Parametricity Tutorial (Part 1)

Edsko de Vries – Saturday, 23 May 2015

all coding types  

A powerful feature of Haskell’s type system is that we can deduce properties of functions by looking only at their type. For example, a function of type

f :: ∀a. a -> a

can only be the identity function: since it must return something of type a, for any type a, the only thing it can do is return the argument of type a that it was given (or crash). Similarly, a function of type

f :: ∀a. a -> a -> a

can only do one of two things: either return the first argument, or return the second. This kind of reasoning is becoming more and more important with the increasing use of types such as this definition of a “lens”:

type Lens s t a b = forall f. Functor f => (a -> f b) -> s -> f t

Since a lens is just a function of a particular type, the only thing we can conclude about such a function is whatever we can deduce from its type.

To reason about the properties of functions based only on their types we make use of the theory of parametricity, which tells how to derive a so-called “free theorem” from a type. This blog post is a tutorial on how to do this; it won’t explain why the theory works, only how it works. A Haskell practitioner’s guide to parametricity, if you will.

This is a two-part blog post. In part 1 (this post) we will cover the basics: constant types, functions and polymorphism (over types of kind *).

In part 2 we will deal with more advanced material: type constructor (types of kind * -> *), type classes, polymorphism over type constructors and type constructor classes.

The Basics

The main theorem of parametricity is the following:

if f :: t then f ℛ(t) f

When t is a closed type, ℛ(t) is a relation between two terms of type t (we shall see later that the type of ℛ is actually slightly more general). In words, parametricity states that any term f of type t is related to itself by ℛ(t). Don’t worry if this all looks incredibly abstract! We shall see lots and lots of examples.

Constant types (types of kind *)

For any constant type C, the relation ℛ(C) is the identity relation. In other words,

     x ℛ(C) x'
iff  x ≡ x'

(We will use ≡ throughout to mean mathematical equality, to distinguish it from Haskell’s equality function (==).)

Let’s see an example. Suppose that x :: Int. Then parametricity tells us that

     x ℛ(Int) x
iff  x ≡ x

I.e., it tells us that x is equal to itself. Not very insightful! Intuitively, this makes sense: if all we know about x is that it is an integer, we cannot tell anything about its value.

TOOLING. Many of the examples in this blog post (though sadly not all) can also be auto-derived by one of two tools. On the #haskell IRC channel we can ask lambdabot to derive free theorems for any types not involving type classes or type constructor classes. If you ask

@free x :: Int

lambdabot will reply with

x = x

(I recommend starting a private conversation with lambdabot so you avoid spamming the whole #haskell channel.)

Alternatively, you can also try the online free theorem generator. This free theorem generator is a bit more precise than lambdabot (which takes some shortcuts sometimes), and supports type classes, but cannot work with type constructors (lambdabot can work with unknown type constructors but not with quantification over type constructors, unfortunately).

Functions

For functions we map related arguments to related results:

     f ℛ(A -> B) f'
iff  forall x, x'.
       if x ℛ(A) x' then f x ℛ(B) f' x'

(The types of x and x' here depend on what precisely A is; see The type of ℛ, below.)

Example. Suppose f :: Int -> Bool. By parametricity

     f ℛ(Int -> Bool) f
iff  forall x :: Int, x' :: Int.
       if x ℛ(Int) x' then f x ℛ(Bool) f x'
-- both Int and Bool are constant types
iff  forall x :: Int, x' :: Int.
       if x ≡ x' then f x ≡ f x'
-- simplify
iff  f ≡ f

Again, not very informative. Parametricity doesn’t tell us anything about functions between constant types. Time to look at something more interesting!

Polymorphism (over types of kind *)

The definition for polymorphic values is

     f ℛ(∀a. t) f'
iff  forall A, A', a :: A ⇔ A'.
       f@A ℛ(t) f'@A'           -- where 'a' can occur free in t

That is, whenever we pick two types A and A', and some relation a between A and A', the function f@A obtained by instantiating the type variable by A must be related to the function f'@A' obtained by instantiating the type variable by A'. In what follows we will write explicit type instantiation like this only if the type is not clear from the context; specifically, we will omit it when we supply arguments to the function.

The type of ℛ.

∀ab. a -> b -> a is an example of a closed type: all type variables are bound by a universal quantifier. An open type is a type with free type variables such as ∀b. a -> b -> a or even a -> b -> a. (Note that this distinction is harder to see in Haskell where universal quantifiers are often implicit. We will not follow that convention in this article.)

We said in the introduction that if t is a closed type, ℛ(t) relates two terms of type t. As we saw, in order to be able to give a meaning to open types we need a mapping from any free variable a to a relation a :: A ⇔ A'. In this article we somewhat informally maintain this mapping simply by using the same name for the type variable and the relation.

Given two relations a :: A ⇔ A' and b :: B ⇔ B', ℛ(a -> b -> a) relates terms of type A -> B -> A with terms of type A' -> B' -> A'. It is important to realize that ℛ can therefore relate terms of different types. (For a more precise treatment, see my Coq formalization of a part of this blog post.)

The interpretation of ℛ for a free type variable a is defined in terms of the corresponding relation:

     x ℛ(a) x'     -- the type variable
iff  (x, x') ∈ a    -- the relation

Example: ∀a. a -> a

Let’s consider a number of examples, starting with an f :: ∀a. a -> a:

     f ℛ(∀a. a -> a) f
-- parametricity
iff  forall A, A', a :: A ⇔ A'.
       f@A ℛ(a -> a) f@A'
-- definition for function types
iff  forall A, A', a :: A ⇔ A', x :: A, x' :: A'.
       if x ℛ(a) x' then f x ℛ(a) f x'

It might not be immediately evident from the last line what this actually allows us to conclude about f, so let’s look at this a little closer. A function g is a special kind of relation, relating any argument x to g x; since the property holds for any kind of relation a : A ⇔ A', it must also hold for a function a⃯ :: A -> A':

     forall x, x'.
       if x ℛ(a⃯) x' then f x ℛ(a⃯) f x'
-- x ℛ(a⃯) x' iff a⃯ x ≡ x'
iff  forall x :: A, x' :: A'.
       if a⃯ x ≡ x' then a⃯ (f x) ≡ f x'
-- simplify
iff  forall x :: A,
       a⃯ (f x) ≡ f (a⃯ x)

We can apply this result to show that any f :: ∀a. a -> a must be the identity function: picking a⃯ = const x, we get const x (f x) ≡ f (const x x), i.e. x ≡ f x, as required.

NOTE. We are doing fast and loose reasoning in this tutorial and will completely ignore any totality issues. See Automatically Generating Counterexamples to Naive Free Theorems, or the associated web interface, for some insights about what wrong conclusions we can draw by ignoring totality.

Example: ∀a. a -> a -> a

Intuitively, there are only two things a function f :: ∀a. a -> a -> a can do: it can either return its first argument, or it can return its second argument. What does parametricity tell us? Let’s see:

     f ℛ(∀a. a -> a -> a) f
iff  forall A, A', a :: A ⇔ A'.
       f@A ℛ(a -> a -> a) f@A'
-- applying the rule for functions twice
iff  forall A, A', a :: A ⇔ A', x :: A, x' :: A', y :: A, y' :: A'.
       if x ℛ(a) x', y ℛ(a) y' then f x y ℛ(a) f x' y'

Let’s again specialize the last line to pick a function a⃯ :: A -> A' for the relation a:

     forall x :: A, x' :: A', y :: A, y' :: A'.
       if x ℛ(a⃯) x', y ℛ(a⃯) y' then f x y ℛ(a⃯) f x' y'
-- a⃯ is a function
iff  forall x :: A, y :: A.
       if a⃯ x ≡ x' and a⃯ y ≡ y' then a⃯ (f x y) ≡ f x' y'
-- simplify
iff  a⃯ (f x y) = f (a⃯ x) (a⃯ y)

So parametricity allows us to “push in” or “distribute” a⃯ over f. The fact that f must either return its first or its second argument follows from parametricity, but not in a completely obvious way; see the reddit thread How to use free theorems.

## Example: ∀ab. a -> b

Other than undefined (which we are ignoring), there can be no function f :: ∀ab. a -> b. Let’s suppose that one did exist; what does parametricity tell us?

     f ℛ(∀ab. a -> b) f
-- applying the rule for universal quantification, twice
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'.
       f@A,B ℛ(a -> b) f@A',B'
-- applying the rule for functions
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B', x :: A, x' :: A'.
       if x ℛ(a) x' then f x ℛ(b) f x'

Picking two functions a⃯ :: A -> A' and b⃯ :: B -> B' for a and b, we get

b⃯ . f = f . a⃯

It’s not too difficult to derive contradiction from this (remember that you can pick any two functions a⃯ and b⃯ between any types of your choosing). Hence, such a function cannot exist.

Example: ∀ab. (a -> b) -> a -> b

The only thing a function of this type can do is apply the supplied function to the supplied argument (alternatively, if you prefer, this must be the identity function). Let’s spell this example out in a bit of detail because it is our first example of a higher order function.

     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', x :: A, x' :: A'.
       if g ℛ(a -> b) g' and x ℛ(a) x' then f g x ℛ(b) f g' x'

Let’s expand what that premise g ℛ(a -> b) g' means:

     g ℛ(a -> b) g'
iff  forall y :: A, y' :: A'.
       if y ℛ(a) y' then g y ℛ(b) g' y'

For the special case that we pick functions a⃯ :: A -> A' and b⃯ :: B -> B' for a and b, that premise collapses to

     forall y :: A, y' :: A'.
       if y ℛ(a⃯) y' then g y ℛ(b⃯) g' y'
-- a⃯ and b⃯ are functions
iff  forall y :: A, y' :: A'.
       if a⃯ y ≡ y' then b⃯ (g y) ≡ g' y'
-- simplify
iff  forall y :: A.
       b⃯ (g y) ≡ g' (a⃯ y)
-- simplify (extensionality)
iff  b⃯ . g ≡ g' . a⃯

So that the free theorem for f :: ∀ab. (a -> b) -> a -> b becomes

if b⃯ . g ≡ g' . a⃯ then b⃯ . f g ≡ f g' . a⃯

Picking b⃯ = const g, g' = g, and a⃯ = id (verify that the premise holds) we get that indeed g ≡ f g, as expected.

Useful shortcut. This pattern is worth remembering:

      g ℛ(a⃯ -> b⃯) g'
iff   b⃯ . g ≡ g' . a⃯
whenever a⃯ and b⃯ are function(al relation)s.

Example: ∀ab. (∀c. c -> String) -> a -> b -> String

A function f :: ∀ab. (∀c. c -> String) -> a -> b -> String is not only higher order, but has a rank-2 type: it insists that the function it is given is itself polymorphic. This makes it possible to write, for example

f g x y = g x ++ g y

Note that since x and y have different types, it is important that g is polymorphic. What is the free theorem for f?

     f ℛ(∀ab. (∀c. c -> String) -> a -> b -> String) f
-- apply rule for polymorphism, twice
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'.
       f ℛ((∀c. c -> String) -> a -> b -> String) f
-- apply rule for functions three times, and simplify ℛ(String) to (≡)
iff  forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'.
     forall g :: ∀c. c -> String, g' :: ∀c. c -> String.
     forall x :: A, x' :: A', y :: B, y' :: B'.
       if
         g ℛ(∀c. c -> String) g', x ℛ(a) x', y ℛ(b) y'
       then
         f g x y ≡ f g' x' y'

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

forall A, A', B, B', a⃯ :: A -> A', b⃯ :: B -> B'.
forall g :: ∀c. c -> String, g' :: ∀c. c -> String.
forall x :: A, y :: B.
  if
    g ℛ(∀c. c -> String) g'
  then
    f g x y ≡ f g' (a⃯ x) (b⃯ y)

But that is somewhat surprising, because it seems to say that the values of x and y cannot matter at all! What is going on? Expanding the first premise:

     g ℛ(∀c. c -> String) g'
iff  forall C, C', c :: C ⇔ C',
       g ℛ(c -> String) g'
iff  forall C, C', c :: C ⇔ C', z :: C, z' :: C'.
       if z ℛ(c) z' then g z = g' z'

Let’s stop for a moment to ponder what this requirement for g and g' really says: given any relation c, and any elements z and z' that are related by c—in other words, for any z and z' at all—we must have that g z and g' z' give us equal results. This means that g and g' must be constant functions, and the same constant function at that. As a consequence, for any function f of the above type, f g must itself be constant in x and y. In part two we will see a more useful variation which uses the Show type class.

Incidentally, note that this quantification over an arbitrary relation c is a premise to the free theorem, not a conclusion; hence we cannot simply choose to consider only functions c.

TOOLING. Indeed, if you enter

(forall c. c -> String) -> a -> b -> String
in the online free theorem generator you will see that it first gives the free theorem using relations only, and then says it will reduce all “permissible” relation variables to functions; in this example, that is all relations except for c; lambdabot doesn’t make this distinction and always reduces relations to functions, which is not correct.

To be continued

Part 2 covers type constructors, type classes and type constructor classes. Meanwhile, here are some links to papers on the subject if you want to read more.