YES

TRS:

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

max/plus interpretations on N:

  f_A(x1) = max{11, 11}
  f#_A(x1) = max{5, 3}
  0_A = 1
  0#_A = 4
  s_A(x1) = max{6, 4 + x1}
  s#_A(x1) = max{2, x1}

precedence:

  s > f > 0