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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          g#_A(x1) = x1
          c_A(x1,x2) = x2
          s_A(x1) = x1
          g_A(x1) = ((1,0),(0,0)) x1
          f_A(x1) = ((1,0),(1,1)) x1 + (2,1)
          d_A(x1) = (1,1)
          rand_A(x1) = x1 + (1,0)
  
    precedence:
    
      s > g# = c = g = f = 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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1) = x1
          c_A(x1,x2) = x1 + (1,2)
          s_A(x1) = x1
          g_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          f_A(x1) = x1 + (2,1)
          d_A(x1) = (1,1)
          rand_A(x1) = x1 + (1,1)
  
    precedence:
    
      f = d > c > f# = s = g > rand
    
    partial status:
  
      pi(f#) = [1]
      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.