YES

TRS:

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

max/plus interpretations on N:

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

precedence:

  + > s > sum > 0