In the theory, the rule looks like this:
A : \Type0    p : isSet A
----------------------------------
      F(A,p) : \Set0

and we also add an equivalence between A and F(A,p). In the actual
implementation, you can do this using \use \level construction. If you can
prove that some \data or \record satisfies isSet (or, more generally, that
it is an n-type), then you can put this proof in \use \level function
corresponding to this definition and it will be put in the corresponding
universe. So, you can define \data F(A,p) with one constructor with one
parameter of type A and put it in \Set0.

Regards,
Valery Isaev


сб, 10 авг. 2019 г. в 12:47, Michael Shulman <shulman@sandiego.edu>:

> On Thu, Aug 8, 2019 at 2:56 AM Valery Isaev <valery.isaev@gmail.com>
> wrote:
> > You can say that they are hidden in the background, but I prefer to
> think about this in a different way. I think about \Set0 as a strict
> subtype of \Type0. In comparison, the type \Sigma (A : \Type0) (isSet A) is
> only homotopically embeds into \Type0. It is equivalent to \Set0, but not
> isomorphic to it. In particular, this means that every type in \Set0
> satisfies isSet and every type in \Type0 which satisfies isSet is
> equivalent to some type in \Set0, but not necessarily belongs to \Set0
> itself. So, if we have (1), we also have (2) and we do not have (3). It may
> be true that A is a 2-type, which means that there is a type A' : \2-Type 1
> equivalent to A, but A itself does not belong to \2-Type 1.
>
> How do you ensure that "every type in \Type0 which satisfies isSet is
> equivalent to some type in \Set0"?  Is it just an axiom?
>
> Also, since \Prop "has no predicative level", does this property
> applied to \Prop imply that propositional resizing holds?
>

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