21 Aug
2007
21 Aug
'07
10:21 p.m.
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