Proof using.
intros l1; elim l1; simpl in |- *; auto.
intros l2 l3 l4 a b H; right; left; exists l3; auto.
intros a l H l2 l3 l4 a0 b; case l3; simpl in |- *.
case l; simpl in |- *.
intros H0; right; right; injection H0; split; auto.
apply f_equal2 with (f := cons (A:=A)); auto.
intros b0 l0 H0; left; exists l0; injection H0; intros;
repeat apply f_equal2 with (f := cons (A:=A)); auto.
intros b0 l0 H0; case (H l2 l0 l4 a0 b); auto.
injection H0; auto.
intros (l5, HH1); left; exists l5; apply f_equal2 with (f := cons (A:=A));
auto; injection H0; auto.
intros [H1| (H1, H2)]; auto.
right; right; split; auto; apply f_equal2 with (f := cons (A:=A)); auto;
injection H0; auto.
Qed.

Theorem same_length_ex :
forall (a : A) l1 l2 l3,
length (l1 ++ a :: l2) = length l3 ->
exists l4 : _,
(exists l5 : _,
(exists b : B,
length l1 = length l4 /\ length l2 = length l5 /\ l3 = l4 ++ b :: l5)).