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]) 
:xs') ℛ([a]) (y:ys') (xand xs' ℛ([a]) ys'iff x ℛ(a) x'
For the special case that
a is a function
a⃗ :: A -> A', this amounts to saying that
map a⃗ 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:
Fbe a functor. Then for all relations
a :: A ⇔ A',
b :: B ⇔ B'and functions
f :: A -> Band
g :: A' -> B', such that
f ℛ(a -> b) g:where we overload
forall xs :: F A, xs' :: F A'. if xs ℛ(F a) xs' then F f xs ℛ(F b) F g xs'
Fto also mean the “map” function associated with
F. (Provided that the
Functortype class instance for
F fshould be the same as
(If we had the precise rules for algebraic data types we would be able to prove this characterization for any specific functor
Intuitively, think about
xs' as two containers of the same shape with elements related by
a, and suppose we have a pair of functions
g which map
a-related arguments to
b-related results. Then the characterization states that if we apply function
f to the elements of
g to the elements of
xs', we must end up with two containers of the same shape with elements related by
For the special case that
b are functions (and
F is a functor), the mapping relations characterization simply says that
if xs ℛ(F a⃗) xs' then F f xs ℛ(F b⃗) F g xs' -- simplify if F a⃗ xs ≡ xs' then F f xs ℛ(F b⃗) F g xs' -- simplify F b⃗ (F f xs) ≡ F g (F a⃗ xs) -- functoriality F (b⃗ . f) xs ≡ F (g . a⃗) xs
which follows immediately from the premise that
b⃗ . f ≡ g . a⃗ (which in turn is a consequence of
f ℛ(a⃗ -> b⃗) 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
Fis not a functor, then
F amay not be a functional relation even if
For example, let
a⃗ :: A -> A', and take
F(a) = a -> a. Then
F a⃗) g f ℛ(-- expand definition -> a⃗) g iff f ℛ(a⃗ -- rule for functions forall x :: A, x' :: A'. iff if x ℛ(a⃗) x' then f x ℛ(a⃗) g x' -- simplify (a⃗ is a function) forall x :: A. iff a⃗ (f x) ≡ g (a⃗ x)
TakingGiven a function
a :: Int -> Int ; a x = 0, this would relate two functions
f, g :: Int -> Intwhenever
0 ≡ g 0; it is clear that this is not a functional relation between
a⃗ :: A -> A',
F a⃗is a function
F A -> F A'when
Fis a functor, or a function
F A' -> F Aif
Fis 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.
∀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
. (a -> b) -> [a] -> [b]) f f ℛ(∀ab-- apply rule for polymorphism, twice forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'. iff @A,B ℛ((a -> b) -> [a] -> [b]) f@A',B' f-- apply rule for functions, twice forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'. iff 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
a⃗ :: A -> A' and
b⃗ :: B -> B', we get
forall A, A', B, B', a⃗ :: A -> A', b⃗ :: B -> B'. forall g :: A -> B, g' :: A' -> B', xs :: [A], xs' :: [A']. if g ℛ(a⃗ -> b⃗) g', xs ℛ([a⃗]) xs' then f g xs ℛ([b⃗]) f g' xs' -- simplify forall A, A', B, B', a⃗ :: A -> A', b⃗ :: B -> B'. iff forall g :: A -> B, g' :: A' -> B'. if b⃗ . g ≡ g' . a⃗ then map b⃗ . f g ≡ f g' . map a⃗
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
∀a. F a -> G a
Consider a funtion
f :: ∀a. F a -> G a, polymorphic in
a but between fixed (constant) type constructors
G; for example, a function of type
∀a. Maybe a -> [a] fits this pattern. What can we tell about
. F a -> G a) f f ℛ(∀aforall A, A', a :: A ⇔ A'. iff @A ℛ(F a -> G a) f@A' fforall A, A', a :: A ⇔ A', x :: F A, x' :: F A'. iff if x ℛ(F a) x' then f x ℛ(G a) f x'
For the special case where we pick a function
a⃗ :: A -> A' for
a, this is equivalent to
forall A, A', a⃗ :: A -> A'. G a⃗ . f == f . F a⃗
For the categorically inclined, this means that polymorphic functions must be natural transformations.
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
. C a => t) f' f ℛ(∀ aforall A, A', a :: A ⇔ A' iff A, A' instances of C and a respects C. such that @A ℛ(t) f'@A' f
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
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 forall x :: A, x' :: A', y :: A, y' :: A'. iff if x ℛ(a) x', y ℛ(a) y' then x == y ℛ(Bool) x' == y' -- Bool is a constant type, simplify forall x :: A, x' :: A', y :: A, y' :: A'. iff if x ℛ(a) x', y ℛ(a) y' then x == y ≡ x' == y'
For the special case where we pick a function
a⃗ :: A -> A', the function respects Eq iff
forall x :: A, y :: A. == y ≡ a⃗ x == a⃗ y x
I.e., the function maps
(==)-equal arguments to
In the following we will write
forall A, A', a :: A ⇔ A' A, A' instances of C and a respects C.such that
more concisely as
forall C(A), C(A'), C(a) :: A ⇔ A'.
∀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 = if x == y then y else x f x y g :: Nat -> Nat Nat n) = Nat (n + 1)g (
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.
∀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)
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):
. C f => t) g' g ℛ(∀fforall C(F), C(F'), C(f) :: F ⇔ F'. iff @F ℛ(t) g'@F' g
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
fmap ℛ(∀ab. (a -> b) -> f a -> f b) fmap forall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'. iff @A,B ℛ((a -> b) -> f a -> f b) fmap@A,B fmapforall A, A', B, B', a :: A ⇔ A', b :: B ⇔ B'. iff 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'
∀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:
. Functor f => f Int -> f Int) g g ℛ(∀fforall Functor(F), Functor(F'), Functor(f) :: F ⇔ F'. iff @F ℛ(f Int -> f Int) g@F' gforall Functor(F), Functor(F'), Functor(f) :: F ⇔ F'. iff 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
f⃗ :: F -> F' (with
F' type constructors) to mean
f⃗ :: ∀ab. (a -> b) -> (F a -> F' b). Then we can specialize the free theorem to
forall Functor(F), Functor(F'), Functor(f⃗) :: F -> F'. iff forall x :: F Int, x' :: F' Int. if x ℛ(f⃗ Int) x' then g x ℛ(f⃗ Int) g x' -- `f⃗` is a function; recall that `Int` as a relation is the identity: forall Functor(F), Functor(F'), Functor(f⃗) :: F -> F'. iff id . g ≡ g . f⃗ id f⃗
for any Functor-respecting
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
Functor we have to prove that
if g ℛ(a -> b) g', x ℛ(f⃗ a) x' then fmap g x ℛ(f⃗ 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
= k ⚬ F(a)f(a)
where we use
k also as a relation. Then
x ℛ(f a) x'. x ℛ(k) i and i ℛ(F(a)) x' iff ∃i-- k is a function F(a)) x' iff k x ℛ(-- by the Functor Characterization F g (k x) ℛ(F b) F g' x' iff -- naturality F g x) ℛ(F b) F g' x' iff k (-- use k as a relation again F g x ℛ(k) k (F g x) ℛ(F b) F g' x' iff -- 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 fmap g x ℛ(f b) fmap g' x'iff
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(f⃗) :: F -> F'. id . g ≡ g . f⃗ id f⃗
for any higher order function which respects
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.
∀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
. Functor f => (B -> f B) -> f A) g g ℛ(∀fforall Functor(F), Functor(F'), Functor(f) : F ⇔ F'. iff @F ℛ((B -> f B) -> f A) g@F' gforall Functor(F), Functor(F'), Functor(f) : F ⇔ F'. iff 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
f⃗ :: ∀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 -> f⃗ B) l' then g l ℛ(f⃗ A) g l' forall Functor(F), Functor(F'), Functor(f) : F ⇔ F'. iff forall l :: B -> F B, l' :: B -> F' B. if f⃗ id . l ≡ l' . id then f⃗ id (g l) ≡ g l' -- simplify forall Functor(F), Functor(F'), Functor(f) : F ⇔ F'. iff forall l :: B -> F B. id (g l) ≡ g (f⃗ id . l) f⃗
f⃗; we can now apply the same reasoning as we did in the previous section, and give the following free theorem instead:
. l)k (g l) ≡ g (k
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.
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).
Thanks to Auke Booij on
#haskell for his helpful feedback on both parts of this blog post.