Free applicative functors in Coq

Posted on July 14, 2019

This is a long presentation of a short formalization of free applicative functors. In summary, we prove the functor laws, and the associativity law of applicative composition (or “composition law”).

The full Coq code is available here.

The highlight of this formalization is that it is axiom-free, and the (only!) two main functions are defined by simple structural induction, where previous attempts1 have relied on parametricity axioms and induction on an auxiliary measure. The proof scripts are short (5 lines each, 10 if you really like to spread out tactics), and will be mostly omitted from this post.

In spite of the apparent simplicity, the formalization involves crucial design decisions that are detailed in this post, and making the wrong choice at any point would jeopardize the highlighted features.

Admitting functional extensionality would certainly solve most of the difficulties discussed here. It is perhaps the least controversial axiom to admit, which also indicates that the subtleties at play may be hard to catch, even by diving deep into the details. At any rate, I hope this will help to better your understanding of the language Coq (or rather, Gallina), and dependent types in general.

An important point to take away is that = in Coq (eq) is not “equality” as we usually think of it. Deciding when to use eq or pointwise eq (or something else) is a subtle balancing game, and even after the choice is made and the proofs are done, it is perhaps still not clear what the equalities mean.

The file starts with those two lines which mean that the variables f, a, b, c have these types wherever they’re bound.

Implicit Type f : Type -> Type.
Implicit Type a b c : Type.

Free applicative functor

Here is the definition of the free applicative functor:

Inductive Free f a : Type :=
| Pure : a -> Free f a
| Ap {e} : f e -> Free f (e -> a) -> Free f a
.

(* Make the type arguments implicit. *)
Arguments Pure {f a}.
Arguments Ap {f a e}.

In other words, terms of type Free f a are lists of values ti of type f ei with heterogeneous result types ei (i.e., each value can have a different ei), terminated by a function which takes those results to produce a single value v, of type a. Such a term looks like this, where t1, …, tn are terms, x1, …, xn are variables to bind their results, and v is a term which may use x1, …, xn:

Ap t1 (Ap t2 (... (Ap tn (Pure (
fun x1     x2  ...     xn => v))) ...))

Comparison with other versions

This is the same definition as the “default” one offered by the free library by Edward Kmett in Haskell. The library also includes two other definitions (Free.Fast, Free.Final) with much better asymptotic complexity.

Interestingly, the original paper which introduced free applicative functors, by Paolo Capriotti and Ambrus Kaposi, mentions it first (under the name FreeAL), but it is soon replaced with a different one (FreeA) where, in the Ap constructor, the arrow type b -> a goes under f instead of Free f (there is also a mostly inconsequential matter of left or right nesting):

Inductive FreeA f a : Type :=
| Pure : a -> FreeA f a
| Ap {b} : f (b -> a) -> FreeA f b -> FreeA f a
.

The reason given for choosing FreeA is that its Functor instance (i.e., fmap) and Applicative instance (i.e., pure and, most importantly, (<*>)) are easier to define. That is certainly the case for fmap, since it doesn’t need recursion, it simply peels off the first action and uses the Functor instance of f. However, ap uses induction on size to ensure termination; this complication is easily isolated on paper, but would be exacerbated in Coq.

Furthermore, proving that FreeA is an applicative functor (i.e., it satisfies the laws) requires an assumption of parametricity which sadly does not hold in Coq. The conclusion of the paper explains this and one possible solution. Another way would be to accept that eq is too fine a relation for this purpose, and instead work with an ad-hoc but coarser notion of equivalence for which the parametricity assumption is admissible, or even provable. Admittedly, equivalence relations come with their own set of proof-engineering problems down the line.

One more difference is that Free is freer than FreeA: Free f is an applicative functor for any indexed type f : Type -> Type, not only functors. It is the same difference as that between the “free monad” and the “freer monad” (it’s worth mentioning that both are free monads in the categorical sense, just in different categories). In Coq, Free, as opposed to FreeA, simplifies the formalization because there is one fewer assumption to keep track of in proofs.

See also Flavours of free applicative functors, by Roman Cheplyaka.

Functor

As a warm up exercise, let’s first show that Free f is a functor: we implement map : (a -> b) -> Free f a -> Free f b and show that it commutes with function composition.

The definition of map is straightforward. In the Ap case, the subterm ts : Free f (e -> a) has a different type from the parameter ts0 : Free f a, and that is taken care of by adapting the function argument of map.

Fixpoint map {f a b} (h : a -> b) (ts0 : Free f a) : Free f b :=
match ts0 with
| Pure x => Pure (h x)
| Ap t ts => Ap t (map (fun i x => h (i x)) ts)
end.

As for the laws, there’s two obvious ways to formulate them; only one leads to provable statements.

If you look up a reference on category theory or some other language’s documentation, you’ll find this:

map id = id              (map_id_classic)
map (k ∘ h) = map k ∘ map h   (map_map_classic)

… with the usual definitions id = (fun x => x) and k ∘ h = (fun x => k (h x)).

It’s a trap!

If we try to prove that, we are immediately stuck, there is simply no way forward.

We can in fact show that no such proof exists, i.e., there are no closed terms of type map id = id2. Terms in Coq have normal forms of the same type (normalization and type preservation); the only possible closed normal form of type _ = _ is eq_refl, but for it to be well-typed the two sides of the equality must be convertible3, and of course map id is not convertible to id. Note that the “closed term” assumption is key: we can prove nontrivial equations by eliminating variables in the context, and this argument shows that we’re stuck if the context is empty.

That is a metalinguistic argument which cannot be carried out within Coq. It would mean proving the negation of that equality, which cannot be because that equality is in fact admissible. See also this discussion of the same topic on SO.

Those equations don’t hold, does that mean Free f is not a functor? The mismatch is actually in the notation =. When we write = on paper, we pretend that there is such a thing as “equality” and that it has obvious properties. In mechanized formalization, the way I like to think about it is that there is simply no canonical “equality”. When we write = in Coq, it means eq by default, an indexed type defined in Coq (i.e., not a primitive), and whether or not it matches our mental picture of “equality” is a subjective matter. Other definitions may also be worthy of the name “equality” depending on the situation, and they don’t even have to be expressible as types in Coq (convertibility for example). To take another language, Idris overloads the symbol = so it may denote heterogeneous equality if the context requires it. This cements my conviction that equality is merely a social construct.

So instead, we will show that the equations hold pointwise. For all u : Free f a:

map id u = u                 [map_id]
map (k ∘ h) u = map k (map h u)   [map_map]

Often, the = symbol on paper can be translated to an equivalence relation in Coq, which has no reason to be eq a priori. In this case, we are interpreting = from the original laws as pointwise eq, with its definition unfolded here. It is also sometimes called extensional equality for functions, but that terminology can become confusing when codomains of functions start carrying their own relations, while “pointwise eq” is always precise.

Definition pointwise_eq {a b} (i j : a -> b) : Prop :=
forall x, i x = j x.

Infix "=" := pointwise_eq : fun_scope.
Delimit Scope fun_scope with fun.

In Coq, we can redefine symbols such as =, which would allow us to write something familiar like the following; the %fun annotations open a scope in which = desugars to pointwise eq instead of plain eq:

(map id = id)%fun              [map_id_pointwise]
(map (k ∘ h) = map k ∘ map h)%fun   [map_map_pointwise]

Looking at the categorical definition of a functor, once we reject “equality” as a primitive notion, we have to review the very definition of a category, which makes use of = ostensibly:

id ∘ k = k ∘ id = k
(i ∘ j) ∘ k = i ∘ (j ∘ k)

The definition of a category should now specify the equivalence relation between arrows.4 The definition of a functor also needs to be adjusted: it must not only “commute with composition”, it must also “preserve equivalences”, although this is trivial to prove if the domain category uses eq as the equivalence relation.

In this case, map makes Free f : Type -> Type a functor from the category where objects are types, arrows are functions, and arrow equivalence is eq, to the category with the same objects and arrows, but arrow equivalence is pointwise_eq.

What if we change the domain category to also use pointwise_eq as the equivalence relation? Things break again: map doesn’t send pointwise-equal functions to pointwise-equal functions. Something fishy is going on, but let’s postpone that for another day.

So much for warming up.

Applicative

Free is a type of lists, with a few ornaments. In particular, ap is list concatenation, if we ignore the a and b ornaments.

ap : Free f (a -> b) -> Free f a -> Free f b`,

That tells us the general shape of the function, but these ornaments still matter a whole lot to the whole function.

Let’s first try implementing ap. Here, by induction on u. In the Pure case, we have h : a -> b, and us : Free f a, these types strongly hint at map.

Fixpoint ap {f a b} (ts0 : Free f (a -> b)) (us : Free f a)
: Free f b :=
match ts0 with
| Pure h => map h v
| Ap t ts => Ap r _
end.

The Ap case is puzzling… In the underscore (also called “typed hole”), we expect to find Free f (e -> b), for some fresh type e (which is the result type of r). We have ts : Free f (e -> a -> b) and we can’t directly apply ap to it with us : Free f a. There is a cunning plan: we can flip us to map flip us : Free f (a -> e -> b), and ap that with us.

Definition flip {a b c} (f : a -> b -> c) : b -> a -> c :=
fun x y => f y x.

Fixpoint ap {f a b} (ts0 : Free f (a -> b)) (us : Free f a)
: Free f b :=
match ts0 with
| Pure h => map h v
| Ap t ts => Ap t (ap (map flip ts) us)
end.

(* But nope! *)

Nice try, but the Fixpoint police is watching: we can only ap to a subterm of ts0, and map flip ts is not one. In the match branch with Ap t ts, its subterms are t and ts. t : f e doesn’t have the right type, so it only makes sense to apply ap to ts, but we don’t want to do that, which leaves us right back where we started.

There are several workarounds, and I think the simplest one is to implement liftA2 instead.

liftA2 : (a -> b -> c) -> (Free f a -> Free f b -> Free f c)

The problem with ap was that it imposes a somewhat arbitrary structure on the type indices of its arguments and result: a -> b, a, b. In contrast, liftA2 keeps the three indices loosely related through a separate function a -> b -> c, so we have much more flexibility to write the recursive application. Indeed, in the Ap case, the definition of liftA2 follows the structure of list concatenation (++, app) quite clearly, we only need to fill the hole.

liftA2 h (Ap u us) vs = Ap u (liftA2 _ us vs)
app (cons u us) vs = cons u (app us vs)

And the hole is uniquely determined by its context.

(* Given this context *)
h : a -> b -> c
============================= (* The Type-Tetris game *)
(e -> a) -> b -> (e -> c)
(* Implement that goal *)

This leads to the following definition:

Fixpoint liftA2 {f a b c}
(h : a -> b -> c) (ts0 : Free f a) (us : Free f b)
: Free f c :=
match ts0 with
| Pure x => map (h x) us
| Ap t ts => Ap t (liftA2 (fun i y z => h (i z) y) ts us)
(* Type-Tetris here   *)
end.

The Ap case doesn’t need any call to map, that simplifies the implementation, and thus the proofs.

Applicative law: associativity

The most interesting applicative functor law here is the associativity law:

liftA2 _ (liftA2 _ ts us) vs = liftA2 _ ts (liftA2 _ us vs)

But what goes in the underscores? We will try different things. There are two points to take into consideration for choosing the right theorem, they are perhaps self-evident: the difficulty of the proof, and the generality of the theorem. Despite appearances, these objectives don’t necessarily conflict with each other: the proof is most likely going to happen by induction over ts, so some generalization is required to get a suitably strong induction hypothesis. Often, that generalization is a perfectly fine theorem on its own: we get a general theorem, and by design the first step of the proof is induction.

On one end of the spectrum, a first candidate is to put the results of ts, us, and vs in tuples. The concreteness makes it easy to understand.

liftA2 snocpair (liftA2 pair ts us) vs
=
liftA2 conspair ts (liftA2 pair us vs)

[liftA2_liftA2_tuple]    <- name of the equation for reference

where
pair = (fun x y => (x, y))
snocpair = (fun '(x, y) z => (x, y, z))
conspair = (fun x '(y, z) => (x, y, z))

Sadly, that doesn’t pass the test of “is this a good induction hypothesis?”. Moreover, in practice most occurrences of liftA2 in the wild don’t use pair, which makes that equation hard to apply. In that practical sense, it lacks generality, although theoretically it is fully expressive in the presence of a suitable “naturality lemma” which is useful to prove at any rate.

Another approach is to put variables in the left hand side, and then figure out what right hand side matches.

liftA2 k (liftA2 h ts us) vs = liftA2 _ ts (liftA2 _ us vs)

On the right hand side, we need one function to combine the results of us and vs, and another one to combine that with the result of ts. There is very little to work with from the left hand side, which is unfortunate because we are so constrained, but also fortunate because we are so constrained. This looks like a canonical answer: wrap us and vs in a pair so we can freely rearrange the results at the next level.

liftA2 k (liftA2 h ts us) vs
= liftA2 (fun x '(y, z) => k (h x y) z) ts (liftA2 pair us vs)

[liftA2_liftA2_simple]

That is an easy proposition to prove by induction, modulo a few details5. Regarding generality, that equation also looks good because it can be used to rewrite any expression with two left-associated liftA2 into a right-associated one. By now it’s clear I am nitpicky about this stuff: what about right-to-left rewriting?

A symmetric formulation

Symmetry is often something worth pondering in mathematics, both in ideas and in notations. It’s not necessarily bad to lack symmetry: liftA2 does not have a symmetric definition, privileging one decreasing argument ts over the other us. Nevertheless, breaking or restoring symmetry can teach us new insights.

In this case, a symmetric equation is obtained using different variables for the function arguments on both sides.

liftA2 i (liftA2 h ts us) vs = liftA2 j ts (liftA2 k us vs)

[liftA2_liftA2]

Now, we need an extra assumption to relate h, i, j, k6. From the structure of the above equation, or from the types of these functions, it would be reasonable to suggest the following equation, for all x, y, and z:

i (h x y) z = j x (k y z)   [hyp_0]

So, the full theorem would look like this, to give you an idea of the actual Coq code. Most of the space is taken by the signatures for the parameters, I don’t know whether we should be sad about it, but the last two lines are the important ones.

Theorem almost_liftA2_liftA2 {f a1 a2 a3 b12 b23 c}
(h : a1 -> a2 -> b12) (i : b12 -> a3 -> c)
(j : a1 -> b23 -> c) (k : a2 -> a3 -> b23)
(ts : Free f a1) (us : Free f a2) (vs : Free f a3)

: forall (hyp_0 : forall x y z, i (h x y) z = j x (k y z)),

liftA2 i (liftA2 h ts us) vs = liftA2 j ts (liftA2 k us vs).

Before induction, we should revert just about every parameter except ts (actually, only about half of them, yet nothing goes wrong if there’s more). Such generalization is par for the course in formal proofs by induction.

We’ll skip the base case (ts = Pure x), but suffice it to say that the gist of it can be extracted into separate theorems about the interaction of liftA2 with map. I like to call them “naturality lemmas” because they show that liftA2 is a natural transformation (map_liftA2, liftA2_map in the Coq source).

In the inductive case (ts = Ap t ts'), to use the induction hypothesis, we instantiate h, i and j with a fun expression of three parameters, as found in the definition of liftA2: fun l y z => h (l z) y (and replace h with i and j, respectively); since all the functions in that equation are applied to two arguments, we will be left with one fun on each side. The induction hypothesis gives us the following obligation to prove, for all x', y', z':

(fun z0 => i (h (x' z0) y') z') = (fun z0 => j (x' z0) (k y' z'))

As you might have guessed, the hyp_0 assumption, i.e., i (h x y) z = j x (k y z) (with x := x' z0, y := y', z := z'), cannot be applied to prove that. Theoretically, the same argument as the one given earlier applies to explain that no axiom-free proof is possible. Practically, we can see that the subexpression we are trying to rewrite using hyp_0 contains a bound variable, z0. Indeed, the proof principle used for rewriting, which is a theorem m = n -> P m -> P n, implicitly requires that the expressions m and n have no variables bound in the “context” P.

With a lot of hand-waving, the idea is that the equation i (h x y) z = j x (k y z) has too many free variables: to use that equation, we must instantiate x, y, z, and the trouble starts when those instantiations contain bound variables. If that is indeed the problem, we can try to bind these variables so we don’t have to instantiate them. A normal reaction is to think that idea doesn’t even make sense. But hey, it works. Thus we bind x, y, and z with fun on both sides we get the following equation hyp_1 to replace hyp_0.

(fun x y z => i (h x y) z) = (fun x y z => j x (k y z))

[hyp_1]

Now this is the theorem we can actually prove.

Theorem liftA2_liftA2 {f a1 a2 a3 b12 b23 c}
(h : a1 -> a2 -> b12) (i : b12 -> a3 -> c)
(j : a1 -> b23 -> c) (k : a2 -> a3 -> b23)
(ts : Free f a1) (us : Free f a2) (vs : Free f a3)

: forall (hyp_1 : (fun x y => i (h x y)) = (fun x y z => j x (k y z))),

liftA2 i (liftA2 h ts us) vs = liftA2 j ts (liftA2 k us vs).

After replaying the first steps of the proof, the goal given to us by the induction hypothesis looks like this:

(fun x' y' z' z0 => i (h (x' z0) y') z')
=
(fun x' y' z' z0 => j (x' z0) (k y' z'))

At a glance, things don’t look any better if you’re used to thinking that equations of the form (fun ...) = (fun ...) require functional extensionality. But here we have specifically fixed our assumption hyp_1 to work around that. We only need to make the two functions of hyp_1 appear in the goal, which is a little exercise in β-expansion.

(fun x' y' z' z0 =>
(fun x y z => i (h x y) z) (x' z0) y' z')
=
(fun x' y' z' z0 =>
(fun x y z => j x (k y z)) (x' z0) y' z')

Actually, β-expansion in Coq can be tedious to do with tactics (I can’t see a concise way to do that, short of spelling out the desired goal). Instead, we can manipulate the assumption hyp_1 to obtain a proposition which β-reduces to the goal. Start by representing the context common to both sides of the β-expanded goal as a function:

fun l => (fun x' y' z' z0 => l (x' z0) y' z')

Then use the f_equal lemma to apply that function to both sides of hyp_1. After β-reduction, we obtain exactly the goal.

(* ... *)
apply (@f_equal _ _ (fun l => (fun x' y' z' z0 => l (x' z0) y' z'))) in hyp_1.
apply hyp_1.
Qed.

(* Apply directly in the term and solve the goal with the result. *)
apply (@f_equal _ _ (fun l => (fun x' y' z' z0 => l (x' z0) y' z')) _ _ hyp_1).
Qed.

The associativity of liftA2 is now proved. Phew.

Corollaries

The first two version of associativity we gave, liftA2_liftA2_tuple and liftA2_liftA2_simple, are immediate corollaries of liftA2_liftA2: a proof for both of them is literally apply liftA2_liftA2; auto.. The converse is not difficult, but still a little less obvious. Although the statement of liftA2_liftA2 is not so easy to come up with, its proof is very similar to the “more intuitive” liftA2_liftA2_simple.

Given liftA2, defining ap is as easy as liftA2 (fun k x => k x), and the associativity law expressed in terms of ap is a piece of cake using liftA2_liftA2 and naturality (liftA2_map in the source).

What does the theorem really mean?

Hopefully I’ve made my point that eq is actually much stronger than it may seem at first. The relation is hard to prove (e.g., map id = id is unprovable when = is eq, but provable when = is pointwise_eq), and as an assumption, for example hyp_1, it implies a lot.

Let’s take another look at hyp_1. How might we prove such an equation, in order to use liftA2_liftA2?

(fun x y z => i (h x y) z) = (fun x y z => j x (k y z))

The corollaries above provide some concrete examples. For example, to prove liftA2_liftA2_tuple, we use the following instantiation.

i := fun xy z => (fst xy, snd xy, z)
j := fun x yz => (x, fst yz, snd yz)
h := fun x y => (x, y)   (* h = k = pair *)
k := fun x y => (x, y)

Then hyp_1 simplifies to this, which is trivial by reflexivity.

(fun x y z => (x, y, z)) = (fun x y z => (x, y, z))

In fact, reflexivity is basically the only way to prove any equation of the form hyp_1. Thus hyp_1 could be said to be roughly an encoding of the fact that i (h x y) z is convertible to j x (k y z), which flies in the face of the common wisdom that convertibility is not expressible in Coq!

So liftA2_liftA2 requires a “proof of convertibility”, if we may say so, which is quite a lot to ask for. For example, setting all four of i, j, h, and k to add : nat -> nat -> nat, i.e., + (or some other not-too-trivial associative operation), we get this unprovable statement:

(fun x y z => (x + y) + z) = (fun x y z => x + (y + z))

As it turns out, the conclusion we would get from that with liftA2_liftA2 is also unprovable:

liftA2 add (liftA2 add ts us) vs = liftA2 add ts (listA2 add us vs)

Consequently, liftA2_liftA2 makes a strong assumption which is not easily weakened.

Still, programmers familiar with applicative functors would expect that last equation above to hold: liftA2 add should be associative. When did we go astray? The root of the problem is again eq. Our informal intuition of “equality” is extensional. When are two Free computations the same? When they perform the same actions and produce the same results.

Concretely, a Free term is a list of Ap nodes, terminated by a Pure node holding a function.

Ap t1 (Ap t2 (... (Ap tn (Pure (
fun x1     x2  ...     xn => v))) ...))

When we equate two such terms, such as we do above in the functor and applicative laws, this implies equating the functions at the end of the lists, and that spells trouble. It would be more intuitive to equate them pointwise.

(fun x1 ... xn => v) = (fun x1 ... xn => w)   (* [eq]uality *)
forall x1 ... xn, v = w                       (* Pointwise [eq]uality *)

An extensional version of equality for Free terms is nontrivial to define and prove to be an equivalence relation, but with it, we would be able to show that liftA2 add is associative.

Introducing that relation opens up a whole new set of problems, especially regarding what we mean when we say that Free f is an applicative functor (the categories of interest are different), but let’s stop at this because I don’t want to make this post much longer.

Conclusion

To summarize the main results formalized in this post:

• the definition of the free applicative functor as an inductive type, which is the same as what can be found in existing literature and libraries, but in Coq;

• a proof that Free f is a functor, which made us refine the notion of categories to 2-categories;

• a definition of liftA2 with a proof of its associativity, whose statement is not even obvious. From that we can derive ap and derive the laws in its terms.

However, that is not a complete demonstration that the “free applicative functor” is worthy of the name. To finish it, the interested reader is invited to investigate the following points:

• the identity laws of applicative functors;

• Free is indeed the free applicative functor: Free (not Free f) must be a functor satisfying a certain universal property.

Many thanks to Georgy Lukyanov for the discussion on free applicative and selective functors which sparked this piece.

1. Free Applicative Functors, by Paolo Capriotti and Ambrus Kaposi, in MSFP 2014 (arxiv) (previously submitted to ICFP 2013, to get the timeline right).↩︎

2. map id hides some type arguments: @map f a a id, so we’re not actually in an empty context. This can be addressed with a longer proof, or by specializing it @map (fun x => x) nat nat id, since a general proof map id = id would imply this special case.↩︎

3. “Convertible” roughly means “β-equivalent”, and anyone would understand if you use these words as synonyms, but β is actually only a subset of convertibility. There are more rules to unfold definitions and to handle constructs other than fun: match, let, fix, cofix.↩︎

4. We’re inches away from from 2-categories and higher category theory.↩︎

5. Instead of pattern-matching on the pair, one should use projections fst and snd.↩︎

6. About the choice of variable names, these are the four letters after f and g, which are already taken (of course g must be in the same category as f); h looks like k, and i looks like j, so we can fine-tune the symmetry further in notations; the alphabet is a fortunate thing here.↩︎