YES

1 decompositions
 #1 -----------
   3: +(0(),x2) -> x2
   4: +(s(x1),x2) -> s(+(x1,x2))
   5: sum(0()) -> 0()
   6: sum(s(x1)) -> +(s(x1),sum(x1))
   7: +(+(x1,x2),x3) -> +(x1,+(x2,x3))
   8: +(x1,x2) -> +(x2,x1)

@Jouannaud and Kirchner's criterion
--- R
   3: +(0(),x2) -> x2
   4: +(s(x1),x2) -> s(+(x1,x2))
   5: sum(0()) -> 0()
   6: sum(s(x1)) -> +(s(x1),sum(x1))
   7: +(+(x1,x2),x3) -> +(x1,+(x2,x3))
   8: +(x1,x2) -> +(x2,x1)

--- S
   3: +(0(),x2) -> x2
   4: +(s(x1),x2) -> s(+(x1,x2))
   5: sum(0()) -> 0()
   6: sum(s(x1)) -> +(s(x1),sum(x1))
   7: +(+(x1,x2),x3) -> +(x1,+(x2,x3))
   8: +(x1,x2) -> +(x2,x1)


NOTE: input TRS is reduced

original is
   1: +(x1,0()) -> x1
   2: +(x1,s(x2)) -> s(+(x1,x2))
   3: +(0(),x2) -> x2
   4: +(s(x1),x2) -> s(+(x1,x2))
   5: sum(0()) -> 0()
   6: sum(s(x1)) -> +(s(x1),sum(x1))
   7: +(+(x1,x2),x3) -> +(x1,+(x2,x3))
   8: +(x1,x2) -> +(x2,x1)

reduced to
   3: +(0(),x2) -> x2
   4: +(s(x1),x2) -> s(+(x1,x2))
   5: sum(0()) -> 0()
   6: sum(s(x1)) -> +(s(x1),sum(x1))
   7: +(+(x1,x2),x3) -> +(x1,+(x2,x3))
   8: +(x1,x2) -> +(x2,x1)