Maybe I'm missing something, but can't you just define:
(≡) f g = (f &&& g) >>^ (==)
(You could also define it as: (≡) f g x = f x == g x)
And use ≡ at will?
Why do you need an instance or a newtype?
Ivan
On Wed, 31 May 2023 at 04:25, Olaf Klinke
Dear Cafe,
The expression
\x -> f x == g x
is a testable property, as long as values for x can be randomly generated. For clarity I'd prefer a point-free style, e.g.
f ≡ g
Are there extensions to QuickCheck that let me write this? The QuickCheck package itself does not seem to contain such an operator. My current work-around is a newtype ExtensionalEquality a b that holds two functions of type (a -> b) and a Testable instance for it. But I've got a hunch that I re-invented some wheel here. (My ExtensionalEquality is isomorphic to Refl (a -> b) (a -> b) but Refl ist conceptually about type equality, not term equality.)
Thanks Olaf
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