Proof using.
intros.
induction H.
- constructor.
- eapply RT1n_step; eauto.
Qed.

Lemma refl_trans_1n_trace_n1_ind :
forall (step : step_relation) (P : A -> A -> list trace -> Prop),
(forall x, P x x []) ->
(forall x x' x'' tr1 tr2,
refl_trans_1n_trace step x x' tr1 ->
step x' x'' tr2 ->
P x x' tr1 ->
refl_trans_1n_trace step x x'' (tr1 ++ tr2) ->
P x x'' (tr1 ++ tr2)) ->
forall x y l,
refl_trans_1n_trace step x y l -> P x y l.