YES

TRS:

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

max/plus interpretations on N:

  f_A(x1) = max{0, x1}
  f#_A(x1) = max{1, -2 + x1}
  s_A(x1) = max{5, 8 + x1}
  s#_A(x1) = max{5, 4}
  p_A(x1) = max{0, -8 + x1}
  p#_A(x1) = max{0, 2}
  0_A = 1
  0#_A = 0

precedence:

  f > s = 0 > p