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Home/ Questions/Q 7998301
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T15:10:55+00:00 2026-06-04T15:10:55+00:00

Well i’m just wondering why does it work, where prolog loose variable. ?- \+

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Well i’m just wondering why does it work, where prolog loose variable.

?- \+ \+ member(X,[a]),X=b.
X=b.

Where X=a. automagicly vanished.

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  1. Editorial Team
    Editorial Team
    2026-06-04T15:10:56+00:00Added an answer on June 4, 2026 at 3:10 pm

    \+/1 is not “not” in the logical sense, but is implemented through “negation as failure”. That means, \+ Goal succeeds iff Goal fails.

    Think of \+/1 as being implemented as:

    \+(Goal) :- Goal -> fail ; true.
    

    As you can see, in both cases can no variable in Goal become bound. In the “if” branch, any binding will be undone by backtracking, and in the “else” branch, no variable will have been become bound anyway.

    In your example, member(X,[a] succeeds by binding X to a. This will have \+ member(X,[a]) fail, and so \+\+ member(X,[a]) succeeds because of this failure. Due to the intermediate failure, X will not be bound to a.

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