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In this paper, we have studied generating sets of the complete semigroups defined by X-semilattices of the class
Σ
_{2}(
*X*, 4).

Let X be an arbitrary nonempty set and D be a nonempty set of subsets of the set X. If D is closed under the union, then D is called a complete X-semilattice of unions. The union of all elements of the set D is denoted by the symbol D ⌣ .

Let B X be the set of all binary relations on X. It is well known that B X is a semigroup.

Let f be an arbitrary mapping from X into D. Then we denote a binary relation

α f = ∪ x ∈ X ( { x } × f ( x ) ) for each f. The set of all such binary relations is denoted

by B X ( D ) . It is easy to prove that B X ( D ) is a semigroup with respect to the product operation of binary relations. This semigroup B X ( D ) is called a complete semigroup of binary relations defined by an X-semilattice of unions D. This structure was comprehensively investigated in Diasamidze [

y α = { x ∈ X | y α x } , Y α = ∪ y ∈ Y y α ,

V ( D , α ) = { Y α | Y ∈ D } , X ∗ = { Y | ∅ ≠ Y ⊆ X } Y T α = { y ∈ X | y α = T } , V ( X ∗ , α ) = { Y α | ∅ ≠ Y ⊆ X } D t = { Z ′ ∈ D | t ∈ Z ′ } , B 0 = { α ∈ B X ( D ) | V ( X ∗ , α ) = D }

Let D = { D ⌣ , Z 1 , Z 2 , ⋯ , Z m − 1 } be finite X-semilattice of unions and C ( D ) = { P 0 , P 1 , P 2 , ⋯ , P m − 1 } be the family of pairwise nonintersecting subsets of X. If φ = ( D ⌣ Z 1 ⋯ Z m − 1 P 0 P 1 ⋯ P m − 1 ) is a mapping from D on C ( D ) , then the equalities D ⌣ = P 0 ∪ P 1 ∪ P 2 ∪ ⋯ ∪ P m − 1 and Z i = P 0 ∪ ∪ T ∈ D \ D Z φ ( T ) are valid. These equalities are called formal.

Let D be a complete X-semilattice of unions α ∈ B X . Then a representation

of a binary relation α of the form α = ∪ T ∈ V ( X ∗ , α ) ( Y T α × T ) is called quasinormal.

Let P 0 , P 1 , P 2 , ⋯ , P m − 1 be parameters in the formal equalities, β ∈ B X ( D ) , β ¯ 2

be mapping from X \ D ⌣ to D . Then β ¯ = ∪ i = 0 m − 1 ( P i × ∪ t ∈ P i t β ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

is called subquasinormal represantation of β . It can be easily seen that the following statements are true.

a) β ¯ ∈ B X ( D ) .

b) ∪ i = 0 m − 1 ( P i × ∪ t ∈ P i t β ) ⊆ β and β = β ¯ for some β ¯ 2 .

c) Subquasinormal represantation of β is quasinormal.

d) β ¯ 1 = ( P 0 P 1 ⋯ P m − 1 P 0 β ¯ P 1 β ¯ ⋯ P m − 1 β ¯ ) is mapping from C ( D ) on D ∪ { ∅ } .

β ¯ 1 and β ¯ 2 are respectively called normal and complement mappings for β .

Let α ∈ B X ( D ) . If α ≠ δ ∘ β for all δ , β ∈ B X ( D ) \ { α } then α is called external element. Every element of the set B 0 = { α ∈ B X ( D ) | V ( X ∗ , α ) = D } is an external element of B X ( D ) .

Theorem 1. [

Corollary 1. [

It is known that the set of all external elements is subset of any generating set of B X ( D ) in [

In this work by symbol Σ 2.2 ( X ,4 ) we denote all semilattices D = { Z 3 , Z 2 , Z 1 , D ⌣ } of the class Σ 2 ( X ,4 ) which the intersection of minimal elements Z 3 ∩ Z 2 = ∅ . This semilattices graphic is given in

D ⌣ = P 1 ∪ P 2 ∪ P 3 Z 1 = P 2 ∪ P 3 Z 2 = P 1 ∪ P 3 Z 3 = P 2 (1)

Let δ , β ¯ ∈ B X ( D ) . If quasinormal representation of binary relation δ has a form δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ ) then

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ )

We denote the set

B 32 = { α ∈ B X ( D ) | V ( X ∗ , α ) = { Z 3 , Z 2 , D ⌣ } } B 21 = { α ∈ B X ( D ) | V ( X ∗ , α ) = { Z 2 , Z 1 , D ⌣ } } B 31 = { α ∈ B X ( D ) | V ( X ∗ , α ) = { Z 3 , Z 1 } } B ˜ 32 = { α ∈ B 32 | α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) , Y 3 α ∪ Y 2 α = X , Y 3 α ∩ Y 2 α = ∅ } B ˜ 21 = { α ∈ B 21 | α = ( Y 2 α × Z 2 ) ∪ ( Y 1 α × Z 1 ) , Y 2 α ∪ Y 1 α = X , Y 2 α ∩ Y 1 α = ∅ }

It is easy to see that

B 0 ∩ B 32 = B 0 ∩ B 21 = B 0 ∩ B 31 = B 21 ∩ B 32 = B 31 ∩ B 32 = B 21 ∩ B 31 = ∅ .

Lemma 2. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) . Then following statements are true for the sets B 0 , B 32 , B ˜ 32 .

a) If α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 1 α , Y 0 α ∉ ∅ , then α is product of some elements of the set B 0 .

b) If β 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 2 ) , then ( B 0 ∘ β 0 ) ∪ B ˜ 32 = B 32 .

c) If σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) , then ( B 0 ∘ σ 1 ) ∪ B ˜ 21 = B 21 .

d) If σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) , then B 32 ∘ σ 1 = B 21 .

e) If σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) , then B 32 ∘ σ 0 = B 31 .

f) Every element of the set B 32 is product of elements of the set B 0 ∪ B ˜ 32 .

g) Every element of the set B 21 is product of elements of the set B 0 ∪ B ˜ 32 ∪ { σ 1 } .

Proof. It will be enough to show only a, b and g. The rest can be similarly seen.

a. Let α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 1 α , Y 0 α ∉ { ∅ } , δ , β ¯ ∈ B 0 . Then quasinormal representation of δ has a form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 1 δ , Y 0 δ ∉ { ∅ } . We suppose that

β ¯ = ( P 2 × Z 3 ) ∪ ( P 1 × Z 2 ) ∪ ( P 3 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 2 Z 3 Z 1 ) is normal mapping for β ¯ and β ¯ 2 is com-

plement mapping of the set X \ D ⌣ on the set D ˜ . So, β ¯ ∈ B 0 since V ( X ∗ , β ¯ ) = D . From the equalities (2.1) and definition of β ¯

Z 3 β ¯ = P 2 β ¯ = Z 3 Z 2 β ¯ = ( P 1 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 3 β ¯ = Z 2 ∪ Z 1 = D ⌣ Z 1 β ¯ = ( P 2 ∪ P 3 ) β ¯ = P 2 β ¯ ∪ P 3 β ¯ = Z 3 ∪ Z 1 = Z 1 D ⌣ β ¯ = ( P 1 ∪ P 2 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 2 β ¯ ∪ P 3 β ¯ = Z 2 ∪ D ⌣ ∪ Z 1 = D ⌣

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × D ⌣ ) ∪ ( Y 2 δ × D ⌣ ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × D ⌣ ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × D ⌣ ) = α .

b. Let α ∈ B 0 ∘ β 0 ∪ B ˜ 32 . Then α ∈ B 0 ∘ β 0 or α ∈ B ˜ 32 . If α ∈ B 0 ∘ β 0 then α = δ ∘ β 0 for some δ ∈ B 0 . In this case we have

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 1 δ , Y 0 δ ∉ { ∅ } . Also

α = δ ∘ β 0 = ( Y 3 δ × Z 3 β 0 ) ∪ ( Y 2 δ × Z 2 β 0 ) ∪ ( Y 1 δ × Z 1 β 0 ) ∪ ( Y 0 δ × D ⌣ β 0 ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × D ⌣ ) ∈ B 32 \ B ˜ 32

is satisfied. So, we have ( B 0 ∘ β 0 ) ∪ B ˜ 32 ⊆ B 32 . On the other hand, if α ∈ B ˜ 32 ⊆ B 32 then ( B 0 ∘ β 0 ) ∪ B ˜ 32 ⊆ B 32 is satisfied. Conversely, if α ∈ B 32 then quasinormal representation of α has a form

α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 0 α × D ⌣ )

where Y 3 α , Y 2 α , Y 0 α ∉ { ∅ } or Y 3 α , Y 2 α ∉ { ∅ } and Y 0 α = ∅ . We suppose that Y 3 α , Y 2 α ∉ { ∅ } . In this case, we have

δ ∘ β 0 = ( Y 3 δ × Z 3 β 0 ) ∪ ( Y 2 δ × Z 2 β 0 ) ∪ ( Y 0 δ × Z 1 β 0 ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ( Y 0 δ × D ⌣ ) = α

for δ = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 0 α × Z 1 ) ∈ B 0 . So, we have B 32 ⊆ ( B 0 ∘ β 0 ) ∪ B ˜ 32 . Now suppose that Y 3 α , Y 2 α ∉ { ∅ } and Y 0 α = ∅ . In this case, we have α ∈ B ˜ 32 ⊆ ( B 0 ∘ β 0 ) ∪ B ˜ 32 . So, ( B 0 ∘ β 0 ) ∪ B ˜ 32 = B 32 .

g. From the statement c, we have that ( B 0 ∘ β 0 ) ∪ B ˜ 32 = B 32 where β 0 ∈ B ˜ 32 by definition of β 0 . Thus, every element of the set B 32 is product of elements of the set B 0 ∪ B ˜ 32 .

Lemma 3. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) . If | X \ D ⌣ | ≥ 1 then the following statements are true.

a) If α = X × D ⌣ then α is product of elements of the set B 0 .

b) If α = X × Z 1 then α is product of elements of the set B 0 .

c) If α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) for some Y 3 α , Y 1 α ∉ ∅ , then α is product of elements of the B 0 .

d) If α = ( Y 3 α × Z 3 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 0 α ∉ ∅ , then α is product of elements of the B 0 .

e) If α = ( Y 2 α × Z 2 ) ∪ ( Y 0 α × D ⌣ ) for some Y 2 α , Y 0 α ∉ ∅ , then α is product of elements of the B 0 .

f) If α = ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 1 α , Y 0 α ∉ ∅ , then α is product of elements of the B 0 .

Proof. c. Let quasinormal representation of α has a form α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) where Y 3 δ , Y 1 δ ∉ { ∅ } . By definition of the semilattice D, | X | ≥ 3 . We suppose that | Y 3 α | ≥ 1 and | Y 1 α | ≥ 2 . In this case, we suppose that

β ¯ = ( P 2 × Z 3 ) ∪ ( ( P 1 ∪ P 3 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 3 Z 1 ) is normal mapping for β ¯ and β ¯ 2 is comple-

ment mapping of the set X × D ⌣ on the set D ˜ \ { Z 3 , Z 1 } = { Z 2 } (by suppose | X \ D ⌣ | ≥ 1 ). So, β ¯ ∈ B 0 since V ( X ∗ , β ¯ ) = D . Also, Y 3 δ = Y 3 α and Y 2 δ ∪ Y 1 δ ∪ Y 0 δ = Y 1 δ since | Y 3 δ | ≥ 1 , | Y 2 δ | ≥ 1 , | Y 1 δ | ≥ 1 , | Y 0 δ | ≥ 0 . From the equalities (2.1) and definition of β ¯ we obtain that

Z 3 β ¯ = P 2 β ¯ = Z 3 Z 2 β ¯ = ( P 1 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 1 = Z 1 Z 1 β ¯ = ( P 2 ∪ P 3 ) β ¯ = P 2 β ¯ ∪ P 3 β ¯ = Z 3 ∪ Z 1 = Z 1 D ⌣ β ¯ = ( P 1 ∪ P 2 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 2 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 3 ∪ Z 1 = Z 1

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × Z 1 ) = ( Y 3 δ × Z 3 ) ∪ ( ( Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 1 ) = α

Now, we suppose that | Y 3 α | ≥ 2 and | Y 1 α | ≥ 1 . In this case, we suppose that

β ¯ = ( ( P 2 ∪ P 3 ) × Z 3 ) ∪ ( P 1 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 3 Z 3 ) is normal mapping for β ¯ and β ¯ 2 is com-

plement mapping of the set X × D ⌣ on the set D ˜ \ { Z 3 , Z 1 } = { Z 2 } (by suppose | X \ D ⌣ | ≥ 1 ). So, β ¯ ∈ B 0 since V ( X ∗ , β ¯ ) = D . Also, Y 3 δ ∪ Y 1 δ = Y 3 α and Y 2 δ ∪ Y 0 δ = Y 1 α since | Y 3 δ | ≥ 1, | Y 2 δ | ≥ 1, | Y 1 δ | ≥ 1, | Y 0 δ | ≥ 0 . From the equalities (2.1) and definition of β ¯ we obtain that

Z 3 β ¯ = P 2 β ¯ = Z 3 Z 2 β ¯ = ( P 1 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 3 = Z 1 Z 1 β ¯ = ( P 2 ∪ P 3 ) β ¯ = P 2 β ¯ ∪ P 3 β ¯ = Z 3 ∪ Z 3 = Z 3 D ⌣ β ¯ = ( P 1 ∪ P 2 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 2 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 3 ∪ Z 3 = Z 1

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × Z 3 ) ∪ ( Y 0 δ × Z 1 ) = ( ( Y 3 δ ∪ Y 1 δ ) × Z 3 ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × Z 1 ) = α

Lemma 4. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) , σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) and σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . If X = D ⌣ then the following statements are true

a) If α = ( Y 3 α × Z 3 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 0 α ∉ { ∅ } , then α is product of elements of the B 0 ∪ B 32 .

b) If α = ( Y 2 α × Z 2 ) ∪ ( Y 0 α × D ⌣ ) for some Y 2 α , Y 0 α ∉ { ∅ } , then α is product of elements of the B 32 ∪ { σ 1 } .

c) If α = ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 1 α , Y 0 α ∉ { ∅ } , then α is product of elements of the B 32 ∪ { σ 0 , σ 1 } .

Proof. First, remark that Z 3 σ 0 = Z 3 , Z 2 σ 0 = D ⌣ σ 0 = Z 1 , Z 3 σ 1 = Z 1 , Z 2 σ 1 = Z 2 , D ⌣ σ 1 = D ⌣ .

a. Let α = ( Y 3 α × Z 3 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 0 α ∉ ∅ . In this case, we suppose that

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 0 δ × D ⌣ )

and

β 1 = ( Z 3 × Z 3 ) ∪ ( ( Z 1 \ Z 3 ) × Z 1 ) ∪ ( ( X \ Z 1 ) × D ⌣ )

where Y 3 δ , Y 2 δ ∉ { ∅ } . It is easy to see that δ ∈ B 32 and β 1 is generating by elements of the B 0 by statement b of Lemma 2. Also, Y 3 δ = Y 3 α and Y 2 δ ∪ Y 0 δ = Y 0 α since Z 3 β ¯ = Z 3 , Z 2 β ¯ = D ⌣ β ¯ = D ⌣ and | Y 3 δ | ≥ 1, | Y 2 δ | ≥ 1, | Y 0 δ | ≥ 0 . So, α is product of elements of the B 0 ∪ B 32 . □

Lemma 5. Let

D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 )

and

σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) .

If | X \ D ⌣ | ≥ 1 then S 1 = B 0 ∪ B ˜ 32 ∪ { σ 1 } is an irreducible generating set for the semigroup B X ( D ) .

Proof. First, we must prove that every element of B X ( D ) is product of elements of S 1 . Let α ∈ B X ( D ) and

α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ )

where Y 3 α ∪ Y 2 α ∪ Y 1 α ∪ Y 0 α = X and Y 3 α ∩ Y 2 α = ∅ , ( 0 ≤ i ≠ j ≤ 3 ) . We suppose

that | V ( X ∗ , α ) | = 1 . Then we have V ( X ∗ , α ) ∈ { { Z 3 } , { Z 2 } , { Z 1 } , { D ⌣ } } . If

V ( X ∗ , α ) ∈ { { Z 3 } , { Z 2 } , { Z 1 } } then α = X × Z 3 or α = X × Z 2 or α = X × Z 1 . Quasinormal representations of δ , β 1 , β 2 and β 3 has form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ ) β 1 = ( D ⌣ × Z 3 ) ∪ ( ( X \ D ⌣ ) × Z 2 ) β 2 = ( D ⌣ × Z 2 ) ∪ ( ( X \ D ⌣ ) × Z 1 ) β 3 = ( D ⌣ × Z 1 ) ∪ ( ( X \ D ⌣ ) × Z 2 )

where Y 3 δ , Y 2 δ , Y 1 δ ∉ { ∅ } . So, δ ∈ B 0 , β 1 ∈ B ˜ 32 and β 2 , β 3 ∈ B 21 since | X \ D ⌣ | ≥ 1 . From the definition of δ , β 1 , β 2 and β 3 we obtain that

δ ∘ β 1 = ( Y 3 δ × Z 3 β 1 ) ∪ ( Y 2 δ × Z 2 β 1 ) ∪ ( Y 1 δ × Z 1 β 1 ) ∪ ( Y 0 δ × D ⌣ β 1 ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 3 ) ∪ ( Y 1 δ × Z 3 ) ∪ ( Y 0 δ × Z 3 ) = ( Y 3 δ ∪ Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 3 = X × Z 3

δ ∘ β 2 = ( Y 3 δ × Z 3 β 2 ) ∪ ( Y 2 δ × Z 2 β 2 ) ∪ ( Y 1 δ × Z 1 β 2 ) ∪ ( Y 0 δ × D ⌣ β 2 ) = ( Y 3 δ × Z 2 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 2 ) ∪ ( Y 0 δ × Z 2 ) = ( Y 3 δ ∪ Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 2 = X × Z 2

δ ∘ β 3 = ( Y 3 δ × Z 3 β 3 ) ∪ ( Y 2 δ × Z 2 β 3 ) ∪ ( Y 1 δ × Z 1 β 3 ) ∪ ( Y 0 δ × D ⌣ β 3 ) = ( Y 3 δ × Z 1 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × Z 1 ) = ( Y 3 δ ∪ Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 1 = X × Z 1

That means, X × Z 1 , X × Z 2 and X × Z 3 are generated by B 0 ∪ B ˜ 32 , B 0 ∪ B 21 and B 0 ∪ B 21 respectively. By using statement g and h of Lemma 3, we have X × Z 1 , X × Z 2 and X × Z 3 are generated by B 0 ∪ B ˜ 32 ∪ { σ 1 } . On the other hand, if V ( X ∗ , α ) = { D ⌣ } then α = X × D ⌣ By using statement a of Lemma 3, we have α is product of some elemets of B 0 .

So, S 1 is generating set for the semigroup B X ( D ) . Now, we must prove that S 1 = B 0 ∪ B ˜ 32 ∪ { σ 1 } is irreducible. Let α ∈ S 1 .

If α ∈ B 0 then α ≠ σ ∘ τ for all σ , τ ∈ B X ( D ) \ { α } from Lemma 2. So, α ≠ σ ∘ τ for all σ , τ ∈ S 1 \ { α } . That means, α ∉ B 0 .

If α ∈ B ˜ 32 then the quasinormal representation of α has form α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) for some Y 3 α , Y 2 α ∉ ∅ . Let α = δ ∘ β for some δ , β ∈ S 1 \ { α } .

We suppose that δ ∈ B 0 \ { α } and β ∈ S 1 \ { α } . By definition of B 0 , quasinormal representation of δ has form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 2 δ , Y 1 δ ∉ { ∅ } . By using Z 3 ⊂ Z 1 ⊂ D ⌣ and Z 2 ⊂ D ⌣ we have Z 3 β and Z 2 β are minimal elements of the semilattice { Z 3 β , Z 2 β , Z 1 β , D ⌣ β } . Also, we have

( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) = α = δ ∘ β = ( Y 3 δ × Z 3 β ) ∪ ( Y 2 δ × Z 2 β ) ∪ ( Y 1 δ × Z 1 β ) ∪ ( Y 0 δ × D ⌣ β )

Since Z 3 and Z 2 are minimal elements of the semilattice { Z 3 , Z 2 , D ⌣ } , this equality is possible only if Z 3 = Z 3 β , Z 2 = Z 2 β or Z 3 = Z 2 β , Z 2 = Z 3 β . By using formal equalities and P 3 β , P 2 β , P 1 β ∈ D , we obtain

Z 3 = Z 3 β = P 2 β and Z 2 = Z 2 β = P 1 β = P 3 β Z 2 = Z 3 β = P 2 β and Z 3 = Z 2 β = P 1 β = P 3 β

respectively. Let Z 3 = P 2 β and Z 2 = P 1 β = P 3 β . If β ¯ is sub-quasinormal representation of β then δ ∘ β = δ ∘ β ¯ and

β ¯ = ( ( P 1 ∪ P 3 ) × Z 2 ) ∪ ( P 2 × Z 3 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 2 Z 3 Z 2 ) is normal mapping for β ¯ and β ¯ 2 is com-

plement mapping of the set X × D ⌣ on the set D ˜ = { Z 3 , Z 2 , Z 1 } . From formal equalities, we obtain

β ¯ = ( Z 2 × Z 2 ) ∪ ( Z 3 × Z 3 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) ) ∈ S 1 \ { α }

and by using Z 1 ∩ Z 2 ≠ ∅ , Z 3 ∪ Z 2 = D and | Y 1 δ ∪ Y 0 δ | ≥ 1 , we have

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × D ⌣ ) ≠ α

This contradicts with α = δ ∘ β . So, δ ∉ B 0 \ { α } .

Now, we suppose that δ ∈ B ˜ 32 \ { α } and β ∈ S 1 \ { α } . Similar operations are applied as above, we obtain δ ∉ B ˜ 32 \ { α } .

Now, we suppose that δ = σ 1 and β ∈ S 1 \ { α } . Similar operations are applied as above, we obtain δ ≠ σ 1 .

That means α ≠ δ ∘ β for any α ∈ B ˜ 32 and δ , β ∈ S 1 \ { α } .

If α = σ 1 , then by the definition of σ 1 , quasinormal representation of α has a form α = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . Let α = δ ∘ β for some δ , β ∈ S 1 \ { σ 1 } .

We suppose that δ ∈ B 0 \ { σ 1 } and β ∈ S 1 \ { σ 1 } . By definition of B 0 , quasinormal representation of δ has form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 2 δ , Y 1 δ ∉ { ∅ } . By using Z 3 ⊂ Z 1 ⊂ D ⌣ and Z 2 ⊂ D ⌣ we have Z 3 β and Z 2 β are minimal elements of the semilattice { Z 3 β , Z 2 β , Z 1 β , D ⌣ β } . Also, we have

( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) = α = δ ∘ β = ( Y 3 δ × Z 3 β ) ∪ ( Y 2 δ × Z 2 β ) ∪ ( Y 1 δ × Z 1 β ) ∪ ( Y 0 δ × D ⌣ β )

From Z 2 and Z 1 are minimal elements of the semilattice { Z 2 , Z 1 , D ⌣ } , this equality is possible only if Z 2 = Z 3 β , Z 1 = Z 2 β or Z 2 = Z 2 β , Z 1 = Z 3 β . By using formal equalities, we obtain

Z 2 = Z 3 β = P 2 β and Z 1 = Z 2 β = P 1 β ∪ P 3 β Z 1 = Z 3 β = P 2 β and Z 2 = Z 2 β = P 1 β = P 3 β

respectively. Let Z 2 = P 2 β and Z 1 = P 1 β ∪ P 3 β where P 1 β , P 3 β ∈ { Z 3 , Z 1 } . Then subquasinormal representation of β has one of the form

β ¯ 1 = ( P 1 × Z 3 ) ∪ ( P 2 × Z 2 ) ∪ ( P 3 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 2 = ( P 3 × Z 3 ) ∪ ( P 2 × Z 2 ) ∪ ( P 1 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 3 = ( P 2 × Z 2 ) ∪ ( ( P 1 ∪ P 3 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where

β ¯ 1 1 = ( ∅ P 1 P 2 P 3 ∅ Z 3 Z 2 Z 1 ) , β ¯ 1 2 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 2 Z 3 ) , β ¯ 1 3 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 2 Z 1 )

are normal mapping for β ¯ , β ¯ 2 is complement mapping of the set X × D ⌣ on the set D ˜ = { Z 3 , Z 2 , Z 1 } and δ ∘ β = δ ∘ β ¯ i . From formal equalities, we obtain

β ¯ 1 = ( ( Z 2 \ Z 1 ) × Z 3 ) ∪ ( ( Z 1 \ Z 2 ) × Z 2 ) ∪ ( ( Z 2 \ Z 1 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 2 = ( ( Z 2 ∩ Z 1 ) × Z 3 ) ∪ ( ( Z 1 \ Z 2 ) × Z 2 ) ∪ ( ( Z 2 \ Z 1 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 3 = ( ( Z 1 \ Z 2 ) × Z 2 ) ∪ ( Z 2 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

and by using | Y 1 δ ∪ Y 0 δ | ≥ 1 , we have

δ ∘ β ¯ 1 = δ ∘ β ¯ 2 = δ ∘ β ¯ 3 = ( Y 3 δ × Z 3 β ¯ 1 ) ∪ ( Y 2 δ × Z 2 β ¯ 1 ) ∪ ( Y 1 δ × Z 1 β ¯ 1 ) ∪ ( Y 0 δ × D ⌣ β ¯ 1 ) = ( Y 3 δ × Z 2 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × Z 2 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × D ⌣ ) ≠ α

This contradicts with α = δ ∘ β . So, δ ∉ B 0 \ { σ 1 } .

Now, we suppose that δ ∈ B ˜ 32 \ { σ 1 } and β ∈ S 1 \ { σ 1 } . Similar operations are applied as above, we obtain δ ∉ B ˜ 32 \ { σ 1 } .

That means α ≠ δ ∘ β for any α ∈ B ˜ 32 and δ , β ∈ S 1 \ { α } . □

Lemma 6. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) , σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) and σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . If X = D ⌣ then S 2 = B 0 ∪ B ˜ 32 ∪ { σ 0 , σ 1 } is irreducible generating set for the semigroup B X ( D ) .

Theorem 7. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) , σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) and σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . If X is a finite set and | X | = n then the following statements are true

a) If | X \ D ⌣ | ≥ 1 then | B 0 ∪ B ˜ 32 ∪ { σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 2

b) If X = D ⌣ then | B 0 ∪ B ˜ 32 ∪ { σ 0 , σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 1

Proof. Let

S n = { φ i | φ i : M = { 1,2, ⋯ , n } → M = { 1,2, ⋯ , n } , onetoonemapping }

be a group, φ i 1 , φ i 2 , ⋯ , φ i m ∈ S n ( m ≤ n ) and Y φ 1 , Y φ 2 , ⋯ , Y φ m be partitioning of

X. It is well known that k n m = | { Y φ 1 , Y φ 2 , ⋯ , Y φ m } | = ∑ i = 1 m ( − 1 ) m + i ( i − 1 ) ! ( m − i ) ! . If m = 2 , 3 , 4

then we have

k n 2 = 2 n − 1 − 1 k n 3 = 1 2 ⋅ 3 n − 1 − 2 n − 1 + 1 2 k n 4 = 1 6 ⋅ 4 n − 1 − 1 2 ⋅ 3 n − 1 + 1 2 ⋅ 2 n − 1 − 1 6

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

2 ⋅ k n 2 = 2 n − 2 (2)

If Y φ 1 , Y φ 2 , Y φ 3 are any three elements of partitioning of X and β ¯ = ( Y φ 1 × T 1 ) ∪ ( Y φ 2 × T 2 ) ∪ ( Y φ 3 × T 3 ) where T 1 , T 2 , T 3 are pairwise different elements of D, then the number of different binary relations β ¯ of semigroup B X ( D ) is equal to

6 ⋅ k n 3 = 3 n − 3 ⋅ 2 n + 3 (3)

If Y φ 1 , Y φ 2 , Y φ 3 , Y φ 4 are any four elements of partitioning of X and β ¯ = ( Y φ 1 × T 1 ) ∪ ( Y φ 2 × T 2 ) ∪ ( Y φ 3 × T 3 ) ∪ ( Y φ 4 × T 4 ) where T 1 , T 2 , T 3 , T 4 are pairwise different elements of D, then the number of different binary relations β ¯ of semigroup B X ( D ) is equal to

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

Let α ∈ B 0 . Quasinormal represantation of α has form

α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ )

where Y 3 α , Y 2 α , Y 1 α ∉ { ∅ } . Also, Y 3 α , Y 2 α , Y 1 α or Y 3 α , Y 2 α , Y 1 α , Y 0 α are partitioning of X for | X | ≥ 4 . By using Equations (2.3) and (2.4) we obtain

| B 0 | = 4 n − 3 n + 1 + 3 ⋅ 2 n − 1

Let α ∈ B ˜ 32 . Quasinormal represantation of α has form α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) where Y 3 α , Y 2 α ∉ { ∅ } . Also, Y 3 α , Y 2 α are partitioning of X. By using (2.2) we obtain

| B ˜ 32 | = 2 n − 2

So, we have

| B 0 ∪ B ˜ 32 ∪ { σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 2 | B 0 ∪ B ˜ 32 ∪ { σ 0 , σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 1

since B 0 ∩ B ˜ 32 = B 0 ∩ { σ 0 , σ 1 } = B ˜ 32 ∩ { σ 0 , σ 1 } = ∅ . □

Sincere thanks to Prof. Dr. Neşet AYDIN for his valuable suggestions.

Albayrak, B., Givradze, O. and Partenadze,^{ }G. (2018) Generating Sets of the Complete Semigroups of Binary Relations Defined by Semilattices of the Class . Applied Mathematics, 9, 17-27. https://doi.org/10.4236/am.2018.91002