Qed.
Hint Resolve refl_l_egal.
Section more_about_words.
Variable f : Elt -> Elt.
Let wef := Word_ext f.
(* Lemma wef_cons : (a:Elt)(u:Word)(wef (cons a u))=(cons (f a) (wef u)). Proof [a:Elt][u:Word](refl_equal Word (wef (cons a u))). *)

Lemma wef_append :
forall u v : Word, wef (Append u v) = Append (wef u) (wef v).