Brandon S. Allbery KF8NH wrote:
On Aug 21, 2007, at 22:13 , Twan van Laarhoven wrote:
Other rules that could be interesting are:
> forall a b. fromInteger a + fromInteger b = fromInteger (a + b)
I don't think this will work, a and b have to be the same type.
They are of the same type, both are Integers,
> forall a b :: Integer.
> ((fromInteger (a::Integer)) + (fromInteger b)) :: Num n => n
> =
> (fromInteger (a + b :: Integer)) :: Num n => n
Twan
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