YES

TRS:

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

max/plus interpretations on N:

  +_A(x1,x2) = max{3, 2 + x1, 1 + x2}
  +#_A(x1,x2) = max{3, 6, 1}
  0_A = 1
  0#_A = 0
  s_A(x1) = max{5, 3}
  s#_A(x1) = max{0, 2}

precedence:

  + > 0 = s