YES

TRS:

  sum(0()) -> 0()
  sum(s(x)) -> +(sum(x),s(x))
  sum1(0()) -> 0()
  sum1(s(x)) -> s(+(sum1(x),+(x,x)))

max/plus interpretations on N:

  sum_A(x1) = max{5, 3 + x1}
  sum#_A(x1) = max{5, 3 + x1}
  0_A = 0
  0#_A = 0
  s_A(x1) = max{6, 4 + x1}
  s#_A(x1) = max{6, 4 + x1}
  +_A(x1,x2) = max{0, x1, x2}
  +#_A(x1,x2) = max{0, x1, x2}
  sum1_A(x1) = max{3, 1 + x1}
  sum1#_A(x1) = max{3, 1 + x1}

precedence:

  0 > sum = s = sum1 > +