YES

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

  f(0,1,x) -> f(s(x),x,x)
  f(x,y,s(z)) -> s(f(0,1,z))

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#(0,1,x) -> f#(s(x),x,x)
  f#(x,y,s(z)) -> f#(0,1,z)

and Q consists of the rules:

  f(0,1,x) -> f(s(x),x,x)
  f(x,y,s(z)) -> s(f(0,1,z))
  rand(x) -> x
  rand(x) -> rand(s(x))

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

  f#_A((x1,y1),(x2,y2),(x3,y3)) = (x3, y3)
  0_A = (1, 1)
  1_A = (1, 1)
  s_A((x1,y1)) = (x1, y1 + 1)
  f_A((x1,y1),(x2,y2),(x3,y3)) = (x3, y3)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  f#(x,y,s(z)) -> f#(0,1,z)

We remove them from the set of dependency pairs.

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

  f#_A((x1,y1),(x2,y2),(x3,y3)) = (x1, 0)
  0_A = (1, 1)
  1_A = (1, 1)
  s_A((x1,y1)) = (0, 2)
  f_A((x1,y1),(x2,y2),(x3,y3)) = (x1 + 1, y1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  f#(0,1,x) -> f#(s(x),x,x)

We remove them from the set of dependency pairs.

No dependency pair remains.