I\'ve been trying to write a simple code, that would behave in this manner:
| ?- hasCoppiesOf(X,[a,b,a,b,a,b,a,b]).
X = [a,b] ? ;
X = [a,b,a,b] ? ;
X = [a,b,a,b
Okay, I got your problem, you want to eliminate the repetitions.
hasCoppiesOf(A,[]).
hasCoppiesOf([H1|T1], [H1|T2]) :-
append(T1, [H1], X),
hasCoppiesOf([H1|T1], X, T2).
hasCoppiesOf(A, A, B) :-
hasCoppiesOf(A, B),!. %Change here, place a cut after the termination.
hasCoppiesOf(A, [H1|T1], [H1|T2]) :-
append(T1, [H1], X),
hasCoppiesOf(A, X, T2).
This is the change that you need to make.
hasCoppiesOf(A, A, B) :- hasCoppiesOf(A, B),!.
A Cut '!' terminates the unwanted backtracking and thereby repetitions.