YES

Problem:
 app(nil(),xs) -> nil()
 app(cons(x,xs),ys) -> cons(x,app(xs,ys))

Proof:
 DP Processor:
  DPs:
   app#(cons(x,xs),ys) -> app#(xs,ys)
  TRS:
   app(nil(),xs) -> nil()
   app(cons(x,xs),ys) -> cons(x,app(xs,ys))
  Subterm Criterion Processor:
   simple projection:
    pi(app#) = 0
   problem:
    DPs:
     
    TRS:
     app(nil(),xs) -> nil()
     app(cons(x,xs),ys) -> cons(x,app(xs,ys))
   Qed