[Haskell-cafe] Free theorem for `forall z. (A,z) -> (B,z)`?

Joachim Breitner mail at joachim-breitner.de
Mon Jul 23 15:33:39 UTC 2018

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
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

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


   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.


Joachim Breitner
  mail at joachim-breitner.de
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