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