YES

1 decompositions
 #0 -----------
   1: +(0(),y) -> y
   2: +(s(0()),y) -> s(+(0(),y))
   3: +(x,y) -> +(y,x)
   4: s(s(x)) -> x

@Jouannaud and Kirchner's criterion
--- R
   1: +(0(),y) -> y
   2: +(s(0()),y) -> s(+(0(),y))
   3: +(x,y) -> +(y,x)
   4: s(s(x)) -> x

--- S
   1: +(0(),y) -> y
   2: +(s(0()),y) -> s(+(0(),y))
   3: +(x,y) -> +(y,x)
   4: s(s(x)) -> x