I am working upon formalization of groups theory in coq. I have 2 files:
groups.v - contains definitions and theorems for groups
groups_Z.v - contains theorems and definitions for Z group.
groups.v:
Require Import Coq.Setoids.Setoid.
Require Import Coq.Lists.List.
Require Import PeanoNat.
Class Semigroup G : Type :=
{
mult : G -> G -> G;
assoc : forall x y z:G,
mult x (mult y z) = mult (mult x y) z
}.
Class Monoid G `{Hsemi: Semigroup G} : Type :=
{
e : G;
left_id : forall x:G, mult e x = x;
}.
Class Group G `{Hmono: Monoid G} : Type :=
{
inv : G -> G;
left_inv : forall x:G, mult (inv x) x = e;
}.
Declare Scope group_scope.
Infix "*" := mult (at level 40, left associativity) : group_scope.
Open Scope group_scope.
Section Group_theorems.
Parameter G: Type.
Context `{Hgr: Group G}.
(* More theorems follow *)
Fixpoint pow (a: G) (n: nat) {struct n} : G :=
match n with
| 0 => e
| S n' => a * (pow a n')
end.
Notation "a ** b" := (pow a b) (at level 35, right associativity).
End Group_theorems.
Close Scope group_scope.
groups_Z.v:
Add LoadPath ".".
Require Import groups.
Require Import ZArith.
Open Scope group_scope.
Section Z_Groups.
Parameter G: Type.
Context `{Hgr: Group G}.
Definition pow_z (a: groups.G) (z: Z) : G :=
match z with
| Z0 => e
| Zpos x => pow a (Pos.to_nat x)
| Zneg x => inv (pow a (Pos.to_nat x))
end.
Notation "a ** b" := (pow_z a b) (at level 35, right associativity).
End Z_groups.
Close Scope group_scope.
The attempt to define pow_z fails with message:
The term "pow a (Pos.to_nat x)" has type "groups.G" while it is
expected to have type "G".
If we use the different signature: Definition pow_z (a: G) (z: Z) : G
instead of Definition pow_z (a: groups.G) (z: Z) : G.
then it gives another error:
The term "a" has type "G" while it is expected to have type
"groups.G".
How to fix this?
In Coq, the command Parameter G : Type declares a global constant, which is akin to axiomatizing the existence of an abstract Type G : Type. From a theoretical point of view, this should be ok as this axiom is trivially realizable, but I think you meant Variable G : Type to denote a local variable instead.
The errors messages of Coq follow from there because you declare two global constants named G, one in each module. As soon as the second one is declared, the first one is designated by groups.G by Coq (it's the shortest name that disambiguates this constant from others). Now pow operates on and returns a groups.G, while you require pow_z returns a G (which in file groups_Z.v at this location means groups_Z.G, and is different from groups.G).
NB: Group theory has been developed several times in Coq, and if you want to do anything else than experimenting with the system, I would advise you work on top of existing libraries. For example the mathematical components library has a finite group library.
I changed Parameter G: Type. to Variable G: Type in both files and pow_z definition to this:
Definition pow_z (a: G) (z: Z) : G :=
match z with
| Z0 => e
| Zpos x => pow G a (Pos.to_nat x)
| Zneg x => inv (pow G a (Pos.to_nat x))
end.
Related
Definitions
I'm working on formalizing a typed lambda calculus in Coq, and to keep the notation manageable I've come to rely a lot on coercions. However, I've been running into some difficulties which seem odd.
Right now I'm trying to work with the following types:
type: A descriptor of an allowable type in the language (like function, Unit, etc...)
var: A variable type, defined as nat
VarSets: set of vars
Judgement: a var/my_type pair
ty_ctx: Lists of judgements.
ctx_join: Pairs of ty_ctx's describing disjoint sets of variables
The actual definitions are all given below, except for ctx_join which is given in the next block
(* Imports *)
Require Import lang_spec.
From Coq Require Import MSets.
Require Import List.
Import ListNotations.
Module VarSet := Make(Nat_as_OT).
Inductive Judgement : Type :=
| judge (v : var) (t : type)
.
Definition ty_ctx := (list Judgement).
Definition disj_vars (s1 s2 : VarSet.t) := VarSet.Empty (VarSet.inter s1 s2).
Often I'd like to make statements like "this var does not appear in the set of vars bound by ty_ctx", and to that end I've set up a bunch of coercions between these types below.
(* Functions to convert between the different types listed above *)
Fixpoint var_to_varset (v : var) : VarSet.t :=
VarSet.singleton v.
Coercion var_to_varset : var >-> VarSet.t.
Fixpoint bound_variables (g : ty_ctx) : VarSet.t :=
match g with
| nil => VarSet.empty
| cons (judge v _) g' =>VarSet.union (VarSet.singleton v) (bound_variables g')
end.
Coercion bound_variables : ty_ctx >-> VarSet.t.
Inductive ctx_join :=
| join_single (g : ty_ctx)
| join_double (g1 g2 : ty_ctx)
(disjoint_proof : disj_vars g1 g2)
.
Fixpoint coerce_ctx_join (dj : ctx_join) : ty_ctx :=
match dj with
| join_single g => g
| join_double g1 g2 _ => g1 ++ g2
end.
Coercion coerce_ctx_join : ctx_join >-> ty_ctx.
Fixpoint coerce_judgement_to_ty_ctx (j : Judgement) : ty_ctx :=
cons j nil.
Coercion coerce_judgement_to_ty_ctx : Judgement >-> ty_ctx.
You'll notice that the definition of ctx_join relies on coercing its arguments from ty_ctx to VarSet.
I've drawn up the conversion hierarchy just to make things clear
The Problem
I'd like to define an inductive type with the following constructor
Inductive expr_has_type : ty_ctx -> nat -> type -> Prop :=
(* General Expressions *)
| ty_var (g : ty_ctx) (x : var) (t : type) (xfree : disj_vars x g)
: expr_has_type (join_double (judge x t) g xfree) x t
.
The problem is that when I do, I get the following error:
Error:
In environment
expr_has_type : ty_ctx -> nat -> type -> Prop
g : ty_ctx
x : var
t : type
xfree : disj_vars x g
The term "xfree" has type "disj_vars x g" while it is expected to have type
"disj_vars (judge x t) g" (cannot unify "VarSet.In a (VarSet.inter (judge x t) g)" and
"VarSet.In a (VarSet.inter x g)").
However, if I change the type of xfree to disj_vars (VarSet.singleton x) g, then the definition works fine! This seems very odd, as disj_vars is defined only on VarSets, and so it seems like x should automatically be converted toVarSet.singleton x since that's how the coercion is set up.
Even weirder is the fact that if I don't set up the coercion from vars to varsets, then Coq correctly complains about applying dis_vars to a var instead of a VarSet. So the coercion is definitely doing something
Can someone explain to me why the first definition fails? Given the coercions I've set up, to me it like all the definitions above should be equivalent
Note
Changing the type of xfree to disj_vars (judge x t) g also fixes the error. This seems odd too, since to be able to apply disj_vars to j := (judge x t), it first needs to be coerced to a ty_ctx via cons j nil, then to a VarSet via bound_variables, which should produce a VarSet containing only x (which is equivalent to VarSet.singleton x?). So this coercion chain seems to go off without a hitch, while the other one fails even though it's simpler
If you use Set Printing Coercions., the error message will be much more informative about the problem:
The term "xfree" has type "disj_vars (var_to_varset x) (bound_variables g)"
while it is expected to have type
"disj_vars (bound_variables (coerce_judgement_to_ty_ctx (judge x t)))
(bound_variables g)"
The problem is that the coercion of x into a VarSet.t is equal to Var.singleton x, while the coercion in judge reduces to VarSet.union (VarSet.singleton x) VarSet.empty. While these two are propositionally equal, they are not judgmentally equal, so as far as Coq is concerned, the term it came up with is ill-typed.
Let's define two helper types:
Inductive AB : Set := A | B.
Inductive XY : Set := X | Y.
Then two other types that depend on XY and AB
Inductive Wrapped : AB -> XY -> Set :=
| W : forall (ab : AB) (xy : XY), Wrapped ab xy
| WW : forall (ab : AB), Wrapped ab (match ab with A => X | B => Y end)
.
Inductive Wrapper : XY -> Set :=
WrapW : forall (xy : XY), Wrapped A xy -> Wrapper xy.
Note the WW constructor – it can only be value of types Wrapped A X and Wrapped B Y.
Now I would like to pattern match on Wrapper Y:
Definition test (wr : Wrapper Y): nat :=
match wr with
| WrapW Y w =>
match w with
| W A Y => 27
end
end.
but I get error
Error: Non exhaustive pattern-matching: no clause found for pattern WW _
Why does it happen? Wrapper forces contained Wrapped to be A version, the type signature forces Y and WW constructor forbids being A and Y simultaneously. I don't understand why this case is being even considered, while I am forced to check it which seems to be impossible.
How to workaround this situation?
Let's simplify:
Inductive MyTy : Set -> Type :=
MkMyTy : forall (A : Set), A -> MyTy A.
Definition extract (m : MyTy nat) : nat :=
match m with MkMyTy _ x => S x end.
This fails:
The term "x" has type "S" while it is expected to have type "nat".
wat.
This is because I said
Inductive MyTy : Set -> Type
This made the first argument to MyTy an index of MyTy, as opposed to a parameter. An inductive type with a parameter may look like this:
Inductive list (A : Type) : Type :=
| nil : list A
| cons : A -> list A -> list A.
Parameters are named on the left of the :, and are not forall-d in the definition of each constructor. (They are still present in the constructors' types outside of the definition: cons : forall (A : Type), A -> list A -> list A.) If I make the Set a parameter of MyTy, then extract can be defined:
Inductive MyTy (A : Set) : Type :=
MkMyTy : A -> MyTy A.
Definition extract (m : MyTy nat) : nat :=
match m with MkMyTy _ x => S x end.
The reason for this is that, on the inside, a match ignores anything you know about the indices of the scrutinee from the outside. (Or, rather, the underlying match expression in Gallina ignores the indices. When you write a match in the source code, Coq tries to convert it into the primitive form while incorporating information from the indices, but it often fails.) The fact that m : MyTy nat in the first version of extract simply did not matter. Instead, the match gave me S : Set (the name was automatically chosen by Coq) and x : S, as per the constructor MkMyTy, with no mention of nat. Meanwhile, because MyTy has a parameter in the second version, I actually get x : nat. The _ is really a placeholder this time; it is mandatory to write it as _, because there's nothing to match, and you can Set Asymmetric Patterns to make it disappear.
The reason we distinguish between parameters and indices is because parameters have a lot of restrictions—most notably, if I is an inductive type with parameters, then the parameters must appear as variables in the return type of each constructor:
Inductive F (A : Set) : Set := MkF : list A -> F (list A).
(* ^--------^ BAD: must appear as F A *)
In your problem, we should make parameters where we can. E.g. the match wr with Wrap Y w => _ end bit is wrong, because the XY argument to Wrapper is an index, so the fact that wr : Wrapper Y is ignored; you would need to handle the Wrap X w case too. Coq hasn't gotten around to telling you that.
Inductive Wrapped (ab : AB) : XY -> Set :=
| W : forall (xy : XY), Wrapped ab xy
| WW : Wrapped ab (match ab with A => X | B => Y end).
Inductive Wrapper (xy : XY) : Set := WrapW : Wrapped A xy -> Wrapper xy.
And now your test compiles (almost):
Definition test (wr : Wrapper Y): nat :=
match wr with
| WrapW _ w => (* mandatory _ *)
match w with
| W _ Y => 27 (* mandatory _ *)
end
end.
because having the parameters gives Coq enough information for its match-elaboration to use information from Wrapped's index. If you issue Print test., you can see that there's a bit of hoop-jumping to pass information about the index Y through the primitive matchs which would otherwise ignore it. See the reference manual for more information.
The solution turned out to be simple but tricky:
Definition test (wr : Wrapper Y): nat.
refine (match wr with
| WrapW Y w =>
match w in Wrapped ab xy return ab = A -> xy = Y -> nat with
| W A Y => fun _ _ => 27
| _ => fun _ _ => _
end eq_refl eq_refl
end);
[ | |destruct a]; congruence.
Defined.
The issue was that Coq didn't infer some necessary invariants to realize that WW case is ridiculous. I had to explicitly give it a proof for it.
In this solution I changed match to return a function that takes two proofs and brings them to the context of our actual result:
ab is apparently A
xy is apparently Y
I have covered real cases ignoring these assumptions, and I deferred "bad" cases to be proven false later which turned to be trivial. I was forced to pass the eq_refls manually, but it worked and does not look that bad.
Inductive ty: Set :=
| I
| O.
Definition f (x: ty) (y: ty): nat :=
if x = y then 0 else 1.
I want the function f to compare two terms of type ty but it does not compile and I see this error:
The term x = y has type Prop which is not a (co-)inductive type.
You need to prove that equality is decidable for ty (which can be done automatically using decide equality) and then use that definition in the if ... then ... else ... statement. Concretely:
Inductive ty: Set :=
| I
| O.
Definition ty_eq_dec : forall (x y : ty), { x = y } + { x <> y }.
Proof.
decide equality.
Defined.
Definition f (x: ty) (y: ty): nat :=
if ty_eq_dec x y then 0 else 1.
You can use match to compare the elements of inductive data types.
Definition f x y := match x,y with I, I | O, O => 0 | _,_ => 1 end.
decide equality is a more general tactic and works for infinite sets, but it is good to know that it is match that is doing the real work.
Motivation: I am attempting to study category theory while creating a Coq formalization of the ideas I find in whatever textbook I follow. In order to make this formalization as simple as possible, I figured I should identify objects with their identity arrow, so a category can be reduced to a set (class, type) of arrows X with a source mapping s:X->X, target mapping t:X->X, and composition mapping product : X -> X -> option X which is a partial mapping defined for t f = s g. Obviously the structure (X,s,t,product) should follow various properties. For the sake of clarity, I am spelling out the formalization I chose below, but there is no need to follow it I think in order to read my question:
Record Category {A:Type} : Type := category
{ source : A -> A
; target : A -> A
; product: A -> A -> option A
; proof_of_ss : forall f:A, source (source f) = source f
; proof_of_ts : forall f:A, target (source f) = source f
; proof_of_tt : forall f:A, target (target f) = target f
; proof_of_st : forall f:A, source (target f) = target f
; proof_of_dom: forall f g:A, target f = source g <-> product f g <> None
; proof_of_src: forall f g h:A, product f g = Some h -> source h = source f
; proof_of_tgt: forall f g h:A, product f g = Some h -> target h = target g
; proof_of_idl: forall a f:A,
a = source a ->
a = target a ->
a = source f ->
product a f = Some f
; proof_of_idr: forall a f:A,
a = source a ->
a = target a ->
a = target f ->
product f a = Some f
; proof_of_asc:
forall f g h fg gh:A,
product f g = Some fg ->
product g h = Some gh ->
product fg h = product f gh
}
.
I have no idea how practical this is and how far it will take me. I see this as an opportunity to learn category theory and Coq at the same time.
Problem: My first objective was to create a 'Category' which would resemble as much as possible the category Set. In a set theoretic framework, I would probably consider the class of triplets (a,b,f) where f is a map with domain a and range a subset of b. With this in mind I tried:
Record Arrow : Type := arrow
{ dom : Type
; cod : Type
; arr : dom -> cod
}
.
So that Arrow becomes my base type on which I could attempt building a structure of category. I start embedding Type into Arrow:
Definition id (a : Type) : Arrow := arrow a a (fun x => x).
which allows me to define the source and target mappings:
Definition domain (f:Arrow) : Arrow := id (dom f).
Definition codomain (f:Arrow) : Arrow := id (cod f).
Then I move on to defining a composition on Arrow:
Definition compose (f g: Arrow) : option Arrow :=
match f with
| arrow a b f' =>
match g with
| arrow b' c g' =>
match b with
| b' => Some (arrow a c (fun x => (g' (f' x))))
| _ => None
end
end
end.
However, this code is illegal as I get the error:
The term "f' x" has type "b" while it is expected to have type "b'".
Question: I have the feeling I am not going to get away with this, My using Type naively would take me to some sort of Russel paradox which Coq will not allow me to do. However, just in case, is there a way to define compose on Arrow?
Your encoding does not work in plain Coq because of the constructive nature of the theory: it is not possible to compare two sets for equality. If you absolutely want to follow this approach, Daniel's comment sketches a solution: you need to assume a strong classical principle to be able to check whether the endpoints of two arrows match, and then manipulate an equality proof to make Coq accept the definition.
Another approach is to have separate types for arrows and objects, and use type dependency to express the compatibility requirement on arrow endpoints. This definition requires only three axioms, and considerably simplifies the construction of categories:
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
Record category : Type := Category {
obj : Type;
hom : obj -> obj -> Type;
id : forall {X}, hom X X;
comp : forall X Y Z, hom X Y -> hom Y Z -> hom X Z;
(* Axioms *)
idL : forall X Y (f : hom X Y), comp id f = f;
idR : forall X Y (f : hom X Y), comp f id = f;
assoc : forall X Y Z W
(f : hom X Y) (g : hom Y Z) (h : hom Z W),
comp f (comp g h) = comp (comp f g) h
}.
We can now define the category of sets and ask Coq to automatically prove the axioms for us.
Require Import Coq.Program.Tactics.
Program Definition Sets : category := {|
obj := Type;
hom X Y := X -> Y;
id X := fun x => x;
comp X Y Z f g := fun x => g (f x)
|}.
(This does not lead to any circularity paradoxes, because of Coq's universe mechanism: Coq understands that the Type used in this definition is actually smaller than the one used to define category.)
This encoding is sometimes inconvenient due to the lack of extensionality in Coq's theory, because it prevents certain axioms from holding. Consider the category of groups, for example, where the morphisms are functions that commute with the group operations. A reasonable definition for these morphisms could be as follows (assuming that there is some type group representing groups, with * denotes multiplication and 1 denotes the neutral element).
Record group_morphism (X Y : group) : Type := {
mor : X -> Y;
mor_1 : mor 1 = 1;
mor_m : forall x1 x2, mor (x1 * x2) = mor x1 * mor x2
}.
The problem is that the properties mor_1 and mor_m interfere with the notion of equality for elements of group_morphism, making the proofs for associativity and identity that worked for Sets break. There are two solutions:
Adopt extra axioms into the theory so that the required properties still go through. In the above example, you would need proof irrelevance:
proof_irrelevance : forall (P : Prop) (p q : P), p = q.
Change the category axioms so that the identities are valid up to some equivalence relation specific to that category, instead of the plain Coq equality. This approach is followed here, for example.
Let's say I have again a small problem with my datatype with an existential quantified component. This time I want to define when two values of type ext are equal.
Inductive ext (A: Set) :=
| ext_ : forall (X: Set), option X -> ext A.
Fail Definition ext_eq (A: Set) (x y: ext A) : Prop :=
match x with
| ext_ _ ox => match y with
| ext_ _ oy => (* only when they have the same types *)
ox = oy
end
end.
What I'd like to do is somehow distinguish between the cases where the existential type is actually same and where it's not. Is this a case for JMeq or is there some other way to accomplish such a case distinction?
I googled a lot, but unfortunately I mostly stumbled upon posts about dependent pattern matching.
I also tried to generate a (boolean) scheme with Scheme Equality for ext, but this wasn't successful because of the type argument.
What I'd like to do is somehow distinguish between the cases where the existential type is actually same and where it's not.
This is not possible as Coq's logic is compatible with the univalence axiom which says that isomorphic types are equal. So even though (unit * unit) and unit are syntactically distinct, they cannot be distinguished by Coq's logic.
A possible work-around is to have a datatype of codes for the types you are interested in and store that as an existential. Something like this:
Inductive Code : Type :=
| Nat : Code
| List : Code -> Code.
Fixpoint meaning (c : Code) := match c with
| Nat => nat
| List c' => list (meaning c')
end.
Inductive ext (A: Set) :=
| ext_ : forall (c: Code), option (meaning c) -> ext A.
Lemma Code_eq_dec : forall (c d : Code), { c = d } + { c <> d }.
Proof.
intros c; induction c; intros d; destruct d.
- left ; reflexivity.
- right ; inversion 1.
- right ; inversion 1.
- destruct (IHc d).
+ left ; congruence.
+ right; inversion 1; contradiction.
Defined.
Definition ext_eq (A: Set) (x y: ext A) : Prop.
refine(
match x with | #ext_ _ c ox =>
match y with | #ext_ _ d oy =>
match Code_eq_dec c d with
| left eq => _
| right neq => False
end end end).
subst; exact (ox = oy).
Defined.
However this obviously limits quite a lot the sort of types you can pack in an ext. Other, more powerful, languages (e.g. equipped with Induction-recursion) would give you more expressive power.