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16 lines
460 B
Idris
16 lines
460 B
Idris
%hide List.filter
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filter : (p : a -> Bool) -> 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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_ | True = x :: filter p xs
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_ | False = filter p xs
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filterFilter : (p : a -> Bool) -> (xs : List a) ->
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filter p (filter p xs) === filter p xs
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filterFilter p [] = Refl
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filterFilter p (x :: xs) with (p x) proof eq
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_ | False = filterFilter p xs
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_ | True with (p x)
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_ | True = cong (x ::) (filterFilter p xs)
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