1 8 2 7 3 6 4 5

5 4 6 3 7 2 8 1

9 8 3 10 7 3 11 6 3 12 6 2

13 6 1 14 5 1 15 4 1

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

, , , , , , , , ,

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

′= ∅ ′= ∅ ′= ∅ ′= ∅

′= ∅ ′ = ∅ ′ = ∅ ′ = ∅

′ = ∅ ′ = ∅ ′ = ∅

   

then

( )

3 15

( )

1

i i

R∗ Q R D =

′

=



(3)

(see remark page 5 in [1] and Theorem 1).

{

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 the semigroup B_X

( )

D such that each element satisfies the condition c) of Theorem 1. Then

( )

( ) 5( )3

( )

8 3

1 ,

i j k

i j k M Q

R∗ Q R D R D R D

= ∈

′ ′ ′

=

∑

−

∑

∩

where

( )

{

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

}

(5)

Proof. Let Di′= ∅

{

, ,Y Yi′ ′′i

}

(

∅ ⊂ ⊂Yi′ Yi′′

)

be arbitrary element of the set Q3ϑXI and

α

∈R D

( )

i′ . Then

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

(

Y_∅α× ∅ ∪

) (

Y_Tα×T

) (

∪ Y_Tα_′×T′

)

for some

, ,

T T′∈D ∅ ⊂ ⊂T T′, YT ,YT

{ }

α α

′∉ ∅ and by statement c) of Theorem 1 satisfies the conditions

T i

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



. From this and by the condition Y_∅α∪Y_Tα ⊇Y_i′, Y_Tα∩ ≠ ∅Y_i′ , Y_Tα′ ∩ ≠ ∅Y_i′′ we have

, , ,

T i T i T

Y∅α∪Yα ⊇Y Y′ α ∩ ≠ ∅Y′ Yα′ ∩ ≠ ∅D



i.e. α∈R D

( )

′_j , where Dj′= ∅

{

, ,Y Di′

}



. 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, ∈M Q₁

( )

₃ , where

( )

{

( ) (

) (

)

}

1 3 9,1 , 10, 2 , 11, 3 , 12, 3 , 13, 3 , 14, 4 , 15, 5 . M Q =

Therefore the following equality holds

( )

3 8

( )

1 .

i i

R∗ Q R D =

′

=



(4)

Now, let Di′= ∅

{

, ,Y Di

}

,D′j = ∅

{

,Y Dj,

}

⊂

{

D D₁′ ′, ₂, ,D₈′

}

 

 , D_i′≠D′_j and α∈R D

( )

_i′ ∩R D

( )

′_j . Then for the binary relation α we have

, , ,

, , .

T i T i T

T j T j T

Y Y Y Y Y Y D Y Y Y Y Y Y D

α α α α

′ ∅

∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

 

From the last condition it follows that Y_∅α∪Y_Tα ⊇ ∪Y_i Y_j.

1) Y_i∪Y_j=D. Then we have, that

(

Y∅α∪Y_Tα

)

∩Y_Tα′ ⊇

(

Y_i∪Y_j

)

∩Y_Tα′ =D∩Y_Tα′ ≠ ∅



. But the inequality

(

Y 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 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′= ∅

{

, ,Y Di

}

,D′j= ∅

{

,Y Dj,

}

,Dk′= ∅

{

,Yi∪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

, , ,

T i T i T

T j T j T

α α α α

′ ∅

∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

 

and

(

)

, , ,

, ,

T i T i T

T j T j T

T i j T i j T

Y Y Y Y Y Y D

Y Y Y Y Y Y Y Y D

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

, , , .

T i j T i T j T

Y∅α∪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 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

( )

{

(

) (

)

}

(6)

3 ) Di′= ∅

{

, ,Y Di

}

, ,D′j= ∅

{

Y Dj,

}

, ,Dk′= ∅

{

Y Dk,

}

, ,Dt′= ∅

{

Yi∪ ∪Yj Y Dk,

}

⊂

{

D D₁′ ′, ₂, ,D₈′

}

   

 , D_p′ ≠D_q′,

{

}

, , , ,

p q∈ i j k t , p≠q, α∈R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ and α∈R D

( )

_i′ ∩R D

( )

′_j ∩R D

( )

_k′ ∩R D

( )

_t′ are true. Then we have

, , ,

, ,

T i T i T

T j T j T

T k T k T

Y Y Y Y Y Y D

α α α α

′ ∅

∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

  

and

(

)

, , ,

, ,

T i T i T

T j T j T

T k T k T

T i j k T i j k T

Y Y Y Y Y Y D

Y Y Y Y Y Y Y Y Y Y D

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

, , , , .

T i j k T i T j T k T

Y∅α∪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 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 (2) and by 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 5. Let D′ = ∅

{

, ,Y D

}

, D′′= ∅

{

,Y D′, 

}

, 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∅α× ∅ ∪

) (

Y_Tα×T

)

∪

(

Y₀α×D

)



for some T∈D, ∅ ⊂ ⊂T D and YT ,Y₀

{ }

α α_{∉ ∅} _,_then

_α

_∈_{R D}

( )

_′ _∩_{R D}

( )

_′′ _iff 0

, , .

T T

Y∅α∪Yα ⊇Y Y′ α∩ ≠ ∅Y Yα∩ ≠ ∅D



Proof. If

α

∈R D

( )

′ ∩R D

( )

′′ , then by statement c) of theorem 1 we have

0 0

, , ;

, , .

T T

α α α α

∅

∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅

′ ′

∪ ⊇ ∩ ≠ ∅ ∩ ≠ ∅



 (5)

From the last condition we have

0

, ,

T T

Y∅α ∪Yα ⊇Y Y′ α∩ ≠ ∅Y Yα∩ ≠ ∅D



(6) since Y′ ⊇Y by assumption. On the other hand, if the conditions of (6) holds, then the conditions of (5) follow,

i.e.

α

∈R D

( )

′ ∩R D

( )

′′ . □

Lemma 6. Let D′= ∅

{

, ,Y D

}

,D′′= ∅

{

,Y D′,

}

⊂

{

D D₁′ ′, ₂,,D₈′

}

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

( )

\

(

)

(

\ \

)

\

15 2Y Y 2Y 1 3D Y 2D Y 3X D. R D′ ∩R D′′ = ⋅ ′ ⋅ − ⋅ ′ − ′ ⋅

  

Proof. Let D′= ∅

{

, ,Y D

}

,D′′= ∅

{

,Y D′, 

}

⊂

{

D D₁′ ′, ₂,,D₈′

}

, where Y′ ⊇Y . Assume that

( )

R D R D

(7)

(

Y

) (

YT T

)

(

Y0 D

)

α α α

α= _∅ × ∅ ∪ × ∪ ×  for some T∈D, ∅ ⊂ ⊂T D and Y_Tα,Y₀α∉ ∅

{ }

. Then according to Lemma 5, we have

0

, , ,

T T

Y∅α ∪Yα ⊇Y Y′ α∩ ≠ ∅Y Yα ∩ ≠ ∅D



 (7) Further, let f_α be a mapping from X to the semilattice D satisfying the conditions f_α

( )

t =t

α

for all

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



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

{

Y D Y′, \ ′,X \D

}

is an empty set, and

(

\

) (

\

)

Y′∪ D Y ′ ∪ X D = X . We are going to find properties of the maps f₁_α, f₂_α, f₃_α. 1) t∈Y′. Then by the properties of D we have Y Y YT

α α ∅

′ ⊆ ∪ , i.e., t Y YT

α α ∅

∈ ∪ and t

α

∈ ∅

{

,T

}

by definition of the sets Y∅α and YT

α

. Therefore f₁α

( ) {

t ∈ ∅,T

}

for all t∈Y′. By suppose we have that

T

Yα∩ ≠ ∅Y , i.e. t′ =

α

T for some t′∈Y. Therefore f₁α

( )

t′ =T for some t′∈Y. 2) t∈D Y\ ′. Then by properties of D we have D Y\ D X Y YT Y₀

α α α

∅

′ ⊆ ⊆ = ∪ ∪

 

, i.e., t∈Y∅α∪Y_Tα∪Y₀α and tα ∈ ∅

{

, ,T D

}

by definition of the sets Y∅α , YT

α _and 0

Yα . Therefore f₃_α

( )

t ∈ ∅

{

, ,T D

}

for all \

t∈D Y ′. By suppose we have, that Y₀α∩ ≠ ∅D , i.e. t′′ =α D for some t′′∈D. If t′′∈Y′. Then

T

t′′∈ ⊆Y′ Y∅α∪Yα. Therefore t′′ ∈ ∅

α

{

,T

}

by definition of the set Y∅α and YT

α

. We have contradiction to the equality t′′ =α D. Therefore f₃_α

( )

t′′ =D for some t′′∈D Y\ ′.

3) t∈X\D. Then by definition quasinormal representation binary relation α and by property of D we have 0

\ T

t∈X D⊆X =Y∅α∪Yα ∪Yα



, i.e. tα ∈ ∅

{

, ,T D

}

by definition of the sets Y ,YT

α α

∅ and Y0 α

. Therefore

( )

{

}

4 , ,

f_α t ∈ ∅T D for all t∈X \D. Therefore for every binary relation

α

∈R D

( )

′ ∩R D

( )

′′ there exists ordered system

(

f₁_α,f₂_α,f₃_α

)

. It is obvious that for disjoint binary relations there exists disjoint ordered systems. Further, let

{

}

{

}

{

}

1: , , 2: \ , , , 3: \ , , f Y′→ ∅T f D Y ′→ ∅T D f X D→ ∅T D

be such mappings, which satisfy the conditions: f t₁

( ) {

∈ ∅,T

}

for all t∈Y′ and f t₁

( )

′ =T′ for some t′∈Y; f₂

( )

t ∈ ∅

{

, ,T D

}

for all t∈D Y\ ′ and f₂

( )

t′′ =D for some t′′∈D Y\ ′; f₃

( )

t ∈ ∅

{

, ,T D

}

for all

\

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

( )

1 9

2

3

, if \ , , if \ , , if \ . 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 =

_

_∈ × , Y∅β =

{

t t|

β

= ∅

}

, YT

{

t t| T

}

β ₌

_β

₌

and Y₀β =

{

t t| β =D

}

. Then binary relation

β

may be represented by

(

Y

) (

YT T

)

(

Y0 D

)

β β β

β = _∅ × ∅ ∪ × ∪ × 

and satisfy the conditions:

0

, ,

T T

Y∅β ∪Yβ ⊇Y Y′ β ∩ ≠ ∅Y Yβ ∩ ≠ ∅D



(By suppose f t1

( )

′ =T′ for some t′∈Y and f2

( )

t′′ =D



for some t′′∈D Y\ ′), i.e., by lemma 5 we have that

β

∈R D

( )

′ ∩R D

( )

′′ . Therefore for every binary relation

α

∈R D

( )

′ ∩R D

( )

′′ and ordered system

(

f1α,f2α,f3α

)

there exists one to one mapping. By Lemma 1 and by Theorem 1 in [1] the number of the mappings f₁_α,f₂_α,f₃_α,f₄_α are respectively:

( \ )

(

)

\ \ \

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

  

Note that the number 2Y Y′\ ⋅

(

2Y − ⋅1

)

(

3D Y\ ′ −2D Y\ ′

)

⋅3X D\

  

(8)

( )

\

(

)

(

\ \

)

\ 15 2Y Y 2Y 1 3D Y 2D Y 3X D. R D′ ∩R D′′ = ⋅ ′ ⋅ − ⋅ ′ − ′ ⋅

  

□

Therefore, we obtain:

( )

(

)

(

)

( )

(

)

(

)

( )

(

)

(

)

( )

(

)

(

)

( )

3 3

3 8 8

3 3

3 7 7

3 3

3 6 6

2 2

2 6 6

1 \ \ \ \ 1 6 \ \ \ \ 2 6 \ \ \ \ 3 6 \ \ \ \ 3 7 3 8

15 2 2 1 3 2 3 ,

15 2

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

Z

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

′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅            

(

)

(

)

( )

(

)

(

)

( )

(

)

(

)

1 1 6 6 1 1

1 5 5

1 1

1 4 4

\ \ \ \ \ \ \ \ 4 8 \ \ \ \ 5 8

2 1 3 2 3 ,

15 2 2 1 3 2 3 ,

15 2 2 1 3 2 3 .

D Z D Z X D

Z Z

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

R D R D R D R D

⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅ ′ ∩ ′ = ⋅ ⋅ − ⋅ − ⋅          (8)

{

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

( )

c) of Theorem 1. Then

( )

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

8 8 8 7 7 7 6 6 6 5 5 5 4 4 4 3 3 3 2 2 \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

15 2 1 3 2 3

15 2 1 3 2

D Z D Z X D

Z

D Z D Z X D

Z

D Z D Z X D

Z

D Z D Z X D

Z

D Z D Z X D

Z

D Z D Z X D

Z

D Z D Z

Z

R∗ Q = ⋅ − ⋅ − ⋅

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

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

2 1 1 1 3 3

3 8 8

3 3

3 7 7

3 3

3 6 6

2 2

2 6 6

1 6 6

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

15 2 1 3 2 3

15 2 2 1 3 2 3

22 2 2 1 3

X D

D Z D Z X D

Z

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

D

Z Z Z

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

(

)

(

)

(

)

(

)

(

)

1 1 1 1

1 5 5

1 1

1 4 4

\ \ \ \ \ \ \ \ \ \ 2 3

15 2 2 1 3 2 3

15 2 2 1 3 2 3 .

Z D Z X D

D Z D Z X D

Z Z Z

D Z D Z X D

Z Z Z

(9)

Proof. Let Z₉= ∅. Then the given Lemma immediately follows from Lemma 4 and from the Equalities (3).

□

Now let binary relation α of the semigroup B_X

( )

D satisfy the condition d) of Theorem 1 (see diagram 4 of the Figure 1). In this case we have Q₄= ∅

{

, ,T T D′, 

}

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

{

} {

}

{

}

{

} {

}

4 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

= ∅ ∅ ∅ ∅

∅ ∅ ∅

   

  

ϑ

(

Q Q₄, ₄

)

=1 and Ω

( )

Q₄ =7. If

{

}

{

}

{

}

{

}

{

}

{

}

{

}

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

( )

7

( )

4 1

i i

R Q∗ R D =

′

=

_

(9)

(see Definition [1], Definition 4 and [1], Theorem 2).

{

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

}

∈ Σ₁

(

X,10

)

and Z₉= ∅. Let

( )

4

( )

D such that each element satisfies the condition d) of Theorem 1. Then

( )

(

) (

)

(

)

(

) (

)

(

)

(

) (

)

(

)

(

) (

)

(

)

(

) (

)

3 3

8 3 8 3 8

3 3

7 3 7 3 7

3 3

6 3 6 3 6

2 2

6 2 6 2 6

6 1 6 1 6

\ \ \

\ \

4

\ \ \

\ \

\ \ \

\ \

\ \ \

\ \

\

\ \

7 2 1 3 2 4 3 4

7 2 1 3 2 4

D Z D Z X D

Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z

D Z

Z Z Z Z Z

R∗ Q = ⋅ − ⋅ − ⋅ − ⋅

+ ⋅ − ⋅ − ⋅ − ⋅

+ ⋅ − ⋅ − ⋅

            

(

)

(

) (

)

(

)

(

) (

)

(

)

1 1

5 1 5 1 5

1 1

4 1 4 1 4

\ \

\ \ \

\ \

\ \ \

\ \

3 4

7 2 1 3 2 4 3 4

7 2 1 3 2 4 3 4 .

D Z X D

D Z D Z X D

Z Z Z Z Z

D Z D Z X D

Z Z Z Z Z

− ⋅

+ ⋅ − ⋅ − ⋅ − ⋅

       

Proof. Let Z₉ = ∅. Then the given Lemma immediately follows from ([1], Lemma 10). □ Now let binary relation α of the semigroup B_X

( )

D satisfy the condition e) of Theorem 1 (see diagram 5 of the Figure 1). In this case we have Q₅ = ∅

{

, ,T T T′, ∪T′

}

, where T T, ′∈D and T T\ ′ ≠ ∅ and

\

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

{

} {

}

{

} {

}

{

}

{

}

{

} {

}

{

}

{

} {

}

{

}

{

} {

}

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

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

, , , , , , , , , ,

XI

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

ϑ = ∅ ∅ ∅ ∅ ∅

∅ ∅ ∅ ∅ ∅

∅ ∅ ∅

  

   

 

_{

_{} {}

_}

{

}

3 2 3 1 2 1

, , , , , , , , , ,

, , , .

D Z Z D Z Z D Z Z D

∅ ∅

∅

  



[image:9.595.196.441.365.541.2]

(10)

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

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

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

, , , ,

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

′= ∅ ′= ∅ ′= ∅ ′= ∅

′= ∅ ′ = ∅ ′ = ∅ ′ = ∅

′ = ∅ ′

   

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

14 6 7 3 15 7 5 16 5 7 17 7 4 18 4 7 19 7 2 20 2 7 21 7 1 22 1 7 23 6 5 1 24 5 6 1 25 6 4 1 26 4

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

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

, , , , ,

D Z Z Z 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 Z

′ ′

= ∅ = ∅ = ∅

′ = ∅ ′ = ∅ ′ = ∅ ′ = ∅

′ = ∅ ′ = ∅

 

   

 

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

{

}

6 1 27 5 4 1 28 4 5 1 29 5 3 30 3 5 31 5 2 32 2 5 33 4 3 34 3 4 35 4 2 36 2 4 37 3 2 38 2 3

, , , , , , , , , , ,

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

, , , , , , , ,

Z Z D Z Z Z D Z Z Z 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

′ = ∅ ′ = ∅

′ = ∅ ′ = ∅ ′ = ∅ ′ = ∅

′ = ∅ ′ = ∅ ′

   

 

_{

_}

_{

_}

{

}

{

}

39 3 1 40 1 3 41 2 1 42 1 2

, , , , , , , ,

, , , , , , ,

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

′

= ∅ = ∅

′ = ∅ ′ = ∅

 

then

( )

5 42

( )

1

i i

R∗ Q R D =

′

=



(10)

(see [1], Definition 4 and [1], Theorem 1).

{

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

}

∈ Σ₁

(

X,10

)

and Z₉= ∅. Let

( )

5

( )

D such that each element satisfies the condition e) of Theorem 1. Then

( )

( ) 4

( )

42 5

1 ,

,

i k j

i k j M

R∗ Q R D R D R D

′

= ∈

′ ′ ′

=

∑

−

∑

∩

where

( ) ( ) (

{

) (

)

(

) (

)

(

) (

)

(

) (

)

(

)

4 5 3, 9 , 3,11 , 4,10 , 4,12 , 5,11 , 5, 30 , 5, 39 , 6,12 , 6, 29 , 6, 40 , 7,11 , 7, 34 , 7, 39 , 8,12 , 8, 33 , 8, 40 , 9, 37 , 10, 38 , 11, 39 , 12, 40 , 13,19 , 13, 21 , 14, 20 , 14, 22 , 15, 21 , 15, 30 , 15, 39 , 16, 22 , 16, 29 , 16, 40 , 17, 21 , 17, 34 , 17, 39 , 18, 22 , 18, 33 , 18

M′ Q =

(

) (

)

(

) (

)

(

) (

)

}

, 40 , 19, 37 , 20, 38 , 21, 39 , 22, 40 , 23, 30 , 23, 32 , 23, 39 , 23, 41 , 24, 29 , 24, 31 , 25, 34 , 25, 36 , 26, 33 , 26, 35 , 29, 40 , 30, 39 , 31, 42 , 32, 41 , 33, 40 , 34, 39 , 35, 42 , 36, 41 .

Proof. Let Z₉= ∅. Then the given Lemma immediately follows from ([1], Lemma 13). □

Lemma 10. Let D_i′= ∅

{

, , ,Y Y Y_i _i′ _i∪Y_i′

}

and Dj′= ∅

{

,Y Y Yj, j′, j∪Yj′

}

be arbitrary elements of the set

{

D D1′ ′, 2,,D′42

}

, where Di′≠D′j, Yj ⊇Yi and Yj′⊇Yi′. Then the following equality holds

( )

(

\( )

)

(

\

)

(

\( )

)

(

\

)

\( )

21 2Yj Yi Yj 2Y Yi j 1 2Yj Yi Yj 2Y Yi j 1 4X Yj Yj .

i j

R D′ ∩R D′ = ⋅ ∪ ′ ⋅ ′ − ⋅ ′ ′∪ ⋅ ′ − ⋅ ∪ ′

Proof. Let Z₉= ∅. Then the given Lemma immediately follows from definition semilattice D and by ([1], Lemma 13). □

{

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

}

∈ Σ₁

(

X,10

)

and Z₉= ∅. Let

( )

5

( )

e) of Theorem 1. Then

( )

_{( )}

( )

4

42

5 i1 i k j, M k j

(11)

( )

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

8 7 7 8 3 8 6 6 8 3

8 5 5 8 8 4 4 8

8 2 2 8 8 1 1 8

7 6 6 7 3

42 \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

42 2 1 2 1 4 42 2 1 2 1 4

42 2 1 2 1 4

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

i i

X D X D

Z Z Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

Z Z Z Z X Z

R D = ′ = ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ +

∑

   

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

)

7 5 5 7

7 4 4 7 7 2 2 7

7 1 1 7 6 5 5 6 1

6 4 4 6 1 5 4 4 5 1

5 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

42 2 1 2 1 4

42 2 1 2 1 4 42 2 1 2 1 4

42 2 1

X D

Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

X D

Z Z Z Z Z Z Z Z X Z

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

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

(

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

3 5 5 2 2 5

4 3 3 4 4 2 2 4

3 2 2 3 3 1 1 3

2 1 1 2

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

2 1 4 42 2 1 2 1 4

42 2 1 2 1 4 42 2 1 2 1 4

42 2 1 2 1 4 .

X D X D

Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

X D

Z Z Z Z

⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅        and ( )

( )

(

)

(

)

(

)

(

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

4

8 2 2 3 6 8 8 1 1 3 6 8

8 1 5 8 8 5 5 3

8 1 5 3 8 1 4 8

8 , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

42 2 1 2 2 1 4 42 2 1 2 2 1 4

42 2 1 2 1 4 42 2 1 2 1 4

42 2

k j

k j M

X D X D

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

X D X D

Z Z Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

Z

R D R D

′ ∈ ′ ∩ ′ = ⋅ − ⋅ ⋅ − ⋅ + ⋅ − ⋅ ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ + ⋅

∑

     

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

)

(

)

(

)

(

)

(

) (

)

4 4 3 8 1 4 3

8 2 2 3 8 1 1 3

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

7 1 5 7

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

1 2 1 4 42 2 1 2 1 4

42 2 1 2 1 4 42 2 1 2 1 4

42 2 1 2 2 1 4 42 2 1 2 2 1 4

42 2 1 2 1 4 42

X D X D

Z Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

X D X D

X D

Z Z Z Z

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

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

7 5 5 3

7 1 5 3 7 1 4 7

7 4 4 3 7 1 4 3

7 2 2 3 7 1 1 3

3 1 6

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

2 1 2 1 4

42 2 1 2 1 4 42 2 1 2 1 4

42 2 2

X D

Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

Z Z Z

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

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

(

) (

)

5 5 3 2 1 6 5 5 2

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

5 3 3 1 5 2 2 1

4 3 3 1

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

1 2 1 4 42 2 2 1 2 1 4

42 2 2 1 2 1 4 42 2 2 1 2 1 4

42 2 1 2 1 4 42 2 1 2 1 4

42 2 1 2 1 4

X D X D

Z Z Z Z Z Z Z Z Z

X D X D

Z Z Z Z Z Z Z Z

X

Z Z Z Z

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

(

4\ 2

) (

2\1

)

\

42 2 1 2 1 4 .