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Julian Ospald 2015-04-20 15:06:54 +02:00
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@ -2,7 +2,7 @@ To cut the story short, the abstract solution looks like this:
\begin{haskellcode}
fold :: b -> (a -> b -> b) -> [a] -> b
fold z f [] = z
fold z f (x:xs) = f x (fold z f xs)
fold z f (x:xs) = x `f` (fold z f xs)
\end{haskellcode}
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