Use the right fold definition Set The derived Haskell definition is thereforeEach step prepends the next element to the accumulator. This is a left fold derivation from list concatenation, rather than merely a proposed implementation.
xs ++ ys = foldr (:) ys xs, the right-identity law xs ++ [] = xs, and the finite-list involution law reverse (reverse xs) = xs. The last two laws are the additional results used here. Part (b) now givesxs ++ ys
= foldr (:) ys xs
= foldl (flip (:)) ys (reverse xs)ys = [] to obtain xs = foldl (flip (:)) [] (reverse xs). Substitute reverse xs for xs and simplify the double reversal:reverse xs
= foldl (flip (:)) [] (reverse (reverse xs))
= foldl (flip (:)) [] xsreverse :: [a] -> [a]
reverse = foldl (flip (:)) [] Articles by others on the same topic
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