Here is a pure version using dif/2 which implements sound inequality. dif/2 is offered by B-Prolog, YAP-Prolog, SICStus-Prolog and SWI-Prolog.
firstdup(E, [E|L]) :-
member(E, L).
firstdup(E, [N|L]) :-
non_member(N, L),
firstdup(E, L).
member(E, [E|_L]).
member(E, [_X|L]) :-
member(E, L).
non_member(_E, []).
non_member(E, [F|Fs]) :-
dif(E, F),
non_member(E, Fs).
The advantages are that it can also be used with more general queries:
?- firstdup(E, [A,B,C]).
E = A, A = B ;
E = A, A = C ;
E = C,
B = C,
dif(A, C) ;
false.
Here, we get three answers: A is the duplicate in the first two answers, but on two different grounds: A might be equal to B or C. In the third answer, B is the duplicate, but it will only be a duplicate if C is different to A.
To understand the definition, read :- as an arrow ← So what is on the right-hand side is what you know and on the left is what you conclude. It is often in the beginning a bit irritating to read predicates in that direction, after all you might be tempted to follow "the thread of execution". But often this thread leads to nowhere – it gets too complex to understand.
The first rule reads:
Provided E is an element of list L we conclude that [E|L] has E as first duplicate.
The second rule reads:
Provided E is the first duplicate of L (don't panic here and say we don't know that ...) and provided that N is not an element of L we conclude that [N|L] has E as first duplicate.
The member/2 predicate is provided in many Prolog systems and non_member(X,L) can be defined as maplist(dif(X),L). Thus one would define firstdup/2 rather as:
firstdup(E, [E|L]) :-
member(E, L).
firstdup(E, [N|L]) :-
maplist(dif(N), L),
firstdup(E, L).
