I'm trying to further my understanding of Prolog, and how it handles unification. In this case, how it handles unification with lists.
This is my knowledgebase;
member(X, [X|_]).
member(X, [_|T]):- member(X, T).
If I'm understanding the process correctly. If member(X, [X|_]) is not true, then it moves into the recursive rule, and if X is in list T, then [_|T] is unified with T.
So what happens to the anonymous variable in my recursive predicate? Does it get discarded? I'm having difficulty understanding the exact unification process with lists, as [_|T] is two variables, rather than one. I'm just trying to figure out how the unification process works precisely with lists.
Assume that _ is Y
member(X, [Y|T]):- member(X, T).
Then this is True regardless Y. Now you are "returning" member(X, T). In other words, you are discarding Y and "returning" member(X, T).
_ means, whatever it is, ignore that variable.
The _ is just like any other variable, except that each one you see is treated as a different variable and Prolog won't show you what it unifies with. There's no special behavior there; if it confuses you about the behavior, just invent a completely new variable and put it in there to see what it does.
In your case, your function check if a given element exists on a list, so, you take first element of the list, check if is equal, if not, you discard that element and moves on.