Prolog中的递归succ加法

Question

add(0,Y,Y).
add(succ(X),Y,succ(Z)) :-
        add(X,Y,Z).    

I don't understand how this one works. If I run the query:

add(succ(succ(succ(0))), succ(succ(0)), R)

How does it remove succ from the first argument?

If I run it:

X=succ(succ(succ(0)))
Y=succ(succ(0))
R=?

This satisfied the body. We input them into the head:

add(succ(X),Y,succ(Z)) %% becomes
add(succ(succ(succ(0))),succ(succ(0)),succ(R))

Ie it looks to me as if it just adds succ, not removes them from the first argument? I saw that there's another post about this exact question but it does not make sense to me.

Answer 1

When you call the goal add(succ(succ(succ(0))), succ(succ(0)), R) and Prolog tries to execute the clause with the head add(succ(X),Y,succ(Z)) , unification happens. Unification is one of the central concepts of Prolog, and it's important to understand it well. Unification is also the operation that is performed by the = operator. What happens when the head of the clause is executed is exactly the same as happens when you execute the following equivalent query:

?- add(succ(succ(succ(0))), succ(succ(0)), R) = add(succ(X),Y,succ(Z)).

It is important to understand that unification is not just "unpacking" or "building up" terms, but both at the same time .

It can "extract" information from terms:

?- X = f(a), Y = f(Z), X = Y.
X = Y, Y = f(a),
Z = a.

Here we found that a is the argument of the f functor in X .

It can "construct" terms:

?- X = f(Y), Y = g(h).
X = f(g(h)),
Y = g(h).

Here we "inserted" the term g(h) into the "hole" Y in X .

It can also do both at the same time to more than one term:

?- X = f(a, B), Y = f(A, b), X = Y.
X = Y, Y = f(a, b),
B = b,
A = a.

Here unification "unpacked" both f terms, then "filled in" a hole in each of them.

Now, armed with this knowledge, what does the unification in your example do?

?- add(succ(succ(succ(0))), succ(succ(0)), R) = add(succ(X),Y,succ(Z)).
R = succ(Z),
X = Y, Y = succ(succ(0)).

In particular, when succ(succ(succ(0))) is unified with succ(X) , this will bind X to succ(succ(0)) . X is not bound to the entire term succ(succ(succ(0))) but only to an "inner" part. This is what removes one level of succ from the argument.