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-minimal dependency 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-minimal dependency 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{1, x1}
          s_A(x1) = max{1, x1}
          f_A(x1) = x1
          |0|_A = 2
          rand_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        s = f = |0| > f# > rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(f) = [1]
        pi(|0|) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = 0
          s_A(x1) = max{5, x1 - 1}
          f_A(x1) = max{0, x1 - 1}
          |0|_A = 8
          rand_A(x1) = 0
    
      precedence:
      
        s = f > |0| > f# = rand
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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


-- Reduction pair.

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

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 2
          s_A(x1) = x1
          f_A(x1) = 0
          |0|_A = 0
          rand_A(x1) = x1 + 1
    
      precedence:
      
        f# = s = f = |0| = rand
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{0, x1 - 19}
          s_A(x1) = max{17, x1 + 12}
          f_A(x1) = 19
          |0|_A = 6
          rand_A(x1) = 16
    
      precedence:
      
        s = f > |0| > rand > f#
      
      partial status:
    
        pi(f#) = []
        pi(s) = [1]
        pi(f) = []
        pi(|0|) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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