YES

We show the termination of R/S, where R is

  p(0,y) -> y
  p(s(x),y) -> s(p(x,y))

and S is:

  p(x,y) -> p(x,s(y))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  p#(s(x),y) -> p#(x,y)

and Q consists of the rules:

  p(0,y) -> y
  p(s(x),y) -> s(p(x,y))
  p(x,y) -> p(x,s(y))
  p#(x,y) -> p#(x,s(y))

The weakly monotone algebra (N^2, >_lex) with

  p#_A((x1,y1),(x2,y2)) = (x1 + x2, y1)
  s_A((x1,y1)) = (x1, y1 + 1)
  p_A((x1,y1),(x2,y2)) = (x1 + x2 + 1, y1 + 1)
  0_A = (1, 1)

strictly orients the following dependency pairs:

  p#(s(x),y) -> p#(x,y)

We remove them from the set of dependency pairs.

No dependency pair remains.