[Haskell-cafe] instance Eq (a -> b)

Ashley Yakeley ashley at semantic.org
Wed Apr 14 02:42:07 EDT 2010

On Wed, 2010-04-14 at 16:11 +1000, Ivan Miljenovic wrote:
> but the only way you can "prove" it in
> Haskell is by comparing the values for the entire domain  (which gets
> computationally expensive)...

It's not expensive if the domain is, for instance, Bool.

Ashley Yakeley

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