YES

We show the termination of the relative TRS R/S:

  R:
  f(|0|()) -> s(|0|())
  f(s(|0|())) -> s(|0|())
  f(s(s(x))) -> f(f(s(x)))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimaldependency pair problem (P, R), where P consists of

p1: f#(s(s(x))) -> f#(f(s(x)))
p2: f#(s(s(x))) -> f#(s(x))

and R consists of:

r1: f(|0|()) -> s(|0|())
r2: f(s(|0|())) -> s(|0|())
r3: f(s(s(x))) -> f(f(s(x)))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the non-minimaldependency pair problem (P, R), where P consists of

p1: f#(s(s(x))) -> f#(f(s(x)))
p2: f#(s(s(x))) -> f#(s(x))

and R consists of:

r1: f(|0|()) -> s(|0|())
r2: f(s(|0|())) -> s(|0|())
r3: f(s(s(x))) -> f(f(s(x)))
r4: rand(x) -> x
r5: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1) = x1
      s_A(x1) = x1 + (0,2,0,2)
      f_A(x1) = (0,3,1,1)
      |0|_A() = (0,0,0,1)
      rand_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,1,1,0)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.