[Haskell-cafe] Re: forall & ST monad

Heinrich Apfelmus apfelmus at quantentunnel.de
Tue Feb 17 07:28:18 EST 2009

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')



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