YES

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

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

and S is:

  rand(x) -> x
  rand(x) -> rand(s(x))

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

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

and Q consists of the rules:

  f(0,y) -> 0
  f(s(x),y) -> f(f(x,y),y)
  rand(x) -> x
  rand(x) -> rand(s(x))

The weakly monotone algebra (N^3, >_lex) with

  f#_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + y2, z1 + z2)
  s_A((x1,y1,z1)) = (x1, y1 + 2, z1 + 1)
  f_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1 + 1, z1 + 2)
  0_A = (1, 1, 1)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

No dependency pair remains.