CODEWITHSUNDEEP
prologmember
daniel451
daniel451

Reputation: 11002

Prolog: true / false output of member predicate?

I am wondering why the member/2 predicate of Prolog delivers multiple alternatives (via backtracking?!), if true was already unified for the output.

For example member(1, [1,2,3]). delivers the following output:

true ;
false.

Why does member return false after it already found out that the atom 1 is indeed a member of the list [1,2,3]?

Even more confusing to me is the following output:

?- member(1, [1,2,3,1]).
true ;
true.

Upvotes: 0

Views: 1499

Answers (1)

Scott Hunter
Scott Hunter

Reputation: 49920

In your first example, you ask it to prove member(1,[1,2,3]).; since it can, it reports true. When you enter ;, you are asking to find another way to prove that query; since it cannot, it reports false.

In the second case, the first true is because it found one of the 1's in the list; the second is because it found the second. If you had hit ; again, it would have come back with false, as it had no other ways to prove the query. (Note: As @WillNess points out, you don't actually get the chance to hit ; again; this is probably due to the implementation of member being such that Prolog knows there are no remaining alternatives. If the list did not end with a 1, you would be able to hit ; again.)

Upvotes: 5

Related Questions