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