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