More determinism for `memberd/2`?

Question

Many systems provide a pure and efficient implementation of member/2. In particular, no choice point is left open for:

?- member(b,[a,b]).
true.         


        

Answer 1

First, we rename memberd to memberd_old for the sake of clarity.

Then, we implement memberd_new/2, which uses lagging and 1st argument indexing to prevent the creation of the useless choicepoint at the end of the list.

memberd_new(E,[X|Xs]) :-
   memberd_new_aux(Xs,X,E).

% auxiliary predicate to enable first argument indexing
memberd_new_aux([],E,E).
memberd_new_aux([X1|Xs],X0,E) :-
   if_(E=X0, true, memberd_new_aux(Xs,X1,E)).

Let's compare member/2 (SWI-Prolog builtin predicate), memberd_old/2, and memberd_new/2!

First, a ground query:

?- member(a,[a,a]).
true ;
true.                       % BAD!

?- memberd_old(a,[a,a]).
true.

?- memberd_new(a,[a,a]).
true.

Next, another ground query:

?- member(a,[a,b]).
true ;                      % BAD!
false.

?- memberd_old(a,[a,b]).
true.

?- memberd_new(a,[a,b]).
true.

Now, a query having multiple distinct solutions:

?- member(X,[a,b]).
X = a ;
X = b.

?- memberd_old(X,[a,b]).
X = a ;
X = b ;                     % BAD!
false.

?- memberd_new(X,[a,b]).
X = a ;
X = b.

Edit

The implementation of memberd_new/2 presented here is deprecated.

I recommend using the newer implementation shown in this answer.