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[Jason Gross <jasongross9 AT gmail.com>]

> Axiom OrTrueTrueRelevant : (or_introl I <> or_intror I).

> Axiom UIPNatNat : forall (f:nat ->nat) (p1 p2 : f=f), p1=p2.

Yes, that's safe (or, at least, if it's unsafe, the proof of contradiction would need impredicativity, and it seems unlikely that this would interact with impredicativity in such a way); it's validated by considering the predicative subset of Coq, replacing [Prop] with [Set] (and bumping the other universe levels appropriately), and considering the model in sets.

The claim I'm making is that if there are two elements of a Prop that can be distinguished in a Prop context (i.e., if you could find distinct elements were you to put the inductive in [Type] rather than [Prop]), then having distinguishable elements of [or] gives you distinguishable elements of that type.  The fully general statement would probably say something about W-types in Type and W-types in Prop, and say that if you have a W-type in Type with two provably distinct elements, and you have [or_introl I <> or_intror I], then you can find two elements of the same W-type in Prop that are provably distinct.

On Sun, Apr 24, 2016 at 12:21 PM, Abhishek Anand <abhishek.anand.iitg AT gmail.com> wrote:

Thanks, Jason. I understand your proof for the True \/ True case.

However, I don't see a proof of the full generality of your claim.

For example, for any f:nat->nat, I think proof irrelevance of f=f is not provable. However, unlike for True \/ True, I do not have any 2 elements in f=f that match can distinguish between.

More formally, is it safe to simultaneously assume the following?

Axiom OrTrueTrueRelevant : (or_introl I <> or_intror I).

Axiom UIPNatNat : forall (f:nat ->nat) (p1 p2 : f=f), p1=p2.

-- Abhishek

http://www.cs.cornell.edu/~aa755/

On Sun, Apr 24, 2016 at 10:25 AM, Jason Gross <jasongross9 AT gmail.com> wrote:

Note that having *any* prop with two distinguishable elements is inconsistent with proof irrelevance for any props for which it is not already provable: when you're in a prop context, you can distinguish elements of, e.g., [or].  Since you can map two elements of, e.g., [True \/ True] to non equal things, you can't have proof irrelevance for [or].

On Sun, Apr 24, 2016, 10:13 AM Jonathan Leivent <jonikelee AT gmail.com> wrote:

On 04/24/2016 12:22 AM, Abhishek Anand wrote:
> Your proof essentially shows that Erasable_inj implies proof relevance of
> Foo, and is thus inconsistent with assuming proof irrelevance of Foo :
> (No was a Yes? Sorry if I am misunderstanding something.)

I believe that is correct.


>
> Inductive Erasable(A : Set) : Prop :=
>    erasable: A -> Erasable A.
>
> Arguments erasable [A] _.
>
> Axiom Erasable_inj : forall {A : Set}{a b : A}, erasable a=erasable b ->
> a=b.
>
> Inductive Foo : Prop :=
>    foo: nat -> Foo.
>
> Definition foo2erasable(f : Foo) : Erasable nat :=
>    match f with
>      | foo x => erasable x
>    end.
>
> Lemma FooRelevant : foo 1 <> foo 2.
> Proof.
>    intros H.  apply f_equal with (f := foo2erasable) in H.
>    simpl in H.
>    apply Erasable_inj in H.
>    discriminate H.
> Qed.
>
>
> Do you need to assume Erasable_inj in its full generality? or just some
> specific instances for some sets?
> If only certain instances suffice, and if there was a way to easily
> understand the implications of such instances, it may still be possible to
> safely have some instances of proof irrelevance.
> More formally, if we assume  @Erasable_inj A for some set A, is there are a
> simple characterization of the Props (members of sort Prop) whose proof
> relevance become implied?
>

Unfortunately, one of the more important cases of Erasable_inj I use is
when the Set is list A for arbitrary A.  That arbitrary A would seem to
imply that there is no way to restrict this proof-relevance domino effect.

Thanks for the help.

-- Jonathan