[Haskell-cafe] Re: Propositional logic question

[Dave Tapley]

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,