[Haskell-cafe] Re: Propositional logic question

[Eric]

This seems rather complicated! What about this:

     A => (B => C)
=      { X => Y   ==  ¬X \/ Y }
    ¬A \/ (¬B \/ C)
=      {associativity}
    (¬A \/ ¬B) \/ C
=      { DeMorgan }
    ¬(A /\ B) \/ C
=      { X => Y   ==  ¬X \/ Y }
    A /\ B   =>  C

E.

Dave Tapley wrote:
> Whoops, okay after two lines (thanks to oerjan) on #haskell I realise
> that yes, it is as easy as it should have been.
>
> For completeness:
>
> [A /\ B]1
> ------------ (/\ E1)      [A => (B => C)]2
>      A
> ------------------------------------------------- (=> E)
>                                               B => C
> [A /\ B]1
> ------------ (/\ E2)
>      B                                      B => C
> ------------------------------------------------- (=> E)
>                                                         C
>                                             ------------------ (=> I)1
>                                             (A /\ B) => C
>
> ------------------------------------------------ (=> I)2
>                                             (A => (B => C)) => ((A /\ 
> B) => C)
>
>
> Learning is fun :)
>
> Dave,
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