SWI-Prolog sort predicate not working - prolog

Question

I just made up a program, doing following task: "Get elements, which values are equal to their indexes".
Here is the code:
% get element's index
get_index([Element|_], Element, 0).
get_index([_|T], Element, Index) :-
get_index(T, Element, Index1),
Index is Index1+1.
index_equals_to_element(List, List2) :-
member(X, List),
get_index(List, X, Index),
Index =:= X,
append([], [X], List2).
It works pretty well. But there is one problem. For list [0, 3, 2, 4, 0] my predicate index_equals_to_element returns [0, 2, 0].
Okay, let it happen. But when I'm trying to output only unique elements, I'm getting the same list without any changes. Example:
?- index_equals_to_element([0, 3, 2, 4, 0], List).
% Outputs [0, 2, 0]
?- sort(List, List2).
% Outputs [0, 2, 0] either, when expected [0, 2]
It's very strange for me, because this works fine:
?- sort([0, 2, 1, 0], List).
% Outputs [0, 1, 2].
Why sort doesn't work only with the list, generated by my predicate?

Answer 1

A simple solution is:
index_equals_to_element(List1, List2) :-
% assume that list position index starts at 0
index_equals_to_element(List1, 0, List2).
index_equals_to_element([], _, []).
index_equals_to_element([X| Xs], Index, List2) :-
NextIndex is Index + 1,
( X == Index ->
List2 = [X| Tail],
index_equals_to_element(Xs, NextIndex, Tail)
; index_equals_to_element(Xs, NextIndex, List2)
).
Example call:
?- index_equals_to_element([0, 3, 2, 4, 0], List).
List = [0, 2].
I suggest you study it by using the trace feature of your Prolog system by typing the query:
?- trace, index_equals_to_element([0, 3, 2, 4, 0], List).
Step trough the execution until is the predicate definition is clear for you.

Answer 2

Your index_equals_to_element([0, 3, 2, 4, 0], List). doesn't output [0, 2, 0] as you claim, but gives three answers [0], [2] and [0]:
?- index_equals_to_element([0, 3, 2, 4, 0], List).
List = [0] ;
List = [2] ;
List = [0] ;
false.
You can use findall to get what you want:
?- findall(X, index_equals_to_element([0, 3, 2, 4, 0], [X]), List).
List = [0, 2, 0].
Update. Here is what I think a better implementation of index_equals_to_element/2:
index_equals_to_element(List, List2) :-
index_equals_to_element(List, 0, List2).
index_equals_to_element([], _, []).
index_equals_to_element([X | Rest], I, Rest2) :-
Inext is I + 1,
index_equals_to_element(Rest, Inext, NewRest),
( X =:= I ->
Rest2 = [X | NewRest]
;
Rest2 = NewRest
).
Test run:
?- index_equals_to_element([0, 3, 2, 4, 0], List).
List = [0, 2].
?- index_equals_to_element([0, 1, 2, 2, 4, 5], List).
List = [0, 1, 2, 4, 5].

Answer 3

The other answers are best for learning the nuts and bolts of Prolog. But here's a more concise (but also easier to grok) solution using the higher-order predicate findall/3 and nth0/3 from the SWI-Prolog library(lists):
elements_equal_to_index(List, Elements) :-
findall(Index, nth0(Index, List, Index), Elements).
Edit:
As #Paulo Moura pointed out in a comment, the above answer is only equivalent to the others offered here if all argument are instantiated. I.e., if the above encounters a free variable in the list, I will bind that variable to its index in the list instead of rejecting it as an unsatisfactory element. The addition of a test for strong equality between the index and the list element should make the answer conform:
elements_equal_to_index(List, Elements) :-
findall( Index,
( nth0(Index, List, Elem),
Elem == Index ),
Elements
).