I have a question about a problem in Yet Another Haskell Tutorial (problem 7.1). My answers seems to disagree with Hal's, and it fact something looks wrong in Hal's answer (maybe it's an error in his paper). The problem is to write the following function in "point-free" style: func5 f list = foldr (\x y -> f(y,x)) 0 list Here's how I approached it. It's easy to drop 'list' by the eta-reduction (using partial application): func5 f = foldr (\x y -> f(y,x)) 0 But how to get rid of f? First I reasoned that f is a function of a tuple. That is, it is not curried. So to curry it: func5 f = foldr (\x y -> (curry f) y x) 0 The second argument to foldr is obviously flipping the arugments, so func5 f = foldr (flip $ curry f) 0 Now I want to use the eta-reduction again, but I have to transform this expression into something that takes f as its last argument instead of the 0. I can use flip again, this time on foldr: func5 f = flip foldr 0 (flip $ curry f) Now f can be dropped: THE FINAL ANSWER: func5 = flip foldr 0 . flip. curry This works, in a couple of my test cases. Now Hal gives this as the answer: func5 = foldr (uncurry $ flip f) 0 The first problem is that there's no argument f, though he refers to it. So maybe he meant func5 f = foldr (uncurry $ flip f) 0 But more problems. He's applying flip to f, but f takes only one argument (a 2-tuple). Then he's uncurry-ing it, but I thought it needed currying, not uncurry-ing. Can anyone figure out what Hal is up to, or does it look like a simple mistake? Thanks, Mike