Hi everyone, this is my first post: I am a ph.d. student from Italy who is learning the wonderful world of Haskell :) I have encountered a problem and I cannot find a way to get past it (or even to begin to understand it). I wish to define a simple type class that defines how to convert a type function into its argument (like going from a dumb constructor like data T a = T a into a itself): class Convert rec where convert :: rec a -> a now when I try to use the conversion operator class (CNum n, HasField n (a -> (b,rec a)) l, Convert rec) => HasMethod n l a b rec where (..!) :: l -> n -> (a -> (b,a)) instance (CNum n, HasField n (a -> (b,rec a)) l, Convert rec) => HasMethod n l a b rec where l ..! n = let m = l .! n in (\x -> let (y,v) = m x in (y,convert v)) in what looks to me as a straightforward use, the compiler complains that : *References> :load Main.hs [1 of 5] Compiling Records ( Records.hs, interpreted ) [2 of 5] Compiling References ( References.hs, interpreted ) [3 of 5] Compiling Methods ( Methods.hs, interpreted ) Methods.hs:25:14: Could not deduce (HasField n (a -> (b, rec a)) l) from the context (HasMethod n l a b rec1, CNum n, HasField n (a -> (b, rec1 a)) l, Convert rec1) arising from a use of `.!' at Methods.hs:25:14-19 Possible fix: add (HasField n (a -> (b, rec a)) l) to the context of the instance declaration or add an instance declaration for (HasField n (a -> (b, rec a)) l) In the expression: l .! n In the definition of `m': m = l .! n In the expression: let m = l .! n in (\ x -> let (y, v) = ... in (y, convert v)) Failed, modules loaded: References, Records. *References> I apologize for what might look like a huge dumping of code, but I have no idea what might have caused this to further refine the code or to write a more focused example...