Stopping infinite loop in Prolog - list

Question

I'm trying to return distinct items in a list. I've come up with the following code. I seem to run into a problem when checking the members of the list using member(X,Y), and it runs into an infinite loop.
get_unique(Y, [H|T]) :-
memberchk(H, Y)-> get_unique(Y, T) ; get_unique([H|T], T).
get_unique(_, []) :- true.
final_list(X, L) :-
get_unique(Y, L), member(X,Y).
There are two input cases:
?- final_list(dog,[dog,cat,bat,dog]).
true;
true;
true;
true;
(infinitely)
In this case I would expect just
true;
false.
The other input case:
?- final_list(X,[dog,cat,bat,dog]).
X = dog;
X = cat;
X = bat;
true ;
true ;
true ;
true ;
true ;
true ;
(infinitely)
In this case I would expect just
X = dog;
X = cat;
X = bat;
false.
I tried to do some debugging and I think I can see that member(X,Y) is generating the infinite loop -- tracing steps seen below. I don't know how to stop this, such that member stops checking once it returns bat.
X = bat ;
Redo: (9) lists:member(_6632, [dog, cat, bat|_6882]) ? creep
Exit: (9) lists:member(_6632, [dog, cat, bat, _6632|_6888]) ? creep
Exit: (8) final_list(_6632, [dog, cat, bat, dog]) ? creep
true ;
Redo: (9) lists:member(_6632, [dog, cat, bat, _6886|_6888]) ? creep
Exit: (9) lists:member(_6632, [dog, cat, bat, _6886, _6632|_6894]) ? creep
Exit: (8) final_list(_6632, [dog, cat, bat, dog]) ? creep
true ;
Redo: (9) lists:member(_6632, [dog, cat, bat, _6886, _6892|_6894]) ? creep
Exit: (9) lists:member(_6632, [dog, cat, bat, _6886, _6892, _6632|_6900]) ? creep
Exit: (8) final_list(_6632, [dog, cat, bat, dog]) ? creep
true ;
Redo: (9) lists:member(_6632, [dog, cat, bat, _6886, _6892, _6898|_6900]) ? creep
Exit: (9) lists:member(_6632, [dog, cat, bat, _6886, _6892, _6898, _6632|...]) ? creep
Exit: (8) final_list(_6632, [dog, cat, bat, dog]) ? creep
true ;
Thank you!

Answer 1

Try get_unique/2 alone:
?- get_unique(Y,[dog,cat,bat,dog]).
Y = [dog,cat,bat|_A].
% ^^^
Thus, Y is only a partial list, but not a (real) list. It comprises all lists with three or more elements.
The second clause should read get_unique([], []). to avoid this problem. And also the first needs some rearrangement:
get_unique(Y, [H|T]) :-
( memberchk(H, T) -> get_unique(Y, T) ; Y = [H|Y1], get_unique(Y1, T) ).
% ^^ ^^^^^^^^^^ ^^
get_unique([], []).
But still, your definition makes some odd assumptions:
?- final_list(X,[dog,Cat]).
X = dog, Cat = dog.
So it forces variables to become one specific value.
There is no clean way out of this, except by reconsidering member/2. See this answer for a clean solution. Here is the answer memberd/2 produces:
?- memberd(X,[dog,Cat]).
X = dog
; X = Cat, dif(Cat,dog)
; false.
So the answer says: X = dog as before and additionally, X = Cat but only if Cat is different to dog.