YES

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

  R:
  g(c(x,s(y))) -> g(c(s(x),y))
  f(c(s(x),y)) -> f(c(x,s(y)))
  f(f(x)) -> f(d(f(x)))
  f(x) -> x

  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#(c(x,s(y))) -> g#(c(s(x),y))
p2: f#(c(s(x),y)) -> f#(c(x,s(y)))
p3: f#(f(x)) -> f#(d(f(x)))

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x
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#(c(x,s(y))) -> g#(c(s(x),y))

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x
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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1) = max{5, x1}
        c_A(x1,x2) = max{x1 + 6, x2 + 12}
        s_A(x1) = max{2, x1}
        g_A(x1) = 0
        f_A(x1) = max{31, x1 + 16}
        d_A(x1) = max{31, x1}
        rand_A(x1) = max{5, x1 + 3}
  
    precedence:
    
      c = f > s > g# > g = d = rand
    
    partial status:
  
      pi(g#) = [1]
      pi(c) = [2]
      pi(s) = [1]
      pi(g) = []
      pi(f) = []
      pi(d) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: f#(c(s(x),y)) -> f#(c(x,s(y)))

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x
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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 5
        c_A(x1,x2) = max{x1 + 4, x2 + 4}
        s_A(x1) = x1
        g_A(x1) = max{5, x1 + 2}
        f_A(x1) = x1 + 6
        d_A(x1) = max{6, x1}
        rand_A(x1) = x1 + 2
  
    precedence:
    
      f# = c = s = g = f = d = rand
    
    partial status:
  
      pi(f#) = [1]
      pi(c) = [1, 2]
      pi(s) = [1]
      pi(g) = []
      pi(f) = [1]
      pi(d) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.