Generating Set of the Complete Semigroups of Binary Relations

Applied Mathematics, 2016, 7, 98-107 Published Online January 2016 in SciRes. http://www.scirp.org/journal/am http://dx.doi.org/10.4236/am.2016.71009

Generating Set of the Complete Semigroups of Binary Relations Yasha Diasamidze1, Neset Aydin2, Ali Erdoğan3 1

Shota Rustavelli University, Batumi, Georgia Çanakkale Onsekiz Mart University, Çanakkale, Turkey 3 Hacettepe University, Ankara, Turkey 2

Received 16 December 2015; accepted 25 January 2016; published 28 January 2016 Copyright © 2016 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

Abstract Difficulties encountered in studying generators of semigroup BX ( D ) of binary relations defined by a complete X-semilattice of unions D arise because of the fact that they are not regular as a rule, which makes their investigation problematic. In this work, for special D, it has been seen that the semigroup BX ( D ) , which are defined by semilattice D, can be generated by the set

{

(

)

}

B= α ∈ BX ( D ) V X ∗ , α = D .

Keywords Semigroups, Binary Relation, Generated Set, Generators

1. Introduction

 Theorem 1. Let D = D, Z1 , Z 2 , , Z m −1

{

}

be some finite X-semilattice of unions and C ( D ) = { P0 , P1 , P2 , , Pm −1}

be the family of sets of pairwise nonintersecting subsets of the set X.  If φ is a mapping of the semilattice D on the family of sets C ( D ) which satisfies the condition ϕ D = P0 any i 1, 2, , m − 1 and Dˆ Z = D \ DT , then the following equalities are valid: and ϕ ( Z i ) = Pi for =  D = P0 ∪ P1 ∪ P2 ∪  ∪ Pm −1 , (1) ϕ (T ) . Z= P ∪

( )

i

0



T ∈Dˆ Zi

How to cite this paper: Diasamidze, Y., Aydin, N. and Erdoğan, A. (2016) Generating Set of the Complete Semigroups of Binary Relations. Applied Mathematics, 7, 98-107. http://dx.doi.org/10.4236/am.2016.71009

Y. Diasamidze et al.

In the sequel these equalities will be called formal. It is proved that if the elements of the semilattice D are represented in the form 1, then among the parameters = Pi ( i 0,1, 2, , m − 1) there exist such parameters that cannot be empty sets for D. Such sets Pi ( 0 < i ≤ m − 1) are called basis sources, whereas sets Pi ( 0 ≤ j ≤ m − 1) which can be empty sets too are called completeness sources. It is proved that under the mapping ϕ the number of covering elements of the pre-image of a basis source is always equal to one, while under the mapping ϕ the number of covering elements of the pre-image of a completeness source either does not exist or is always greater than one (see [1], Chapter 11). Some positive results in this direction can be found in [2]-[6]. Let P0 , P1 , P2 , , Pm −1 be parameters in the formal equalities, β ∈ BX ( D ) and m −1



=i 0





β =   Pi × t β  ∪ m −1



=i 0



t∈Pi

 ({t ′} × t ′β ) .



 t ′∈ X \ D



(





β =   Pi × t β  ∪ ( X \ D ) × D t∈Pi



(2)

)

(3)

The representation of the binary relation β of the form β and β will be called subquasinormal and maximal subquasinormal. If β and β are the subquasinormal and maximal subquasinormal representations of the binary relation β , then for the binary relations β and β the following statements are true: a) β , β ∈ BX ( D ) ; b) β ⊆ β ⊆ β ; c) the subquasinormal representation of the binary relation β is quasinormal; d) if P1  Pm −1   P β1 =  0 ,  P0 β P1 β  Pm −1 β  then β1 is a mapping of the family of sets C ( D ) = { P0 , P1 , P2 , , Pm −1} in the X-semilattice of unions  D = D, Z1 , Z 2 , , Z m −1 .   e) if β 2 : X \ D → D is a mapping satisfying the condition β 2 ( t ′ ) = t ′β for all t ′ ∈ X \ D , then

{

m −1



=i 0



}



β =   Pi × t β  ∪ 

t∈Pi

 ({t ′} × β 2 ( t ′) ) .

 t ′∈ X \ D

2. Results Proposition 2. Let α , β ∈ BX ( D ) . Then

α=  β α=  β α  β .

(

)

Proof. It is easy to see the inclusion α  β ⊆ α  β ⊆ α  β holds, since β ⊆ β ⊆ β . If x α  β y  m −1 z  i 0 Pi ×  t∈P t β ( x, y ∈ X ) , then xα z β y for some z ∈ X . So, z ∉ X \ D since z ∈ xα ∈ D and=

(

(

)

Then z Pk ×  t∈P t β y for some k k

( 0 ≤ k ≤ m − 1)

(

i

)) y .

i.e. z ∈ Pk and y ∈ z β ⊆  t∈P t β . For the last condition k

follows that z β y . We have xα z β y and x (α  β ) y . Therefore, the inclusion α  β ⊆ α  β is true. Of this  and by inclusion α  β ⊆ α  β ⊆ α  β follows that the equality α=  β α=  β α  β holds.  Corollary 1. If α , δ , β ∈ BX ( D ) and β ⊆ δ ⊆ β , then α=  β α=  δ α  β . Proof. We have β ⊆ δ ⊆ β and α  β ⊆ α  δ ⊆ α  β . Of this follows that α=  β α=  δ α  β since   α β =α β .  Let the X-semilattice D = Z 4 , Z 3 , Z 2 , Z1 , D of unions given by the diagram of Figure 1. Formal equalities of the given semilattice have a form:

{

}

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Figure 1. Diagram of D.

 D = P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 , Z1 = P0 ∪ P2 ∪ P3 ∪ P4 , Z 2 = P0 ∪ P1 ∪ P3 ∪ P4 , Z 3 = P0 ∪ P1 ∪ P2 ∪ P4 ,

(4)

Z= P0 ∪ P2 . 4

The parameters P1, P2, P3 are basis sources and the parameters P0 , P4 are completeness sources, i.e. X ≥ 3 . Example 3. Let X = {1, 2,3, 4,5, 6, 7} , P1 = {1, 4} , P2 = {2,5} , P3 = {3, 6} , P0 = P4 = ∅ . Then for the for-

mal equalities of the semilattice D follows that Z 4 = {2,5} , Z 3 = {1, 2, 4,5} , Z 2 = {1,3, 4, 6} , Z1 = {2,3,5, 6} ,  D = {1, 2,3, 4,5, 6} , D = {{2} , {1, 2} , {1,3} , {2,3} , {1, 2,3}} , and

β = ({1} × {2,3,5, 6} ) ∪ ({2} × {1,3, 4, 6} ) ∪ ({3} × {1, 2, 4,5} ) ∪ ({4,5, 6, 7} × {2,5} ) . Then we have:

= t β t∈P1

2,3,5, 6} ,  t β {1, 2,3, 4,5, 6} ,  t β {1, 2, 4,5} , = {= t∈P2

tβ = tβ =

t∈P0

t∈P3

∅;

t∈P4

β=

({1, 4} × {2,3,5, 6}) ∪ ({2,5} × {1, 2,3, 4,5, 6}) ∪ ({3, 6} × {1, 2, 4,5}) ∪ ({7} × {2,5}) ;

β =

({1, 4} × {2,3,5, 6}) ∪ ({2,5} × {1, 2,3, 4,5, 6}) ∪ ({3, 6} × {1, 2, 4,5}) ∪ ({7} × {1, 2,3, 4,5, 6, 7}) .

 Theorem 4. Let the X-semilattice D = Z 4 , Z 3 , Z 2 , Z1 , D of unions given by the diagram of Figure 1,  B= α ∈ BX ( D ) | V ( X ∗ , α ) = D and X \ D ≥ 3 . Then the set B is generating set of the semigroup BX ( D ) .  Proof. It is easy to see that X ≥ 6 since P1 , P2 , P3 ∉ {∅} and X \ D ≥ 3 . Now, let α be any binary relation of the semigroup BX ( D ) ; α1 , α 2 , , α n ∈= B , α α1  α 2  ⋅⋅⋅  α n and = β α 2  α 3  ⋅⋅⋅  α n . Then the ( is subquasinormal representation of a binary relation β ) is true. By equality α α= β α β =   β 1 1 assumption α1 ∈ BX ( D ) , i.e. the quasinormal representation of a binary relation α1 have a form  α1 = Y4α1 × Z 4 ∪ Y3α1 × Z 3 ∪ Y2α1 × Z 2 ∪ Y1α1 × Z1 ∪ Y0α1 × D .

{

{

}

(

}

) (

) (

) (

) (

)

Of this follows that



α = α1  β = (Y4α1 × Z 4 ) ∪ (Y3α1 × Z 3 ) ∪ (Y2α1 × Z 2 ) ∪ (Y1α1 × Z1 ) ∪ (Y0α1 × D )  β  = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1 β ∪ Y0α1 × D β .

(

) (

) (

) (

) (

)

For the binary relation α we consider the following case. a) Let V ( X ∗ , α ) = 1 . Then α= X × T , where T ∈ D . By element T we consider the following cases:  1. T ≠ D . In this case suppose that P



P

β1 =  0 1 ∅ T

and β 2 are mapping of the set X \ D on the set

P2 T

P3 T

P4   ∅

{Z 4 , Z3 , Z 2 , Z1} \ {T } . Then

100

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β = ( P0 × ∅ ) ∪ ( P1 × T ) ∪ ( P2 × T ) ∪ ( P3 × T ) ∪ ( P4 × ∅ ) ∪

Y. Diasamidze et al.

 ({t ′} × β 2 ( t ′) ) ,

 t ′∈ X \ D



α1 = (Y4α1 × Z 4 ) ∪ (Y3α1 × Z 3 ) ∪ (Y2α1 × Z 2 ) ∪ (Y1α1 × Z1 ) ∪ (Y0α1 × D ) ,

(6)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (6) and (5) we have:

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ T = T ; Z 3 β =( P0 ∪ P1 ∪ P2 ∪ P4 ) β =P0 β ∪ P1 β ∪ P2 β ∪ P4 β =∅ ∪ T ∪ T ∪ ∅ = T; Z 2 β =( P0 ∪ P1 ∪ P3 ∪ P4 ) β =P0 β ∪ P1 β ∪ P3 β ∪ P4 β =∅ ∪ T ∪ T ∪ ∅ = T; Z1 β =( P0 ∪ P2 ∪ P3 ∪ P4 ) β =P0 β ∪ P2 β ∪ P3 β ∪ P4 β =∅ ∪ T ∪ T ∪ ∅ = T;  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1 β ∪ P2 β ∪ P3 β ∪ P4 β =∅ ∪ T ∪ T ∪ T ∪ ∅ = T ;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1 β ∪ Y0α1 × D β

( = (Y

α1

4

) ( ) ( ) ( × T ) ∪ (Y × T ) ∪ (Y × T ) ∪ (Y × T ) ∪ (Y α1

3

α1

α1

2

1

α1

0

) ( ×T ) = X ×T

)

α1 α1 α1 α1 since Y4α1 ∪ X.  Y3 ∪ Y2 ∪ Y1 ∪ Y0 = 2. T = D . In this case suppose that

P D

β1 =  0

P1

P2

P3

Z1

Z2

Z3

P4   Z4 

 and β 2 are mapping of the set X \ D in the set D. Then  β = ( P0 × D ) ∪ ( P1 × Z1 ) ∪ ( P2 × Z 2 ) ∪ ( P3 × Z 3 ) ∪ ( P4 × Z 4 ) ∪

 ({t ′} × β 2 ( t ′) ) ,

 t ′∈ X \ D

 α1 = (Y4α1 × Z 4 ) ∪ (Y3α1 × Z 3 ) ∪ (Y2α1 × Z 2 ) ∪ (Y1α1 × Z1 ) ∪ (Y0α1 × D ) ,

(7)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (7) and (5) we have:  Z 4 β =( P0 ∪ P2 ) β =P0 β ∪ P2 β =D;  Z 3 β = ( P0 ∪ P1 ∪ P2 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P4 β = D;  Z 2 β = ( P0 ∪ P1 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P3 β ∪ P4 β = D;  Z1β = ( P0 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = D;   Dβ = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P3 β ∪ P4 β = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1β ∪ Y0α1 × D β      = Y4α1 × D ∪ Y3α1 × D ∪ Y2α1 × D ∪ Y1α1 × D ∪ Y0α1 × D  = X × D.

( (

b) V ( X ∗ , α ) = 2 . Then

) ( ) (

) (

) (

) (

) (

) (

) ( )

)

    V X ∗ , α ∈ {Z 4 , Z1} , {Z 4 , Z 3 } , Z 4 , D , Z 3 , D , Z 2 , D , Z1 , D

(

(

) {

{

)

}{

}{

} { }} V ( X , α ) consider the following

∗ since V X ∗ , α is X-semilattice of unions. For the semilattice of unions cases. 1. Let V X ∗ , α = {Z 4 , T } , where, T ∈ {Z1 , Z 3 } . Then binary relation α has representation of the form

(

)

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α = (Y4α × Z 4 ) ∪ (YTα × T ) . In this case suppose that  P0 ∅

β1 = 

P1

P2

P3

T

Z4

T

P4   ∅

 and β 2 are mapping of the set X \ D on the set D \ {Z 4 , T } . Then

( P0 × ∅ ) ∪ ( P1 × T ) ∪ ( P2 × Z 4 ) ∪ ( P3 × T ) ∪ ( P4 × ∅ ) ∪   ({t ′} × β 2 ( t ′ ) ) ,

= β

t ′∈ X \ D



α1 = (Y4 × Z 4 ) ∪ (Y3 × Z 3 ) ∪ (Y2 × Z 2 ) ∪ (Y1 × Z1 ) ∪ (Y0 × D ) , α1

α1

α1

α1

α1

(8)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , Y4α1 = Y4α and Y3α1 ∪ Y2α1 ∪ Y1α1 ∪ Y0α1 = Y0α , then it is easy to see, that

α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (8) and (5) we have: Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ Z 4 = Z 4 ;

( P0 ∪ P1 ∪ P2 ∪ P4 ) β =

Z3 β =

P0 β ∪ P1 β ∪ P2 β ∪ P4 β

= ∅ ∪ T ∪ Z4 ∪ ∅ = T ;

( P0 ∪ P1 ∪ P3 ∪ P4 ) β =

Z2 β =

P0 β ∪ P1 β ∪ P3 β ∪ P4 β

T; =∅ ∪ T ∪ T ∪ ∅ = Z1 β =

( P0 ∪ P2 ∪ P3 ∪ P4 ) β

= P0 β ∪ P2 β ∪ P3 β ∪ P4 β

= ∅ ∪ Z4 ∪ T ∪ ∅ = T ;  Dβ =

( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β =

P0 β ∪ P1 β ∪ P2 β ∪ P3 β ∪ P4 β

T; = ∅ ∪ T ∪ Z4 ∪ T ∪ ∅ =

α = α1  β  = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1 β ∪ Y0α1 × D β

(

) (

(

) (

(

) ((

) (

) (

) (

) (

) (

) (

= Y4α1 × Z 4 ∪ Y3α1 × T ∪ Y2α1 × T ∪ Y1α1 × T ∪ Y0α1 × T

) ) (

)

)

) (

)

= Y4α1 × Z 4 ∪ Y3α1 ∪ Y2α1 ∪ Y1α1 ∪ Y0α1 × T = Y4α × Z 4 ∪ Y0α × T .

 2. Let V X ∗ , α = T , D , where, T ∈ {Z 4 , Z 3 , Z 2 , Z1} . Then binary relation α has representation of the  form α = YTα × T ∪ Y0α × D . In this case suppose that

(

(

) { ) (

}

)

P

P2 P3 P4    T D ∅   are mapping of the set X \ D on the set D \ T , D . Then   = β ( P0 × ∅ ) ∪ P1 × D ∪ ( P2 × T ) ∪ P3 × D ∪ ( P4 × ∅ ) ∪ P1  D

β1 =  0 ∅

and β 2

(

{ (

)

α1 = (Y4 × Z 4 ) ∪ (Y3 × Z 3 ) ∪ (Y2 × Z 2 ) α1

α1

α1

}

)

 ({t ′} × f ( t ′) ) ,

 t ′∈ X \ D

 ∪ Y1 × Z1 ∪ Y0 × D ,

(

α1

) (

α1

)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , Y4α1 = YTα and Y3α1 ∪ Y2α1 ∪ Y1α1 ∪ Y0α1 = Y0α , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (9) and (5) we have:

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Y. Diasamidze et al.

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ T = T ;

  Z 3 β = ( P0 ∪ P1 ∪ P2 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P4 β = ∅ ∪ D ∪ T ∪ ∅ = D;    Z2 β = ∅∪ D∪ D∪∅ = D; ( P0 ∪ P1 ∪ P3 ∪ P4 ) β =P0 β ∪ P1β ∪ P3 β ∪ P4 β =   Z1β = ( P0 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = ∅ ∪ T ∪ D ∪ ∅ = D;  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P3 β ∪ P4 β    = ∅ ∪ D ∪T ∪ D ∪∅ = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1β ∪ Y0α1 × Dβ     = Y4α1 × T ∪ Y3α1 × D ∪ Y2α1 × D ∪ Y1α1 × D ∪ Y0α1 × D   = Y4α1 × T ∪ Y3α1 ∪ Y2α1 ∪ Y1α1 ∪ Y0α1 × D = YTα × T ∪ Y0α × D .

( ( (

) ( ) ( ) ((

c) V ( X ∗ , α ) = 3 . Then

) (

) (

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

)

)

      V X ∗ , α ∈ Z 4 , Z1 , D , Z 4 , Z 3 , D , Z 4 , Z 2 , D , Z1 , Z 2 , D , Z1 , Z 3 , D , Z 2 , Z 3 , D

since cases.

) {{

( V ( X ,α ) ∗

}{

}{

}{

}{

is X-semilattice of unions. For the semilattice of unions

}}

}{ V ( X ,α ) ∗

consider the following

 1. Let V X ∗ , α = Z 4 , Z1 , D . Then binary relation α has representation of the form  α = Y4α × Z 4 ∪ Y1α × Z1 ∪ Y0α × D . In this case suppose that

(

(

) (

) {

) (

}

)

 P0 ∅

β1 = 

P1

P2

P3

Z2

Z4

Z1

P4   ∅

 and β 2 are mapping of the set X \ D on the set D \ {Z 4 , Z 2 , Z1} . Then

( P0 × ∅ ) ∪ ( P1 × Z 2 ) ∪ ( P2 × Z 4 ) ∪ ( P3 × Z1 ) ∪ ( P4 × ∅ ) ∪   ({t ′} × f ( t ′ ) ) ,

= β

α1 = (

Y4α1

) (

× Z4 ∪

Y3α1

) (

× Z3 ∪

Y2α1

) (

× Z2 ∪

Y1α1

) (

× Z1 ∪

t ′∈X \ D

Y0α1

 ×D ,

)

(10)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , Y4α1 = Y4α , Y1α1 = Y1α and Y3α1 ∪ Y2α1 ∪ Y0α1 = Y0α , then it is easy to see, that

α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (10) and (5) we have: Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ Z 4 = Z 4 ;

 Z3 β = P0 β ∪ P1β ∪ P2 β ∪ P4 β = ∅ ∪ Z2 ∪ Z4 ∪ ∅ = D; ( P0 ∪ P1 ∪ P2 ∪ P4 ) β =  Z 2 β = ( P0 ∪ P1 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P3 β ∪ P4 β = ∅ ∪ Z 2 ∪ Z1 ∪ ∅ = D; Z1β = ( P0 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = ∅ ∪ Z 4 ∪ Z1 ∪ ∅ = Z1 ;  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P3 β ∪ P4 β  = ∅ ∪ Z 2 ∪ Z 4 ∪ Z1 ∪ ∅ = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1β ∪ Y0α1 × Dβ    = Y4α1 × Z 4 ∪ Y3α1 × D ∪ Y2α1 × D ∪ Y1α1 × Z1 ∪ Y0α1 × D  = Y4α1 × Z 4 ∪ Y1α1 × Z1 ∪ Y3α1 ∪ Y2α1 ∪ Y0α1 × D  = Y4α × Z 4 ∪ Y1α × Z1 ∪ Y0α × D .  2. Let V X ∗ , α = Z 4 , Z 3 , D . Then binary relation α has representation of the form

( ( ( (

(

) {

) ( ) ( ) ( ) ( }

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

103

)

Y. Diasamidze et al.

 α = (Y4α × Z 4 ) ∪ (Y3α × Z 3 ) ∪ (Y0α × D ) . In this case suppose that  P0 ∅

P1 Z3

β1 = 



P2 Z4

P4   ∅

P3 Z2

and β 2 are mapping of the set X \ D on the set D \ {Z 4 , Z 3 , Z 2 } . Then

= β

( P0 × ∅ ) ∪ ( P1 × Z3 ) ∪ ( P2 × Z 4 ) ∪ ( P3 × Z 2 ) ∪ ( P4 × ∅ ) ∪   ({t ′} × f ( t ′ ) ) ,

α1 = (

Y4α1

) (

× Z4 ∪

Y3α1

) (

× Z3 ∪

Y2α1

) (

× Z2 ∪

Y1α1

) (

× Z1 ∪

t ′∈X \ D

Y0α1

 ×D ,

)

(11)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , Y4α1 = Y4α , Y3α1 = Y3α and Y2α1 ∪ Y1α1 ∪ Y0α1 = Y0α , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (11) and (5) we have:

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ Z 4 = Z 4 ;

( P0 ∪ P1 ∪ P2 ∪ P4 ) β =

Z3 β =

P0 β ∪ P1 β ∪ P2 β ∪ P4 β

= ∅ ∪ Z3 ∪ Z 4 ∪ ∅ = Z3 ;

( P0 ∪ P1 ∪ P3 ∪ P4 ) β =

Z2 β =

P0 β ∪ P1 β ∪ P3 β ∪ P4 β

 = ∅ ∪ Z 3 ∪ Z 2 ∪ ∅ = D; Z1 β =

( P0 ∪ P2 ∪ P3 ∪ P4 ) β

 Dβ =

( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β =

= P0 β ∪ P2 β ∪ P3 β ∪ P4 β  = ∅ ∪ Z4 ∪ Z2 ∪ ∅ = D; P0 β ∪ P1 β ∪ P2 β ∪ P3 β ∪ P4 β

 = ∅ ∪ Z3 ∪ Z 4 ∪ Z 2 ∪ ∅ = D;

α = α1  β  = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1 β ∪ Y0α1 × D β

( = (Y

α1

4

× Z4

) ( ) ∪ (Y

α1

3

× Z3

) ( ) ( ) (    ) ∪ (Y × D ) ∪ (Y × D ) ∪ (Y × D ) α1

α1

2

)

α1

1

0

 = Y4α1 × Z 4 ∪ Y3α1 × Z 3 ∪ Y2α1 ∪ Y1α1 ∪ Y0α1 × D

(

) (

(

) (

= Y4α × Z 4 ∪ Y3α × Z 3

) (( ) ∪ (Y

α

0

) )

 ×D .

)

 3. Let V X ∗ , α = Z 4 , Z 2 , D . Then binary relation α has representation of the form  α = Y4α × Z 4 ∪ Y2α × Z 2 ∪ Y0α × D . In this case suppose that

(

(

) (

) {

) (

}

)

 P0 ∅

β1 = 



P1 Z2

P2 Z4

P3 Z2

P4   ∅

and β 2 are mapping of the set X \ D on the set D \ {Z 4 , Z 2 } . Then

β = ( P0 × ∅ ) ∪ ( P1 × Z 2 ) ∪ ( P2 × Z 4 ) ∪ ( P3 × Z 2 ) ∪ ( P4 × ∅ ) ∪

 ({t ′} × f ( t ′) ) ,

 t ′∈X \ D



α1 = (Y4α1 × Z 4 ) ∪ (Y3α1 × Z 3 ) ∪ (Y2α1 × Z 2 ) ∪ (Y1α1 × Z1 ) ∪ (Y0α1 × D ) ,

(12)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , Y4α1 = Y4α , Y2α1 = Y2α and Y3α1 ∪ Y1α1 ∪ Y0α1 = Y0α , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (12) and (5) we have:

104

Y. Diasamidze et al.

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ Z 4 = Z 4 ;

 Z3 β = P0 β ∪ P1β ∪ P2 β ∪ P4 β = ∅ ∪ Z2 ∪ Z4 ∪ ∅ = D; ( P0 ∪ P1 ∪ P2 ∪ P4 ) β = Z2 β = P0 β ∪ P1β ∪ P3 β ∪ P4 β = ∅ ∪ Z2 ∪ Z2 ∪ ∅ = Z2 ; ( P0 ∪ P1 ∪ P3 ∪ P4 ) β =  Z1β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = ∅ ∪ Z4 ∪ Z2 ∪ ∅ = D; ( P0 ∪ P2 ∪ P3 ∪ P4 ) β =  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P3 β ∪ P4 β  = ∅ ∪ Z2 ∪ Z4 ∪ Z2 ∪ ∅ = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1β ∪ Y0α1 × Dβ    = Y4α1 × Z 4 ∪ Y3α1 × D ∪ Y2α1 × Z 2 ∪ Y1α1 × D ∪ Y0α1 × D  = Y4α1 × Z 4 ∪ Y2α1 × Z 2 ∪ Y3α1 ∪ Y1α1 ∪ Y0α1 × D  = Y4α × Z 4 ∪ Y2α × Z 2 ∪ Y0α × D .  4. Let V X ∗ , α = Z1 , Z 2 , D . Then binary relation α has representation of the form  α = Y1α × Z1 ∪ Y2α × Z 2 ∪ Y0α × D . In this case suppose that

) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ) ( ) (( ) ) ) ( ) ( ) ( ) { } ) ) ( ) ( ( ( ( (

(

P

P4   Z1 ∅   are mapping of the set X \ D on the set D \ {Z 2 , Z1} . Then P

P

2 β1 =  0 1 ∅ Z Z  1 2

and β 2

)

= β

P3

( P0 × ∅ ) ∪ ( P1 × Z1 ) ∪ ( P2 × Z 2 ) ∪ ( P3 × Z1 ) ∪ ( P4 × ∅ ) ∪   ({t ′} × f ( t ′ ) ) ,

α1 = (

Y4α1

) (

× Z4 ∪

Y3α1

) (

× Z3 ∪

Y2α1

) (

× Z2 ∪

Y1α1

) (

× Z1 ∪

t ′∈X \ D

Y0α1

 ×D ,

)

(13)

where Y4α1 , Y3α1 , Y2α1 , Y1α1 ∉ {∅} , Y2α1 = Y1α , Y4α1 = Y2α and Y3α1 ∪ Y1α1 ∪ Y0α1 = Y0α , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (13) and (5) we have:

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ Z 2 = Z 2 ;

 Z 3 β = ( P0 ∪ P1 ∪ P2 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P4 β = ∅ ∪ Z1 ∪ Z 2 ∪ ∅ = D; Z2 β = P0 β ∪ P1β ∪ P3 β ∪ P4 β = ∅ ∪ Z1 ∪ Z1 ∪ ∅ = Z1 ; ( P0 ∪ P1 ∪ P3 ∪ P4 ) β =  Z1β = ( P0 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = ∅ ∪ Z 2 ∪ Z1 ∪ ∅ = D;  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P3 β ∪ P4 β  = ∅ ∪ Z1 ∪ Z 2 ∪ Z1 ∪ ∅ = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1β ∪ Y0α1 × Dβ    = Y4α1 × Z 2 ∪ Y3α1 × D ∪ Y2α1 × Z1 ∪ Y1α1 × D ∪ Y0α1 × D  = Y2α1 × Z1 ∪ Y4α1 × Z 2 ∪ Y3α1 ∪ Y1α1 ∪ Y0α1 × D  = Y1α × Z1 ∪ Y2α × Z 2 ∪ Y0α × D .

) ( ) ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ) ( ) (( ) ) ) ( ) ( )  5. Let V ( X , α ) = {Z , Z , D} . Then binary relation α has representation of the form  α = (Y × Z ) ∪ (Y × Z ) ∪ (Y × D ) . In this case suppose that ( ( ( (

∗

1

α 1

1

α 3

3

α

3

0

 P0 ∅

β1 = 

P1 Z3

105

P2 Z1

P3 Z3

P4   ∅

Y. Diasamidze et al.



and β 2 are mapping of the set X \ D on the set D \ {Z 3 , Z1} . Then

= β

( P0 × ∅ ) ∪ ( P1 × Z3 ) ∪ ( P2 × Z1 ) ∪ ( P3 × Z3 ) ∪ ( P4 × ∅ ) ∪   ({t ′} × f ( t ′ ) ) ,

α1 = (

Y4α1

) (

× Z4 ∪

Y3α1

) (

× Z3 ∪

Y2α1

) (

× Z2 ∪

Y1α1

) (

× Z1 ∪

t ′∈X \ D

Y0α1

 ×D ,

)

(14)

∉ {∅} , Y4 = Y1 , Y2 = Y3 and Y3 ∪ Y1 ∪ Y0 = where Y0α , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (14) and (5) we have: Y4α1 , Y3α1 , Y2α1 , Y1α1

α1

α1

α

α1

α

α1

α1

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∩ Z1 = Z1 ;

 Z 3 β = ( P0 ∪ P1 ∪ P2 ∪ P4 ) β = P0 β ∪ P1 β ∪ P2 β ∪ P4 β = ∅ ∪ Z 3 ∪ Z1 ∪ ∅ = D;

Z2 β = P0 β ∪ P1 β ∪ P3 β ∪ P4 β = ∅ ∪ Z3 ∪ Z3 ∪ ∅ = Z3 ; ( P0 ∪ P1 ∪ P3 ∪ P4 ) β =  Z1 β = ( P0 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = ∅ ∪ Z1 ∪ Z 3 ∪ ∅ = D;  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1 β ∪ P2 β ∪ P3 β ∪ P4 β  = ∅ ∪ Z 3 ∪ Z1 ∪ Z 3 ∪ ∅ = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1 β ∪ Y0α1 × D β    = Y4α1 × Z1 ∪ Y3α1 × D ∪ Y2α1 × Z 3 ∪ Y1α1 × D ∪ Y0α1 × D  = Y4α1 × Z1 ∪ Y2α1 × Z 3 ∪ Y3α1 ∪ Y1α1 ∪ Y0α1 × D  = Y1α × Z1 ∪ Y3α × Z 3 ∪ Y0α × D .  6. Let V X ∗ , α = Z 2 , Z 3 , D . Then binary relation α has representation of the form  α = Y2α × Z 2 ∪ Y3α × Z 3 ∪ Y0α × D . In this case suppose that  P P1 P2 P3 P4  β1 =  0   ∅ Z3 Z 2 Z3 ∅ 

( ( ( (

(

) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (( ) ) ) ( ) ( ) ( ) { } ) ( ) ( )

) ( )

)



and β 2 are mapping of the set X \ D on the set D \ {Z 3 , Z 2 } . Then

β =

( P0 × ∅ ) ∪ ( P1 × Z3 ) ∪ ( P2 × Z 2 ) ∪ ( P3 × Z3 ) ∪ ( P4 × ∅ ) ∪   ({t ′} × f ( t ′ ) )

α1 = (

Y4α1

) (

× Z4 ∪

Y3α1

) (

× Z3 ∪

Y2α1

) (

× Z2 ∪

Y1α1

) (

× Z1 ∪

t ′∈X \ D

Y0α1

 ×D ,

)

(15)

∉ {∅} , Y4 = Y2 , Y2 = Y3 and Y3 ∪ Y1 ∪ Y0 = where Y0α , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (15) and (5) we have: Y4α1 , Y3α1 , Y2α1 , Y1α1

α1

α

α1

α

α1

α1

α1

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ Z 2 = Z 2 ;

 Z 3 β = ( P0 ∪ P1 ∪ P2 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P4 β = ∅ ∪ Z 3 ∪ Z 2 ∪ ∅ = D; Z2 β = P0 β ∪ P1β ∪ P3 β ∪ P4 β = ∅ ∪ Z3 ∪ Z3 ∪ ∅ = Z3 ; ( P0 ∪ P1 ∪ P3 ∪ P4 ) β =  Z1β = ( P0 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = ∅ ∪ Z 2 ∪ Z 3 ∪ ∅ = D;  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P3 β ∪ P4 β  = ∅ ∪ Z3 ∪ Z 2 ∪ Z3 ∪ ∅ = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1β ∪ Y0α1 × Dβ    = Y4α1 × Z 2 ∪ Y3α1 × D ∪ Y2α1 × Z 3 ∪ Y1α1 × D ∪ Y0α1 × D  = Y4α1 × Z 2 ∪ Y2α1 × Z 3 ∪ Y3α1 ∪ Y1α1 ∪ Y0α1 × D  = Y2α × Z 2 ∪ Y3α × Z 3 ∪ Y0α × D .

( ( ( (

) ( ) ( ) ( ) (

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

106

)

 7. Let V X ∗ , α = Z 4 , Z 3 , Z1 , D . Then binary relation α has representation of the form  α = Y4α × Z 4 ∪ Y3α × Z 3 ∪ Y1α × Z1 ∪ Y0α × D . In this case suppose that

(

(

) (

) {

) (

}

) (

Y. Diasamidze et al.

)

P

P

1 β1 =  0 Z ∅ 3 

P2

P3

Z4

Z1

P4   ∅

 and β 2 are mapping of the set X \ D on the set D \ {Z 4 , Z 3 , Z1} . Then = β

( P0 × ∅ ) ∪ ( P1 × Z3 ) ∪ ( P2 × Z 4 ) ∪ ( P3 × Z1 ) ∪ ( P4 × ∅ ) ∪   ({t ′} × f ( t ′ ) ) ,

α1 = (

Y4α1

) (

× Z4 ∪

Y3α1

) (

× Z3 ∪

Y2α1

) (

× Z2 ∪

Y1α1

) (

× Z1 ∪

t ′∈X \ D

Y0α1

 ×D ,

)

(16)

∉ {∅} , Y4 = Y4 , Y3 = Y3 , Y1 = Y1 and Y2 ∪ Y0α1 = where Y0α , then it is easy to see, that α1 , β ∈ B since V= ( D, α1 ) V= ( D, β ) D . From the formal equality and equalities (16) and (5) we have: Y4α1 , Y3α1 , Y2α1 , Y1α1

α1

α

α1

α

α1

α

α1

Z 4 β = ( P0 ∪ P2 ) β = P0 β ∪ P2 β = ∅ ∪ Z 4 = Z 4 ; Z 3 β = ( P0 ∪ P1 ∪ P2 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P4 β = ∅ ∪ Z 3 ∪ Z 4 ∪ ∅ = Z 3 ;  Z 2 β = ( P0 ∪ P1 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P3 β ∪ P4 β = ∅ ∪ Z 3 ∪ Z1 ∪ ∅ = D; Z1β = ( P0 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P2 β ∪ P3 β ∪ P4 β = ∅ ∪ Z 4 ∪ Z1 ∪ ∅ = Z1 ;  D β = ( P0 ∪ P1 ∪ P2 ∪ P3 ∪ P4 ) β = P0 β ∪ P1β ∪ P2 β ∪ P3 β ∪ P4 β  = ∅ ∪ Z 3 ∪ Z 4 ∪ Z1 ∪ ∅ = D;

α = α1  β

 = Y4α1 × Z 4 β ∪ Y3α1 × Z 3 β ∪ Y2α1 × Z 2 β ∪ Y1α1 × Z1β ∪ Y0α1 × Dβ   = Y4α1 × Z 4 ∪ Y3α1 × Z 3 ∪ Y2α1 × D ∪ Y1α1 × Z1 ∪ Y0α1 × D  = Y4α1 × Z 4 ∪ Y3α1 × Z 3 ∪ Y1α1 × Z1 ∪ Y2α1 ∪ Y0α1 × D  = Y4α × Z 4 ∪ Y3α × Z 3 ∪ Y1α × Z1 ∪ Y0α × D .

( ( ( (

) ( ) ( ) ( ) (

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

)



References [1]

Diasamidze, Ya. and Makharadze, Sh. (2013) Complete Semigroups of Binary Relations. Cityplace Kriter, CountryRegion Turkey, 520 p.

[2]

Davedze, M.Kh. (1968) Generating Sets of Some Subsemigroups of the Semigroup of All Binary Relations in a Finite Set. Proc. A. I. Hertzen Leningrad State Polytechn. Inst., 387, 92-100. (In Russian)

[3]

Davedze, M.Kh. (1971) A Semigroup Generated by the Set of All Binary Relations in a Finite Set. XIth All-Union Algebraic Colloquium, Abstracts of Reports, Kishinev, 193-194. (In Russian)

[4]

Davedze, M.Kh. (1968) Generating Sets of the Subsemigroup of All Binary Relations in a Finite Set. Doklady AN BSSR, 12, 765-768. (In Russian)

[5]

Givradze, O. (2010) Irreducible Generating Sets of Complete Semigroups of Unions BX ( D ) Defined by Semilattices of Class

[6]

∑ ( X , 4 ) . Proceedings of the International Conference “Modern Algebra and Its Aplications”, Batumi. 2

Givradze, O. (2011) Irreducible Generating Sets of Complete Semigroups of Unions BX ( D ) Defined by Semilattices   of Class in Case, When X = D and D \ Z 3 > 1 . Proceedings of the International Conference “Modern Algebra and Its Aplications”, Batumi.

107