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Proof.

From the proposition above.

Example 14.14.

There exists a principal funcoid which is not a complemented element of the

lattice of funcoids.

Proof.

I will prove that quasi-complement of the funcoid id

FCD

(

N

)

is not its complement. We have:

¡

id

FCD

(

N

)

=

c

2

FCD

(

N

;

N

)

j

c

id

FCD

(

N

)

 

w

"

N

f

FCD

"

N

f

g j

2

N

;

"

N

f

FCD

"

N

f

id

FCD

(

N

)

 

=

G

f"

N

f

FCD

"

N

f

g j

2

N

=

/

g

=

"

FCD

(

N

;

N

)

G

ff

g  f

g j

2

N

=

/

g

=

"

FCD

(

N

;

N

)

(

N

N

n

id

N

)

(used corollary

6.107

). But by proved above

¡

id

FCD

(

N

)

u

id

FCD

(

N

)

=

/ 0

F

(

N

)

:

[TODO: Stronger conjecture:

(

"

FCD

f

)

=

f

for every binary relation

f

.]

Example 14.15.

There exists a funcoid

h

such that GR

h

is not a lter.

Proof.

Consider the funcoid

h

=

id

(

N

)

FCD

. We have (from the proof of proposition

14.12

that

f

2

GR

h

and

g

2

GR

h

, but

f

\

g

=

; 2

/

GR

h

.

Example 14.16.

There exists a funcoid

h

=

/ 0

FCD

(

A

;

B

)

such that

(

RLD

)

out

h

= 0

RLD

(

A

;

B

)

.

Proof.

Consider

h

=

id

(

N

)

FCD

. By proved above

h

=

f

u

g

where

f

=

id

FCD

(

N

)

=

"

FCD

(

N

;

N

)

id

N

,

g

=

"

FCD

(

N

;

N

)

(

N

N

n

id

N

)

.

We have id

N

;

N

N

n

id

N

2

GR

h

.

So

(

RLD

)

out

h

=

d

"

RLD

(

N

;

N

)

GR

h

v "

RLD

(

N

;

N

)

(

id

N

\

(

N

N

n

id

N

)) = 0

RLD

(

N

;

N

)

; and thus

(

RLD

)

out

h

= 0

RLD

(

N

;

N

)

.

Example 14.17.

There exists a funcoid

h

such that

(

FCD

)(

RLD

)

out

h

=

/

h

.

Proof.

It follows from the previous example.

Example 14.18.

(

RLD

)

in

(

FCD

)

f

=

/

f

for some convex reloid

f

.

Proof.

Let

f

=

id

RLD

(

N

)

. Then

(

FCD

)

f

=

id

FCD

(

N

)

. Let

a

be some non-trivial ultralter on

N

.

Then

(

RLD

)

in

(

FCD

)

f

w

a

RLD

a

v

id

RLD

(

N

)

and thus

(

RLD

)

in

(

FCD

)

f

v

f

.

Example 14.19.

There exist composable funcoids

f

and

g

such that

(

RLD

)

out

(

g

f

) =

/ (

RLD

)

out

g

(

RLD

)

out

f :

Proof.

Take

f

=

id

(

N

)

FCD

and

g

= 1

F

(

N

)

FCD

"

N

f

g

for some

2

N

. Then

(

RLD

)

out

f

= 0

RLD

(

N

;

N

)

and thus

(

RLD

)

out

g

(

RLD

)

out

f

= 0

RLD

(

N

;

N

)

.

We have

g

f

= (

N

)

FCD

"

N

f

g

.

Let's prove

(

RLD

)

out

((

N

)

FCD

"

N

f

g

) = (

N

)

RLD

"

N

f

g

.

[TODO: Separate proposition

asserting

(

RLD

)

in

(

FCD

"

F

f

g

) = (

RLD

)

out

(

FCD

"

F

f

g

) =

RLD

"

F

f

g

. Does it hold for

every (co)complete funcoid?]

Really:

(

RLD

)

out

((

N

)

FCD

"

N

f

g

) =

d

"

RLD

(

N

;

N

)

GR

((

N

)

FCD

"

N

f

g

) =

d

"

RLD

(

N

;

N

)

(

K

 f

g

)

j

K

2

(

N

)

 

.

F

2

GR

d

"

RLD

(

N

;

N

)

(

K

 f

g

)

j

K

2

(

N

)

 

,

F

2

GR

(

d

f"

N

K

j

K

2

(

N

)

RLD

"

N

f

g

)

for every

F

2

P

(

N

N

)

. Thus

d

"

RLD

(

N

;

N

)

(

K

 f

g

)

j

K

2

(

N

)

 

=

d

f"

N

K

j

K

2

(

N

)

RLD

"

N

f

g

= (

N

)

RLD

"

N

f

g

.

Counter-examples about funcoids and reloids

173