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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x1 + x2
        s_A(x1) = x1
        f_A(x1,x2) = x1
        |0|_A() = 1
        rand_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 3
        f_A(x1,x2) = 2
        |0|_A() = 1
        rand_A(x1) = 0
    

The next rules are strictly ordered:

  p1, p2

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