[Haskell-cafe] Re: Propositional logic question
Eric
eeoam at ukfsn.org
Mon Jun 25 16:11:58 EDT 2007
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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>
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