I believe R J was consciously restricting himself to finite lists in the original post.
2009/3/18 MigMit <[email protected]> > More interesting: > > foldl (flip const) whatever (repeat 1 ++ [1,2,3]) > > Daniel Fischer wrote on 18.03.2009 15:17: > > Am Mittwoch, 18. März 2009 13:10 schrieb Daniel Fischer: >> >>> Now prove the >>> >>> Lemma: >>> >>> foldl g e (ys ++ zs) = foldl g (foldl g e ys) zs >>> >>> for all g, e, ys and zs of interest. >>> (I don't see immediately under which conditions this identity could >>> break, >>> maybe there aren't any) >>> >> >> Of course, hit send and you immediately think of >> >> foldl (flip const) whatever (undefined ++ [1,2,3]) >> _______________________________________________ >> Haskell-Cafe mailing list >> [email protected] >> http://www.haskell.org/mailman/listinfo/haskell-cafe >> >> _______________________________________________ > Haskell-Cafe mailing list > [email protected] > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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