Parametricity Tutorial (Part 2): Type constructors and type classes

Friday, 14 August 2015, by Edsko de Vries.
Filed under coding.

This is part 2 of a two-part series on parametricity.

In part 1 we covered the basics: constant types, functions and polymorphism (over types of kind *). In this post we will deal with more advanced material: type constructors, type classes, polymorphism over type constructors and type constructor classes.

Type constructors (types of kind * -> *)

Before considering the general case, let’s think about lists. Given a :: A ⇔ A', two lists xs :: [A] and ys :: [A'] are related iff their elements are related by a; that is,

[] ℛ([a]) []

and

     (x:xs') ℛ([a]) (y:ys')
iff  x ℛ(a) x'  and  xs' ℛ([a]) ys'

For the special case that a is a function a⃯ :: A -> A', this amounts to saying that map a⃯ xs ≡ ys.

You can imagine a similar relation F a exists for any type constructor F. However, we will not give a general treatment of algebraic data types in this blog post. Doing this would require giving instances for products and sums (which is fine), but also for (least) fixed points, and that would take us much too far afield.

Thankfully, we will not need to be quite so precise. Instead, it will only require the following characterization:

Characterization: Functors.

Let F be a functor. Then for all relations a :: A ⇔ A', b :: B ⇔ B' and functions f :: A -> B and g :: A' -> B', such that f ℛ(a -> b) g:

forall xs :: F A, xs' :: F A'.
  if    xs ℛ(F a) xs'
  then  F f xs ℛ(F b) F g xs'

where we overload F to also mean the “map” function associated with F. (Provided that the Functor type class instance for F is correct, F f should be the same as fmap f.)

(If we had the precise rules for algebraic data types we would be able to prove this characterization for any specific functor F.)

Intuitively, think about xs and xs' as two containers of the same shape with elements related by a, and suppose we have a pair of functions f and g which map a-related arguments to b-related results. Then the characterization states that if we apply function f to the elements of xs and g to the elements of xs', we must end up with two containers of the same shape with elements related by b.

For the special case that a and b are functions (and F is a functor), the mapping relations characterization simply says that

    if xs ℛ(F a⃯) xs' then F f xs ℛ(F b⃯) F g xs'
-- simplify
    if F a⃯ xs ≡ xs' then F f xs ℛ(F b⃯) F g xs'
-- simplify
    F b⃯ (F f xs) ≡ F g (F a⃯ xs)
-- functoriality
    F (b⃯ . f) xs ≡ F (g . a⃯) xs

which follows immediately from the premise that b⃯ . f ≡ g . a⃯ (which in turn is a consequence of f ℛ(a⃯ -> b⃯) g), so the mapping relations characterization is trivially satisfied (provided that the mapping of relations correspond to the functor map in the case for functions).

Technical note. When we use parametricity results, we often say something like: “specializing this result to functions rather than relations…”. It is important to realize however that if F is not a functor, then F a may not be a functional relation even if a is.

For example, let a⃯ :: A -> A', and take F(a) = a -> a. Then

     f ℛ(F a⃯) g
-- expand definition
iff  f ℛ(a⃯ -> a⃯) g
-- rule for functions
iff  forall x :: A, x' :: A'.
       if x ℛ(a⃯) x' then f x ℛ(a⃯) g x'
-- simplify (a⃯ is a function)
iff  forall x :: A.
       a⃯ (f x) ≡ g (a⃯ x)

Taking a :: Int -> Int ; a x = 0, this would relate two functions f, g :: Int -> Int whenever 0 ≡ g 0; it is clear that this is not a functional relation between f and g.

Given a function a⃯ :: A -> A', F a⃯ is a function F A -> F A' when F is a functor, or a function F A' -> F A if F is a contravariant functor. We will not consider contravariant functors further in this blog post, but there is an analogous Contravariant Functor Characterization that we can use for proofs involving contravariant functors.

 Example: ∀ab. (a -> b) -> [a] -> [b]

This is the type of Haskell’s map function for lists of course; the type of map doesn’t fully specify what it should do, but the elements of the result list can only be obtained from applying the function to elements of the input list. Parametricity tells us that

     f ℛ(∀ab. (a -> b) -> [a] -> [b]) f
-- apply rule for polymorphism, twice
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'.
       f@A,B ℛ((a -> b) -> [a] -> [b]) f@A',B'
-- apply rule for functions, twice
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'.
     forall g :: A -> B, g' :: A' -> B', xs :: [A], xs' :: [A'].
       if g ℛ(a -> b) g', xs ℛ([a]) xs' then f g xs ℛ([b]) f g' xs'

Specializing to functions a⃯ :: A -> A' and b⃯ :: B -> B', we get

     forall A, A', B, B', a⃯ :: A -> A', b⃯ :: B -> B'.
     forall g :: A -> B, g' :: A' -> B', xs :: [A], xs' :: [A'].
       if g ℛ(a⃯ -> b⃯) g', xs ℛ([a⃯]) xs' then f g xs ℛ([b⃯]) f g' xs'
-- simplify
iff  forall A, A', B, B', a⃯ :: A -> A', b⃯ :: B -> B'.
     forall g :: A -> B, g' :: A' -> B'.
       if b⃯ . g ≡ g' . a⃯ then map b⃯ . f g ≡ f g' . map a⃯

As an aside, Functor instances should satisfy two laws:

• map id ≡ id
• map f . map g ≡ map (f . g)

It turns out that the second property follows from the first by parametricity; see The free theorem for fmap.

Example: ∀a. F a -> G a

Consider a funtion f :: ∀a. F a -> G a, polymorphic in a but between fixed (constant) type constructors F and G; for example, a function of type ∀a. Maybe a -> [a] fits this pattern. What can we tell about f?

     f ℛ(∀a. F a -> G a) f
iff  forall A, A', a :: A ⇔ A'.
       f@A ℛ(F a -> G a) f@A'
iff  forall A, A', a :: A ⇔ A', x :: F A, x' :: F A'.
       if x ℛ(F a) x' then f x ℛ(G a) f x'

For the special case where we pick a function a⃯ :: A -> A' for a, this is equivalent to

forall A, A', a⃯ :: A -> A'.
  G a⃯ . f == f . F a⃯

For the categorically inclined, this means that polymorphic functions must be natural transformations.

Type classes

Now that we’ve covered the basics, it’s time to consider some more advanced language features. We will first consider qualified types, such as ∀a. Eq a => a -> a -> a.

The rule for a qualified type is

     f ℛ(∀ a. C a => t) f'
iff  forall A, A', a :: A ⇔ A'
     such that A, A' instances of C and a respects C.
       f@A ℛ(t) f'@A'

What does it mean for a relation a :: A ⇔ A' to respect a type class C? Every type class introduces a new constraint on relations defined by the members of the type class. Let’s consider an example; Haskell’s equality type class is defined by

class Eq a where
  (==) :: a -> a -> Bool

(Let’s ignore (/=) for simplicity’s sake.). Then a relation a respects Eq, written Eq(a), iff all class members are related to themselves. For the specific case of Eq this means that

     (==) ℛ(a -> a -> Bool) (==)
-- rule for functions, twice
iff  forall x :: A, x' :: A', y :: A, y' :: A'.
       if x ℛ(a) x', y ℛ(a) y' then x == y ℛ(Bool) x' == y'
-- Bool is a constant type, simplify
iff  forall x :: A, x' :: A', y :: A, y' :: A'.
       if x ℛ(a) x', y ℛ(a) y' then x == y ≡ x' == y'

For the special case where we pick a function a⃯ :: A -> A', the function respects Eq iff

forall x :: A, y :: A.
  x == y ≡ a⃯ x == a⃯ y

I.e., the function maps (==)-equal arguments to (==)-equal results.

Syntactic convention.

In the following we will write

forall A, A', a :: A ⇔ A'
such that A, A' instances of C and a respects C.

more concisely as

forall C(A), C(A'), C(a) :: A ⇔ A'.

Example: ∀a. Eq a => a -> a -> a

We already considered the free theorem for functions f :: ∀ a. a -> a -> a:

g (f x y) = f (g x) (g y)

Is this free theorem still valid for ∀a. Eq a => a -> a -> a? No, it’s not. Consider giving this (admittedly somewhat dubious) definition of natural numbers which considers all “invalid” natural numbers to be equal:

newtype Nat = Nat Int
  deriving (Show)

instance Eq Nat where
  Nat n == Nat n' | n <= 0, n' <= 0 = True
                  | otherwise       = n == n'

If we define

f :: forall a. Eq a => a -> a -> a
f x y = if x == y then y else x

g :: Nat -> Nat
g (Nat n) = Nat (n + 1)

then for x ≡ Nat (-1) and y ≡ Nat (-2) we have that g (f x y) ≡ Nat (-1) but f (g x) (g y) ≡ Nat 0. Dubious or not, free theorems don’t assume anything about the particular implementation of type classes. The free theorem for ∀a. Eq a => a -> a -> a however only applies to functions g which respect Eq; and this definition of g does not.

 Example: ∀ab. (Show a, Show b) => a -> b -> String

We promised to look at this type when we considered higher rank types above. If you go through the process, you will find that the free theorem for functions f of this type is

f x y = f (g x) (h y)

for any Show-respecting functions g and h. What does it mean for a function to respect Show? Intuitively it means that the function can change the value of its argument but not its string representation:

show (g x) = show x

Type constructor classes

Type constructor classes are classes over types of kind * -> *; a typical example is

class Functor f where
  fmap :: ∀ab. (a -> b) -> f a -> f b

The final rule we will discuss is the rule for universal quantification over a qualified type constructor (universal quantification over a type constructor without a qualifier is rarely useful, so we don’t discuss it separately):

     g ℛ(∀f. C f => t) g'
iff  forall C(F), C(F'), C(f) :: F ⇔ F'.
       g@F ℛ(t) g'@F'

If F and F' are type constructors rather than types (functions on types), f :: F ⇔ F' is a relational action rather than a relation: that is, it is a function on relations. As before, C(f) means that this function must respect the type class C, in much the same way as for type classes. Let’s consider what this means for the example of Functor:

     fmap ℛ(∀ab. (a -> b) -> f a -> f b) fmap
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'.
       fmap@A,B ℛ((a -> b) -> f a -> f b) fmap@A,B
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'.
     forall g :: A -> B, g' :: A' -> B', x :: F A, x' :: F' A'.
       if    g ℛ(a -> b) g', x ℛ(f a) x'
       then  fmap g x ℛ(f b) fmap g' x'

Example: ∀f. Functor f => f Int -> f Int

Intuitively, a function g :: ∀f. Functor f => f Int -> f Int can take advantage of the Int type, but only by applying fmap; for example, when we apply g to a list, the order of the list should not matter. Let’s derive the free theorem for functions of this type:

     g ℛ(∀f. Functor f => f Int -> f Int) g
iff  forall Functor(F), Functor(F'), Functor(f) :: F ⇔ F'.
       g@F ℛ(f Int -> f Int) g@F'
iff  forall Functor(F), Functor(F'), Functor(f) :: F ⇔ F'.
     forall x :: F Int, x' :: F' Int.
       if x ℛ(f Int) x' then g x ℛ(f Int) g x'

As before, we can specialize this to higher order functions, which are special cases of relational actions. Let’s use the notation f⃯ :: F -> F' (with F and F' type constructors) to mean f⃯ :: ∀ab. (a -> b) -> (F a -> F' b). Then we can specialize the free theorem to

iff  forall Functor(F), Functor(F'), Functor(f⃯) :: F -> F'.
     forall x :: F Int, x' :: F' Int.
       if x ℛ(f⃯ Int) x' then g x ℛ(f⃯ Int) g x'
-- `f⃯` is a function; recall that `Int` as a relation is the identity:
iff  forall Functor(F), Functor(F'), Functor(f⃯) :: F -> F'.
       f⃯ id . g ≡ g . f⃯ id

for any Functor-respecting f⃯.

Example continued: further specializing the free theorem

The free theorem we saw in the previous section has a very useful special case, which we will derive now. Recall that in order to prove that a higher order function f⃯ respects Functor we have to prove that

if g ℛ(a -> b) g', x ℛ(f⃯ a) x' then fmap g x ℛ(f⃯ b) fmap g' x'

As in the higher rank example, this is a proof obligation (as opposed to the application of a free theorem), so that we really have to consider relations a :: A ⇔ A' and b :: B ⇔ B' here; it’s not sufficient to consider functions only.

We can however derive a special case of the free theorem which is easier to use. Take some arbitrary polymorphic function k :: ∀a. F a -> F' a, and define the relational action f :: F ⇔ F' by

f(a) = k ⚬ F(a)

where we use k also as a relation. Then

     x ℛ(f a) x'
iff  ∃i. x ℛ(k) i and i ℛ(F(a)) x'
-- k is a function
iff  k x ℛ(F(a)) x'
-- by the Functor Characterization
iff  F g (k x) ℛ(F b) F g' x'
-- naturality
iff  k (F g x) ℛ(F b) F g' x'
-- use k as a relation again
iff  F g x ℛ(k) k (F g x) ℛ(F b) F g' x'
-- pick k (F g x) as the intermediate
then F g x ℛ(f b) F g' x'
-- if we assume that fmap is the "real" functor instance
iff  fmap g x ℛ(f b) fmap g' x'

In the previous section we derived that the free theorem for g :: ∀f. Functor f => f Int -> f Int was

forall Functor(F), Functor(F'), Functor(f⃯) :: F -> F'.
  f⃯ id . g ≡ g . f⃯ id

for any higher order function which respects Functor. The f we defined above is a higher order function provided that a if a function, and we just proved that it must respect functor. The identity relation is certainly a function, so we can specialize the free theorem to

k . g ≡ g . k

for any polymorphic function k (no restrictions on k). As a special case, this means that we must have

reverse . g ≡ g . reverse

formalizing the earlier intuition that when we apply such a function to a list, the order of the list cannot matter.

 Example: ∀f. Functor f => (B -> f B) -> f A

As our last example, we will consider higher-order functions of type g :: ∀f. Functor f => (B -> f B) -> f A. The free theorem for such functions is

     g ℛ(∀f. Functor f => (B -> f B) -> f A) g
iff  forall Functor(F), Functor(F'), Functor(f) : F ⇔ F'.
       g@F ℛ((B -> f B) -> f A) g@F'
iff  forall Functor(F), Functor(F'), Functor(f) : F ⇔ F'.
     forall l :: B -> F B, l' :: B -> F' B.
       if l ℛ(B -> f B) l' then g l ℛ(f A) g l'

Specializing to higher order functions f⃯ :: ∀ab. (a -> b) -> F a -> F' b (rather than a relational action f), we get

     forall Functor(F), Functor(F'), Functor(f) : F ⇔ F'.
     forall l :: B -> F B, l' :: B -> F' B.
       if l ℛ(B -> f⃯ B) l' then g l ℛ(f⃯ A) g l'
iff  forall Functor(F), Functor(F'), Functor(f) : F ⇔ F'.
     forall l :: B -> F B, l' :: B -> F' B.
       if f⃯ id . l ≡ l' . id then f⃯ id (g l) ≡ g l'
-- simplify
iff  forall Functor(F), Functor(F'), Functor(f) : F ⇔ F'.
     forall l :: B -> F B.
       f⃯ id (g l) ≡ g (f⃯ id . l)

for any Functor respecting f⃯; we can now apply the same reasoning as we did in the previous section, and give the following free theorem instead:

k (g l) ≡ g (k . l)

for any polymorphic function (that is, natural transformation) k :: ∀a. F a -> F' a and function l :: B -> F B. This property is essential when proving that the above representation of a lens is isomorphic to a pair of a setter and a getter; see Functor is to Lens as Applicative is to Biplate, Section 4, for details.

Conclusions

Parametricity allows us to formally derive what we can conclude about a function by only looking at its type. We’ve covered a lot of material in this tutorial, but there is a lot more out there still. If you want to know more, here are some additional references.

• Free Theorems Involving Type Constructor Classes is a very accessible paper by Janis Voigtländer that describes how to deal with type constructor classes in free theorems, and gives lots of examples involving monads. If you prefer, you can also watch the video of the presentation. Janis also has a slide set with an introduction to free theorems and some of the underlying theory.
• Proofs for Free: Parametricity for dependent types by Jean-Philippe Bernardy, Patrik Jansson and Ross Paterson extends the theory of parametricity to dependent types.
• Although it’s not about parametricity per se, A Representation Theorem for Second-Order Functionals by Mauro Jaskelioff and Russell O’Connor gives a general (categorical) result that can be used to derive properties of types such as van Laarhoven lenses (sadly, the theorem does not seem to cover prisms).

Acknowledgements

Thanks to Auke Booij on #haskell for his helpful feedback on both parts of this blog post.