W

=

fh

R

i

X

j

R

2

up

r

g

is?? a generalized lter base?? (No, it isn't, up

r

isn't a lter.)

Let

P

0

; P

1

2

W

. Then

P

0

=

h

R

0

i

X

,

P

1

=

h

R

1

i

X

Let

F

2

up

(

f

r

)

. Then ??

F

2

up

T

f

f

"

FCD

R

j

R

2

up

r

g

.

So

f

r

T

f

f

"

FCD

R

j

R

2

up

r

g

X

[

f

"

FCD

R

]

Y ,

[

"

FCD

R

]

/

h

f

¡

1

iY

??
Let

W

=

fh"

FCD

R

i

a

j

R

2

up

r

g

.

W

is a generalized lter base??. Really If

X

;

Y 2

W

. Then

X

=

h"

FCD

X

i

a

and

Y

=

h"

FCD

Y

i

a

for some

X

2

up

r

,

Y

2

up

r

.

\

f

f

"

FCD

R

j

R

2

up

r

g

a

=

\

fh

f

"

FCD

R

i

a

j

R

2

up

r

g

=

(because

fh"

FCD

R

i

a

j

R

2

up

r

g

is a g.f.b.??)

h

f

i

\

fh"

FCD

R

i

a

j

R

2

up

r

g

=

h

f

ih

r

i

a

Counter-example attempt: Let

r

=

id

FCD

. ??

Corollary 34.

g

f

=

"

FCD

(

Src

f

;

Dst

g

)

(

G

F

)

j

F

2

up

f ; G

2

up

g

.

Proof.

x

"

FCD

(

Src

f

;

Dst

g

)

(

G

F

)

z

, 9

y

2

atoms Dst

f

: (

x

[

"

F

]

y

^

y

[

"

G

]

z

)

x

T

"

FCD

(

Src

f

;

Dst

g

)

(

G

F

)

j

F

2

up

f ; G

2

up

g

z

, 8

F

2

up

f ; G

2

up

g

:

x

"

FCD

(

Src

f

;

Dst

g

)

(

G

F

)

z

, 8

F

2

up

f ; G

2

up

g

9

y

2

atoms Dst

f

:

¡

x

"

FCD

(

Src

f

;

Dst

g

)

F

y

^

y

"

FCD

(

Src

f

;

Dst

g

)

G

z

Conjecture 35.

Let

f

be a set,

F

be the set of f.o. on

f

,

P

be the set of principal f.o. on

f

,

let

n

be an index set. Consider the ltrator

(

F

n

;

P

n

)

. Then if

f

is a multifuncoid of the form

P

n

,

then

E

f

is a multifuncoid of the form

F

n

.

Proof.

(

val

E

f

)

i

L

= ?? =

f

X

2

A

i

j 8

K

2

up

L

:

K

[ f

(

i

;

X

)

g 2

f

g

=

f

X

2

A

i

j 8

K

2

up

L

:

X

2

(

val

f

)

i

K

g

=

f

X

2

A

i

j 8

K

2

up

L

:

K

[ f

(

i

;

X

)

g 2

f

g

= ?? =

f

X

2

A

i

j

L

[ f

(

i

;

X

)

g 2

E

f

g

X

2

(

val

E

f

)

i

L

,

??

,

L

[ f

(

i

;

X

)

g 2

f

A

[

B

2

(

val

E

f

)

i

L

= ?? =

L

[ f

(

i

;

A

[

B

)

g 2

f

,

(

futher trivial

)

??

(

val

E

f

)

i

L

=

f

X

2

A

i

j

L

[ f

(

i

;

X

)

g 2

E

f

g

=

f

X

2

A

i

j

up

(

L

[ f

(

i

;

X

)

g

)

f

g

= ?? =

f

X

2

A

i

j

up

L

X

f

g

=

f

X

2

A

i

j 8

K

2

up

L; x

2

up

X

:

K

[ f

(

i

;

x

)

g 2

f

g , f

X

2

A

i

j 8

K

2

up

L;

x

2

up

X

:

x

2

(

val

f

)

i

K

g

=

f

X

2

A

i

j 8

K

2

up

L

:

up

X

(

val

f

)

i

K

g

.

[TODO: The same formula as

below only with other variable names.] [TODO: Correct order of coords.]

up

(

L

[ f

(

i

;

X

)

g

)

f

, 8

K

2

up

(

L

[ f

(

i

;

X

)

g

):

K

2

f

, 8

P

2

up

L; X

0

2

up

X

:

P

[ f

(

i

;

X

0

)

g 2

f

, 8

P

2

up

L; X

0

2

up

X

:

X

0

2

(

val

f

)

i

P

, 8

P

2

up

L

:

up

X

(

val

f

)

i

P

Thus

(

val

E

f

)

i

L

=

f

X

2

A

i

j 8

P

2

up

L

:

up

X

(

val

f

)

i

P

g

=

X

2

A

i

j

up

X

T

P

2

up

L

(

val

f

)

i

P

.

[TODO: First try to prove for the binary case.]

A

[

B

2

(

val

E

f

)

i

L

,

X

2

A

i

j

up

(

A

[

B

)

T

P

2

up

L

(

val

f

)

i

P

,

X

2

A

i

j

up

A

\

up

B

T

P

2

up

L

(

val

f

)

i

P

(the lemma does not work, try to use a lter base)

A

[

B

2

(

val

E

f

)

i

L

, 8

P

2

up

L

:

up

(

A

[

B

)

(

val

f

)

i

P

, 8

P

2

up

L

: (

up

A

(

val

f

)

i

P

_

up

B

(

val

f

)

i

P

)

( 8

P

2

up

L

: (

up

A

(

val

f

)

i

P

_ 8

P

2

up

L

:

up

B

(

val

f

)

i

P

)

(used the lemma).

[TODO: Reverse implication.]

8

P

2

up

L

: (

up

A

(

val

f

)

i

P

_

up

B

(

val

f

)

i

P

)

) 8

P

2

up

L

: (

up

A

\

up

B

(

val

f

)

i

P

)

,

8

P

2

up

L

:

up

(

A

[

B

)

(

val

f

)

i

P

.

Thus??

A

[

B

2

(

val

E

f

)

i

L

,

A

2

(

val

E

f

)

i

L

_

B

2

(

val

E

f

)

i

L

.

Consider

f

\

RLD

S

S

=

S

h

f

\

RLD

i

S

.

1. If

f

is not required to be complete this formula fails even for set-valued

S

.

2. Let

f

is complete. Then it fails for specically choosen

S

.

Rest

11