YES

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

  R:
  g(s(x)) -> f(x)
  f(|0|()) -> s(|0|())
  f(s(x)) -> s(s(g(x)))
  g(|0|()) -> |0|()

  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: g#(s(x)) -> f#(x)
p2: f#(s(x)) -> g#(x)

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()
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: g#(s(x)) -> f#(x)
p2: f#(s(x)) -> g#(x)

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()
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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = x1
            s_A(x1) = x1
            f#_A(x1) = x1
            g_A(x1) = x1 + (1,1)
            f_A(x1) = x1 + (1,1)
            |0|_A() = (1,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        g = f = rand > g# = s > f# > |0|
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = [1]
        pi(f#) = [1]
        pi(g) = [1]
        pi(f) = [1]
        pi(|0|) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((0,0),(1,0)) x1 + (5,1)
            s_A(x1) = ((0,0),(1,0)) x1 + (3,0)
            f#_A(x1) = x1 + (4,2)
            g_A(x1) = (5,7)
            f_A(x1) = (4,6)
            |0|_A() = (2,1)
            rand_A(x1) = (1,1)
    
      precedence:
      
        g > f > s > f# > g# = |0| = rand
      
      partial status:
    
        pi(g#) = []
        pi(s) = []
        pi(f#) = []
        pi(g) = []
        pi(f) = []
        pi(|0|) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  (no SCCs)