Hi Conal, Am Sonntag, den 22.07.2018, 21:29 -0700 schrieb Conal Elliott:
Suppose `g :: forall z. (A,z) -> (B,z)`. Is it necessarily the case that `g = first f` for some `f :: A -> B` (where `first f (a,b) = (f a, b)`), perhaps as a free theorem for the type of `g`?
there used to be a free theorem calculator at http://www-ps.iai.uni-bonn.de/ft but it seems to be down :-( There is a shell client, ftshell, lets see what it says… it does not build with ghc-8.0 any more :-( Ah, but lambdabot understands @free: <nomeata> @free g :: forall z. (A,z) -> (B,z) <lambdabot> $map_Pair $id f . g = g . $map_Pair $id f Which is basically what you are writing here:
Note that `(A,)` and `(B,)` are functors and that `g` is a natural transformation, so `g . fmap h == fmap h . g` for any `h :: u -> v`. Equivalently, `g . second h == second h . g` (where `second h (a,b) = (a, h b)`).
Doesn’t that already answer the question? Let’s try to make it more rigorous. I define f :: A -> B f a = fst (g (a, ()) and now want to prove that g = first f, using the free theorem… which is tricky, because the free theorem is actually not as strong as it could be; it should actually be phrased in terms of relations relation, which might help with the proof. So let me try to calculate the free theorem by hand, following https://www.well-typed.com/blog/2015/05/parametricity/ g is related to itself by the relation [forall z. (A,z) -> (B,z)] and we can calculate g [forall z. (A,z) -> (B,z)] g ↔ ∀ z z' Rz, g [(A,z) -> (B,z)] g -- Rz relates z and z' ↔ ∀ z z' Rz, ∀ p p', p [(A,z)] p' → g p [(B,z)] g p' ↔ ∀ z z' Rz, ∀ p p', fst p = fst p' ∧ snd p Rz snd p' → fst (g p) = fst (g p') ∧ snd (g p) Rz snd (g p') ↔ ∀ z z' Rz, ∀ a x x', x Rz x' → fst (g (a,x)) = fst (g (a,x')) ∧ snd (g (a,x)) Rz snd (g (a,x')) We can use this to show ∀ (a,x :: z). fst (g (a,x)) = f a using (this is the tricky part that requires thinking) z := z, z' := (), Rz := λ_ _ -> True, a := a, x := x, x' := () as then the theorem implies fst (g (a,x)) = fst (g (a, ()) = f a. We can also use it to show ∀ (a,x :: z). snd (g (a,x)) = x using z := z, z' := z, Rz := const (= x), a := a, x := x, x' := x as then the theorem implies, under the true assumption x Rz z' (snd (g (a,x))) Rz (snd (g (a,x'))) ↔ snd (g (a,x)) = x And from ∀ (a,x :: z). fst (g (a,x)) = f a ∀ (a,x :: z). snd (g (a,x)) = x we can conclude that g = first f as intended. Cheers, Joachim -- Joachim Breitner mail@joachim-breitner.de http://www.joachim-breitner.de/