background image

Proof

X

[

g

f

]

Y ⇔ Y ∩ h

g

f

i X 6

= 0

F

(Dst

g

)

⇔ Y ∩ h

g

i h

f

i X 6

= 0

F

(Dst

g

)

h

f

i X

[

g

]

Y ⇔ X

([

g

]

◦ h

f

i

)

Y

for every

X ∈

F

(Src

f

),

Y ∈

F

(Dst

g

). [

g

f

]=

(

f

1

g

1

)

1

=

f

1

g

1

1

=

(

f

1

g

1

)

1

=

g

1

1

[

f

].

The following theorem is a variant for funcoids of the statement (which

defines compositions of relations) that

x

(

g

f

)

z

⇔ ∃

y

(

x f y

y g z

) for every

x

and

z

and every binary relations

f

and

g

.

Theorem 12

For every small sets

A

,

B

,

C

and

f

FCD

(

A

;

B

)

,

g

FCD

(

B

;

C

)

and

X ∈

F

(

A

)

,

Z ∈

F

(

C

)

X

[

g

f

]

Z ⇔ ∃

y

atoms 1

F

(

B

)

: (

X

[

f

]

y

y

[

g

]

Z

)

.

Proof

y

atoms 1

F

(

B

)

: (

X

[

f

]

y

y

[

g

]

Z

)

⇔ ∃

y

atoms 1

F

(

B

)

:

Z ∩ h

g

i

y

6

= 0

F

(

C

)

y

∩ h

f

i X 6

= 0

F

(

B

)

⇔ ∃

y

atoms 1

F

(

B

)

: (

Z ∩ h

g

i

y

6

= 0

F

(

C

)

y

⊆ h

f

i X

)

⇒ Z ∩ h

g

i h

f

i X 6

= 0

F

(

C

)

⇔ X

[

g

f

]

Z

.

Reversely, if

X

[

g

f

]

Z

then

h

f

i X

[

g

]

Z

, consequently exists

y

atoms

h

f

i X

such that

y

[

g

]

Z

; we have

X

[

f

]

y

.

Theorem 13

For every small sets

A

,

B

,

C

1.

f

(

g

h

) =

f

g

f

h

for

g, h

FCD

(

A

;

B

)

and

f

FCD

(

B

;

C

)

;

2.

(

g

h

)

f

=

g

f

h

f

for

g, h

FCD

(

B

;

C

)

and

f

FCD

(

A

;

B

)

.

Proof

I will prove only the first equality because the other is analogous.

For every

X ∈

F

(

A

),

Z ∈

F

(

C

)

X

[

f

(

g

h

)]

Z

⇔ ∃

y

atoms 1

F

(

B

)

: (

X

[

g

h

]

y

y

[

f

]

Z

)

⇔ ∃

y

atoms 1

F

(

B

)

: ((

X

[

g

]

y

∨ X

[

h

]

y

)

y

[

f

]

Z

)

⇔ ∃

y

atoms 1

F

(

B

)

: (

X

[

g

]

y

y

[

f

]

Z ∨ X

[

h

]

y

y

[

f

]

Z

)

⇔ ∃

y

atoms 1

F

(

B

)

: (

X

[

g

]

y

y

[

f

]

Z

)

∨ ∃

y

atoms 1

F

(

B

)

: (

X

[

h

]

y

y

[

f

]

Z

)

⇔ X

[

f

g

]

Z ∨ X

[

f

h

]

Z

⇔ X

[

f

g

f

h

]

Z

.

Conjecture 2

g

f

=

FCD

(Src

f

;Dst

g

)

(

G

F

)

|

F

up

f, G

up

g

 

for

every composable funcoids

f

and

g

.

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