YES

TRS:

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

max/plus interpretations on N:

  g_A(x1) = max{3, 1 + x1}
  g#_A(x1) = max{5, 4 + x1}
  0_A = 5
  0#_A = 1
  s_A(x1) = max{2, 4 + x1}
  s#_A(x1) = max{1, 3}
  f_A(x1) = max{5, 4}
  f#_A(x1) = max{2, 0}

precedence:

  g > s > f > 0