YES

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

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

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