RE: Implicit 'forall' in data declarations

| On a related note, these are also apparently allowed (in 6.10.4): | f :: forall a. (Eq a => a -> a) -> a -> a | -- the Eq context prevents the function from ever being called. That's not true. E.g. f ((==) True) True works fine. | g :: forall a. Ord a => (Eq a => a -> a) -> a -> a | -- the Eq context is effectively ignored That's a bit more true, because Ord a implies Eq a, but still not really. It still says that you must pass evidence for equality to g's argument. Simon