[Haskell-beginners] Trying to prove Applicative is superclass of Functor, etc

[Dániel Arató]

For a drunk person you're getting these equivalences surprisingly right. :)

>> Ah true, I heard about this Dániel. But then it would be generalized to
> all classes, or just those three ones?

It would make sense for it to work for any class hierarchy, but
honestly I just don't know and I've got an old GHC on my laptop so I
can't check. Maybe someone who actually knows will come along and
enlighten us. ¯\_(ツ)_/¯

>> Anyway, trying the same with Applicative and its *sub*class Monad:
>>> pure = return

That's right.

>>> mf <*> ma =
>>    let mg = mf >>= \f -> return (return . f)
>>    in mg >>= \g -> (ma >>= g)

>> Is there an easier solution? It's easy enough, but not as trivial as for
> Applicative => Functor.

I've got (an alpha-equivalent of) the following on a scrap piece of paper:
mf <*> mx = mf >>= \f -> mx >>= \x -> return . f $ x

It reads nice and fluent.

>> Why do we write constraints like that:
>>> Functor f => (a -> b) -> f a -> f b
>> or:
>>> Functor f => Applicative f

That's a very good question. I agree that it would make a bit more
sense if the arrows were reversed.
I think it helps to read
"Functor f => Applicative f where ..."
as
"if you've got a Functor then you can have an Applicative if you just
implement these functions"
as opposed to
"if f is a Functor that implies that f is an Applicative" (wrong).

Daniel