Implement the member predicate as a one-liner

Question

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 code can also be used to enumerate the elements of a given list:
```
?- member(X, [a, b]).
X = a ;
X = b ;
fail.
```
- And it can be used to “enumerate” all lists which contain a given element:
```
?- member(a, X).
X = [a|_G246] ;
X = [_G245, a|_G249] ;
X = [_G245, _G248, a|_G252] ;
...
```
- Replacing = by == in the above code makes it a lot less flexible: it would immediately fail on member(X, [a]) and cause a stack overflow on member(a, X) (tested with SWI-Prolog version 5.6.57).