Variable vector : Type -> Nat -> Type
Axiom N0 (n : Nat) : n + 0 = n
Theorem V0 (T : Type) (n : Nat) : (vector T (n + 0)) = (vector T n) :=
   Congr (Refl (vector T)) (N0 n)
Variable f (n : Nat) (v : vector Int n) : Int
Variable m : Nat
Variable v1 : vector Int (m + 0)
(*
   The following application will fail because (vector Int (m + 0)) and (vector Int m)
   are not definitionally equal.
*)
Check f m v1
(*
   The next one succeeds using the "casting" operator.
   We can do it, because (V0 Int m) is a proof that
   (vector Int (m + 0)) and (vector Int m) are propositionally equal.
   That is, they have the same interpretation in the lean set theoretic
   semantics.
*)
Check f m (cast (V0 Int m) v1)