YES

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

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