YES

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

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