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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.