[Haskell] Announce: revamped free theorems generator

Jean-Philippe Bernardy bernardy at chalmers.se
Tue Jun 22 10:02:00 EDT 2010

On Tue, Jun 22, 2010 at 1:32 PM, Janis Voigtländer
<jv at informatik.uni-bonn.de> wrote:

>> I need to read his paper again for the proof idea. Maybe I'll find a
>> counter example then.
> In personal communication, Jean-Philippe added that for your type "(Bool
> -> a) -> a" the statement that any such function is either ($True) or
> ($False) does *not* follow from the results in that paper. So I gather
> that you will find neither proof nor counterexample for the more general
> statement in that paper either. (It's worth reading nevertheless!)

Actually I had misread the type. Transforming

forall a. (Bool -> a) -> a



is a direct consequence of our main theorem.
In fact it is a consequence of the simpler theorem:

forall a. (F a -> a) -> a = μ F


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