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8.2. COMPOSITION OF RELOIDS

185

Theorem

1005

.

For every sets

A

,

B

,

C

if

g, h

RLD

(

A, B

) then

1

.

f

(

g

t

h

) =

f

g

t

f

h

for every

f

RLD

(

B, C

);

2

. (

g

t

h

)

f

=

g

f

t

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

t

f

h

=

RLD

l

F

G

F

up

f, G

up

g

t

RLD

l

F

H

F

up

f, H

up

h

=

RLD

l

(

F

1

G

)

t

RLD

(

F

2

H

)

F

1

, F

2

up

f, G

up

g, H

up

h

=

RLD

l

(

F

1

G

)

t

(

F

2

H

)

F

1

, F

2

up

f, G

up

g, H

up

h

.

Obviously

RLD

l

(

F

1

G

)

t

(

F

2

H

)

F

1

, F

2

up

f, G

up

g, H

up

h

w

RLD

l

(((

F

1

u

F

2

)

G

)

t

((

F

1

u

F

2

)

H

))

F

1

, F

2

up

f, G

up

g, H

up

h

=

RLD

l

(

F

G

)

t

(

F

H

)

F

up

f, G

up

g, H

up

h

=

RLD

l

F

(

G

t

H

)

F

up

f, G

up

g, H

up

h

.

Because

G

up

g

H

up

h

G

t

H

up(

g

t

h

) we have

RLD

l

F

(

G

t

H

)

F

up

f, G

up

g, H

up

h

w

RLD

l

F

K

F

up

f, K

up(

g

t

h

)

=

f

(

g

t

h

)

.

Thus we have proved

f

g

t

f

h

w

f

(

g

t

h

). But obviously

f

(

g

t

h

)

w

f

g

and

f

(

g

t

h

)

w

f

h

and so

f

(

g

t

h

)

w

f

g

t

f

h

. Thus

f

(

g

t

h

) =

f

g

t

f

h

.

Theorem

1006

.

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

.