What does the s() predicate do in Prolog?

Question

I have been trying to learn Prolog, and am totally stumped on what the predicate s() does. I see it used often and there is so little resources on the internet about Prolog that I cannot find an answer.

Ex.

    /* sum(Is,S) is true if S is the sum of the list of integers Is.           */
    sum([],0).
    sum([0|Is],S):-sum(Is,S).
    sum([s(I)|Is], s(Z) ):-sum([I|Is],Z).

Teledu

What version of prolog interpreter do you use? – Feingold

I use: SWI-Prolog version 5.10.1 for amd64 – Teledu

Answer 1

s/1 does not do anything in itself, and it's not really a predicate. They are just terms, a representation of the successor of their argument. So, s(0) is used to represent the successor of 0 (i.e. 1), s(s(0)) is used to represent the successor of s(0) (i.e. 2), and so on and so forth. They are so widespread in Prolog because Prolog is quite fine a language to perform symbolic computation, whereas even simple arithmetic operations feel clunky, meaning that they are not seamlessly integrated with the programming paradigm.

Saks

This explains, very well, why the output of my programs using s/1 were so strange. Thank you very much, great explanation! – Teledu

Answer 2

s/1 stands for successor. It's used to represent numbers in a logically accessible ways.

Moralize

That makes sense thank you very much! However when I try run a program count that uses s/1 I get strange output in the form of: ?- mycount(a,[a,a],N). N = s(s(0)) ; N = s(0) ; N = s(0) ; N = 0 ; false. – Teledu

Answer 3

It is Prolog-implementation specific. It refers to a successor-predicate, see this for some more info

Intermeddle

s(X) is not implementation specific. It's an arbitrary functor. Try t(X) or succ(X) in its place. – Havener