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
