Hi,

Sorry for resurrecting an old thread, I just wanted to document that Frobenius J can be derived from Paulin-Mohring J without Pi, Sigma or universes: https://github.com/akaposi/hiit-signatures/blob/master/formalization/FrobeniusJDeriv.agda

This was observed by Rafael Bocquet and András Kovács in October 2019, but maybe other people have known about this.

We mention this briefly in our paper on HIIT signatures where Paulin-Mohring J can be used on previous equality constructors but we don't have Pi, Sigma or (usual) universes: https://lmcs.episciences.org/6100

Cheers,

Ambrus

2018. július 13., péntek 13:05:47 UTC+2 időpontban Valery Isaev a következőt írta:

2018-07-13 13:38 GMT+03:00 Peter LeFanu Lumsdaine <p.l.l...@gmail.com>:

On Thu, Jul 12, 2018 at 7:07 PM Valery Isaev <valer...@gmail.com> wrote:

Hi Peter,

I've been thinking about such eliminators lately too. It seems that they are derivable from ordinary eliminator for most type-theoretic constructions as long as we have identity types and sigma types.

Thankyou — very nice observation, and (at least to me) quite surprising!

I was surprised too. The case of coproducts is especially unexpected.

I mean a strong version of Id:

Side note: this is I think more widely known as the Paulin-Mohring or one-sided eliminator for Id-types; the HoTT book calls it based path-induction.

The fact that the Frobenius version is strictly stronger is known in folklore, but not written up anywhere I know of. One way to show this is to take any non right proper model category (e.g. the model structure for quasi-categories on simplicial sets), and take the model of given by its (TC,F) wfs; this will model the simple version of Id-types but not the Frobenius version.

Are you sure this is true? It seems that we can interpret the strong version of J even in non right proper model categories. Then the argument I gave above shows that the Frobenius version is also definable.

Ah, yes — there was a mistake in the argument I had in mind. In that case, do we really know for sure that the Frobenius versions are really strictly stronger?

I don't know how to prove this, but it seems that we can't even define transport without Frobenius J. Also, if we do have transport and we know that types of the form \Sigma (x : A) Id(a,x) are contractible, then the "one-sided J" is definable, so if we want to prove that the Frobenius version does not follows from the non-Frobenius, we need to find a model where either transport is not definable or \Sigma (x : A) Id(a,x) is not contractible.

–p.