Why are Monoidal and Applicative laws telling us the same thing? - haskell

Question

I have learnt about Monoidal being an alternative way to represent Applicative not so long ago. There is an interesting question on Typeclassopedia:
(Tricky) Prove that given your implementations from the first exercise [pure and (<*>) written down using unit and (**) and the other way around], the usual Applicative laws and the Monoidal laws stated above are equivalent.
Here are these classes and laws:
-- A note from https://wiki.haskell.org/Typeclassopedia#Alternative_formulation:
-- In this and the following laws, ≅ refers to isomorphism rather than equality.
-- In particular we consider (x,()) ≅ x ≅ ((),x) and ((x,y),z) ≅ (x,(y,z)).
-- Monoidal.
class Functor f => Monoidal f where
unit :: f ()
(**) :: f a -> f b -> f (a,b)
-- unit ** v ≅ v - Left Identity.
-- u ** unit ≅ u - Right Identity.
-- u ** (v ** w) ≅ (u ** v) ** w - Associativity.
-- Applicative.
class Functor f => Applicative f where
pure :: a -> f a
infixl 4 <*>, ...
(<*>) :: f (a -> b) -> f a -> f b
...
-- pure id <*> v = v - Identity.
-- pure f <*> pure x = pure (f x) - Homomorphism.
-- u <*> pure y = pure ($ y) <*> u - Interchange.
-- u <*> (v <*> w) = pure (.) <*> u <*> v <*> w - Composition.
Writing down combinators using others is no big deal:
unit = pure ()
f ** g = (,) <$> f <*> g = liftA2 (,) f g
pure x = const x <$> unit
f <*> g = uncurry ($) <$> (f ** g)
Here is my understanding of why the laws are telling us the same thing:
u <*> pure y = pure ($ y) <*> u -- Interchange: Applicative law.
The first thing we shall notice is that ($ y) ≅ y (more formally: (y -> a) -> a ≅ y). Having that in mind, Interchange law simply tells us that (a, b) ≅ (b, a).
pure id <*> v = v -- Identity: Applicative law.
I reckon id to be something of a unit itself as it is the only inhabitant of type forall a. a -> a. Therefore, this law gives us the Left Identity:
unit ** v = v -- Left Identity: Monoidal law.
Now we can use that (a, b) ≅ (b, a) to write the Right Identity down:
u ** unit = u -- Right Identity: Monoidal law.
The Composition law:
u <*> (v <*> w) = pure (.) <*> u <*> v <*> w -- Composition: Applicative law.
I reckon this law to tell the same thing as Associativity for Monoidal:
u ** (v ** w) ≅ (u ** v) ** w
That is, (a, (b, c)) ≅ ((a, b), c). Applicative just adds a layer of application.
So, we have covered all of the Monoidal laws. I believe there is no need to do it the other way around as we are going to use the same isomorphisms. But one could have noticed something odd - we did not use the Homomorphism Applicative law:
pure f <*> pure x = pure (f x)
I tried understanding Homomorphism in terms of the Naturality free theorem for Monoidal:
fmap (g *** h) (u ** v) = fmap g u ** fmap h v
But it seems odd as Homomorphism does not deal with side-effects, yet Naturality works with them just fine.
So, I have 3 questions:
Is my reasoning right?
Where does Homomorphism stand in this picture?
How can we understand the Naturality free theorem in terms of Applicative?

Answer 1

Just dumping this here for now... wanted to discuss this but I already spent way to long implementing it: it's a Coq proof script that shows the equivalence in an absolutely waterproof way.
Require Import Coq.Program.Basics.
Require Import Coq.Init.Datatypes.
Require Import Coq.Init.Notations.
Notation "f ∘ g" := (compose f g).
Class Functor (F: Type -> Type) : Type :=
{ fmap : forall {x} {y}, (x->y) -> (F x->F y)
; fmap_id : forall x, #fmap x x id = id
; fmap_compose : forall {x} {y} {z} (f: y->z) (g: x->y)
, fmap (f∘g) = fmap f ∘ fmap g
}.
Lemma fmap_twice {F} `{Functor F} {x} {y} {z} (f: y->z) (g: x->y) (xs: F x)
: fmap (f∘g) xs = fmap f (fmap g xs).
Proof.
rewrite fmap_compose. now compute.
Qed.
Definition parallel {a} {b} {c} {d} (f: a->c) (g: b->d)
: (a*b) -> (c*d) := fun xy => match xy with
| (x,y) => (f x, g y)
end.
Notation "f *** g" := (parallel f g) (at level 40, left associativity).
Definition rassoc {a} {b} {c} : ((a*b)*c) -> (a*(b*c))
:= fun xyz => match xyz with | ((x,y),z) => (x,(y,z)) end.
Definition tt_ {a} (x:a) := (tt, x).
Definition _tt {a} (x:a) := (x, tt).
Class Monoidal F `{Functor F} : Type :=
{ funit : F unit
; fzip : forall {a} {b}, F a -> F b -> F (a*b)
; left_identity : forall {a} (v: F a)
, fzip funit v = fmap tt_ v
; right_identity : forall {a} (v: F a)
, fzip v funit = fmap _tt v
; associativity : forall {a} {b} {c} (u: F a) (v: F b) (w: F c)
, fzip u (fzip v w) = fmap rassoc (fzip (fzip u v) w)
; naturality : forall {a} {b} {c} {d}
(g: a->c) (h: b->d) (u: F a) (v: F b)
, fmap (g***h) (fzip u v) = fzip (fmap g u) (fmap h v)
}.
Notation "u ** v" := (fzip u v) (at level 40, left associativity).
Lemma naturalityL {F} `{Monoidal F} {a} {b} {c}
(f: a->c) (u: F a) (v: F b)
: fmap (f***id) (fzip u v) = fzip (fmap f u) v.
Proof.
assert (v = fmap id v) as ->. { now rewrite fmap_id. }
rewrite <- naturality.
assert (v = fmap id v) as <-. { now rewrite fmap_id. }
now trivial.
Qed.
Lemma naturalityR {F} `{Monoidal F} {a} {b} {c}
(f: b->c) (u: F a) (v: F b)
: fmap (id***f) (fzip u v) = fzip u (fmap f v).
Proof.
assert (u = fmap id u) as ->. { now rewrite fmap_id. }
rewrite <- naturality.
assert (u = fmap id u) as <-. { now rewrite fmap_id. }
now trivial.
Qed.
Definition to {a} {b} (y: a) (f: a->b) := f y.
Class Applicative F `{Functor F} : Type :=
{ pure : forall {a}, a -> F a
; app : forall {a} {b}, F (a->b) -> F a -> F b
; identity : forall {a} (v: F a)
, app (pure id) v = v
; homomorphism : forall {a} {b} (f: a->b) (x: a)
, app (pure f) (pure x) = pure (f x)
; interchange : forall {a} {b} (u: F (a->b)) (y: a)
, app u (pure y) = app (pure (to y)) u
; composition : forall {a} {b} {c}
(u: F (b->c)) (v: F (a->b)) (w: F a)
, app u (app v w) = app (app (app (pure compose) u) v) w
; appFtor : forall {a} {b} (g: a->b) (x: F a)
, fmap g x = app (pure g) x
}.
Notation "fs <*> xs" := (app fs xs) (at level 40, left associativity).
Require Import Coq.Program.Tactics.
Require Import Coq.Logic.FunctionalExtensionality.
Definition apl {a} {b} (fx: (a->b)*a)
:= match fx with |(f,x) => f x end.
Program Instance MonoidalIsApplicative {F} `{Monoidal F}
: Applicative F
:= { pure := fun {a} (x: a) => fmap (const x) funit
; app := fun {a} {b} (fs: F (a->b)) (xs: F a)
=> fmap apl (fzip fs xs) }.
Next Obligation. (* identity *)
rewrite <- naturalityL.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
rewrite -> fmap_id.
now compute.
Qed.
Next Obligation. (* homomorphism *)
rewrite <- naturality.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
now compute.
Qed.
Next Obligation. (* interchange *)
rewrite <- naturalityL.
rewrite <- naturalityR.
repeat (rewrite <- fmap_twice).
rewrite -> right_identity.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
now compute.
Qed.
Next Obligation. (* composition *)
rewrite <- naturalityR.
rewrite -> associativity.
repeat (rewrite <- naturalityL).
rewrite -> left_identity.
repeat (rewrite <- naturalityL).
repeat (rewrite <- fmap_twice).
f_equal. (* This part is just about *)
unfold compose. (* convincing Coq that two *)
apply functional_extensionality. (* functions are equal, it *)
intro x. (* has nothing to do with *)
destruct x as ((btc, atb), a0). (* applicative or monoidal *)
now compute. (* functors, specifically. *)
Qed.
Next Obligation. (* appFtor *)
rewrite <- naturalityL.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
now compute.
Qed.
Lemma fmapPure {F} `{Applicative F} {a} {b}
(f: a->b) (x: a) : fmap f (pure x: F a) = pure (f x).
Proof.
rewrite -> appFtor.
now apply homomorphism.
Qed.
Lemma fmapBracket {F} `{Applicative F} {a} {b} {c} {d}
(f: c->d) (g: a->b->c) (xs: F a) (ys: F b)
: fmap f (fmap g xs<*>ys) = fmap (fun x y => f (g x y)) xs <*> ys.
Proof.
repeat (rewrite -> appFtor).
rewrite -> composition.
rewrite -> homomorphism.
rewrite -> composition.
repeat (rewrite -> homomorphism).
now compute.
Qed.
Lemma fmap_both {F} `{Applicative F} {a} {b} {c} {d}
(f: a->c->d) (g: b->c) (xs: F a) (ys: F b)
: fmap f xs <*> fmap g ys = fmap (fun x y => f x (g y)) xs <*> ys.
Proof.
repeat (rewrite -> appFtor).
rewrite -> composition.
repeat (rewrite <- appFtor).
rewrite <- fmap_twice.
rewrite -> interchange.
rewrite -> appFtor.
rewrite -> composition.
repeat (rewrite -> homomorphism).
rewrite <- appFtor.
now compute.
Qed.
Definition tup {a} {b} (x:a) (y:b) : (a*b) := (x,y).
Program Instance ApplicativeIsMonoidal {F} `{Applicative F}
: Monoidal F
:= { funit := pure tt
; fzip := fun {a} {b} (u: F a) (v: F b)
=> fmap tup u <*> v }.
Next Obligation. (* left_identity *)
repeat (rewrite -> appFtor).
rewrite -> homomorphism.
now compute.
Qed.
Next Obligation. (* right_identity *)
repeat (rewrite -> appFtor).
rewrite -> interchange.
rewrite -> composition.
repeat (rewrite -> homomorphism).
now compute.
Qed.
Next Obligation. (* associativity *)
repeat (rewrite -> fmapBracket).
rewrite -> composition.
repeat (rewrite <- appFtor).
rewrite <- fmap_twice.
rewrite -> fmap_both.
now compute.
Qed.
Next Obligation. (* naturality *)
rewrite -> fmap_both.
rewrite <- fmap_twice.
rewrite -> fmapBracket.
now compute.
Qed.
Compiled with Coq 8.9.1.

Answer 2

We have
-- Monoidal.
class Functor f => Monoidal f where
unit :: f ()
(**) :: f a -> f b -> f (a,b)
-- unit ** v ≅ v - Left Identity.
-- u ** unit ≅ u - Right Identity.
-- u ** (v ** w) ≅ (u ** v) ** w - Associativity.
-- Applicative,
class Functor f => Applicative f where
pure :: a -> f a
infixl 4 <*>
(<*>) :: f (a -> b) -> f a -> f b
-- pure id <*> v = v - Identity.
-- pure f <*> pure x = pure (f x) - Homomorphism.
-- u <*> pure y = pure ($ y) <*> u - Interchange.
-- u <*> (v <*> w) = pure (.) <*> u <*> v <*> w - Composition.
Implementation 1. Applicative --> Monoidal
unit = pure ()
xs ** ys = pure (,) <*> xs <*> ys
Implementation 2. Monoidal --> Applicative
pure x = const x <$> unit
fs <*> xs = uncurry ($) <$> (fs ** xs)
Now prove Monoidal Laws given Applicative Laws and Implementation 1:
Left Identity. unit ** v ≅ v
unit ** v = pure () ** v
= pure (,) <*> pure () <*> v
= pure (\x -> (,) () x) <*> v
= pure (\x -> (() , x)) <*> v
= pure (() ,) <*> v
≅ pure id <*> v
= v
Right Identity. u ** unit ≅ u
u ** unit = u ** pure ()
= pure (,) <*> u <*> pure ()
= pure ($ ()) <*> (pure (,) <*> u) -- u <*> pure y = pure ($ y) <*> u
-- u <*> (v <*> w) = pure (.) <*> u <*> v <*> w
= pure (.) <*> pure ($ ()) <*> pure (,) <*> u
= pure ((.) ($ ())) <*> pure (,) <*> u
= pure ((.) ($ ()) (,)) <*> u
= pure (\x -> (.) ($ ()) (,) x) <*> u
= pure (\x -> ($ ()) ((,) x)) <*> u
= pure (\x -> (,) x ()) <*> u
= pure (\x -> (x , ())) <*> u
= pure (, ()) <*> u
≅ pure id <*> u
= u
Associativity. u ** (v ** w) ≅ (u ** v) ** w
u ** (v ** w) = ......
You should be able to continue this. I hope I didn't make any mistakes here, but if I did, correct them.

Answer 3

Following Will Ness's advice:
Here is what we get of Homomorphism (besides the laws spoken of about, I used the law specifying how Applicative should relate to Functor: fmap g x = pure g <*> x.)
pure f <*> pure x =
= uncurry ($) <$> ((,) <$> (pure f) <*> (pure x)) =
= (uncurry ($) .) <$> ((,) <$> (pure f)) <*> (pure x) =
= ((uncurry ($) .) . (,) <$> (pure f)) <*> (pure x) =
= (uncurry ($) . (,) f) <$> (pure x) =
= pure $ (uncurry ($) . (,) f) x =
= pure (f x)
So, I guess both Homorphism and fs <*> xs = uncurry ($) <$> (fs ** xs) enable us to perform application on the level of functors.