Re: negotiation by failure- list operations



ne0 wrote:
> the problem with the empty list is just a minor one - its simply
> delete1(_,[],[]).
>
> but thats not what my question was.
>
> The problem is still- how to find out that X isnt in L1 and then to
> throw a YES - can this be solved by a not(member/2) predicate ??

yes - you could use this, although the resulting solution is much less
efficient:

delete1(X,List,List) :-
\+ member(X,List),!.
delete1(X,List,ListWithoutX) :-
delete(X, List, ListWithoutX).

Regards, Brian.

.

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