Antwoorden
 
 Opgave 1:
 
 Welcome to SWI-Prolog (Multi-threaded, Version 5.2.0)
Copyright (c) 1990-2003 University of Amsterdam.
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software,
and you are welcome to redistribute it under certain conditions.
Please visit http://www.swi-prolog.org for details.

For help, use ?- help(Topic). or ?- apropos(Word).

?-  [a,b|X]=[A,B,c].

X = [c]
A = a
B = b ;

No
?- [a,b]=[b,a].

No
?- [a|[b,c]]=[a,b,c].

Yes
?- [a,[b,c]]=[a,b,c].

No
?-  [a,X]=[X,b].

No
?-  [a|[]]=[X].

X = a ;

No
?- [a,b,X,c]=[A,B,Y].

No
?- [H|T]=[[a,b],[c,d]].

H = [a, b]
T = [[c, d]] ;

No
?-  [[X],Y]=[a,b].

No
?-  [X|Y]=[a].

X = a
Y = [] ;

No
?- X =.. [b,Y,[1,2]].

X = b(_G160, [1, 2])
Y = _G160 ;

No
?- X =.. [X1,X2].
ERROR: Arguments are not sufficiently instantiated

?- X is 1, Y is X+1, Y < 3.

X = 1
Y = 2 ;

No
?- Y < 3, X is 1, Y is X+1.
ERROR: Arguments are not sufficiently instantiated
?- X=3, Y=X+1, Z is Y.

X = 3
Y = 3+1
Z = 4 ;

No
?- 


Opgave 2:

a.

plaatje boom:



?- a(X,Y).

X = 1
Y = 1 ;

X = 1
Y = 2 ;

X = 1
Y = 22 ;

No



b.
?-  a(X,Y), !, b(B).

X = 1
Y = 1
B = 1 ;

X = 1
Y = 1
B = 11 ;

No
?- 

Opgave 3: 

set_verschil([],_,[]).
set_verschil([X|S1],S2,S3):-
        member(X,S2), 
        set_verschil(S1,S2,S3).
        
set_verschil([X|S1],S2,[X|S3]):-
       \+ member(X,S2),
       set_verschil(S1,S2,S3).
 
 Met groene cut:
 
set_verschil([],_,[]).
set_verschil([X|S1],S2,S3):-
        member(X,S2), !,
        set_verschil(S1,S2,S3).
        
set_verschil([X|S1],S2,[X|S3]):-
       \+ member(X,S2),
       set_verschil(S1,S2,S3).
 
 
 - Procedurele gedrag = declaratieve gedrag
- cut weghalen geeft dezelfde uitkomsten
 - uitsluitende clauses (member(..) en \+ member(..))
 
 met rode cut
 
set_verschil([],_,[]).
set_verschil([X|S1],S2,S3):-
        member(X,S2), !,
        set_verschil(S1,S2,S3).
        
set_verschil([X|S1],S2,[X|S3]):-
%%%%       \+ member(X,S2),
       set_verschil(S1,S2,S3).
 
 - Procedurele gedrag <> declaratieve gedrag

 
 Opgave 4:
 
 % opgave a

expr(plus(X,Y)) --> basicexpr(X), addterm(Y).
expr(maal(X,Y)) --> basicexpr(X), factorterm(Y).
expr(X) --> basicexpr(X).
basicexpr(1) --> [1].
basicexpr(2) --> [2].
basicexpr(3) --> [3].
addterm(X) --> [+], expr(X).
factorterm(X) --> [*], expr(X).

%opgave b
meaning(plus(X,Y),Z):- meaning(X,MX), 
                      meaning(Y,MY), Z is MX+MY.
meaning(maal(X,Y),Z):- meaning(X,MX), 
                      meaning(Y,MY),
                      Z is MX*MY.
meaning(X,X) :- integer(X).