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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