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 > rand > f# > f > g > |0| > |1|
      
      argument filter:
    
        pi(f#) = [3]
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = 1
        pi(g) = [1, 2]
        pi(f) = 3
        pi(rand) = [1]
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        f#_A(x1,x2,x3) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x3
        |0|_A() = (1,1,1,1)
        |1|_A() = (1,1,1,1)
        s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
        g_A(x1,x2) = (0,0,1,1)
        f_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x3
        rand_A(x1) = (0,0,1,1)
    

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)