YES

TRS:

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

max/plus interpretations on N:

  f_A(x1,x2) = max{1, 4, 1}
  f#_A(x1,x2) = max{0, -2 + x1, 3}
  0_A = 4
  0#_A = 1
  s_A(x1) = max{6, 6 + x1}
  s#_A(x1) = max{0, 0}

precedence:

  f > 0 = s