Sjoerd Visscher-2 wrote
>> class Foo f where
>> foo :: a -> f a
>>
>> data Bar f a = Foo f => Bar {bar :: f a}
>>
>> instance Foo (Bar f) where
>> foo a = Bar (foo a)
>
> No, you can only omit it where you provide Foo f in another way.
Which brings me back to my original question - is there any way that the
type system could be enhanced, so that the compiler "understands" that Bar f
=> Foo f without being told so explicitly every time?
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