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

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

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Hi Jon,

It is not true that Q(T) can be obtained from T simply by replacing
judgmental equalities with propositional ones (I think this should be true
for some theories, but certainly not for all).

So, if we denote the theory with an interval and judgmental rules by T_I
and the theory with propsoitional axioms by T_a, then Q(T_I) is equivalent
to T_I and Q(T_a) is equivalent to T_a, but Q(T_I) is not equivalent to
T_a. Theories T_a and Q(T_a) are actually equivalent to the ordinary MLTT
(since we simply add another contractible type to it).

Functional extensionality is provable in both T_I and Q(T_I). It follows
from the existence of the interval type in T_I and it can be added as an
axiom to Q(T_I). I believe that theories T_I and Q(T_I) are equivalent to
MLTT + functional extensionality, but I don't know how to prove this yet.

Regards,
Valery Isaev


сб, 9 февр. 2019 г. в 05:01, Jon Sterling <jon@jonmsterling.com>:

> Hi Valery,
>
> I'm trying to square what you said with what Anders said. Consider
> extending MLTT with an interval in two different ways:
>
> 1. With judgmental computation rules for the recursor
> 2. With propositional computation axioms for the recursor
>
> I do not expect to obtain a conservativity result for the second version
> over the first version, since the first one derives function
> extensionality, and the second one does not (afaict).
>
> Can you give a bit more detail about how this algebraic power move that
> you are describing works, and whether it applies in this case?
>
> Thanks!
> Jon
>
> On Fri, Feb 8, 2019, at 6:32 PM, Valery Isaev wrote:
> > If you are not interested in computations and convenience of your type
> > theory, then the definitional equality is not essential in the sense
> > that every type theory T is equivalent to a type theory Q(T) with no
> > computational rules. 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, 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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