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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