YES

TRS:

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

max/plus interpretations on N:

  bin_A(x1,x2) = max{6, 5 + x1, -1}
  bin#_A(x1,x2) = max{0, -2 + x1, 0}
  0_A = 4
  0#_A = 1
  s_A(x1) = max{2, 2 + x1}
  s#_A(x1) = max{0, 0}
  +_A(x1,x2) = max{7, 7, x2}
  +#_A(x1,x2) = max{0, -7 + x1, 0}

precedence:

  bin > s = + > 0