YES

TRS:

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

max/plus interpretations on N:

  double_A(x1) = max{1, 2 + x1}
  double#_A(x1) = max{0, 2}
  0_A = 2
  0#_A = 3
  s_A(x1) = max{5, 2}
  s#_A(x1) = max{1, 1}
  +_A(x1,x2) = max{2, 1 + x1, 1 + x2}
  +#_A(x1,x2) = max{1, 1, -4}

precedence:

  double > 0 = + > s