YES

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

Proof:
 DP Processor:
  DPs:
   +#(s(x),s(y)) -> +#(y,0())
   +#(s(x),s(y)) -> +#(s(x),+(y,0()))
  TRS:
   +(0(),y) -> y
   +(s(x),0()) -> s(x)
   +(s(x),s(y)) -> s(+(s(x),+(y,0())))
  Usable Rule Processor:
   DPs:
    +#(s(x),s(y)) -> +#(y,0())
    +#(s(x),s(y)) -> +#(s(x),+(y,0()))
   TRS:
    +(0(),y) -> y
    +(s(x),0()) -> s(x)
   KBO Processor:
    argument filtering:
     pi(0) = []
     pi(+) = [0,1]
     pi(s) = [0]
     pi(+#) = 1
    usable rules:
     +(0(),y) -> y
     +(s(x),0()) -> s(x)
    weight function:
     w0 = 1
     w(+#) = w(s) = w(0) = 1
     w(+) = 0
    precedence:
     s ~ 0 > +# ~ +
    problem:
     DPs:
      
     TRS:
      +(0(),y) -> y
      +(s(x),0()) -> s(x)
    Qed