Use cut in Prolog to define a once_member/2 function

Disclaimer: This is informal and non-assessed coursework to do in my own time. I have tried it myself, failed and am now looking for some guidance.

I am trying to implement a version of the member/2 function which will only return members for a list once.

For example:

| ?- member(X, [1,2,3,1]).
X = 1 ? ;
X = 2 ? ;
X = 3 ? ;
X = 1 ? ;

I would like it to only print out each number a maximum of once.

| ?- once_member(X, [1,2,3,1]).
X = 1 ? ;
X = 2 ? ;
X = 3 ? ;
no

We have been told to do this with the cut '!' operator but I have looked over the notes for my course for cut and more online and yet still can't make it click in my head!

So far I have managed to get:

once_member(E, [E | L]) :- !.
once_member(E, [_, L]) :-
    once_member(E, L).

Which returns 1 and then nothing else, I feel like my cut is in the wrong place and preventing a backtrack for each possible match but I'm really not sure where to go with it next.

I have looked in my course notes and also at: http://www.cs.ubbcluj.ro/~csatol/log_funk/prolog/slides/5-cuts.pdf and Programming in Prolog (Google Books)

Guidance on how to logically apply the cut would be most useful, but the answer might help me figure that out myself.

We have also been told to do another method which uses '\+' negation by failure but hopefully this may be simpler once cut has twigged for me?

Here's an approach that uses a cut in the definition of once_member/2 together with the classic member/2 predicate:

once_member(X,[H|T]) :-
    member(H,T),
    !,
    once_member(X,T).
once_member(H,[H|_]).
once_member(X,[_|T]) :-
    once_member(X,T).

Applied to the example above:

?- once_member(X,[1,2,3,1]).

X = 2 ;

X = 3 ;

X = 1 ;
no

Note: Despite the odd-appearing three clause definition, once_member/2 is last-call/tail-recursive optimization eligible due to the placement of the cut ahead of its first self-invocation.