
W = fhRiX j R 2 up rg is?? a generalized filter base?? (No, it isn't, up r isn't a filter.)
Let P
0
; P
1
2 W . Then P
0
= hR
0
iX, P
1
= hR
1
iX
Let F 2 up(f ◦ r). Then ?? F 2 up
T
ff ◦ "
FCD
R j R 2 up rg.
So f ◦ r ⊇
T
ff ◦ "
FCD
R j R 2 up rg
X [f ◦ "
FCD
R] Y , ["
FCD
R]X / hf
¡1
iY
??
Let W = fh"
FCD
Ria j R 2 up rg. W is a generalized filter base??. Really If X ; Y 2 W . Then
X = h"
FCD
X ia and Y = h"
FCD
Y ia for some X 2 up r, Y 2 up r.
\
ff ◦ "
FCD
R j R 2 up rg
a =
\
fhf ◦ "
FCD
Ria j R 2 up rg = (because fh"
FCD
Ria j R 2 up rg is a g.f.b.??)
hf i
\
fh"
FCD
Ria j R 2 up rg =
hf ihria
Counter-example attempt: Let r = id
Ω
FCD
. ??
Corollary 34. g ◦ f =
T
"
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 , 9y 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 , 8F 2 up f ; G 2 up g:
x
"
FCD(Src f ;Dst g)
(G ◦ F )
z , 8F 2 up f ; G 2 up g9y 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 filtrator ( 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 = ?? = fX 2 A
i
j 8K 2 up L: K [ f(i; X)g 2 f g = fX 2 A
i
j 8K 2 up L:
X 2 (val f )
i
K g = fX 2 A
i
j 8K 2 up L: K [ f(i; X)g 2 f g = ?? = fX 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 = fX 2 A
i
j L [ f(i; X)g 2 E
∗
f g = fX 2 A
i
j up(L [ f(i; X)g) ⊆ f g = ?? =
fX 2 A
i
j up L × X ⊆ f g = fX 2 A
i
j 8K 2 up L; x 2 up X: K [ f(i; x)g 2 f g , fX 2 A
i
j 8K 2 up L;
x 2 up X: x 2 (val f)
i
K g = fX 2 A
i
j 8K 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 , 8K 2 up(L [ f(i; X)g): K 2 f , 8P 2 up L; X
0
2 up X:
P [ f(i; X
0
)g 2 f , 8P 2 up L; X
0
2 up X: X
0
2 (val f)
i
P , 8P 2 up L: up X ⊆ (val f)
i
P
Thus (val E
∗
f)
i
L = fX 2 A
i
j 8P 2 up L: up X ⊆ (val f)
i
P g =
X 2 A
i
j up X ⊆
T
P 2up 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 2up L
(val f )
i
P
,
X 2 A
i
j up A \ up B ⊆
T
P 2up L
(val f)
i
P
(the lemma does not work, try to use a filter base)
A [ B 2 (val E
∗
f)
i
L , 8P 2 up L: up(A [ B) ⊆ (val f)
i
P , 8P 2 up L: (up A ⊆ (val f )
i
P _
up B ⊆ (val f )
i
P ) ( 8P 2 up L: (up A ⊆ (val f)
i
P _ 8P 2 up L: up B ⊆ (val f)
i
P ) (used the lemma).
[TODO: Reverse implication.]
8P 2 up L: (up A ⊆ (val f )
i
P _ up B ⊆ (val f )
i
P ) ) 8P 2 up L: (up A \ up B ⊆ (val f)
i
P ) ,
8P 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
hf \
RLD
iS.
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 specifically choosen S.
Rest 11