YES

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

  R:
  f(|0|(),|1|(),x) -> f(s(x),x,x)
  f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

  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#(|0|(),|1|(),x) -> f#(s(x),x,x)
p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))
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#(|0|(),|1|(),x) -> f#(s(x),x,x)
p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))
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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2,x3) = max{x1 + 4, x2 + 9, x3 + 10}
        |0|_A = 0
        |1|_A = 0
        s_A(x1) = max{3, x1}
        f_A(x1,x2,x3) = max{1, x1 - 2, x2 - 1, x3}
        rand_A(x1) = max{8, x1 + 4}
  
    precedence:
    
      f# = |0| = |1| = s = f = rand
    
    partial status:
  
      pi(f#) = [3]
      pi(|0|) = []
      pi(|1|) = []
      pi(s) = [1]
      pi(f) = [3]
      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#(|0|(),|1|(),x) -> f#(s(x),x,x)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))
r3: rand(x) -> x
r4: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  (no SCCs)