Matching: the built-in predicates = and
\=
Prolog provides a built-in predicate which takes two arguments
and check whether they match. This predicate is
=/2. How does Prolog answer to the following
queries? Then about it first, and then type it into Prolog to
check whether you are right.
?- =(harry,harry).
?- =(harry,'Harry').
?- =(f(a,b),f(a(b))).
Prolog also allows you to use =
as an infix operator. So, instead of writing
=(harry,harry) you can write harry = harry.
Matching in Prolog is a destructive operation because variables get instantiated. So, the terms that are matched may be different after matching to what they were before matching. Here is a sequence of queries that illustrates this.
?- X = harry, Y = hermione.
X = harry
Y = hermione ;
no
Matching the variable X with harry
and the variable Y (a different variable) with hermione
works and instantiates both variables.
?- X = harry, X = hermione. noAfter the first goal
X = harry has been processed,
the variable is instantiated to
harry. harry does not match with
hermione (different atoms), so that the second goal
fails.
The built-in predicate \=/2 works in the opposite
way: Arg1 \= Arg2 is true if
Arg1 = Arg2 fails, and vice
versa. (Note that not all Prolog implementations provide
\=/2. SWI Prolog does, but Sicstus, for example,
doesn't.)
Try some queries involving = and
\=. For example,
?- s(np(pn(dobey)),vp(v(likes),np(pn(harry)))) = s(np(pn(dobey)),v(likes),np(pn(harry))).
?- s(X,vp(v(sings))) = s(np(pn(dobey)),Y).
?- s(X,vp(v(sings))) = s(np(pn(dobey)),vp(X)).
?- father(X) = X.