YES

Problem:
 app(nil(),YS) -> YS
 app(cons(X),YS) -> cons(X)
 from(X) -> cons(X)
 zWadr(nil(),YS) -> nil()
 zWadr(XS,nil()) -> nil()
 zWadr(cons(X),cons(Y)) -> cons(app(Y,cons(X)))
 prefix(L) -> cons(nil())

Proof:
 DP Processor:
  DPs:
   zWadr#(cons(X),cons(Y)) -> app#(Y,cons(X))
  TRS:
   app(nil(),YS) -> YS
   app(cons(X),YS) -> cons(X)
   from(X) -> cons(X)
   zWadr(nil(),YS) -> nil()
   zWadr(XS,nil()) -> nil()
   zWadr(cons(X),cons(Y)) -> cons(app(Y,cons(X)))
   prefix(L) -> cons(nil())
  TDG Processor:
   DPs:
    zWadr#(cons(X),cons(Y)) -> app#(Y,cons(X))
   TRS:
    app(nil(),YS) -> YS
    app(cons(X),YS) -> cons(X)
    from(X) -> cons(X)
    zWadr(nil(),YS) -> nil()
    zWadr(XS,nil()) -> nil()
    zWadr(cons(X),cons(Y)) -> cons(app(Y,cons(X)))
    prefix(L) -> cons(nil())
   graph:
    
   Qed