4.1

Composition of reloids

Definition 44

Reloids

f

and

g

are

composable

when

Dst

f

= Src

g

.

Definition 45

Composition of (composable) reloids is defined by the formula

g

f

=

\ n

RLD

(Src

f

;Dst

g

)

(

G

F

)

|

F

up

f, G

up

g

o

.

Composition of reloids is a reloid.

Theorem 42

(

h

g

)

f

=

h

(

g

f

)

for every composable reloids

f

,

g

,

h

.

Proof

For two nonempty collections

A

and

B

of sets I will denote

A

B

(

K

A

L

B

:

L

K

)

(

K

B

L

A

:

L

K

)

.

It is easy to see that

is a transitive relation.

I will denote

B

A

=

{

L

K

|

K

A, L

B

}

.

Let first prove that for every nonempty collections of relations

A

,

B

,

C

A

B

A

C

B

C.

Suppose

A

B

and

P

A

C

that is

K

A

and

M

C

such that

P

=

K

M

.

K

B

:

K

K

because

A

B

. We have

P

=

K

M

B

C

. Obviously

P

P

. So for every

P

A

C

exist

P

B

C

such that

P

P

; the vice

versa is analogous. So

A

C

B

C

.

up((

h

g

)

f

)

up(

h

g

)

up

f

, up(

h

g

)

(up

h

)

(up

g

). By proven

above up((

h

g

)

f

)

(up

h

)

(up

g

)

(up

f

).

Analogously up(

h

(

g

f

))

(up

h

)

(up

g

)

(up

f

).

So up((

h

g

)

f

)

up(

h

(

g

f

)) what is possible only if up((

h

g

)

f

) =

up(

h

(

g

f

)).

Theorem 43

For every reloid

f

:

1.

f

f

=

RLD

(Src

f

;Dst

f

)

(

F

F

)

|

F

up

f

if

Src

f

= Dst

f

;

2.

f

1

f

=

RLD

(Src

f

;Src

f

)

F

1

F

|

F

up

f

;

3.

f

f

1

=

RLD

(Dst

f

;Dst

f

)

F

F

1

|

F

up

f

.

Proof

I will prove only (1) and (2) because (3) is analogous to (2).

1. It’s enough to show that

F, G

up

f

H

up

f

:

H

H

G

F

. To prove

it take

H

=

F

G

.

2. It’s enough to show that

F, G

up

f

H

up

f

:

H

1

H

G

1

F

. To

prove it take

H

=

F

G

. Then

H

1

H

= (

F

G

)

1

(

F

G

)

G

1

F

.

40