prolog 一元数 - 评估表达式

Question

I am trying to understand Prolog and I came up to following situation. I have defined natural numbers (unary) in following way:

n(0).
n(s(X)) :- nat(X).

Which means that 0 is 0, s(0) is 1, s(s(0)) is 2 etc...

Then I defined predicate add:

add(0, Y, Y) :- nat(Y).
add(s(X), Y, s(Z)) :-
   add(X, Y, Z).

Which adds two unary numbers and result stores to Z.

Now I have following predicate "test" what demonstrates my problem:

test(s(0),0).

Then in interpret I type:

add(s(0),0,R). %result: R = s(0), which is correct

Then i try:

test(add(s(0),0,R), 0).

So the first argument should result in R = s(0), second argument is zero, so the whole expression should be evaluated as true, but prolog says false. I guess that it has something to do with the point, that the add(s(0),0,R) inside the "test" predicate does not evaluate the way I think. Could anyone please explain this to me or eventually provide some link that describes this behaviour? Thanks for any help. Cheers.

Answer 1

No, prolog does not work the way you assume it to. When you ask

?- test(add(s(0),0,R), 0).

prolog tries to find a matching clause. However, there is no matching clause in your database, since s(0) does not match add(s(0),0,R) . Two structures can only match if the have the same functor.

s(0) has the functor s while add(s(0),0,R) has the functor add .