Re: [HoTT] New theorem prover Arend is released

[Michael Shulman <shulman@sandiego.edu>]

In an off-list discussion, Valery clarified to me that \use\level for
\func does not actually put the type in the corresponding universe,
only marks it as having the corresponding homotopy level (and added
some documentation
https://arend-lang.github.io/documentation/language-reference/definitions/level#use-level-for-functions).
However, it also emerged that F(A,p) can also be *defined* for an
arbitrary type A (and an externally fixed n) as

\record F (A : \Type) (p : ofHLevel A n) {
  | in : A
} \where \use \level isntype (A : \Type) (p : ofHLevel A n) : ofHLevel A n => p

and that because records have definitional eta-conversion, this F(A,p)
will actually be definitionally isomorphic to A, rather than just
equivalent to it.

In particular, since \Prop is invariant under predicative level, this
means that (in the current version of Arend) every h-proposition is
definitionally isomorphic to one living in the lowest universe.  This
is not quite as strong as Voeovdsky's propositional resizing rule
(that every h-proposition *itself* lives in the lowest universe), but
I find it about equally semantically dubious, and in particular it's
unclear whether it has any univalent models (not even classically in
simplicial sets).  Perhaps it does; but until such models are known,
it may be better to avoid using this feature for \Prop, and perhaps
add a restriction to Arend preventing it.

On Sun, Aug 11, 2019 at 8:39 AM Michael Shulman <shulman@sandiego.edu> wrote:
>
> On Sun, Aug 11, 2019 at 6:47 AM Valery Isaev <valery.isaev@gmail.com> wrote:
> > I don't remember well, but I think the idea is that you need to prove that there is a trivial cofibration Eq(A,B) -> F(U^I,A,B), where the first object is the object of equivalences between A and B and the second object is the fiber of U^I over A and B. The fact that this map is a weak equivalence is just the univalence axiom. The problem is to show that it is a cofibration and whether this is true or not depends on the definition of Eq(A,B). I don't actually remember whether I finished this proof.
>
> Yes, in a straightforward approach that's what you would need to
> prove, or more precisely that the map Eq -> U^I is a trivial
> cofibration in the slice over UxU (since the lifting has to be done in
> the universal case rather than just over each global element).
> Possibly you could get away with less, since you're only asking to
> recover the action of the given equivalence as a function (rather than
> the entire equivalence data).  Perhaps a clever choice of fibration
> structure in the equivalence extension property would suffice.
>
> > Since F(A,p) is the usual (inductive) data type, you can do everything you can do with other data types. In particular, since it has only one constructor with one parameter A, it is easy to proof that it is equivalent to A.
>
> Okay, it sounds like you're saying that the datatype defined with
> \use\level is F(A,p) and the analogous one defined without \use\level
> is A, while both have their usual rules but are not identical.  In
> this case you have somewhat *more* than an equivalence between F(A,p)
> and A, at least if all datatypes have their usual definitional
> computation rules, since an arbitrary type "F(A,p)" that is only
> *equivalent* to a datatype will only inherit a typal computation rule.
> It seems plausible though that one could construct universes of
> n-types that are strictly closed under a given datatype construction
> that preserves n-types (at least, assuming one can do the same for
> arbitrary universes).
>
> However, the documentation also suggests that \use\level can be
> applied to plain definitions (\func).  How is F(A,p) distinguished
> from A in this case?

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