Ralf Hinze wrote:
> [...] Rewriting the definition to
> 
> > (@@) f g a            =  g a >>= f
> 
> avoids the problem.

I discovered something similar yesterday. Things are even more funny:

> (f @@ g) = \a -> g a >>= f  -- Does not work

> f @@ g = \a -> g a >>= f   -- Does work!

-- 
Sven Panne                                        Tel.: +49/89/2178-2235
LMU, Institut fuer Informatik                     FAX : +49/89/2178-2211
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