Prolog program to find equality of two lists in any order

isSubset([],_).
isSubset([H|T],Y):-
    member(H,Y),
    select(H,Y,Z),
    isSubset(T,Z).
equal(X,Y):-
    isSubset(X,Y),
    isSubset(Y,X).

Try using predicate that checks if one of sets is a permutation of other set:

delete(X, [X|T], T).

delete(X, [H|T], [H|S]):-
    delete(X, T, S).


permutation([], []).

permutation([H|T], R):-
    permutation(T, X), delete(H, R, X).

(Predicate taken from http://www.dreamincode.net/code/snippet3411.htm)

?- permutation([1,2,3],[3,1,2]).
true 

The actual reason for the non-termination that you observed is this: the following clause does not constrain L2 in any way, shape, or form.

equal([H1|T], L2) :- 
   member(H1, L2),
   del(H1, L2, L3),
   equal(T, L3).

So your query ?- equal([1,2,3], X). implies proving the goal member(_, L2) which does not terminate universally. Therefore equal([1,2,3], X) cannot terminate universally, too!

For more information on how to explain non-termination of Prolog code read about failure-slice!

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PS. Looking at the termination problem from a different angle, we see that the non-termination is, in fact, a necessary consequence in this case.

Why? Because you do not constrain the number of multiplicities, which makes the solution set infinite in size. The set cannot be represented by a finite number of answers (provided you do not permit delaying goals).