YES

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

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

  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(x),y) -> f#(f(x,y),y)
p2: f#(s(x),y) -> f#(x,y)

and R consists of:

r1: f(|0|(),y) -> |0|()
r2: f(s(x),y) -> f(f(x,y),y)
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(x),y) -> f#(f(x,y),y)
p2: f#(s(x),y) -> f#(x,y)

and R consists of:

r1: f(|0|(),y) -> |0|()
r2: f(s(x),y) -> f(f(x,y),y)
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. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > |0| > f > f#
      
      argument filter:
    
        pi(f#) = [1, 2]
        pi(s) = 1
        pi(f) = 1
        pi(|0|) = []
        pi(rand) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > |0| > f > f#
      
      argument filter:
    
        pi(f#) = [1, 2]
        pi(s) = 1
        pi(f) = 1
        pi(|0|) = []
        pi(rand) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > |0| > f > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(s) = [1]
        pi(f) = [1]
        pi(|0|) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p2

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