Prolog predicate (\=)/2 ignored - prolog

Question

Getting started with prolog and I am trying to do some simple exercises, however I got stuck pretty early... What I am trying to 'say' is: mike likes anyone if it's a man or a woman and it is not mike. But the X \= mike seems to be ignored:
man(mike).
man(danny).
man(samuel).
man(henry).
woman(samantha).
woman(jane).
woman(betty).
woman(jenny).
likes(mike, X) :-
man(X);
woman(X),
X \= mike.
mike shows up...
?- findall(X, likes(mike, X), L).
L = [mike, danny, samuel, henry, samantha, jane, betty, jenny].
I assume it's because prolog evaluates the rule man(X) first so mike is true. However if I change it to:
likes(mike, X) :-
X \= mike,
man(X);
woman(X).
I will only get the women.
?- findall(X, likes(mike, X), L).
L = [samantha, jane, betty, jenny].
The question is how to do it then? Thanks!
Note: I am using swi-prolog

Answer 1

The problem is operator precedence. Just C/C#/Java/SQL and other procedural languages, in Prolog, logical OR (;) has a different precedence than does logical AND (',').
In most procedural languages, an expression like
A || B && C
Is parsed as
A || ( B && C )
In Prolog, an expression like yours
A ; B , C
is parsed as if written
A ; (B,C)
So your
likes(mike, X) :-
man(X);
woman(X),
X \= mike.
is essentially
likes(mike,X) :- man(X) ; ( woman(X) , X \= mike ) .
You are asserting that Mike likes anyone who is
a man (including himself), OR
a woman who is not mike.
When what you mean was that Mike likes anyone — male or female — other than himself.
So...you need to make the precedence explicit with parentheses,
likes(A,B) :- ( man(B) ; woman(B) ) , A \= B .
or better yet, break your predicate up into 2 clauses and make it generic:
likes(A,B) :- man(B) , A \= B .
likes(A,B) :- woman(B) , A \= B .
Improve things further by making gender an attribute of the entity (a person) rather than a fact in and of itself:
person( mike , male ) .
person( danny , male ) .
person( samuel , male ) .
person( henry , male ) .
person( samantha , female ) .
person( jane , female ) .
person( betty , female ) .
person( jenny , female ) .
Then likes/2 is even simpler, since one can like people of both genders:
likes(A,B) :- person(B,_) , A \= B .

Answer 2

Besides the OR-problem there is another problem. The predicate (\=)/2 is not a constraint, means its essentially not a fully declarative predicate.
The predicate is usually bootstrapped via negation as failure, and negation as failure itself is not fully declarative.
The predicate is bootstrapped from ordinary unification as follows:
X \= Y :- \+ X = Y.
An alternative definition would be:
X \= X :- !, fail.
_ \= _.
If you really want to be able to move around the inequality you should take a Prolog system with constraints and then resort to dif/2.
With dif/2 you can write:
likes(mike, X) :-
(dif(X, mike),
man(X)
; woman(X)).
Or the following:
likes(mike, X) :-
(man(X),
dif(X, mike)
; woman(X)).
And you will get the same results. This is not possible with (\=)/2.
Bye
P.S.: Reasons for the problems with (\=)/2. Negation as failure \+ A sometimes not only acts as ~A, but also as ~exists X1,..,Xn A. The interested reader might like to wade through the admittedly little old, but still applicable:
The Proof Theory of Logic Programs with Negation
Robert Stärk, Bern, 1992
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.29.2745&rep=rep1&type=pdf

Answer 3

The problem is with the OR operator (;) :
man(mike).
man(danny).
man(samuel).
man(henry).
woman(samantha).
woman(jane).
woman(betty).
woman(jenny).
Your solution:
likes(mike, X) :- (man(X);woman(X)),X\=mike.
Another solution (using an equivalent to the OR operator) could be:
likes(mike, X) :- man(X), X\=mike.
likes(mike, X) :- woman(X).
Notice that you don't need an OR operator.