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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1) = x1
        c_A(x1,x2) = x1 + x2 + (1,1)
        s_A(x1) = x1
        g_A(x1) = x1
        f_A(x1) = x1 + (1,0)
        d_A(x1) = x1
        rand_A(x1) = ((1,0),(1,1)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1) = ((0,1),(0,1)) x1
        c_A(x1,x2) = ((0,0),(1,1)) x2 + (1,1)
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        g_A(x1) = ((1,0),(1,1)) x1
        f_A(x1) = (1,1)
        d_A(x1) = ((1,1),(1,1)) x1
        rand_A(x1) = (0,0)
    

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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = x1
        c_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
        s_A(x1) = x1
        g_A(x1) = ((0,1),(1,0)) x1
        f_A(x1) = x1 + (2,1)
        d_A(x1) = (1,1)
        rand_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = ((1,1),(0,0)) x1
        c_A(x1,x2) = ((0,0),(1,1)) x1 + (1,0)
        s_A(x1) = ((0,1),(1,0)) x1 + (1,1)
        g_A(x1) = (0,0)
        f_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        d_A(x1) = (0,2)
        rand_A(x1) = (0,0)
    

The next rules are strictly ordered:

  p1

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