YES

TRS:

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

linear polynomial interpretations on N:

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

precedence:

  s > sum1 > + > sum > 0