20 lines
709 B
Text
20 lines
709 B
Text
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)
|