YES

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

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

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

and Q consists of the rules:

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

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

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

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

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

  f#_A((x1,y1,z1)) = (x1, y1, z1)
  s_A((x1,y1,z1)) = (x1, y1 + 3, z1 + 2)
  f_A((x1,y1,z1)) = (0, 5, 3)
  0_A = (0, 1, 1)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

No dependency pair remains.