YES

TRS:

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

max/plus interpretations on N:

  ack_A(x1,x2) = max{8, -4 + x1, x2}
  ack#_A(x1,x2) = max{1, -11 + x1, -7 + x2}
  0_A = 6
  0#_A = 0
  s_A(x1) = max{3, x1}
  s#_A(x1) = max{1, -6}

precedence:

  ack > s > 0