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Theorem 44

For every small sets

A

,

B

,

C

if

g, h

RLD

(

A

;

B

)

then

1.

f

(

g

h

) =

f

g

f

h

for every

f

RLD

(

B

;

C

)

;

2.

(

g

h

)

f

=

g

f

h

f

for every

f

RLD

(

C

;

A

)

.

Proof

We’ll prove only the first as the second is dual.

By the infinite distributivity law for filters we have

f

g

f

h

=

RLD

(

A

;

C

)

(

F

G

)

|

F

up

f, G

up

g

 

RLD

(

A

;

C

)

(

F

H

)

|

F

up

f, H

up

h

 

=

\ n

RLD

(

A

;

C

)

(

F

1

G

)

∪ ↑

RLD

(

A

;

C

)

(

F

2

H

)

|

F

1

, F

2

up

f, G

up

g, H

up

h

o

=

\ n

RLD

(

A

;

C

)

((

F

1

G

)

(

F

2

H

))

|

F

1

, F

2

up

f, G

up

g, H

up

h

o

.

Obviously

\ n

RLD

(

A

;

C

)

((

F

1

G

)

(

F

2

H

))

|

F

1

, F

2

up

f, G

up

g, H

up

h

o

\ n

RLD

(

A

;

C

)

(((

F

1

F

2

)

G

)

((

F

1

F

2

)

H

))

|

F

1

, F

2

up

f, G

up

g, H

up

h

o

=

\ n

RLD

(

A

;

C

)

((

F

G

)

(

F

H

))

|

F

up

f, G

up

g, H

up

h

o

=

\ n

RLD

(

A

;

C

)

(

F

(

G

H

))

|

F

up

f, G

up

g, H

up

h

o

.

Because

G

up

g

H

up

h

G

H

up(

g

h

) we have

\ n

RLD

(

A

;

C

)

(

F

(

G

H

))

|

F

up

f, G

up

g, H

up

h

o

\ n

RLD

(

A

;

C

)

(

F

K

)

|

F

up

f, K

up(

g

h

)

o

=

f

(

g

h

)

.

Thus we proved

f

g

f

h

f

(

g

h

). But obviously

f

(

g

h

)

f

g

and

f

(

g

h

)

f

h

and so

f

(

g

h

)

f

g

f

h

. Thus

f

(

g

h

) =

f

g

f

h

.

Conjecture 7

If

f

and

g

are reloids, then

g

f

=

[

{

G

F

|

F

atoms

f, G

atoms

g

}

.

Theorem 45

Let

A

,

B

,

C

be sets,

f

RLD

(

A

;

B

)

,

g

RLD

(

B

;

C

)

,

h

RLD

(

A

;

C

)

. Then

g

f

6≍

h

g

6≍

h

f

1

.

Proof

g

f

6≍

h

RLD

(

A

;

C

)

(

G

F

)

|

F

up

f, G

up

g

 

RLD

(

A

;

C

)

up

h

6

=

0

RLD

(

A

;

C

)

RLD

(

A

;

C

)

((

G

F

)

H

)

|

F

up

f, G

up

g, H

up

h

 

6

=

0

RLD

(

A

;

C

)

⇔ ∀

F

up

f, G

up

g, H

up

h

:

RLD

(

A

;

C

)

((

G

F

)

H

)

6

=

41