给定值 x 和 y return 规则名称如果为真

Question

这是我的序言文件。

male(bob).
male(john).

female(betty).
female(dana).

father(bob, john).
father(bob, dana).
mother(betty, john).
mother(betty, dana).

husband(X, Y) :- male(X), mother(Y, Z), father(X, Z).
wife(X, Y) :- female(X), father(Y, Z), mother(X, Z).
son(X, Y) :- male(X), mother(Y, X);female(X), father(Y, X).
daughter(X, Y) :- female(X), mother(Y, X);female(X), father(Y, X).
sister(X, Y) :- female(X), mother(Z, X), mother(Z, Y), X \= Y.
brother(X, Y) :- male(X), mother(Z, X), mother(Z, Y), X \= Y.

我想要一个规则名称，如果它 return对于任何值 x 或 y 都是真的。假设 x = betty 和 y = john.

mother(betty, john). <- 这会满足所以我的规则应该 return 'mother'。类似地，如果任何其他规则或事实对于某个值 x, y 为真，它应该 return 该规则名称。

我怎样才能达到这样的目标？

Answer 1

可能很简单

query_family(P1, P2, P) :-
    % current_predicate(P/2),
    member(P, [father, mother, husband, wife, son, daughter, sister, brother]),
    call(P, P1, P2).

这给了

?- query_family(betty, john, R).
R = mother ;
false.

?- query_family(betty, X, R).
X = john,
R = mother ;
X = dana,
R = mother ;
X = bob,
R = wife ;
X = bob,
R = wife ;
false.

答案后的分号表示'gimme next'

Answer 2

$ swipl
?- ['facts'].
?- setof( Functor,
          Term^(member(Functor, [father, mother, husband, wife, son, daughter, sister, brother]),
           Term =.. [Functor, betty, john],
           once(Term)),
          Answer).

Answer = [mother].
?-

如果您想避免指定感兴趣的函子列表，您可以使用 current_predicate(F/2).

给定值 x 和 y return 规则名称如果为真

Given values x and y return rule name if it is true

prolog