http://rvp74.livejournal.com/ ([identity profile] rvp74.livejournal.com) wrote in [personal profile] deniok 2014-10-05 08:56 am (UTC)

второй почему-то сразу не дался, но решение оказалось еще проще:

ex2 :: (Either a b -> c) -> (a -> c, b -> c)
ex2 h = (h . Left, h . Right)

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