Wolfgang Jeltsch wrote:
> First, I thought so too but I changed my mind. To my knowledge a type
> (forall a. T[a]) -> T' is equivalent to the type exists a. (T[a] -> T'). It’s
> the same as in predicate logic – Curry-Howard in action.
The connection is the other way round, I think.
(exists a. T[a]) -> T' = forall a. (T[a] -> T')
Regards,
apfelmus
--
http://apfelmus.nfshost.com
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