I can't use Ssreflect's rewrite with setoids. Although I don't think this information is relevant to solve the problem, I'm using this formulation of category theory in coq: https://github.com/jwiegley/category-theory.
As far as I know, Ssreflect's rewriting can use user-defined equivalence relations if a suitable RewriteRelation Typeclass instance can be resolved.
Context { x y : obj[C]}.
About RewriteRelation.
(* ∀ A : Type, crelation A → Prop *)
Instance equivrw : (#RewriteRelation (x ~> y) (#equiv (x ~> y) (homset x y))) := {}. (* defined and correctly type-checked *)
About equivrw.
(* RewriteRelation equiv (coq omits implicit arguments) *)
Lemma test_rwrel (a b c : x ~> y) : (#equiv (x ~> y) (homset x y) a b) -> (#equiv (x ~> y) (homset x y) b c) -> (#equiv (x ~> y) (homset x y) a c).
Proof.
move => ->.
(* Fails: not a rewritable relation: (a ≈ b) in rule __top_assumption_ *)
Using the default rewrite tactic instead works (even without manually declaring the instance, because something similar is already declared in the library).
Something funny:
Instance equivrw : (#RewriteRelation False equiv) := {}.
Lemma test_rwrel (a b c : False) : (equiv a b) -> (equiv b c) -> (equiv a c).
Proof.
move => ->.
by [].
Qed.
This works fine.
Why doesn't SSReflect use the RewriteRelation instance I have manually declared to rewrite by user-defined equivalence?
Thanks in advance.
EDIT:
I have figured out why Ssreflect doesn't see my relation. Apparently, you can add relations to be used with Ssreflect's rewrite only with Add Parametric Relation which is documented in the manual, and RewriteRelation instances don't change anything. I have tried to create a fake relation using axioms and adding it with Add Parametric Relation and rewriting works correctly. But now I'm again in trouble. The relation that I want to be able to use has type crelation (x ~> y) (which is (x ~> y) -> (x ~> y) -> Type), but to use Add Parametric Relation I need a term with type relation (x ~> y) (which is (x ~> y) -> (x ~> y) -> Prop. Why two different kinds of relations (relation and crelation) have been defined in the standard library?
How can I convert a crelation to a relation without getting universe inconsistency errors?
And I still can't understand why the above example with False works.
It's not sufficient nor necessary to declare a relation as a RewriteRelation. You must prove properties that ensure that the rewriting is logically valid, the most common one being transitivity (but you often want also reflexivity, and even symmetry as well).
As you have noticed already, the setoid rewriting plugin is actually made of two twin systems, one for relations in Prop, and one for relations in Type (whose module names are prefixed with C). You generally use just one or the other, depending on the sort of your relations.
I'm not sure whether SSReflect changes things, but here is an example usage of setoid rewriting without it:
Require Setoid. to activate the plugin.
Require Import CEquivalence. if you use Type relations, or Require Import Equivalence. if you use Prop relations.
Implement instances for properties of equivalence relations (any of Reflexive, Symmetric, Transitive, or Equivalence for all 3).
You may also have to prove Proper lemmas if the terms you are rewriting are not at the toplevel of the goal.
Note that at step 3, if you do not/cannot prove all three properties, then you will not be allowed to rewrite in all directions.
To the best of my knowledge, commands such as Add Parametric Relation you may find in the documentation are just boilerplate that desugar to these instances. But in my opinion as a user, they add an unnecessary layer of indirection; declaring the instances directly makes things more understandable (especially if you're already familiar with type classes).
Require Setoid. (* 1 *)
Require Import CMorphisms. (* 2 *)
Axiom obj : Type.
Axiom equiv : obj -> obj -> Type.
Axiom cat : forall a b c, equiv a b -> equiv b c -> equiv a c.
(* 3 *)
Instance Transitive_equiv : Transitive equiv.
Proof.
exact cat.
Qed.
(* Example usage *)
Goal forall a b c,
equiv a b -> equiv b c -> equiv a c.
Proof.
intros a b c H1 H2.
rewrite H1.
(* or,
rewrite H2
*)
Related
Is it possible to rewrite something that uses variables from another scope,
such as a function call that uses a variable from a match, fun, or fix ?
For example,
Theorem foo (f : nat -> nat) (rw : forall x, f x = 5) x : match x with
| 0 => 5
| S a => f a
end = 5.
rewrite rw.
(* Error: Found no subterm matching "f ?M160" in the current goal. *)
destruct x; try rewrite rw; apply eq_refl.
Qed.
So, the theorem is provable, but trying to rewrite rw initially fails,
seemingly because a is in another scope. But, the rewrite applies
unconditionally, so it seems like it should apply there too.
Of course, this is a toy example. Assume that, in a real-world scenario,
getting into the scope is a bit more complicated than just a destruct.
From Rewrite tactic fails to find term occurrence within pattern matching it looks like this isn't possible in Coq. So, is it just that it isn't implemented, or does it cause contradictions or allow for bad behavior like smuggling variables out of their scope?
What about harder cases like fix ?
You have likely heard that the Logic of Coq is not powerful enough to derive functional exensionality.
Now what you prove above is a point wise equality, that is you prove that an applied function has a certain value.
A rewrite in the match would correspond to a proof that two unapplied functions are equal: The original match statement (which is a function of x) and the rewritten match statement (also a function of x). This means you would prove a more general result as intermediate step - and Coq's logic is not able to prove this more general result.
Functional extensionality is compatible with Coq's logic, though. So one can add it as axiom and then one can add some setoid instances which allow to setoid_rewrite under binders. But afaik this does not work for rewrites under matches either.
It would also be possible to detect that in the end you prove a point wise equality, do the destruct behind the scenes, do the point wise rewrite and put things together again, but this would work only in rather trivial cases, which I guess are not sufficiently interesting to implement this.
To close here is an example of proving functional extensionality (suggested by the edit by #scubed) with a rewrite under binders - which is enabled by a global type class instance which is itself based on the functional extensionality axiom. So this proves an axiom with itself and it doesn't help for your match case, but it shows that rewriting under binders is equivalent to functional extensionality.
Require Import PeanoNat.
Require Import Setoid.
Require Import Morphisms.
Definition fun1 (x : nat) := x + 1.
Definition fun2 (x : nat) := 1 + x.
Example Fun1EqFun2: fun1 = fun2.
Proof.
unfold fun1, fun2.
Fail setoid_rewrite Nat.add_comm.
Abort.
Require Import FunctionalExtensionality.
(* This is derived from the error message of setoid_rewrite above *)
Global Instance:
forall (A B : Type),
Proper
(#pointwise_relation A B eq ==>
#pointwise_relation A B eq ==> Basics.flip Basics.impl) eq.
Proof.
(* See what all this Setoid stuff actually means ... *)
unfold Proper, pointwise_relation, respectful, Basics.flip, Basics.impl.
intros A B f g fgeq f' g' fg'eq gg'eq.
apply functional_extensionality.
intros x.
rewrite fgeq, fg'eq, gg'eq.
reflexivity.
Qed.
Theorem eq_arg (A B : Type) (f : A -> B) : f = (fun x => f x). reflexivity. Qed.
Lemma functional_extensionality' :
forall (A B : Type), forall (f g : A -> B),
(forall a : A, (f a) = (g a)) -> f = g.
Proof.
intros.
setoid_rewrite eq_arg at 2.
setoid_rewrite H.
reflexivity.
Qed.
Completely new to coq here.
I know about the exists tactic to prove an existential goal, but in this case it wants a function mapping from two sets. What is the syntax for demonstrating such a function?
And if there is no such function how would I disprove this? (I would suppose through a contradiction, but then how would I pose a contradictory hypothesis?)
Context: Trying to work out the proof that all surjective functions have a right inverse.
1 subgoal
A, B : Set
f : A → B
H : ∀ b : B, ∃ a : A, f a = b
______________________________________(1/1)
∃ g : B → A, ∀ b : B, f (g b) = b
Of course, whether or not a function g exists depends on accepting axiom of choice, so where does that come into coq?
I did find this solution:
https://gist.github.com/pedrominicz/0d9004b82713d9244b762eb250b9c808
and the associated reddit post
https://www.reddit.com/r/logic/comments/fxjypn/what_is_not_constructive_in_this_proof/
But I didn't understand it/didn't work for me.
So, what I want to know is:
How do you specify axiom of choice in coq (to prove/disprove this)?
In general, how would I construct a function to provide witness to an existential goal? (I also want to show that all injective functions have a left inverse)
There are several variants of the axiom of choice in the Coq type theory. You can look at the Coq.Logic.ChoiceFacts module for a reasonably comprehensive list of the various formulations and their relative power.
As far as I can tell, your example is equivalent to the axiom of functional choice. One elegant way to phrase and assume it is the following.
Axiom functional_choice : forall (A : Type) (B : A -> Type),
(forall x : A, inhabited (B x)) -> inhabited (forall x : A, B x).
The inhabited type is an inductive box that hides the computational content of a proof in Type into a Prop value that can only be inspected to produce more Prop values. In particular, This axiom is pretty innocuous from the point of view of computation since it only produces values in Prop. There are much more violently non-computational examples of choice like global choice which can be stated as:
Axiom global_choice : forall (A : Type), inhabited A -> A.
This one allows to extract computational content out of thin air.
Here is an answer that is a complete script (tested with coq 8.13.2). Coq by default does not have the axiom of choice loaded, so you need to say explicitly that you are going to work with it.
Require Import ClassicalChoice.
Lemma question (A B : Set) (f : A -> B) :
(forall b, exists a, f a = b) -> exists g, forall b, f (g b) = b.
Proof.
intros H.
apply (choice (fun y x => f x = y)).
exact H.
Qed.
I am trying to use typeclasses for code reuse, but I am getting setoid errors applying parent typeclass axioms in child typeclass theorems. I made a MRE with the following Equality and Addition operations:
Require Import Setoid.
(* Equality *)
Parameter CEq : forall A, A->A->Prop.
Arguments CEq [A] _ _.
Notation "x ¦ y" := (CEq x y) (at level 70, no associativity).
Axiom ceq_reflexivity: forall A, forall a:A, a¦a.
Axiom ceq_symmetry: forall A, forall a b:A, a¦b->b¦a.
Axiom ceq_transitivity: forall A, forall a b c:A, a¦b->b¦c->a¦c.
Add Parametric Relation A : (A) (#CEq A)
reflexivity proved by (#ceq_reflexivity A)
symmetry proved by (#ceq_symmetry A)
transitivity proved by (#ceq_transitivity A)
as ceq_rel.
(* Addition *)
Parameter CAdd: forall A, A->A->A.
Arguments CAdd [A] _ _.
Infix "±" := CAdd (at level 50, left associativity).
The following are the parent and child classes:
(* Parent Typeclass *)
Class CDiscT (CDisc: Set) := {
O: forall CDisc, CDisc;
cdisc_add_neutral:forall CDisc, forall x:CDisc, x±(O CDisc)¦x;
}.
(* Natural Set & Child Typeclass *)
Parameter CNat: Set.
Class CNatT `{CDiscT CNat} := {}.
And here is the failing theorem:
(* Axiom inheritance test *)
Example test `{CNatT}: (O CNat)¦(O CNat)±(O CNat).
Proof.
rewrite <- cdisc_add_neutral. (* Error *)
reflexivity.
Qed.
Here is the error:
Error:
Tactic failure: setoid rewrite failed: Unable to satisfy the following constraints:
In environment:
H : CDiscT CNat
H0 : CNatT
?s : "subrelation (CEq (A:=Prop)) (Basics.flip Basics.impl)"
What is missing here to be able to use CDiscT axioms inside CNatT theorems? Is there a better way to do this?
This may be in part because your example is too simplified, but rewriting right-to-left with cdisc_add_neutral is problematic because the right-hand side x matches anything and can have any type.
The error you get is that Coq is trying to rewrite the whole goal with it, but that would use logical implication impl, which in turn requires that your relation CEq is a subrelation of impl.
You can avoid this by specializing the lemma a bit:
rewrite <- (cdisc_add_neutral CNat) at 1.
You need the at 1 because now the subterm that matches is O CNat, but it occurs three times in your goal. rewrite tries to rewrite all of them by default, which requires Proper instances that are missing here. (You can get those via the Parametric Morphism mechanism described in the manual).
Also, you can rewrite left-to-right:
rewrite cdisc_add_neutral
I want to construct an existential variable interactively. I can't use Grab Existential Variables because I need to fill the existential before I complete the goal.
Minimal exmaple
Here is a minimal example (since it's simple, it has other solutions, but it illustrates my question)
Context (A:Type).
Parameter P Q: A -> Prop.
Definition filter: forall {a}, P a -> A:= fun a Pa=> a.
Lemma my_lemma:
forall a b, Q b -> (Q b -> P a) ->
exists a (H: P a), P (filter H).
Proof.
intros ?? H H0.
do 2 eexists.
At this point, there are two solutions that don't answer my questions: (1) I could run (eauto) and then do Grab Existential Variables, but suppose that eauto doesn't succeed until I instantiate the unification variable; (2) I could just pass the proof term explicitly with instantiate(1:= H0 H) (or even instantiate(1:= ltac:(eauto))), but assume that the proof of the xistential was tedious and we wished to do it interactively.
What else can we do? We can try to use cut or assert, like so:
match goal with
|[|- P (filter ?x)] =>
match type of x with
| ?T => assert (HH:T) by eauto
end
end.
But HH is not in the context of the unification variable, so it can't be instantiated.
instantiate(1:=HH). (* Instance is not well-typed in the environment of ?H. *)
As far as I can tell, the only way to solve this is to use Show Existentials, see the type of the variable copy it by hand, roll back the proof to before the unification was introduced and construct the variable there. In the example it looks like this:
Lemma my_lemma:
forall a b, Q b -> (Q b -> P a) ->
exists a (H: P a), P (filter H).
Proof.
intros ?? H H0.
do 2 eexists.
Show Existentials.
Restart. (* This command restores the proof editing process to the original goal. *)
intros ?? H H0.
assert (HH:P a) by eauto.
eexists; exists HH.
auto.
Qed.
Obviously, I would like to avoid this workflow. So, anyway to turn the existential variables into subgoals?
Your best bet is probably to avoid creating the existential variable as an evar in the first place. You don't have to construct the variable by hand to do this; if you can identify where it's created, you can wrap the offending tactic with unshelve t, which turns all evars created by t into goals. Where this might be difficult is if the relevant tactic is deep inside some automation and difficult to identify or change.
I am thinking about proof irrelevance in COQ.
One provable statement says:
If equality of a type is decidable then there can be only one proof for the equality statement, namely reflexivity.
I wonder if its possible to construct types with more than one equality proof in COQ. Therefore I ask if the following construct is consistent?
(*it is known that f=g is undecidable in COQ *)
Definition f(n:nat) := n.
Definition g(n:nat) := n+0.
Axiom p1: f=g.
Axiom p2: f=g.
Axiom nonirrelevance:p1<>p2.
What me puzzles here is the fact that by introducing p1 I made the equality f=g decidable and therefore it should only have one proof! What is my error in reasoning here?
Is that all a pure COQ behaviour or is it similar in HOTT?
I think you're confusing several things.
The provable statement you are speaking of can be found in https://coq.inria.fr/library/Coq.Logic.Eqdep_dec.html and is
Theorem eq_proofs_unicity A (eq_dec : forall x y : A, x = y \/ x <> y) (x : A) :
forall (y:A) (p1 p2:x = y), p1 = p2.
Now what is quite interesting is the type of eq_dec. First of all, it doesn't even really ask for equality to be decidable, it just asks for it to be true or false which is way less stronger than {x = y} + {x <> y}
Then notice that it doesn't ask this just for the x and y to prove the equality irrevelance of, it ask this property for all functions.
So you would need to prove your contradiction that forall (f g : nat -> nat), f = g \/ f <> g which you cannot. p1 is just a proof that f = g \/ f <> g for your specific f and g.
Notice though that if you could, it would just mean that there is no way to build a system in which you can compare functions and yet have multiple ways which are provably different to check them.
Finally, for P to be undecidable only means that there is no constructible functions over {P} + {~P} yet, it doesn't mean that adding one as an axiom leads to a contradiction. Just adding that in case it wasn't clear.