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/* Opgave 1
{
loves(rose, jack).
loves(jack, rose).
loves(caledon, rose).
happy(rose).
happy(jack).
}
{
loves(rose, jack).
loves(caledon, rose).
}
{
loves(jack, rose).
loves(caledon, rose).
}
{
loves(rose, jack).
loves(jack, rose).
}
{
loves(rose, jack).
loves(jack, rose).
loves(caledon, rose).
happy(jack).
}
{
loves(rose, jack).
loves(jack, rose).
loves(caledon, rose).
happy(rose).
}
{
loves(rose, jack).
loves(jack, rose).
loves(caledon, rose).
}
{
loves(rose, jack).
loves(jack, rose).
happy(jack).
}
{
loves(rose, jack).
loves(jack, rose).
happy(rose).
}
{
loves(rose, jack).
loves(jack, rose).
happy(rose).
happy(jack)
}
Well okay i feel stupid
We say that the universe U_p = {rose, jack, caledon}.
We will then way that the base is:
U_b = { loves(x, y) | x, y \in U_p } \cup { happy(x) | x \in U_p }
Then all the interpretations are.
I = { S | S \subseteq U_b }
*/
/* Opgave 2
loves(rose, jack).
happy(rose)
happy(rose) <= loves(rose, jack),loves(jack,rose)
We know rose is happy, we do not need to check the predicates.
For some reason, kind of TODO.
*/
/* Opgave 3
I_4 og I_5 er modeller for P, hvor I_4 lige har en extra happy(caledon).
Her er I_4 minimal fordi ingen anden model for P er mindre.
*/
/* Opgave 4
M_1 = T_P(Ø) = {god(odin),son(odin,thor),son(odin,baldr),son(thor,mothi),son(thor,magni)}
M_2 = T_P(M_1) = M_1 \cup {demigod(thor),demigod(baldr)}
M_3 = T_P(M_2) = M_2 \cup {mortal(mothi),mortal(magni)}
*/
/* Opgave 5
B -+-> A
D -+-> C
B -+-> A
C ---> A
C -+-> B
D ---> B
D, C -> A, B
D -> C -> B -> A
Det er stratifyable.
|
