Re: [Haskell-cafe] Why is Bool no instance of Num and Bits?

Err, I'm not seeing the danger of this (+) :: forall a. (Num a) => a -> a -> a Doesn't this require the two parameters to be the same instance of Num?

I didn't at first, then I remembered: 1 + True = fromInteger 1 + True And if we have Num for Bool, it type checks. Thanks Neil