[ACCEPTED]-Implement the member predicate as a one-liner-dcg

Accepted answer

Score: 40

Solution:

member(X, [Y|T]) :- X = Y; member(X, T).

Demonstration:

?- member(a, []).
fail.
?- member(a, [a]).
true ;
fail.
?- member(a, [b]).
fail.
?- member(a, [1, 2, 3, a, 5, 6, a]).
true ;
true ;
fail.

How it works:
- We are looking for an occurrence of the first argument, X, in the the second argument, [Y|T].
- The second argument is assumed to be a list. Y matches its head, T matches the tail.
- As a result the predicate fails for the empty list (as it should).
- If X = Y (i.e. X can be unified with Y) then we found X in the list. Otherwise (;) we test whether X is in the tail.
Remarks:
- Thanks to humble coffee for pointing out that using = (unification) yields more flexible code than using == (testing for equality).
- This 7 code can also be used to enumerate the elements 6 of a given list:
```
?- member(X, [a, b]).
X = a ;
X = b ;
fail.
```
- And it can be used to "enumerate" all 5 lists which contain a given element:
```
?- member(a, X).
X = [a|_G246] ;
X = [_G245, a|_G249] ;
X = [_G245, _G248, a|_G252] ;
...
```
- Replacing 4 = by == in the above code makes it a lot less 3 flexible: it would immediately fail on member(X, [a]) and 2 cause a stack overflow on member(a, X) (tested with 1 SWI-Prolog version 5.6.57).

Score: 23

Since you didn't specify what other predicates 2 we're allowed to use, I'm going to try and 1 cheat a bit. :P

member(X, L) :- append(_, [X|_], L).

Score: 7

newmember(X, Xs) :-
   phrase(( ..., [X] ),Xs, _).

With

... --> [] | [_], ... .

Actually, the following definition also 5 ensures that Xs is a list:

member_oflist(X, Xs) :-
   phrase(( ..., [X], ... ), Xs).

The first appearance 4 of above definition of ... is on p. 205, Note 3 1 of

David B. Searls, Investigating the Linguistics 2 of DNA with Definite Clause Grammars. NACLP 1 1989, Volume 1.

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