Can someone provide a complete hand calculation of zip [1,2,3] [4,5,6] using the following definition of zip, based on foldr: zip :: [a] -> [b] -> [(a, b)] zip = foldr f e where e ys = [] f x g [ ] = [] f x g (y : ys) = (x , y) : g ys foldr :: (a -> b -> b) -> b -> ([a] -> b) foldr _ e [] = e foldr f e (x : xs) = f x (foldr f e xs) This implementation of zip produces the expected result [(1, 4), (2, 5), (3, 6)], but I'm unable to do the hand calculation and don't understand why it works. Part of my problem is that "e" is defined as a function that takes one argument, I don't see how that fits in with the usual scheme for foldr, which, as I understand it, is: foldr f e [x1, x2, ...] = f x1 (f x2 (f x3 ...(f xn e)))... Thanks, as always, to all in this great community. _________________________________________________________________ Windows Live™: Life without walls. http://windowslive.com/explore?ocid=TXT_TAGLM_WL_allup_1a_explore_032009