[Haskell-cafe] Re: forall & ST monad

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