YES

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

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

--- S
   1: +(x1,0()) -> x1
   2: +(x1,s(x2)) -> s(+(x1,x2))
   3: +(x1,p(x2)) -> p(+(x1,x2))
   4: +(0(),x2) -> x2
   5: +(s(x1),x2) -> s(+(x1,x2))
   6: +(p(x1),x2) -> p(+(x1,x2))
   7: s(p(x1)) -> x1
   8: p(s(x1)) -> x1
   9: +(+(x1,x2),x3) -> +(x1,+(x2,x3))
  10: +(x1,+(x2,x3)) -> +(+(x1,x2),x3)