Lemma 1

h

f

i

X

=

Dst

f

h

F

i

X

|

F

up

f

for every funcoid

f

and

set

X

P

(Src

f

)

.

Proof

Obviously

h

f

i

X

Dst

f

h

F

i

X

|

F

up

f

.

Let

B

up

h

f

i

X

. Let

F

B

=

X

×

B

((Src

f

)

\

X

)

×

(Dst

f

).

h

F

B

i

X

=

B

.

We have

∅ 6

=

P

X

⇒ h

F

B

i

P

=

B

⊇ h

f

i

P

and

∅ 6

=

P

*

X

⇒ h

F

B

i

P

=

Dst

f

⊇ h

f

i

P

. Thus

h

F

B

i

P

⊇ h

f

i

P

for every set

P

P

(Src

f

) and so

FCD

(Src

f

;Dst

f

)

F

B

f

that is

F

B

up

f

.

Thus

B

up

h

f

i

X

:

B

up

Dst

f

h

F

i

X

|

F

up

f

because

B

up

Dst

f

h

F

B

i

X

.

So

Dst

f

h

F

i

X

|

F

up

f

⊆ h

f

i

X

.

Theorem 9

h

f

i X

=

FCD

(Src

f

;Dst

f

)

F

X

|

F

up

f

for every fun-

coid

f

and

X ∈

F

(Src

f

)

.

Proof

FCD

(Src

f

;Dst

f

)

F

X

|

F

up

f

=

T T

Dst

f

hh

F

ii

up

X

|

F

up

f

=

T T

Dst

f

h

F

i

X

|

X

up

X

|

F

up

f

=

T T

Dst

f

h

F

i

X

|

F

up

f

|

X

up

X

=

Dst

f

h

f

i

X

|

X

up

X

=

h

f

i X

(the lemma used).

Conjecture 1

Every filtrator of funcoids is:

1. with separable core;

2. with co-separable core.

Below it is shown that

FCD

(

A

;

B

) are complete lattices for every small sets

A

and

B

. We will apply lattice operations to subsets of such sets without

explicitly mentioning

FCD

(

A

;

B

).

Theorem 10

FCD

(

A

;

B

)

is a complete lattice (for every small sets

A

and

B

).

For every

R

P

FCD

(

A

;

B

)

and

X

P

A

,

Y

P

B

1.

X

[

S

R

]

Y

⇔ ∃

f

R

:

X

[

f

]

Y

;

2.

h

S

R

i

X

=

h

f

i

X

|

f

R

.

Proof

Accordingly [14] to prove that it is a complete lattice it’s enough to

prove existence of all joins.

2

αX

def

=

h

f

i

X

|

f

R

. We have

α

= 0

F

(Dst

f

)

;

α

(

I

J

) =

h

f

i

(

I

J

)

|

f

R

=

h

f

i

I

∪ h

f

i

J

|

f

R

=

h

f

i

I

|

f

R

h

f

i

J

|

f

R

=

αI

αJ.

16