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. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > d > c > f > g > g#
      
      argument filter:
    
        pi(g#) = 1
        pi(c) = [2]
        pi(s) = 1
        pi(g) = []
        pi(f) = [1]
        pi(d) = 1
        pi(rand) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > d > c > f > g > g#
      
      argument filter:
    
        pi(g#) = 1
        pi(c) = 2
        pi(s) = [1]
        pi(g) = []
        pi(f) = []
        pi(d) = 1
        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. lexicographic path order with precedence:
    
      precedence:
      
        rand > s > d > f > g > f# > c
      
      argument filter:
    
        pi(f#) = 1
        pi(c) = 1
        pi(s) = 1
        pi(g) = []
        pi(f) = 1
        pi(d) = 1
        pi(rand) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > d > g > c > f > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(c) = 1
        pi(s) = [1]
        pi(g) = []
        pi(f) = 1
        pi(d) = 1
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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