YES

We show the termination of R/S, where R is

  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

and S is:

  rand(x) -> x
  rand(x) -> rand(s(x))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  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#(f(x)) -> f#(x)

and Q consists of the rules:

  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
  rand(x) -> x
  rand(x) -> rand(s(x))

The weakly monotone algebra (N^2, >_lex) with

  g#_A((x1,y1)) = (x1, y1)
  c_A((x1,y1),(x2,y2)) = (x2, y2)
  s_A((x1,y1)) = (x1, y1 + 1)
  f#_A((x1,y1)) = (0, 0)
  f_A((x1,y1)) = (x1 + 1, 1)
  d_A((x1,y1)) = (0, 0)
  g_A((x1,y1)) = (0, 0)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  g#(c(x,s(y))) -> g#(c(s(x),y))

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^2, >_lex) with

  f#_A((x1,y1)) = (x1, y1)
  c_A((x1,y1),(x2,y2)) = (x1 + x2, y1)
  s_A((x1,y1)) = (x1, y1)
  f_A((x1,y1)) = (x1 + 1, 1)
  d_A((x1,y1)) = (0, 2)
  g_A((x1,y1)) = (x1, y1)
  rand_A((x1,y1)) = (x1 + 1, y1)

strictly orients the following dependency pairs:

  f#(f(x)) -> f#(d(f(x)))
  f#(f(x)) -> f#(x)

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^2, >_lex) with

  f#_A((x1,y1)) = (x1, y1)
  c_A((x1,y1),(x2,y2)) = (x1, y1)
  s_A((x1,y1)) = (x1, y1 + 1)
  g_A((x1,y1)) = (x1, 0)
  f_A((x1,y1)) = (x1 + 1, 1)
  d_A((x1,y1)) = (0, 2)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

No dependency pair remains.