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:
    
   KBO Processor:
    argument filtering:
     pi(s) = [0]
     pi(+#) = 0
    usable rules:
     
    weight function:
     w0 = 1
     w(+#) = 1
     w(s) = 0
    precedence:
     +# ~ s
    problem:
     DPs:
      
     TRS:
      
    Qed