Re: [HoTT] Re: Why do we need judgmental equality?

[Valery Isaev <valery.isaev@gmail.com>]

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Hi Nicolai,

The notion of a conservative extension is certainly related to the notion
of an equivalence of type theories that I'm referring to (I call it Morita
equivalence). Hofmann defines various properties of the interpretation |-|
: ITT -> ETT, which are weaker than Morita equivalence in general. The
statement of Theorem 3.2.5 is almost equivalent to the notion of Morita
equivalence. We just need to replace "|P'| = P" with "|P'| and P are
propositionally equivalent" and add the condition that, for every \Gamma
|-_I and every |\Gamma| |-_E A there is a type \Gamma |-_I A' such that
|A'| and A are equivalent in the sense of HoTT (the latter condition
follows from Theorem 3.2.5 for ITT and ETT, but not for general type
theories). Then we get precisely the notion of Morita equivalence between
type theories I and E.

Regards,
Valery Isaev


сб, 9 февр. 2019 г. в 04:41, Nicolai Kraus <nicolai.kraus@gmail.com>:

> Hi Valery,
>
> On Fri, Feb 8, 2019 at 11:32 PM Valery Isaev <valery.isaev@gmail.com>
> wrote:
>
>> Now, what do I mean when I say that type theories T and Q(T) are
>> equivalent? I won't give here the formal definition
>>
>
> Would it be correct to say that T is a conservative extension of T, in the
> sense of Martin Hofmann's thesis? Your description sounds a bit like this,
> or do you have something different in mind?
> Nicolai
>
>
>
>> , but the idea is that Q(T) can be interpreted in T and, for every type A
>> of T, there is a type in Q(T) equivalent to A in T and the same is true for
>> terms. This implies that every statement (i.e., type) of Q(T) is provable
>> in Q(T) if and only if it is provable in T and every statement of T has an
>> equivalent statement in Q(T), so the theories are "logically equivalent".
>> Moreover, equivalent theories have equivalent (in an appropriate
>> homotopical sense) categories of models.
>>
>> Regards,
>> Valery Isaev
>>
>>
>> сб, 9 февр. 2019 г. в 00:19, Martín Hötzel Escardó <
>> escardo.martin@gmail.com>:
>>
>>> I would also like to know an answer to this question. It is true that
>>> dependent type theories have been designed using definitional equality.
>>>
>>> But why would anybody say that there is a *need* for that? Is it
>>> impossible to define a sensible dependent type theory (say for the purpose
>>> of serving as a foundation for univalent mathematics) that doesn't mention
>>> anything like definitional equality? If not, why not? And notice that I am
>>> not talking about *usability* of a proof assistant such as the *existing*
>>> ones (say Coq, Agda, Lean) were definitional equalities to be removed. I
>>> don't care if such hypothetical proof assistants would be impossibly
>>> difficult to use for a dependent type theory lacking definitional
>>> equalities (if such a thing exists).
>>>
>>> The question asked by Felix is a very sensible one: why is it claimed
>>> that definitional equalities are essential to dependent type theories?
>>>
>>> (I do understand that they are used to compute, and so if you are
>>> interested in constructive mathematics (like I am) then they are useful.
>>> But, again, in principle we can think of a dependent type theory with no
>>> definitional equalities and instead an existence property like e.g. in
>>> Lambek and Scott's "introduction to higher-order categorical logic". And
>>> like was discussed in a relatively recent message by Thierry Coquand in
>>> this list,)
>>>
>>> Martin
>>>
>>>
>>> On Wednesday, 30 January 2019 11:54:07 UTC, Felix Rech wrote:
>>>>
>>>> In section 1.1 of the HoTT book it says "In type theory there is also a
>>>> need for an equality judgment." Currently it seems to me like one could, in
>>>> principle, replace substitution along judgmental equality with explicit
>>>> transports if one added a few sensible rules to the type theory. Is there a
>>>> fundamental reason why the equality judgment is still necessary?
>>>>
>>>> Thanks,
>>>> Felix Rech
>>>>
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