NO

Problem:
 +(0(),y) -> y
 +(s(0()),y) -> s(y)
 +(s(s(x)),y) -> s(s(+(y,x)))
 +(x,+(y,z)) -> +(+(z,y),x)

Proof:
 Nonconfluence Processor:
  terms: +(y,z) *<- +(0(),+(y,z)) ->* +(+(z,y),0())
  Qed