Prolog member/2 predicate

Question

Why does prolog answer false to: member([5], [2, 5]). ?

Is there a way to work around it?

Any help will be greatly appreciated.

Answer 1

Why does Prolog answer false to: member([5], [2, 5]). ?

Let's ask Prolog why! Download library(diadem) into your working directory and:

?- use_module(diadem).
   true.

?- member([5], [2,5]).? Expl.
   Expl = member([_|_], [2, 5])
;  ...

Not only does it fail but also a generalization fails:instead of [5] that is a list with a single element 5, we have now simply at least one element - no matter which. So we can take the value for Expl as a query which still fails. Something in the remaining goal must thus be the culprit.

?- member([_|_], [2, 5]).
   false.

Note also what was not generalized away: The two elements are still here! If they would be variables, the query would succeed! Generalize the query a bit:

?- member([5], [2, Any]).
   Any = [5].

Now it succeeds!

Answer 2

As described in the SWI Prolog documentation for member/2 :

member( ?Elem , ?List )

True if Elem is a member of List .

The predicate is member . It is not subset or sublist or subsequence . It succeeds if the first argument is a member (that is, an element ) of the list given in the second argument. The element [5] is not a member of the list [2, 5] since the element [5] isn't 2 and it isn't 5. However, [5] would be a member of the list [2, [5]] , so member([5], [2, [5]]) would succeed.

Answer 3

It could be that you are looking for another predicate?
From the SWI Prolog documentation for subset/2 :

subset( +SubSet , +Set )
True if all elements of SubSet belong to Set as well.

It works as expected:

Welcome to SWI-Prolog (threaded, 64 bits, version 7.5.5)
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software.

?- subset([5], [2, 5]).
true.

?- subset([5, 3], [2, 3, 5, 7]).
true.

?- subset([5, 3], [2, 5]).
false.