background image

2.

a

#

b

= (

a

b

)

(

Da

)

.

Proof

1.

(

a

b

)

+(

Da

)

=

\

{

c

Da

|

c

(

a

b

) =

a

}

=

\

{

c

Da

|

c

(

a

b

)

a

}

=

\

{

c

Da

|

(

c

a

)

(

c

b

)

a

}

=

\

{

c

A

|

c

a

c

b

a

}

=

a

\

b.

2.

(

a

b

)

(

Da

)

=

[

{

c

Da

|

c

a

b

= 0

}

=

[

{

c

A

|

c

a

c

a

b

= 0

}

=

[

{

c

A

|

c

a

c

b

= 0

}

=

a

#

b.

Theorem 59

Let

(

F

;

A

)

be a primary filtrator where

A

is a boolean lattice. Let

A ∈

F

. Then for each

X ∈

F

X ∈

Z

(

D

A

)

⇔ ∃

X

A

:

X

=

X

F

A

.

Proof

Let

X

=

X

F

A

where

X

A

. Let also

Y

=

X

F

A

.

Then

X ∩

F

Y

=

X

F

X

F

A

= (

X

A

X

)

F

A

= 0 (used the theorem 29 and

X ∪

F

Y

=

(

X

F

X

)

F

A

= (

X

A

X

)

F

A

= 1

F

A

=

A

(used the theorems 23

and corollary 10). So

X ∈

Z

(

D

A

).

Let

X ∈

Z

(

D

A

). Then exists

Y ∈

Z

(

D

A

) such that

X ∩

F

Y

= 0 and

X ∪

F

Y

=

A

. Then (used the theorem 37) exists

X

up

X

such that

X

F

Y

= 0. We have

X

=

X ∪

F

(

X

F

Y

) =

X

F

(

X ∪

F

Y

) =

X

F

A

.

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