Section SubstPred.
Variable A : Type.
(* Take two covers an substitute the elements of one by the element of the other *)
Inductive subst_pred :
list (btree A) -> list (btree A) -> btree A -> btree A -> Prop :=
| subst_pred_id :
forall (t1 t2 : btree A) (l1 l2 : list (btree A)),
subst_pred (t1 :: nil) (t2 :: nil) t1 t2
| subst_pred_node :
forall (t1 t2 t3 t4 : btree A) (l1 l2 l3 l4 l5 l6 : list (btree A)),
subst_pred l1 l2 t1 t2 ->
subst_pred l3 l4 t3 t4 ->
subst_pred (l1 ++ l3) (l2 ++ l4) (node t1 t3) (node t2 t4).
Hint Resolve subst_pred_id subst_pred_node : core.
Proof using.
intros t1 t2 l1 l2 H; elim H; auto.
Qed.
Proof using.
intros t1 t2 l1 l2 H; elim H; auto.
Qed.

(* The two covers have same length *)
Theorem subst_pred_length :
forall (t1 t2 : btree A) (l1 l2 : list (btree A)),
subst_pred l1 l2 t1 t2 -> length l1 = length l2.