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))
  Subterm Criterion Processor:
   simple projection:
    pi(add#) = 0
   problem:
    DPs:
     
    TRS:
     add(0(),x) -> x
     add(s(x),y) -> s(add(x,y))
   Qed