Free Bicartesian profunctor (with caveat -- see below) a Cartesian profunctor.

Law issues:

FreeBicartesian p can be thought of as a way to add Cocartesian operations on Cartesian p by taking "formal sums" of multplie values of p a b.

Products on sums of multiple values are normalized to sum of products **as if** it satisfy both left and right distribution laws and zero laws. For example, the result of

(p1 +++ p2) *** (q1 +++ q2)

is normalized to

(p1 *** q1) +++ (p1 *** q2) +++ (p2 *** q1) +++ (p2 *** q2)

up to isomorphisms of parameters of these profunctors.

Because there are some profunctors which are both Cartesian and Cocartesian but do not satisfy distributive laws, interpreting FreeBicartesian into such a profunctor might cause a surprising behavior.

For example, Joker [] does not satisfy right distribution, inheriting Alternative [] does not.

>>> import Control.Applicative
>>> let x = [id, id]
>>> let y = [1]; z = [2]
>>> x <*> (y <|> z)
[1,2,1,2]
>>> (x <*> y) <|> (x <*> z)
[1,1,2,2]

With such non-distributive p, foldFree does not preserve the Cartesian operations. The following equation does not have to hold.

-- Not necessarily holds!
foldFree id (ps *** qs)
 == foldFree id ps *** foldFree id qs

It is guaranteed that foldFree f preserves both Cartesian and Cocartesian operations if it is intepreting into a Bicartesian profunctor, in other words both Cartesian p and Cocartesian p which satisfy these additional laws.

• Cocartesian instance is commutative
• Left zero, Right zero, Left distribution, Right distribution

There are no guarantees if any of these conditions are not met.

Interpret a FreeBicartesian p into a Bicartesian profunction q.

It is guaranteed that foldFree f preserves both Cartesian and Cocartesian operations if q is a Bicartesian profunctor. There are no guarantees if any of extra laws to be a Bicartesian are not met.