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29 lines
875 B
Idris
29 lines
875 B
Idris
module WithProof
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%default total
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%hide List.filter
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filter : (p : a -> Bool) -> (xs : List a) -> List a
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filter p [] = []
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filter p (x :: xs) with (p x)
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filter p (x :: xs) | False = filter p xs
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filter p (x :: xs) | True = x :: filter p xs
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filterSquared : (p : a -> Bool) -> (xs : List a) ->
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filter p (filter p xs) === filter p xs
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filterSquared p [] = Refl
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{-
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filterSquared p (x :: xs) with (p x)
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filterSquared p (x :: xs) | False = filterSquared p xs -- easy
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filterSquared p (x :: xs) | True = ?a
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-- argh! stuck on another with-block casing on (p x)!
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-- we could check (p x) again but how do we prove it
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-- can only ever be `True`?!
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-}
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filterSquared p (x :: xs) with (p x) proof eq
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filterSquared p (x :: xs) | False = filterSquared p xs -- easy
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filterSquared p (x :: xs) | True
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= rewrite eq in cong (x ::) (filterSquared p xs)
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