Re: more detailed explanation about forall in Haskell

[Peter Hancock]

This is getting hysterically funny: 

    > *Jan Brosius writes:
    > * > >  > I must put this in the good way;
    > * > >  >
    > * > >  > [forall x . alpha(x)] => alpha(x) is True
    > * > >
    > * > > Yes, by instantiation.
    > * > 
    > * > I disagree.

    > *You disagree with my agreeing with you?

    > About what do you agree with me?

Please guys, you are making clowns of yourselves. 

--
Peter Hancock