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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{9, x1}
          c_A(x1,x2) = max{x1 + 10, x2 + 12}
          s_A(x1) = x1
          g_A(x1) = max{11, x1}
          f_A(x1) = max{5, x1 + 3}
          d_A(x1) = max{4, x1}
          rand_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        s = f > c > g# = g = d > rand
      
      partial status:
    
        pi(g#) = [1]
        pi(c) = [2]
        pi(s) = [1]
        pi(g) = [1]
        pi(f) = []
        pi(d) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = x1 + 3
          c_A(x1,x2) = x2 + 7
          s_A(x1) = max{13, x1 + 4}
          g_A(x1) = max{40, x1 + 20}
          f_A(x1) = 9
          d_A(x1) = max{15, x1 + 6}
          rand_A(x1) = 2
    
      precedence:
      
        s = f = d > g# = c = g = rand
      
      partial status:
    
        pi(g#) = [1]
        pi(c) = [2]
        pi(s) = []
        pi(g) = [1]
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{2, x1}
          c_A(x1,x2) = max{x1 + 6, x2 + 6}
          s_A(x1) = max{3, x1}
          g_A(x1) = max{10, x1 + 5}
          f_A(x1) = x1 + 10
          d_A(x1) = max{0, x1 - 11}
          rand_A(x1) = max{8, x1 + 4}
    
      precedence:
      
        c = s > f# > 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) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = 1
          c_A(x1,x2) = max{x1 + 12, x2 + 6}
          s_A(x1) = max{5, x1}
          g_A(x1) = 13
          f_A(x1) = x1 + 19
          d_A(x1) = 1
          rand_A(x1) = 1
    
      precedence:
      
        f = d > f# = c = s = g = rand
      
      partial status:
    
        pi(f#) = []
        pi(c) = [1]
        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.