How to capture parameters under universal quantification (using Modules? Sections?) - coq

I am wondering how best to handle this sort of situation in Coq:
Suppose I need to define and prove a number of things about an arbitrary structure (for the purposes of this discussion, let's say a set with a binary relation). Of course, I can always supply the set and relation as arguments for each such definition/proof:
Inductive star (X : Set) (R : X -> X -> Prop) := ...
Lemma star_trans (X : Set) (R : X -> X -> Prop) : ...
Naturally, this gets tiresome after a while. What I would like to do is have X and R as parameters locally within some demarcated area of code like so:
Parameter X : Set.
Parameter R : X -> X -> Prop.
Inductive star := ...
Lemma star_trans : ...
in such a way that the definitions and theorems, when used outside this area of code, capture X and R under universal quantification in order to give them the proper type. For instance, Check star should yield star : forall X : Set, (X -> X -> Prop) -> X -> X -> Prop.
I figured that this might be what modules are for, but I can't figure out how to use them for this situation.

This is exactly what the "Section mechanism" does: see https://coq.inria.fr/distrib/current/refman/Reference-Manual004.html#Section .
Section rel_star.
Variables (X : Set) (R : X -> X -> Prop).
Inductive star := ...
Lemma star_trans: ...
End rel_star.

Related

Retrieving constraints from GADT to ensure exhaustion of pattern matching in Coq

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.

Characteristic function of a union

In a constructive setting such as Coq's, I expect the proof of a disjunction A \/ B to be either a proof of A, or a proof of B. If I reformulate this on subsets of a type X, it says that if I have a proof that x is in A union B, then I either have a proof that x is in A, or a proof that x is in B. So I want to define the characteristic function of a union by case analysis,
Definition characteristicUnion (X : Type) (A B : X -> Prop)
(x : X) (un : A x \/ B x) : nat.
It will be equal to 1 when x is in A, and to 0 when x is in B. However Coq does not let me destruct un, because "Case analysis on sort Set is not allowed for inductive definition or".
Is there another way in Coq to model subsets of type X, that would allow me to construct those characteristic functions on unions ? I do not need to extract programs, so I guess simply disabling the previous error on case analysis would work for me.
Mind that I do not want to model subsets as A : X -> bool. That would be unecessarily stronger : I do not need laws of excluded middle such as "either x is in A or x is not in A".
As pointed out by #András Kovács, Coq prevents you from "extracting" computationally relevant information from types in Prop in order to allow some more advanced features to be used. There has been a lot of research on this topic, including recently Univalent Foundations / HoTT, but that would go beyond the scope of this question.
In your case you want indeed to use the type { A } + { B } which will allow you to do what you want.
I think the union of subsets should be a subset as well. We can do this by defining union as pointwise disjunction:
Definition subset (X : Type) : Type := X -> Prop.
Definition union {X : Type}(A B : subset X) : subset X := fun x => A x \/ B x.

Messing around with category theory

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.

How to express subset relation in Coq?

How can I describe in Coq that one set Y is a subset of another set X?
I tested the following:
Definition subset (Y X:Set) : Prop :=
forall y:Y, y:X.
, trying to express that if an element y is in Y, then y is in X. But this generates type errors about y, not surprisingly.
Is there an easy way to define subset in Coq?
Here is how it is done in the standard library (Coq.Logic.ClassicalChoice):
Definition subset (U:Type) (P Q:U->Prop) : Prop := forall x, P x -> Q x.
Unary predicates P and Q represent some subsets of the (universal) set U, so the above definition reads: whenever some x is in P, it is in Q at the same time.
A somewhat similar defintion can be found in Coq.MSets.MSetInterface:
Definition Subset s s' := forall a : elt, In a s -> In a s'.
where In has type elt -> t -> Prop, which means that some element of type elt is a member of a set of type t.

Coq - use Prop (True | False) in if ... then ... else

I'm kind of new to Coq.
I'm trying to implement a generic version of insertion sort. I'm implementing is as a module that takes a Comparator as a parameter. This Comparator implements comparison operators (such as is_eq, is_le, is_neq, etc.).
In insertion sort, in order to insert, I must compare two elements in the input list, and based on the result of the comparison, insert the element into the correct location.
My problem is that the implementations of the comparison operators are type -> type -> prop (i need them to be like this for implementation of other types/proofs). I'd rather not create type -> type -> bool versions of the operators if it can be avoided.
Is there any way to convert a True | False prop to a bool for use in a if ... then ... else clause?
The comparator module type:
Module Type ComparatorSig.
Parameter X: Set.
Parameter is_eq : X -> X -> Prop.
Parameter is_le : X -> X -> Prop.
Parameter is_neq : X -> X -> Prop.
Infix "=" := is_eq (at level 70).
Infix "<>" := (~ is_eq) (at level 70).
Infix "<=" := is_le (at level 70).
Parameter eqDec : forall x y : X, { x = y } + { x <> y }.
Axiom is_le_trans : forall (x y z:X), is_le x y -> is_le y z -> is_le x z.
End ComparatorSig.
An implementation for natural numbers:
Module IntComparator <: Comparator.ComparatorSig.
Definition X := nat.
Definition is_le x y := x <= y.
Definition is_eq x y := eq_nat x y.
Definition is_neq x y:= ~ is_eq x y.
Definition eqDec := eq_nat_dec.
Definition is_le_trans := le_trans.
End IntComparator.
The insertion part of insertion sort:
Fixpoint insert (x : IntComparator .X) (l : list IntComparator .X) :=
match l with
| nil => x :: nil
| h :: tl => if IntComparator.is_le x h then x :: h :: tl else h :: (insert x tl)
end.
(obviously, the insert fixpoint doesn't work, since is_le is returns Prop and not bool).
Any help is appreciated.
You seem to be a bit confused about Prop.
is_le x y is of type Prop, and is the statement x is less or equal to y. It is not a proof that this statement is correct. A proof that this statement is correct would be p : is_le x y, an inhabitant of that type (i.e. a witness of that statement's truth).
This is why it does not make much sense to pattern match on IntComparator.is_le x h.
A better interface would be the following:
Module Type ComparatorSig.
Parameter X: Set.
Parameter is_le : X -> X -> Prop.
Parameter is_le_dec : forall x y, { is_le x y } + { ~ is_le x y }.
In particular, the type of is_le_dec is that of a decision procedure for the property is_le, that is, it returns either a proof that x <= y, or a proof that ~ (x <= y). Since this is a type with two constructors, you can leverage the if sugar:
... (if IntComparator.is_le_dec x h then ... else ...) ...
This is, in some sense, an enhanced bool, which returns a witness for what it is trying to decide. The type in question is called sumbool and you can learn about it here:
http://coq.inria.fr/library/Coq.Init.Specif.html#sumbool
In general, it does not make sense to talk about True or False in executing code.
First, because these live in Prop, which means that they cannot be computationally relevant as they will be erased.
Second, because they are not the only inhabitants of Prop. While true and false are the only values of type bool, which implies you can pattern-match, the type Prop contains an infinite number of elements (all the statements you can imagine), thus it makes no sense to try and pattern-match on a element of type Prop.