Initial and final encodings of free monads

Posted on October 20, 2021

Free monads are often introduced as an algebraic data type, an initial encoding:

data Free f a = Pure a | Free (f (Free f a))

Thanks to that, the term “free monads” tends to be confused with that encoding, even though “free monads” originally refers to a representation-independent idea. Dually, there is a final encoding of free monads:

type Free' f a = (forall m. MonadFree f m => m a)

where MonadFree is the following class:

class Monad m => MonadFree f m where
  free :: f (m a) -> m a

The two types Free and Free' are isomorphic. An explanation a posteriori is that free monads are unique up to isomorphism. In this post, we will prove that they are isomorphic more directly,1 in Coq.

In other words, there are two functions:

fromFree' :: Free' f a -> Free f a
toFree' :: Free f a -> Free' f a

such that, for all u :: Free f a,

fromFree' (toFree' u) = u  -- easy

and for all u :: Free' f a,

toFree' (fromFree' u) = u  -- hard

(Also, these functions are monad morphisms.)

The second equation is hard to prove because it relies on a subtle fact about polymorphism. If you have a polymorphic function forall m ..., it can only interact with m via operations provided as parameters—in the MonadFree dictionary. The equation crashes down if you can perform some kind of case analysis on types, such as isinstanceof in certain languages. This idea is subtle because, how do you turn this negative property “does not use isinstanceof” into a positive, useful fact about the functions of a language?

Parametricity is the name given to such properties. You can get a good intuition for it with some practice. For example, most people can convince themselves that forall a. a -> a is only inhabited by the identity function. But formalizing it so you can validate your intuition is a more mysterious art.

Proof sketch

First, unfolding some definitions, the equation we want to prove will simplify to the following:

foldFree (u @(Free f)) = u @m

where u :: forall m. MonadFree f m => m a is specialized at Free f on the left, at an arbitrary m on the right, and foldFree :: Free f a -> m a is a certain function we do not need to look into for now.

The main idea is that those different specializations of u are related by a parametricity theorem (aka. free theorem).

For all monads m1, m2 that are instances of MonadFree f, and for any relation r between m1 and m2, if r satisfies $CERTAIN_CONDITIONS, then r relates u @m1 and u @m2.

In this case, we will let r relate u1 :: Free f a and u2 :: m a when:

foldFree u1 = u2

As it turns out, r will satisfy $CERTAIN_CONDITIONS, so that the parametricity theorem above applies. This yields exactly the desired conclusion:

foldFree (u @(Free f)) = u @m

It is going to be a gnarly exposition of definitions before we can even get to the proof, and the only reason I can think of to stick around is morbid curiosity. But I had the proof and I wanted to do something with it.2

Formalization in Coq

Imports and setting options
From Coq Require Import Morphisms.

Set Implicit Arguments.
Set Contextual Implicit.

Initial free monads

Right off the bat, the first hurdle is that we cannot actually write the initial Free in Coq. To guarantee that all functions terminate and to prevent logical inconsistencies, Coq imposes restrictions about what recursive types can be defined. Indeed, Free could be used to construct an infinite loop by instantiating it with a contravariant functor f. The following snippet shows how we can inhabit the empty type Void, using only non-recursive definitions, so it’s fair to put the blame on Free:

newtype Cofun b a = Cofun (a -> b)

omicron :: Free (Cofun Void) Void -> Void
omicron (Pure y) = y
omicron (Free (Cofun z)) = z (Free (Cofun z))

omega :: Void
omega = omicron (Free (Cofun omicron))

To bypass that issue, we can tweak the definition of Free into what you might know as the freer monad, or the operational monad. The key difference is that the recursive occurrence of Free f a is no longer under an abstract f, but a concrete (->) instead.

Inductive Free (f : Type -> Type) (a : Type) : Type :=
| Pure : a -> Free f a
| Bind : forall e, f e -> (e -> Free f a) -> Free f a
.

Digression on containers

With that definition, it is no longer necessary for f to be a functor—it’s even undesirable because of size issues. Instead, f should rather be thought of as a type of “shapes”, containing “positions” of type e, and that induces a functor by assigning values to those positions (via the function e -> Free f a here); such an f is also known as a “container”.

For example, the Maybe functor consists of two “shapes”: Nothing, with no positions (indexed by Void), and Just, with one position (indexed by ()). Those shapes are defined by the following GADT, the Maybe container:

data PreMaybe _ where
  Nothing_ :: PreMaybe Void
  Just_ :: PreMaybe ()

A container extends into a functor, using a construction that some call Coyoneda:

data Maybe' a where
  MkMaybe' :: forall a e. PreMaybe e -> (e -> a) -> Maybe' a

data Coyoneda f a where
  Coyoneda :: forall f a e. f e -> (e -> a) -> Coyoneda f a

Freer f a (where Freer is called Free here in Coq) coincides with Free (Coyoneda f) a (for the original definition of Free at the top). If f is already a functor, then it is observationally equivalent to Coyoneda f.

Monad and MonadFree

The Monad class hides no surprises. For simplicity we skip the Functor and Applicative classes. Like in C, return is a keyword in Coq, so we have to settle for another name.

Class Monad (m : Type -> Type) : Type :=
  { pure : forall {a}, a -> m a
  ; bind : forall {a b}, m a -> (a -> m b) -> m b
  }.
(* The braces after `forall` make the arguments implicit. *)

Our MonadFree class below is different than in Haskell because of the switch from functors to containers (see previous section). In the original MonadFree, the method free takes an argument of type f (m a), where the idea is to “interpret” the outer layer f, and “carry on” with a continuation m a. Containers encode that outer layer without the continuation.3

Class MonadFree {f m : Type -> Type} `{Monad m} : Type :=
  { free : forall {x}, f x -> m x }.

(* Some more implicit arguments nonsense. *)
Arguments MonadFree f m {_}.

Here comes the final encoding of free monads. The resemblance to the Haskell code above should be apparent in spite of some funny syntax.

Definition Free' (f : Type -> Type) (a : Type) : Type :=
  forall m `(MonadFree f m), m a.

Type classes in Coq are simply types with some extra type inference rules to infer dictionaries. Thus, the definition of Free' actually desugars to a function type forall m, Monad m -> MonadFree f m -> m a. A value u : Free' f a is a function whose arguments are a type constructor m, followed by two dictionaries of the Monad and MonadFree classes. We specialize u to a monad m by writing u m _ _, applying u to the type constructor m and two holes (underscores) for the dictionaries, whose contents will be inferred via type class resolution. See for example fromFree' below.

While we’re at it, we can define the instances of Monad and MonadFree for the initial encoding Free.

Fixpoint bindFree {f a b} (u : Free f a) (k : a -> Free f b) : Free f b :=
  match u with
  | Pure a => k a
  | Bind e h => Bind e (fun x => bindFree (h x) k)
  end.

Instance Monad_Free f : Monad (Free f) :=
  {| pure := @Pure f
  ;  bind := @bindFree f
  |}.

Instance MonadFree_Free f : MonadFree f (Free f) :=
  {| free A e := Bind e (fun a => Pure a)
  |}.

Interpretation of free monads

To show that those monads are equivalent, we must exhibit a mapping going both ways.

The easy direction is from the final Free' to the initial Free: with the above instances of Monad and MonadFree, just monomorphize the polymorph.

Definition fromFree' {f a} : Free' f a -> Free f a :=
  fun u => u (Free f) _ _.

The other direction is obtained via a fold of Free f, which allows us to interpret it in any instance of MonadFree f: replace Bind with bind, interpret the first operand with free, and recurse in the second operand.

Fixpoint foldFree {f m a} `{MonadFree f m} (u : Free f a) : m a :=
  match u with
  | Pure a => pure a
  | Bind e k => bind (free e) (fun x => foldFree (k x))
  end.

Definition toFree' {f a} : Free f a -> Free' f a :=
  fun u M _ _ => foldFree u.

Equality

In everyday mathematics, equality is a self-evident notion that we take for granted. But if you want to minimize your logical foundations, you do not need equality as a primitive. Equations are just equivalences, where the equivalence relation is kept implicit.

Who even decides what the rules for reasoning about equality are anyway? You decide, by picking the underlying equivalence relation. 4

Here is a class for equality. It is similar to Eq in Haskell, but it is propositional (a -> a -> Prop) rather than boolean (a -> a -> Bool), meaning that equality doesn’t have to be decidable.

Class PropEq (a : Type) : Type :=
  propeq : a -> a -> Prop.

Notation "_ = _" was already used in scope type_scope. [notation-overridden,parsing]

For example, for inductive types, a common equivalence can be defined as another inductive type which equates constructors and their fields recursively. Here it is for Free:

Inductive eq_Free f a : PropEq (Free f a) :=
| eq_Free_Pure x : eq_Free (Pure x) (Pure x)
| eq_Free_Bind p (e : f p) k1 k2
  : (forall x, eq_Free (k1 x) (k2 x)) ->
    eq_Free (Bind e k1) (Bind e k2)
.

(* Register it as an instance of PropEq *)
Existing Instance eq_Free.

Having defined equality for Free, we can state and prove one half of the isomorphism between Free and Free'.

f: Type -> Type
a: Type
u: Free f a

fromFree' (toFree' u) = u

The proof is straightforward by induction, case analysis (which is performed as part of induction), and simplification.

f: Type -> Type
a: Type
u: Free f a

fromFree' (toFree' u) = u
f: Type -> Type
a: Type
a0: a

fromFree' (toFree' (Pure a0)) = Pure a0
f: Type -> Type
a, e: Type
f0: f e
f1: e -> Free f a
H: forall e : e, fromFree' (toFree' (f1 e)) = f1 e
fromFree' (toFree' (Bind f0 f1)) = Bind f0 f1
f: Type -> Type
a: Type
a0: a

Pure a0 = Pure a0
f: Type -> Type
a, e: Type
f0: f e
f1: e -> Free f a
H: forall e : e, fromFree' (toFree' (f1 e)) = f1 e
Bind f0 (fun x : e => foldFree (f1 x)) = Bind f0 f1
all: constructor; auto. Qed.

Equality on final encodings, naive attempts

To state the other half of the isomorphism (toFree' (fromFree' u) = u), it is less obvious what the right equivalence relation on Free' should be. When are two polymorphic values u1, u2 : forall m `(MonadFree f m), m a equal? A fair starting point is that all of their specializations must be equal. “Equality” requires an instance of PropEq, which must be introduced as an extra parameter.

(* u1 and u2 are "equal" when all of their specializations
   (u1 m _ _) and (u2 m _ _) are equal. *)
Definition eq_Free'_very_naive f a (u1 u2 : Free' f a) : Prop :=
  forall m `(MonadFree f m) `(forall x, PropEq (m x)),
    u1 m _ _ = u2 m _ _.

That definition is flagrantly inadequate: so far, a PropEq instance can be any relation, including the empty relation (which never holds), and the Monad instance (as a superclass of MonadFree) might be unlawful. In our desired theorem, toFree' (fromFree' u) = u, the two sides use a priori different combinations of bind and pure, so we expect to rely on laws to be able to rewrite one side into the other.

In programming, we aren’t used to proving that implementations satisfy their laws, so there is always the possibility that a Monad instance is unlawful. In math, the laws are in the definitions; if something doesn’t satisfy the monad laws, it’s not a monad. Let’s irk some mathematicians and say that a lawful monad is a monad that satisfies the monad laws. Thus we will have one Monad class for the operations only, and one LawfulMonad class for the laws they should satisfy. Separating code and proofs that way helps to organize things. Code is often much simpler than the proofs about it, since the latter necessarily involves dependent types.

Class LawfulMonad {m} `{Monad m} `{forall a, PropEq (m a)} : Prop :=
  { Equivalence_LawfulMonad :> forall a, Equivalence (propeq (a := m a))
  ; propeq_bind : forall a b (u u' : m a) (k k' : a -> m b),
      u = u' -> (forall x, k x = k' x) -> bind u k = bind u' k'
  ; bind_pure : forall a (u : m a),
      bind u (pure (a := a)) = u
  ; pure_bind : forall a b (x : a) (k : a -> m b),
      bind (pure x) k = k x
  ; bind_bind : forall a b c (u : m a) (k : a -> m b) (h : b -> m c),
      bind (bind u k) h = bind u (fun x => bind (k x) h)
  }.

The three monad laws should be familiar (bind_pure, pure_bind, bind_bind). In those equations, “=” denotes a particular equivalence relation, which is now a parameter/superclass of the class. Once you give up on equality as a primitive notion, algebraic structures must now carry their own equivalence relations. The requirement that it is an equivalence relation also becomes an explicit law (Equivalence_LawfulMonad), and we expect that operations (in this case, bind) preserve the equivalence (propeq_bind). Practically speaking, that last fact allows us to rewrite subexpressions locally, otherwise we could only apply the monad laws at the root of an expression.

A less naive equivalence on Free' is thus to restrict the quantification to lawful instances:

Definition eq_Free'_naive f a (u1 u2 : Free' f a) : Prop :=
  forall m `(MonadFree f m) `(forall x, PropEq (m x)) `(!LawfulMonad (m := m)),
    u1 m _ _ = u2 m _ _.

That is a quite reasonable definition of equivalence for Free'. In other circumstances, it could have been useful. Unfortunately, it is too strong here: we cannot prove the equation toFree' (fromFree' u) = u with that interpretation of =. Or at least I couldn’t figure out a solution. We will need more assumptions to be able to apply the parametricity theorem of the type Free'. To get there, we must formalize Reynolds’ relational interpretation of types.

Types as relations

The core technical idea in Reynolds’ take on parametricity is to interpret a type t as a relation Rt : t -> t -> Prop. Then, the parametricity theorem is that all terms x : t are related to themselves by Rt (Rt x x is true). If t is a polymorphic type, that theorem connects different specializations of a same term x : t, and that allows us to formalize arguments that rely on “parametricity” as a vague idea.

For example, if t = (forall a, a -> a), then Rt is the following relation, which says that two functions f and f' are related if for any relation Ra (on any types), f and f' send related inputs (Ra x x') to related outputs (Ra (f a x) (f' a' x')).

Rt f f' =
  forall a a' (Ra : a -> a' -> Prop),
  forall x x', Ra x x' -> Ra (f a x) (f' a' x')

If we set Ra x x' to mean “x equals an arbitrary constant z0” (ignoring x', i.e., treating Ra as a unary relation), the above relation Rt amounts to saying that f z0 = z0, from which we deduce that f must be the identity function.

The fact that Rt is a relation is not particularly meaningful to the parametricity theorem, where terms are simply related to themselves, but it is a feature of the construction of Rt: the relation for a composite type t1 -> t2 combines the relations for the components t1 and t2, and we could not get the same result with only unary predicates throughout.5 More formally, we define a relation R[t] by induction on t, between the types t and t', where t' is the result of renaming all variables x to x' in t (including binders). The two most interesting cases are:

In summary:

R[forall a, t](f, f') = forall a a' Ra, R[t](f a)(f' a')
R[a](f, f')           = Ra(f, f')
                        -- Ra should be in scope when a is in scope.
R[t1 -> t2](f, f')    = forall x x', R[t1](x, x') -> R[t2](f x, f' x')

That explanation was completely unhygienic, but refer to Reynolds’ paper or Wadler’s Theorems for free! for more formal details.

For sums (Either/sum) and products ((,)/prod), two values are related if they start with the same constructor, and their fields are related (recursively). This can be deduced from the rules above applied to the Church encodings of sums and products.

Type constructors as relation transformers

While types t : Type are associated to relations Rt : t -> t -> Prop, type constructors m : Type -> Type are associated to relation transformers (functions on relations) Rm : forall a a', (a -> a' -> Prop) -> (m a -> m a' -> Prop). It is usually clear what’s what from the context, so we will often refer to “relation transformers” as just “relations”.

For example, the initial Free f a type gets interpreted to the relation RFree Rf Ra defined as follows. Two values u1 : Free f1 a1 and u2 : Free f2 a2 are related by RFree if either:

We thus have one rule for each constructor (Pure and Bind) in which we relate each field (Ra x1 x2 in RFree_Pure; Rf _ _ _ y1 y2 and RFree Rf Ra (k1 x1) (k2 x2) in RFree_Bind). Let us also remark that the existential type e in Bind becomes an existential relation Re in RFree_Bind.

Inductive RFree {f₁ f₂ : Type -> Type}
    (Rf : forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> f₁ a₁ -> f₂ a₂ -> Prop)
    {a₁ a₂ : Type} (Ra : a₁ -> a₂ -> Prop) : Free f₁ a₁ -> Free f₂ a₂ -> Prop :=
  | RFree_Pure : forall (x₁ : a₁) (x₂ : a₂),
      Ra x₁ x₂ -> RFree Rf Ra (Pure x₁) (Pure x₂)
  | RFree_Bind : forall (e₁ e₂ : Type) (Re : e₁ -> e₂ -> Prop) (y₁ : f₁ e₁) (y₂ : f₂ e₂),
      Rf e₁ e₂ Re y₁ y₂ ->
      forall (k₁ : e₁ -> Free f₁ a₁) (k₂ : e₂ -> Free f₂ a₂),
      (forall (x₁ : e₁) (x₂ : e₂),
        Re x₁ x₂ -> RFree Rf Ra (k₁ x₁) (k₂ x₂)) ->
      RFree Rf Ra (Bind y₁ k₁) (Bind y₂ k₂).

Inductive relations such as RFree, indexed by types with existential quantifications such as Free, are a little terrible to work with out-of-the-box—especially if you’re allergic to UIP. Little “inversion lemmas” like the following make them a bit nicer by reexpressing those relations in terms of some standard building blocks which leave less of a mess when decomposed.

f₁, f₂: Type -> Type
Rf: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> f₁ a₁ -> f₂ a₂ -> Prop
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
u₁: Free f₁ a₁
u₂: Free f₂ a₂

RFree Rf Ra u₁ u₂ -> match u₁ with | Pure a₁ => match u₂ with | Pure a₂ => Ra a₁ a₂ | Bind _ _ => False end | @Bind _ _ e y₁ k₁ => match u₂ with | Pure _ => False | @Bind _ _ e0 y₂ k₂ => exists Re : e -> e0 -> Prop, Rf e e0 Re y₁ y₂ /\ (forall (x₁ : e) (x₂ : e0), Re x₁ x₂ -> RFree Rf Ra (k₁ x₁) (k₂ x₂)) end end
f₁, f₂: Type -> Type
Rf: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> f₁ a₁ -> f₂ a₂ -> Prop
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
u₁: Free f₁ a₁
u₂: Free f₂ a₂

RFree Rf Ra u₁ u₂ -> match u₁ with | Pure a₁ => match u₂ with | Pure a₂ => Ra a₁ a₂ | Bind _ _ => False end | @Bind _ _ e y₁ k₁ => match u₂ with | Pure _ => False | @Bind _ _ e0 y₂ k₂ => exists Re : e -> e0 -> Prop, Rf e e0 Re y₁ y₂ /\ (forall (x₁ : e) (x₂ : e0), Re x₁ x₂ -> RFree Rf Ra (k₁ x₁) (k₂ x₂)) end end
intros []; eauto. Qed.

Type classes, which are (record) types, also get interpreted in the same way. Since Monad is parameterized by a type constructor m, the relation RMonad between Monad instances is parameterized by a relation between two type constructors m1 and m2. Two instances of Monad, i.e., two values of type Monad m for some m, are related if their respective fields, i.e., pure and bind, are related. pure and bind are functions, so two instances are related when they send related inputs to related outputs.

Record RMonad (m₁ m₂ : Type -> Type)
    (Rm : forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> m₁ a₁ -> m₂ a₂ -> Prop)
    `{Monad m₁} `{Monad m₂} : Prop :=
  { RMonad_pure : forall (t₁ t₂ : Type) (Rt : t₁ -> t₂ -> Prop) (x₁ : t₁) (x₂ : t₂),
      Rt x₁ x₂ -> Rm t₁ t₂ Rt (pure x₁) (pure x₂)
  ; RMonad_bind : forall (t₁ t₂ : Type) (Rt : t₁ -> t₂ -> Prop) 
      (u₁ u₂ : Type) (Ru : u₁ -> u₂ -> Prop) (x₁ : m₁ t₁) (x₂ : m₂ t₂),
      Rm t₁ t₂ Rt x₁ x₂ ->
      forall (k₁ : t₁ -> m₁ u₁) (k₂ : t₂ -> m₂ u₂),
      (forall (x₁ : t₁) (x₂ : t₂),
         Rt x₁ x₂ -> Rm u₁ u₂ Ru (k₁ x₁) (k₂ x₂)) ->
      Rm u₁ u₂ Ru (bind x₁ k₁) (bind x₂ k₂)
  }.

MonadFree also gets translated to a relation RMonadFree. Related inputs, related outputs.

Record RMonadFree (f₁ f₂ : Type -> Type)
    (Rf : forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> f₁ a₁ -> f₂ a₂ -> Prop)
    (m₁ m₂ : Type -> Type)
    (Rm : forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> m₁ a₁ -> m₂ a₂ -> Prop)
    `{MonadFree f₁ m₁} `{MonadFree f₂ m₂} : Prop :=
  { RMonadFree_free : forall (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (x₁ : f₁ a₁) (x₂ : f₂ a₂),
      Rf a₁ a₂ Ra x₁ x₂ -> Rm a₁ a₂ Ra (free x₁) (free x₂)
  }.

Note that RMonad and RMonadFree are “relation transformer transformers”, since they take relation transformers as arguments, to produce a relation between class dictionaries.

We can now finally translate the final Free' to a relation. Two values u1 : Free' f1 a1 and u2 : Free' f2 a2 are related if, for any two monads m1 and m2, with a relation transformer Rm, whose Monad and MonadFree instances are related by RMonad and RMonadFree, Rm relates u1 m1 _ _ and u2 m2 _ _.

Definition RFree' {f₁ f₂} Rf {a₁ a₂} Ra (u₁ : Free' f₁ a₁) (u₂ : Free' f₂ a₂) : Prop :=
  forall m₁ m₂ `(MonadFree f₁ m₁) `(MonadFree f₂ m₂) Rm
    (pm : RMonad Rm) (pf : RMonadFree Rf Rm),
    Rm _ _ Ra (u₁ m₁ _ _) (u₂ m₂ _ _).

The above translation of types into relations can be automated by a tool such as paramcoq. However paramcoq currently constructs relations in Type instead of Prop, which got me stuck in universe inconsistencies. That’s why I’m declaring Prop relations the manual way here.

The parametricity theorem says that any u : Free' f a is related to itself by RFree' (for some canonical relations on f and a). It is a theorem about the language Coq which we can’t prove within Coq. Rather than postulate it, we will simply add the required RFree' _ _ u u assumption to our proposition (from_to below). Given a concrete u, it should be straightforward to prove that assumption case-by-case in order to apply that proposition.

These “relation transformers” are a bit of a mouthful to spell out, and they’re usually guessable from the type constructor (f or m), so they deserve a class, that’s a higher-order counterpart to PropEq (like Eq1 is to Eq in Haskell).

Class PropEq1 (m : Type -> Type) : Type :=
  propeq1 : forall a₁ a₂, (a₁ -> a₂ -> Prop) -> m a₁ -> m a₂ -> Prop.

Given a PropEq1 m instance, we can apply it to the relation eq to get a plain relation which seems a decent enough default for PropEq (m a).

Instance PropEq_PropEq1 {m} `{PropEq1 m} {a} : PropEq (m a) := propeq1 eq.

Really lawful monads

We previously defined a “lawful monad” as a monad with an equivalence relation (PropEq (m a)). To use parametricity, we will also need a monad m to provide a relation transformer (PropEq1 m), which subsumes PropEq with the instance just above.6 This extra structure comes with additional laws, extending our idea of monads to “really lawful monads”.

Class Trans_PropEq1 {m} `{PropEq1 m} : Prop :=
  trans_propeq1 : forall a₁ a₂ (r : a₁ -> a₂ -> Prop) x₁ x₁' x₂ x₂',
    x₁ = x₁' -> propeq1 r x₁' x₂ -> x₂ = x₂' -> propeq1 r x₁ x₂'.

Class ReallyLawfulMonad m `{Monad m} `{PropEq1 m} : Prop :=
  { LawfulMonad_RLM :> LawfulMonad (m := m)
  ; Trans_PropEq1_RLM :> Trans_PropEq1 (m := m)
  ; RMonad_RLM : RMonad (propeq1 (m := m))
  }.

Class ReallyLawfulMonadFree f `{PropEq1 f} m `{MonadFree f m} `{PropEq1 m} : Prop :=
  { ReallyLawfulMonad_RLMF :> ReallyLawfulMonad (m := m)
  ; RMonadFree_RLMF : RMonadFree (propeq1 (m := f)) (propeq1 (m := m))
  }.

We inherit the LawfulMonad laws from before. The relations RMonad and RMonadFree, defined earlier, must relate m’s instances of Monad and MonadFree, for the artificial reason that that’s roughly what RFree' will require. We also add a generalized transitivity law, which allows us to rewrite either side of a heterogeneous relation propeq1 r using the homogeneous one = (which denotes propeq1 eq).

It’s worth noting that there is some redundancy here, that could be avoided with a bit of refactoring. That generalized transitivity law Trans_PropEq1 implies transitivity of =, which is part of the claim that = is an equivalence relation in LawfulMonad. And the bind component of RMonad implies propeq_bind in LawfulMonad, so these RMonad and RMonadFree laws can also be seen as generalizations of congruence laws to heterogeneous relations, making them somewhat less artificial than they may seem at first.

Restricting the definition of equality on the final free monad Free' to quantify only over really lawful monads yields the right notion of equality for our purposes, which is to prove the from_to theorem below, validating the isomorphism between Free and Free'.

Instance eq_Free' f `(PropEq1 f) a : PropEq (Free' f a) :=
  fun u₁ u₂ =>
    forall m `(MonadFree f m) `(PropEq1 m) `(!ReallyLawfulMonadFree (m := m)),
      u₁ m _ _ = u₂ m _ _.

Quickly, let’s get the following lemma out of the way, which says that foldFree commutes with bind. We’re really saying that foldFree is a monad morphism but no time to say it properly. The proof of the next lemma will need this, but it’s also nice to look at this on its own.

f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b: Type
u: Free f a
k: a -> Free f b

foldFree (bindFree u k) = bind (foldFree u) (fun x : a => foldFree (k x))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b: Type
u: Free f a
k: a -> Free f b

foldFree (bindFree u k) = bind (foldFree u) (fun x : a => foldFree (k x))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b: Type
a0: a
k: a -> Free f b

foldFree (k a0) = bind (pure a0) (fun x : a => foldFree (k x))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b, e: Type
f0: f e
f1: e -> Free f a
k: a -> Free f b
H2: forall e : e, foldFree (bindFree (f1 e) k) = bind (foldFree (f1 e)) (fun x : a => foldFree (k x))
bind (free f0) (fun x : e => foldFree (bindFree (f1 x) k)) = bind (bind (free f0) (fun x : e => foldFree (f1 x))) (fun x : a => foldFree (k x))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b: Type
a0: a
k: a -> Free f b

foldFree (k a0) = bind (pure a0) (fun x : a => foldFree (k x))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b: Type
a0: a
k: a -> Free f b

bind (pure a0) (fun x : a => foldFree (k x)) = foldFree (k a0)
apply pure_bind with (k0 := fun x => foldFree (k x)).
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b, e: Type
f0: f e
f1: e -> Free f a
k: a -> Free f b
H2: forall e : e, foldFree (bindFree (f1 e) k) = bind (foldFree (f1 e)) (fun x : a => foldFree (k x))

bind (free f0) (fun x : e => foldFree (bindFree (f1 x) k)) = bind (bind (free f0) (fun x : e => foldFree (f1 x))) (fun x : a => foldFree (k x))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b, e: Type
f0: f e
f1: e -> Free f a
k: a -> Free f b
H2: forall e : e, foldFree (bindFree (f1 e) k) = bind (foldFree (f1 e)) (fun x : a => foldFree (k x))

bind (free f0) (fun x : e => foldFree (bindFree (f1 x) k)) = bind (free f0) (fun x : e => bind ((fun x0 : e => foldFree (f1 x0)) x) (fun x0 : a => foldFree (k x0)))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b, e: Type
f0: f e
f1: e -> Free f a
k: a -> Free f b
H2: forall e : e, foldFree (bindFree (f1 e) k) = bind (foldFree (f1 e)) (fun x : a => foldFree (k x))

free f0 = free f0
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b, e: Type
f0: f e
f1: e -> Free f a
k: a -> Free f b
H2: forall e : e, foldFree (bindFree (f1 e) k) = bind (foldFree (f1 e)) (fun x : a => foldFree (k x))
forall x : e, foldFree (bindFree (f1 x) k) = bind (foldFree (f1 x)) (fun x0 : a => foldFree (k x0))
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b, e: Type
f0: f e
f1: e -> Free f a
k: a -> Free f b
H2: forall e : e, foldFree (bindFree (f1 e) k) = bind (foldFree (f1 e)) (fun x : a => foldFree (k x))

free f0 = free f0
reflexivity.
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: forall a : Type, PropEq (m a)
LawfulMonad0: LawfulMonad
a, b, e: Type
f0: f e
f1: e -> Free f a
k: a -> Free f b
H2: forall e : e, foldFree (bindFree (f1 e) k) = bind (foldFree (f1 e)) (fun x : a => foldFree (k x))

forall x : e, foldFree (bindFree (f1 x) k) = bind (foldFree (f1 x)) (fun x0 : a => foldFree (k x0))
auto. Qed.

Finally the proof

Our goal is to prove an equation in terms of eq_Free', which gives us a really lawful monad as an assumption. We open a section to set up the same context as that and to break down the proof into more digestible pieces.

Section ISOPROOF.

Context {f m} `{MonadFree f m} `{PropEq1 m} `{!ReallyLawfulMonad (m := m)}.

As outlined earlier, parametricity will yield an assumption RFree' _ _ u u, and we will specialize it with a relation R which relates u1 : Free f a and u2 : m a when foldFree u1 = u2. However, RFree' actually expects a relation transformer rather than a relation, so we instead define R to relate u1 : Free f a1 and u2 : Free f a2 when propeq1 Ra (foldFree u1) u2, where Ra is a relation given between a1 and a2.

Let R := (fun a₁ a₂ (Ra : a₁ -> a₂ -> Prop) u₁ u₂ => propeq1 Ra (foldFree u₁) u₂).

The following two lemmas are the “$CERTAIN_CONDITIONS” mentioned earlier, that R must satisfy, i.e., we prove that R, via RMonad (resp. RMonadFree), relates the Monad (resp. MonadFree) instances for Free f and m.

f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop

RMonad R
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop

RMonad R
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
x₁: t₁
x₂: t₂
H2: Rt x₁ x₂

R Rt (pure x₁) (pure x₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)
R Ru (bind x₁ k₁) (bind x₂ k₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
x₁: t₁
x₂: t₂
H2: Rt x₁ x₂

R Rt (pure x₁) (pure x₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
x₁: t₁
x₂: t₂
H2: Rt x₁ x₂

R Rt (Pure x₁) (pure x₂)
apply RMonad_RLM; auto.
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)

R Ru (bind x₁ k₁) (bind x₂ k₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)

propeq1 Ru (foldFree (bind x₁ k₁)) (bind x₂ k₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)

foldFree (bind x₁ k₁) = ?x₁'
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)
propeq1 Ru ?x₁' ?x₂
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)
?x₂ = bind x₂ k₂
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)

foldFree (bind x₁ k₁) = ?x₁'
apply foldFree_bindFree.
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)

propeq1 Ru (bind (foldFree x₁) (fun x : t₁ => foldFree (k₁ x))) ?x₂
eapply RMonad_RLM; eauto.
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
t₁, t₂: Type
Rt: t₁ -> t₂ -> Prop
u₁, u₂: Type
Ru: u₁ -> u₂ -> Prop
x₁: Free f t₁
x₂: m t₂
H2: R Rt x₁ x₂
k₁: t₁ -> Free f u₁
k₂: t₂ -> m u₂
H3: forall (x₁ : t₁) (x₂ : t₂), Rt x₁ x₂ -> R Ru (k₁ x₁) (k₂ x₂)

bind x₂ k₂ = bind x₂ k₂
reflexivity. Qed. Context (Rf : PropEq1 f). Context (RMonadFree_m : RMonadFree propeq1 propeq1).
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1

RMonadFree Rf R
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1

RMonadFree Rf R
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

R Ra (free x₁) (free x₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

R Ra (free x₁) (free x₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

propeq1 Ra (foldFree (free x₁)) (free x₂)
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

foldFree (free x₁) = ?x₁'
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂
propeq1 Ra ?x₁' ?x₂
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂
?x₂ = free x₂
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

foldFree (free x₁) = ?x₁'
apply bind_pure.
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

propeq1 Ra (free x₁) ?x₂
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

propeq1 Ra x₁ ?Goal0
eassumption.
f, m: Type -> Type
H: Monad m
H0: MonadFree f m
H1: PropEq1 m
ReallyLawfulMonad0: ReallyLawfulMonad
R:= fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂: forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> Free f a₁ -> m a₂ -> Prop
Rf: PropEq1 f
RMonadFree_m: RMonadFree propeq1 propeq1
a₁, a₂: Type
Ra: a₁ -> a₂ -> Prop
x₁: f a₁
x₂: f a₂
H2: Rf Ra x₁ x₂

free x₂ = free x₂
reflexivity. Qed. End ISOPROOF.

Here comes the conclusion, which completes our claim that toFree'/fromFree' is an isomorphism (we proved the other half to_from on the way here). This equation is under an assumption which parametricity promises to fulfill, but we will have to step out of the system if we want it right now.

f: Type -> Type
Rf: PropEq1 f
a: Type
u: Free' f a

RFree' Rf eq u u -> toFree' (fromFree' u) = u

In the proof, we get the assumption H : RFree' Rf eq u u, which we apply to the above lemmas, RMonad_foldFree and RMonadFree_foldFree, using the specialize tactic. That yields exactly our desired goal.

f: Type -> Type
Rf: PropEq1 f
a: Type
u: Free' f a

RFree' Rf eq u u -> toFree' (fromFree' u) = u
f: Type -> Type
Rf: PropEq1 f
a: Type
u: Free' f a
H: RFree' Rf eq u u
m: Type -> Type
H0: Monad m
H1: MonadFree f m
H2: PropEq1 m
ReallyLawfulMonadFree0: ReallyLawfulMonadFree

toFree' (fromFree' u) H1 = u m H0 H1
f: Type -> Type
Rf: PropEq1 f
a: Type
u: Free' f a
H: RFree' Rf eq u u
m: Type -> Type
H0: Monad m
H1: MonadFree f m
H2: PropEq1 m
ReallyLawfulMonadFree0: ReallyLawfulMonadFree

foldFree (u (Free f) Monad_Free MonadFree_Free) = u m H0 H1
f: Type -> Type
Rf: PropEq1 f
a: Type
u: Free' f a
H: forall (m₁ m₂ : Type -> Type) (H : Monad m₁) (H0 : MonadFree f m₁) (H1 : Monad m₂) (H2 : MonadFree f m₂) (Rm : forall a₁ a₂ : Type, (a₁ -> a₂ -> Prop) -> m₁ a₁ -> m₂ a₂ -> Prop), RMonad Rm -> RMonadFree Rf Rm -> Rm a a eq (u m₁ H H0) (u m₂ H1 H2)
m: Type -> Type
H0: Monad m
H1: MonadFree f m
H2: PropEq1 m
ReallyLawfulMonadFree0: ReallyLawfulMonadFree

foldFree (u (Free f) Monad_Free MonadFree_Free) = u m H0 H1
f: Type -> Type
Rf: PropEq1 f
a: Type
u: Free' f a
m: Type -> Type
H0: Monad m
H1: MonadFree f m
H2: PropEq1 m
H: (fun (a₁ a₂ : Type) (Ra : a₁ -> a₂ -> Prop) (u₁ : Free f a₁) (u₂ : m a₂) => propeq1 Ra (foldFree u₁) u₂) a a eq (u (Free f) Monad_Free MonadFree_Free) (u m H0 H1)
ReallyLawfulMonadFree0: ReallyLawfulMonadFree

foldFree (u (Free f) Monad_Free MonadFree_Free) = u m H0 H1
apply H. Qed.

Conclusion

If you managed to hang on so far, treat yourself to some chocolate.

To formalize a parametricity argument in Coq, I had to move the goalposts quite a bit throughout the experiment:

It could be interesting to see a “really lawful monad” spelled out fully.

Another similar but simpler exercise is to prove the equivalence between initial and final encodings of lists. It probably wouldn’t involve “relation transformers” as much. There are also at least two different variants: is your final encoding “foldr”- or “fold”-based (the latter mentions monoids, the former doesn’t)?

I hope that machinery can be simplified eventually, but given the technical sophistication that is currently necessary, prudence is advised when navigating around claims made “by parametricity”.



  1. Answering Iceland_Jack’s question on Twitter.↩︎

  2. Also an excuse to integrate Alectryon in my blog.↩︎

  3. That idea is also present in Kiselyov and Ishii’s paper.↩︎

  4. Those who do know Coq will wonder, what about eq (“intensional equality”)? It is a fine default relation for first-order data (nat, pairs, sums, lists, ASTs without HOAS). But it is too strong for computations (functions and coinductive types) and proofs (of Props). Then a common approach is to introduce extensionality axioms, postulating that “extensional equality implies intensional equality”. But you might as well just stop right after proving whatever extensional equality you wanted.↩︎

  5. Well, if you tried you would end up with the unary variant of the parametricity theorem, but it’s much weaker than the binary version shown here. n-ary versions are also possible and even more general, but you have to look hard to find legitimate uses.↩︎

  6. To be honest, that decision was a little arbitrary. But I’m not sure making things more complicated by keeping EqProp1 and EqProp separate buys us very much.↩︎