Prolog Successor Arithmetic

Question

I have a knowledge base with the following:

numeral(0).
numeral(s(X)) :- numeral(X).
numeral(X+Y) :- numeral(X), numeral(Y).

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

add2(W+X,Y+Z,R) :- add(W,X,A),add(Y,Z,T),add2(A,T,R).
add2(X+Y,Z,R) :- add(X,Y,A),add2(A,Z,R).
add2(X,Y+Z,R) :- add(Y,Z,A),add2(X,A,R).
add2(X,Y,R) :-  add(X,Y,R).

which evaluates correctly queries such as:

?- add2(s(0)+s(s(0)), s(s(0)), Z).
Z = s(s(s(s(s(0)))))

?- add2(0, s(0)+s(s(0)), Z).
Z = s(s(s(0)))

?- add2(s(s(0)), s(0)+s(s(0)), Z).
Z = s(s(s(s(s(0)))))

However the following query is evaluated to:

?- add2(s(0)+s(0), s(0+s(s(0))), Z).
Z = s(s(s(0+s(s(0))))) .

But the required output is:

?- add2(s(0)+s(0), s(0+s(s(0))), Z).
Z = s(s(s(s(s(0)))))

I know the issue is with the line:

add2(W+X,Y+Z,R) :- add(W,X,A),add(Y,Z,T),add2(A,T,R).

But i just can't figure it out. Any help would be appreciated!

You need to call add2/3 recursively with the first two calls as well. So use add2(W+X,Y+Z,R) :- add2(W,X,A),add2(Y,Z,T),add2(A,T,R).. But if you want a pure Prolog implementation, you will need to alter the code more drastically. — Willem Van Onsem
The form is a little peculiar. Why the extra complication of having in-line addition in two different arguments that are then added together? — lurker

Willem Van Onsem Willem Van Onsem · Accepted Answer · 2017-11-06T17:30:08

I think you make the problem more complex by handling the cases with an add2/3 predicate. You first need to resolve the structure of the first two arguments to something of the shape s(s(...s(0)...)).

In order to do this, we can make an resolve/2 function that looks for (+)/2 terms and recursively works with add/3:

resolve(0,0).
resolve(s(X),s(Y)) :-
    resolve(X,Y).
resolve(X+Y,Z) :-
    resolve(X,RX),
    resolve(Y,RY),
    add(RX,RY,Z).

So now for a grammar:

E -> 0
E -> s(E)
E -> E + E

resolve/2 will convert this to a grammar with:

E -> 0
E -> s(E)

For example:

?- resolve(s(0)+s(0),X).
X = s(s(0)).

?- resolve(s(0+s(s(0))),X).
X = s(s(s(0))).

And now our add2/3 predicate will first resolve/2 the operands, and then add these together:

add2(A,B,C) :-
    resolve(A,RA),
    resolve(B,RB),
    add(RA,RB,C).

The sample queries you then write resolve to:

?- add2(s(0)+s(s(0)), s(s(0)), Z).
Z = s(s(s(s(s(0))))).

?- add2(0, s(0)+s(s(0)), Z).
Z = s(s(s(0))).

?- add2(s(s(0)), s(0)+s(s(0)), Z).
Z = s(s(s(s(s(0))))).

?- add2(s(0)+s(0), s(0+s(s(0))), Z).
Z = s(s(s(s(s(0))))).

Prolog Successor Arithmetic

1 Answers