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A generalization and the proof

We will prove more general statements:

Theorem 1

For an atomistic co-brouwerian lattice

A

and

a, b

A

the following

expressions are always equal:

1.

T

A

z

A

|

a

b

A

z

 

(quasidifference of

a

and

b

);

2.

S

A

z

A

|

z

a

z

A

b

= 0

 

(second quasidifference of

a

and

b

);

3.

S

A

(atoms

A

a

\

atoms

A

b

)

.

Proof

Proof of (1)

=

(3):

a

\

b

=

T

A

z

A

|

a

b

A

z

 

, so it’s enough to prove that

a

\

b

=

A

[

(atoms

A

a

\

atoms

A

b

)

.

Really:

a

\

b

=

 

A

[

atoms

a

!

\

b

=

(theorem 16 in [1])*

A

A

\

b

|

A

atoms

A

a

 

=

A

[

A

if

A

6∈

atoms

A

b

0

if

A

atoms

A

b

|

A

atoms

A

a

=

A

A

|

A

atoms

A

a, A

6∈

atoms

A

b

 

=

A

[

(atoms

A

a

\

atoms

A

b

)

.

* The requirement of theorem 16 that our lattice is complete is superfluous and
can be removed.

Proof of (2)

=

(3):

a

\

b

is defined because our lattice is co-brouwerian. Taking the above into

account, we have

a

\

b

=

[

(atoms

a

\

atoms

b

)

=

z

atoms

a

|

z

A

b

= 0

A

 

.

So

z

atoms

a

|

z

A

b

= 0

A

 

is defined.

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