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. lexicographic path order with precedence:
    
      precedence:
      
        s > f# > g > rand > |1| > |0| > f
      
      argument filter:
    
        pi(f#) = [3]
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = 1
        pi(g) = [1, 2]
        pi(f) = 3
        pi(rand) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f# > |0| > g > rand > s > |1| > f
      
      argument filter:
    
        pi(f#) = [3]
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = 1
        pi(g) = 1
        pi(f) = [3]
        pi(rand) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        rand > g > |1| > |0| > f > s > f#
      
      argument filter:
    
        pi(f#) = [3]
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = [1]
        pi(g) = 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: 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)