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))
  Usable Rule Processor:
   DPs:
    +#(s(x),y) -> +#(x,y)
    +#(s(x),y) -> +#(x,s(y))
   TRS:
    
   CDG Processor:
    DPs:
     +#(s(x),y) -> +#(x,y)
     +#(s(x),y) -> +#(x,s(y))
    TRS:
     
    graph:
     
    Qed