background image

Proof

Our filtrator is with join-closed core.

a

=

S

A

c

A

|

c

A

a

= 0

 

.

But

c

A

a

= 0

⇒ ∃

C

up

c

:

C

A

a

= 0. So

a

=

[

A

C

Z

|

C

A

a

= 0

 

=

[

A

C

Z

|

a

C

 

=

[

A

C

|

C

Z

, a

C

 

=

[

A

C

|

C

up

a

 

=

[

Z

C

|

C

up

a

 

=

\

Z

{

C

|

C

up

a

}

=

\

Z

up

a

=

Cor

a.

(used the theorem 27).

Cor

a

= Cor

a

by the theorem 26.

Corollary 19

If

(

A

;

Z

)

is filtered down-aligned and up-aligned complete lattice

filtrator with finitely meet-closed, separable and co-separable core which is a
complete boolean lattice, then

a

=

a

+

for every

a

A

.

Proof

Comparing two last theorems.

Theorem 62

If

(

A

;

Z

)

is a complete lattice filtrator with join-closed separable

core which is a complete lattice, then

a

Z

for every

a

A

.

Proof

c

A

|

c

A

a

= 0

 

A

Z

|

A

A

a

= 0

 

; consequently

a

S

A

A

Z

|

A

A

a

= 0

 

.

But if

c

c

A

|

c

A

a

= 0

 

then exists

A

Z

such that

A

c

and

A

A

a

= 0 that is

A

A

Z

|

A

A

a

= 0

 

. Consequently

a

S

A

A

Z

|

A

A

a

= 0

 

.

We have

a

=

S

A

A

Z

|

A

A

a

= 0

 

=

S

Z

A

Z

|

A

A

a

= 0

 

Z

.

Theorem 63

If

(

A

;

Z

)

is an up-aligned filtered complete lattice filtrator co-

separable core which is a complete boolean lattice, then

a

+

is dual pseudocom-

plement of

a

, that is

a

+

= min

c

A

|

c

A

a

= 1

 

for every

a

A

.

Proof

Our filtrator is with join-closed core. It’s enough to prove that

a

+

A

a

=

1. But

a

+

A

a

= Cor

a

A

a

Cor

a

A

Cor

a

= Cor

a

Z

Cor

a

= 1 (used the

theorem 24 and the fact that our filtrator is filtered).

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