YES

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

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

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        f#_A(x1) = ((1,0,0,0),(1,1,0,0),(0,1,1,0),(0,1,0,1)) x1
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,0,0)) x1 + (0,2,1,1)
        f_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,0,0)) x1 + (0,2,2,2)
        rand_A(x1) = x1 + (1,0,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > f# > rand > f
      
      argument filter:
    
        pi(f#) = [1]
        pi(s) = []
        pi(f) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p2

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