Reports

Thanks to @brebs and @TessellatingHeckler for your responses. Both of them worked as intended.

I took influence form @brebs solution and implemented an accumulator to basically count the number of occurrences over the pass and unify it with Count in the base case:

fourExactly(X, List) :-
    count_occurrences(X, List, 0, Count),
    Count == 4.

count_occurrences(_, [], Count, Count).
count_occurrences(X, [X|Tail], Acc, Count) :-
    NewAcc is Acc + 1,
    count_occurrences(X, Tail, NewAcc, Count).
count_occurrences(X, [Y|Tail], Acc, Count) :-
    X \= Y,
    count_occurrences(X, Tail, Acc, Count).

Reasons:

Blacklisted phrase (0.5): Thanks
Long answer (-0.5):
Has code block (-0.5):
User mentioned (1): @brebs
User mentioned (0): @TessellatingHeckler
User mentioned (0): @brebs
Self-answer (0.5):
Low reputation (1):

Posted by: Ali A

79115499