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Construction of the free group in coq

February 9, 2020

\[ \newcommand\G{\mathscr{G}} \newcommand\hom{\text{Hom}} \newcommand\M{\mathscr{M}} \newcommand\bet{\rightarrow} \]


While following a course on algebra, I add the unavoidable course on the construction of the free group. The teacher being a pure algebraist, the construction was fully algebraic. It was a complicated piece of mathematics where proving some intuitive result turned out very complicated.

And I could think was that the intuition behind the construction is not that hard, you just have your set set of symbols, you had formal inverses, you take the free monoid and when there is something of the form \(aa^{-1}\) you eliminate it. And it turns out this is the approach described in Wikipedia.

But there usually is a big difference between an intuitive description and a formal one, with the proof of the relevant properties, and sometimes the most intuitive approach is the harder to formalize. I wanted to see if I could formalize the reduction approach using tools from lambda calculus, and see if it was more complicated or not. Turns out that this approach was easy enough, so I present it here.

And it provides me with a perfect exercice to start using Coq again, so I will provide some Coq snippets of the definitions, and some proofs, along the way. For simplicity sake, I will assume function extensionality for those proofs. The final code can be found here.

What’s a free group anyway

Let’s start by defining what’s a free group. Intuitively, a free group over a set \(X\) is the most general group in which one can inject \(X\), and no relations between the injected elements.

Can we make it more formal ? Indeed, the definition is that if you have a set \(X\), the free group \(\G(X)\) over \(X\) is the group such that for every other group \(G\), there is a bijection between \(\hom(\G(X), G)\) and \(G^X\) (the functions from \(X\) to \(G\)).

Intuitively, it means that for every group morphism, knowing the image of the generators of the free group is sufficient. The other direction is the most general part : no matter what images one impose on the generators, it is possible to create a group morphism that respect them.

For those interested in going further, there is a more general notion of free object, which itself comes from the idea of a free fonctor.

The free monoid

The basis of our construction will be the free monoid over a set \(X\) : \(\M(X)\). It is defined using the exact same definition, replacing free group by free monoid and group morphism by monoid morphism.

Let’s formalize that in Coq. First we need to define a monoid structure :

Record Monoid : Type := mkMon {
    type          : Type;
    op            : type -> type -> type;
    empty         : type;
    emptyCorrect  : forall (x : type), op x empty = x /\ op empty x = x;
    associativity : forall (x y z : type), op x (op y z) = op (op x y) z;

Definition is_monoid_morphism (M M' : Monoid) (f : type M -> type M') : Prop :=
    f (empty M) = empty M' /\ forall (x y : type M), f (op M x y) = op M' (f x) (f y).

Record MonoidMorphism (M M' : Monoid) : Type := mkMonMorphism {
    monmor : type M -> type M';
    monmor_correct : is_monoid_morphism M M' monmor;

Now we can define the proposition that tells us what a free monoid is :

Definition is_free_monoid_over (T : Type) (M : Monoid) : Prop :=
    forall (M' : Monoid),
      exists (F : (T -> type M') -> MonoidMorphism M M'),
      exists (G : (type M -> type M') -> (T -> type M')),
           (forall (f : T -> type M'), G (monmor M M' (F f)) = f)
        /\ (forall (m : MonoidMorphism M M'), monmor M M' (F (G (monmor M M' m))) = monmor M M' m).

Now we need to build the free monoid. This is a classic result that I won’t bother proving : the free monoid over a set \(X\) is the set of finite sequences of elements of \(X\), and the operation is the concatenation. If you want more details see on wikipedia or on the nlab.

You can also find the formalised proof in the coq file, for the following definition of free monoid over a type :

Inductive FreeMonT (T : Type) : Type
    := App : T -> FreeMonT T -> FreeMonT T
    | Empty : FreeMonT T

Fixpoint append {T : Type} (x y : FreeMonT T) : FreeMonT T :=
    match x with
    | Empty _   => y
    | App _ h t => App T h (append t y)

Building the free group

Let’s start a bit more seriously now. Let’s review our intuitions : we want need to add formal inverses, and reduce when something is multiplied by its formal inverse. Let’s note our base set \(X\). We define \(X^{-1}\) a set in bijection with \(X\), such that \(X\cup X^{-1} = \emptyset\). If \(a\in X\), we denote its image by the bijection \(a^{-1}\). Those will be our formal inverses.

Now let \(M\) be \(\M(X\uplus X^{-1})\) the free monoid over our initial set and the formal inverses. Here is how to implement something like that in coq :

Inductive WithInv (T : Type) : Type
    := Reg    : T -> WithInv T
    |  ForInv : T -> WithInv T

Now we want a way to say that if, somewhere in the sequence, one find \(aa^{-1}\) or \(a^{-1}a\), they should be removed. But what does it means to removes a subsequence. It sounds more like we’re describing a procedure : start with a sequence, rewrite every reducible pair until there are no more. This sounds a bit like the \(\beta\)-reduction from lambda calculus. Let’s roll with this idea and see where it leads us.

The reduction

We need to define the reduction first :

\[ \left\{\begin{array}{l} \forall x\in X, xx^{-1} \bet \epsilon \\ \forall x\in X, x^{-1}x \bet \epsilon \\ \forall \omega_1, \omega_2, \omega_2', \omega_3\in\M, \omega_2\bet\omega_2' \implies \omega_1\omega_2\omega_3 \bet \omega_1\omega_2'\omega_3 \\ \end{array}\right. \]

The way we’re going to define it in coq is equivalent, but will make it a bit easier to manipulate : given a word, either the redex is at the start, or we reduce the tail of the word. It gives the following definition :

Inductive Reduction (T : Type) : FreeMonT (WithInv T) -> FreeMonT (WithInv T) -> Prop
    := LeftRed  : forall (x : T), forall (tl : FreeMonT (WithInv T)),
                  Reduction T (App (WithInv T) (ForInv T x) (App (WithInv T) (Reg T x) tl)) tl
    |  RightRed : forall (x : T), forall (tl : FreeMonT (WithInv T)),
                  Reduction T (App (WithInv T) (Reg T x) (App (WithInv T) (ForInv T x) tl)) tl
    |  CtxRed   : forall (x : WithInv T), forall (m m' : FreeMonT (WithInv T)),
                  Reduction T m m' -> Reduction T (App (WithInv T) x m) (App (WithInv T) x m')

And now we would like to quotient \(\M(X\uplus X^{-1})\) by an equivalence relationship that says that two terms that have a common reduction are equivalent. This is actually easy because our reduction relation has two very interesting properties.

Strong normalisation

The first thing is that no matter in which order we do the reduction, we will always reach a normal form, that is to say a form that cannot be reduced further.

The intuitive argument is easy : everytime we reduce a word, we strictly reduce its length, so we must stop at some point.

Formally proving that in coq is a bit harder. First we need to define strong normalisation. Here we use the insight that if we define \(x \preceq y\) by \(y \rightarrow x\), we create a new relationship such that \(\rightarrow\) is strongly normalizing if and only if \(\preceq\) is well-founded. So we can just use the well founded module in Coq to get our definition :

Definition Inv (T : Type) (R : T -> T -> Prop) : T -> T -> Prop
    := fun (x y : T) => R y x.

Definition strongly_normalizing (T : Type) (R : T -> T -> Prop) : Prop
    := well_founded (Inv T R).

The proof is then trying to formalize the previous intuition. That is to say we first prove that if we have a relation preserving mapping from one relation to the other, and that the second one is well founded, then the first relation is well founded. Then we prove that length is a relation preserving mapping to the natural with usual order. The fact that they are well founded is proved in the Coq standard library, so we use that to conclude.

Definition monotome_morphism (T T' : Type) (R : T -> T -> Prop)
        (R' : T' -> T' -> Prop) (f : T -> T') : Prop :=
    forall (x y : T), R x y -> R' (f x) (f y).
Theorem preimage_well_founded (T T' : Type) (R : T -> T -> Prop)
        (R' : T' -> T' -> Prop) (f : T -> T') :
    monotome_morphism T T' R R' f -> well_founded R' -> well_founded R.

Lemma monoidLength_monotone (T : Type) :
    monotome_morphism (FreeMonT (WithInv T)) nat
                      (Inv (FreeMonT (WithInv T)) (Reduction T)) lt
                      (monoidLength (WithInv T)).

Theorem reduction_normalizing (T : Type) :
    strongly_normalizing (FreeMonT (WithInv T)) (Reduction T).


Confluence is a sort of weakened determinism. It’s the idea that despite the fact that the reduction is not deterministic, if two terms come from the same original terms, we can reduce them further to reach a common reduction.

Our reduction system has an even stronger property, that is strong confluence. It means that if we have three terms \(\omega\), \(\omega_1\) and \(\omega_2\) such that \(\omega\rightarrow\omega_1\) and \(\omega\rightarrow\omega_2\), then either \(\omega_1 = \omega_2\) or there exists \(\omega'\) such that \(\omega_1\rightarrow\omega'\) and \(\omega_2\rightarrow\omega'\).

This property is actually captured by the following definition (the definition is actually a bit less strong because that’s the official definition, but in out specific case the two cases where \(\omega_1\rightarrow\omega_2\) or vice-versa never happen) :

Definition strongly_confluent (T : Type) (R : T -> T -> Prop) : Prop
     := forall (a b c : T), R a b -> R a c
          -> (b = c) \/ (R b c) \/ (R c b) \/ (exists (d : T), R b d /\ R c d).

Theorem reduction_strongly_confluent' (T : Type) :
    forall (a b c : (FreeMonT (WithInv T))), Reduction T a b -> Reduction T a c
        -> (b = c) \/ (exists (d : (FreeMonT (WithInv T))), Reduction T b d /\ Reduction T c d).

Theorem reduction_strongly_confluent (T : Type) :
    strongly_confluent (FreeMonT (WithInv T)) (Reduction T).

Normal forms

Those two properties, strong normalisation and confluence, means that to test if two terms have a common reduction, we can just reduce them both however we want until they are in normal form, and check the normal form for equality.

This is all well and good, but for now we’ve just defined the normal form as being a term that cannot be reduced further. Surely we can characterize it better ? And indeed, it is just a word such that there now sub word of the form \(xx^{-1}\) or \(x^{-1}x\).

We can just define that naïvely in Coq and it works (we call a word without sub words of the form \(xx^{-1}\) or \(x^{-1}x\) stable, and show it is equivalent to being normal for the reduction) :

Definition normal_form { T : Type } (R : T -> T -> Prop) (x : T) : Prop
    := forall (y : T), R x y -> False.

Inductive is_stable { T : Type } : FreeMonT (WithInv T) -> Prop
    := StableEmpty : is_stable (Empty (WithInv T))
    |  StableSing : forall (x : WithInv T), is_stable (App (WithInv T) x (Empty (WithInv T)))
    |  StableApp : forall (x y : WithInv T), forall (w : FreeMonT (WithInv T)),
            inv x <> y -> is_stable (App (WithInv T) y w)
                -> is_stable (App (WithInv T) x (App (WithInv T) y w)).

Theorem stable_is_normal_form { T : Type } :
    forall (x : FreeMonT (WithInv T)), is_stable x <-> normal_form (Reduction T) x.

Computable reduction

It turns out than under some reasonable assumption of the base type, our reduction also has a very interesting property : it is decidable. To show that we implement a coq function that fully reduce a term and prove it is correct. In order to do so we need the base type to have a decidable equality, which we would automatically have if we had assumed the excluded middle.

Here is the definition of the function, and the theorem proving it is correct :

Fixpoint liftEq { T : Type } (deceq : T -> T -> bool) (x y : WithInv T) : bool
    := match x, y with
     | Reg    _ a, Reg    _ b => deceq a b
     | ForInv _ a, ForInv _ b => deceq a b
     | _,          _          => false

Fixpoint reduce { T : Type } (deceq : T -> T -> bool) (x : FreeMonT (WithInv T)) : FreeMonT (WithInv T) :=
    let witheq : WithInv T -> WithInv T -> bool := liftEq deceq in
    match x with
    | (App _ a w) => let reduced : FreeMonT (WithInv T) := reduce deceq w in
            match reduced with
            | (App _ b w') => if (witheq (inv a) b)
                                then w'
                                else (App (WithInv T) a (App (WithInv T) b w'))
            | Empty _      => App (WithInv T) a (Empty (WithInv T))
    | Empty _ => Empty (WithInv T)

Theorem reduce_is_unique_normal_form { T : Type } (deceq : T -> T -> bool) (decC : DecEqCorrect deceq) :
    forall (x y : FreeMonT (WithInv T)),
        normal_form_of (Reduction T) x y <-> y = reduce deceq x.

Building the quotient

We can now create the candidate group by quotienting by the relation of having a common normal form. First, we need to check that the concatenation operation is compatible with this operation, which is obviously the case. The statement in coq is (where trefl_closure is the reflexive transitive closure of the relation) :

Theorem reduction_compatible_append (T : Type) :
    forall (a b c d : FreeMonT (WithInv T)),
        trefl_closure (Reduction T) a c -> trefl_closure (Reduction T) b d
        -> trefl_closure (Reduction T) (append a b) (append c d).

We would like to quotient in coq, but turns out that taking quotients in coq is hard. So instead we will cheat : we have a canonical representative of every quotient elements, the normal form. So we can define the type of elements that are in normal form, and that will be our quotient.

Record FreeGrpT (T : Type) : Type := mkFreeGrp {
    elem       : FreeMonT (WithInv T);
    elemNormal : is_stable elem;

But there is a problem with that approach. It’s that there can be more than one proof of stability for each element, leading to an equality on FreeGrpT that is not just the equality of the element it stores. There are multiple way to fix that. The first one would be to assume axiom K. I don’t really like this solution since axiom K is incompatible with univalence. Another solution would be to make sure is_stable is a mere proposition, but my knowledge of homotopy type theory is not good enough to make use of that yet. So instead I just added an axiom for that specific case.

Axiom eq_freegrp: forall (T : Type), forall (a b : FreeGrpT T), elem T a = elem T b -> a = b.

Proving it is a free group

As we’ve previously mentioned, the free group can be characterized by a universal property. This is easy enough to define in Coq. The definitions are a bit long, so we refer you to the file FreeGroup.v in the repo for the complete definitions and proofs.

The idea of the proof is to define a function from group morphism from the free group on \(X\) to a group \(G\) and the functions from \(X\) to \(G\) by composing the morphism \(\phi\) to the injection \(\iota\) of \(X\) in \(\G(X)\). This is injective because two morphisms that have the same image have the same image on the image of \(\iota\), and every element of \(\G(X)\) is generated from elements of the image of \(\iota\), so the morphisms are fully determined. To show it is surjective, one can take a function \(f : X \rightarrow G\) and extend it on \(\G(X)\) the naïve way: \(\tilde{f}(x_1^{\epsilon_1}\dots x_n^{\epsilon_n}) = f(x_1)^{\epsilon_1}\dots f(x_n)^{\epsilon_n}\). When restricted along \(\iota\), we get \(f\) again.


We have finally managed to construct the free group in Coq. The informal proof can be described in 2 pages of LaTeX, but the Coq proof took ~900 lines of code. This is probably because I’m still a beginner regarding Coq, but also because there where a few intuitive results I used without proving that are actually not that immediate to formally prove. All things considered it was a fun way to do some Coq again.

I also had to use an axiom, which is a bit disappointing. I’m planning to read more of homotopy type theory, and hopefully I’ll find a way to circumvent the need for that axiom.