
18 Jan
2007
18 Jan
'07
11:45 a.m.
On 18/01/07, Joachim Breitner
(.) :: (b -> c) -> (a -> b) -> a -> c id :: a -> a therefore b = a therefore _|_ :: a -> c
(This is mostly rough guesswork, I might be totally wrong)
That much is right, but remember that just because _|_ has type a -> c doesn't mean it takes a parameter. Bottom can take any type, and I don't think _|_ == \x -> _|_. -- -David House, dmhouse@gmail.com