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