SWI-Prolog Puzzle - prolog

Question

This is the problem :
Victor has been murdered, and Arthur, Bertram, and Carleton are
suspects. Arthur says he did not do it. He says that Bertram was the
victim’s friend but that Carleton hated the victim. Bertram says he
was out of town the day of the murder, and besides he didn’t even know
the guy. Carleton says he is innocent and he saw Arthur and Bertram
with the victim just before the murder. Assuming that everyone–except
possibly for the murderer–is telling the truth, use resolution to
solve the crime.
This is what I wrote in SWI Prolog
% Facts:
p('Arthur'). % suspect
p('Bertram'). % suspect
p('Carleton'). % suspect
p('Victor'). % victim
% Arthur
says('Arthur', i('Arthur')).
says('Arthur', f('Bertram', 'Victor')).
says('Arthur', ht('Carleton', 'Victor')).
% Bertram
says('Bertram', o('Bertram')).
says('Bertram', nk('Bertram', 'Victor')).
% Carleton
says('Carleton', i('Carleton')).
says('Carleton', t('Arthur', 'Victor')).
says('Carleton', t('Bertram', 'Victor')).
% Rules:
holds(X) :- says(Y, X), \+m(Y).
holds(i(X)) :- p(X), \+m(X).
holds(f(X,Y)) :- p(X), p(Y), holds(f(Y,X)).
holds(f(X,Y)) :- p(X), p(Y), \+holds(nk(X,Y)).
holds(o(X)) :- p(X), p(Y), holds(t(X,Y)).
holds(o(X)) :- p(X), \+m(X).
holds(nk(X,Y)) :- p(X), p(Y), \+holds(nk(Y,X)).
holds(nk(X,Y)) :- p(X), p(Y), \+holds(f(X,Y)).
holds(t(X,Y)) :- p(X), p(Y), holds(t(Y,X)).
holds(t(X,Y)) :- p(X), p(Y), p(Z), holds(t(X,Z)), holds(t(Z,Y)).
m(X) :- p(X).
The answer is suppose to be Bertram, but I kept on getting Arthur. Dont know what am I doing wrong.

Answer 1

I'm fairly sure that Rules will be far more simpler than that.
For instance, what does mean m(X) :- p(X)., given that p(X) is always true ? Does Victor have something to say ?
In logic it's essential to stick to Occam's Razor. Programming logic it's not an exception, albeit the term has a more practical connotation - see KISS principle.
I think we can only agree that the murder should be the person that contradicts other two. There is only a fact in question: whether or not a person known Victor.
Then what we know about the crime can be summarized:
t(a) :- k(b), k(c).
t(b) :- \+ k(b).
t(c) :- k(a), k(b).
k(_).
where t(X) stands for X testimony that, and k(X) stands for X known Victor.
We don't really know about k(X), then we must add k(_).
With that, Prolog can suggest:
?- t(X).
X = a ;
X = c.
I.e. only a or b can be true.
EDIT: because Prolog isn't propositive when it came to negation, here is a way to solicit the solution:
m(X) :- member(X, [a,b,c]), \+ t(X).
But let's take a more explicit approach:
Instead of clausal form, that leads to immediate availability of Prolog execution, as shown above, our fact base could also expressed:
say(a, know_victim(b, yes)).
say(a, know_victim(c, yes)).
say(b, know_victim(b, no)).
say(c, know_victim(a, yes)).
say(c, know_victim(b, yes)).
now let's see if some some individual says the opposite of others
liar(I) :-
select(I, [a,b,c], Js),
say(I, Fact),
maplist(negate(Fact), Js).
negate(know_victim(I, X), J) :-
say(J, know_victim(I, Y)),
X \= Y.
yields
?- liar(I).
I = b ;
false.

Answer 2

https://github.com/Anniepoo/prolog-examples
contains several different ways of solving this.