Predicate must be true of all elements in a list


I have a set of facts:

likes(john,mary). likes(mary,robert). likes(robert,kate). likes(alan,george). likes(alan,mary). likes(george,mary). likes(harry,mary). likes(john,alan).

Now I want to write a relation which will check for all element X of an input list if likes(X,A) is true. my relation should return true once if likes(X,A) is true for all element X in my list L. If I try this this:

relat(X) :- member(A,[john,alan,george,harry]), likes(A,X).

but the output is

?- relat(mary). true ; true ; true ; true.

I want to write it such that it returns one true once it found that likes(john,mary),likes(alan,mary),likes(george,mary),likes(harry,mary) all are true. How to approach this problem?


In SWI-Prolog, you can use forall/2:

?- forall(member(A, [john, alan, george, harry]), likes(A, mary)). true. ?- forall(member(A, [john,alan,george,harry,marys_ex]), likes(A, mary)). false.


With standard list processing you can do the following:

helper(X, []). % No one left to check helper(X, [H|L]) :- likes(H, X), helper(X, L). % Check head, then rest relat(X) :- helper(X, [john,alan,george,harry]).


| ?- relat(harry). no | ?- relat(mary). true ? ; no | ?-


Using library(lambda):

liked_byall(X, Ps) :- maplist(X+\P^likes(P,X), Ps).

Equally without lambdas:

liked_byall(X, Ps) :- maplist(liked(X), Ps). liked(X, P) :- likes(P, X).


liked_byall(_X, []). liked_byall(X, [P|Ps]) :- likes(P, X), liked_byall(X, Ps).

With above definitions you can ask even more general questions like "Who is liked by certain persons?"

?- liked_byall(N,[john, alan, george, harry]). N = mary ; false.

With the following definition these general questions are no longer possible.

liked_byall(X, Ps) :- \+ ( member(P, Ps), \+ likes(P, X) ).

This second definition only makes sense if X is ground and Ps is a ground list. We can ensure this as follows:

liked_byall(X, Ps) :- ( ground(X+Ps) -> true ; throw(error(instantiation_error,_)) ), length(Ps,_), \+ ( member(P, Ps), \+ likes(P, X) ).

These extra checks ensure that absurd cases as the following do not succeed:

?- liked_byall(mary, nonlist).

And that otherwise legitimate cases do not produce an incorrect answer:

?- liked_byall(N,[john, alan, george, harry]), N = the_grinch. N = the_grinch.


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