YES

TRS:

  g(s(x)) -> f(x)
  f(0()) -> s(0())
  f(s(x)) -> s(s(g(x)))
  g(0()) -> 0()

max/plus interpretations on N:

  g_A(x1) = max{3, 4 + x1}
  g#_A(x1) = max{3, 4 + x1}
  s_A(x1) = max{0, 2 + x1}
  s#_A(x1) = max{0, 2 + x1}
  f_A(x1) = max{6, 6 + x1}
  f#_A(x1) = max{6, 6 + x1}
  0_A = 2
  0#_A = 2

precedence:

  0 > g = f > s