Consider the set of all rationals with 1 as a numerator, and positive denominator, eg: S = {1/n, n : Nat} this is bounded, enumerable, but infinite. Which makes the whole checking every value bit somewhat, shall we say, difficult. :) So for instance, we want to show f : S -> S f(1/n) = 1/2n and g : S -> S g(1/n) = 1/2 * 1/n would be impossible. Since we would have to check infinitely many values of `n` This, of course, presumes I have understood everything, which seems to be less likely every day. On Apr 14, 2010, at 2:03 AM, Ashley Yakeley wrote:
Why isn't there an instance Eq (a -> b) ?
allValues :: (Bounded a,Enum a) => [a] allValues = enumFrom minBound
instance (Bounded a,Enum a,Eq b) => Eq (a -> b) where p == q = fmap p allValues == fmap q allValues
Of course, it's not perfect, since empty types are finite but not Bounded. One can nevertheless make them instances of Bounded with undefined bounds, and have enumFrom and friends always return the empty list.
It seems one should also be able to write
instance (Bounded a,Enum a) => Traversable (a -> b) where ???
But this turns out to be curiously hard.
-- Ashley Yakeley
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