Checking if unification between two elements is True or False - list

Question

So my problem is this i have this predicate equal(X,Y) which says basically that X is equal to Y and unifies them if they are, as long as they have the same length, but what i want is the predicate to give true or false but instead, it gives me the unifications.
If somebody could help i would appreciate it a lot.
My output:
?- equal([[d, r, a, m, a],[a, m, e, n, o]],[[P11, P21, P31, P41, P51], [a, P23, P33, P43, P53]]).P11 = d,
P21 = r,
P31 = P51, P51 = a,
P41 = P23, P23 = m,
P33 = e,
P43 = n,
P53 = o .
?- equal([[a, m, e, n, o]],[[P11, P21, P31, P41, P51], [a, P23, P33, P43, P53]]).
false.
As you can see the substituitions are fine, like i want those substituitions to happen, but the i dont want them as the output, what i want is whether this is True or False which in this case is True.
Correct output:
?- equal([[d, r, a, m, a],[a, m, e, n, o]],[[P11, P21, P31, P41, P51], [a, P23, P33, P43, P53]]).P11 = d,
true.
?- equal([[a, m, e, n, o]],[[P11, P21, P31, P41, P51], [a, P23, P33, P43, P53]]).
false.
Program:
isSubset([],_).
isSubset([H|T],Y):-
member(H,Y),
select(H,Y,Z),
isSubset(T,Z).
equal(X,Y):-
isSubset(X,Y),
isSubset(Y,X).

Answer 1

You can approach this using Lambda expressions (e.g. with the yall library of SWI-Prolog).
Prolog will yield bindings only for those variables placed within the { } before the query:
?- {P11, P21}/equal([[d, r, a, m, a],[a, m, e, n, o]],[[P11, P21, P31, P41, P51], [a, P23, P33, P43, P53]]).
P11 = d,
P21 = r .
But note that you can also leave the brackets empty and yield no bindings:
?- {}/equal([[d, r, a, m, a],[a, m, e, n, o]],[[P11, P21, P31, P41, P51], [a, P23, P33, P43, P53]]).
true .
This seems to be essentially the functionality you want. You could throw it into a helper rule to make things easier. The following will give you true for each valid relation for equal/2 in the facts:
isEqual(X, Y) :-
{}/equal(X, Y).
If you want to know only if one or more valid relations for equal/2 exists in the facts, then you could add a cut !:
isEqual(X, Y) :-
{}/equal(X, Y),
!.
Output for your query is thus:
?- isEqual([[d, r, a, m, a],[a, m, e, n, o]],[[P11, P21, P31, P41, P51], [a, P23, P33, P43, P53]]).
true.