YES

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

Proof:
 DP Processor:
  DPs:
   +#(s(x),y) -> +#(x,y)
   +#(s(x),y) -> +#(x,s(y))
  TRS:
   +(0(),y) -> y
   +(s(x),y) -> s(+(x,y))
   +(s(x),y) -> +(x,s(y))
  Subterm Criterion Processor:
   simple projection:
    pi(+#) = 0
   problem:
    DPs:
     
    TRS:
     +(0(),y) -> y
     +(s(x),y) -> s(+(x,y))
     +(s(x),y) -> +(x,s(y))
   Qed