YES

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

  R:
  g(x,y) -> x
  g(x,y) -> y
  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: g(x,y) -> x
r2: g(x,y) -> y
r3: f(|0|(),|1|(),x) -> f(s(x),x,x)
r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))
r5: rand(x) -> x
r6: 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: g(x,y) -> x
r2: g(x,y) -> y
r3: f(|0|(),|1|(),x) -> f(s(x),x,x)
r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = max{x1 - 1, x3}
          |0|_A = 2
          |1|_A = 0
          s_A(x1) = max{2, x1}
          g_A(x1,x2) = max{x1, x2 + 1}
          f_A(x1,x2,x3) = max{4, x1 + 3, x3 + 3}
          rand_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        |1| = g = f > f# = |0| = s > rand
      
      partial status:
    
        pi(f#) = [3]
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = [1]
        pi(g) = [1, 2]
        pi(f) = [3]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = x3 + 11
          |0|_A = 13
          |1|_A = 13
          s_A(x1) = max{10, x1 + 9}
          g_A(x1,x2) = max{x1 + 1, x2 + 1}
          f_A(x1,x2,x3) = max{12, x3 + 11}
          rand_A(x1) = 1
    
      precedence:
      
        f# > |0| = |1| = g = f > s > rand
      
      partial status:
    
        pi(f#) = []
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = []
        pi(g) = [1, 2]
        pi(f) = []
        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: g(x,y) -> x
r2: g(x,y) -> y
r3: f(|0|(),|1|(),x) -> f(s(x),x,x)
r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  (no SCCs)