Re: [HoTT] Bishop's work on type theory

[Michael Shulman <shu...@sandiego.edu>]

Of course, if you have higher-order logic with propositional
extensionality, you don't need to use setoids at all, but can instead
define quotients as sets of equivalence classes like in ZF.  But I
suspect Bishop wouldn't have liked that either.

On 5/5/18, Michael Shulman <shu...@sandiego.edu> wrote:
> I think the problem is that it's not consistent about what a
> "proposition" is.  If a "proposition" is a setoid in which all
> elements are equal, then to be consistent, the equality relations of
> other setoids should also be valued in "propositions" of *this* sort,
> not the original collection of "propositions" you started with.
> Otherwise, I think you won't necessarily be able to take the quotient
> of a setoid by a "proposition"-valued equivalence relation, which is
> the whole point of introducing setoids in the first place.  But down
> this route lies infinity.
>
> I only know of three ways to get a well-behaved category of setoids:
>
> 1. Use propositions as types, as in MLTT Type-valued setoids and the
> ex/lex completion.
>
> 2. Define a morphism of setoids to be not an operation respecting
> equality but a total functional relation, as in the tripos-to-topos
> construction and the ex/reg completion.  I personally believe this is
> the correct solution in the most generality, but Bishop-style
> constructivists don't seem to like it.
>
> 3. Assume the axiom of choice, which causes options (1) and (2) to
> coincide.
>
>
> On 5/5/18, Thorsten Altenkirch <Thorsten....@nottingham.ac.uk>
> wrote:
>>
>>
>> On 05/05/2018, 05:27, "homotopyt...@googlegroups.com on behalf of
>> Michael Shulman" <homotopyt...@googlegroups.com on behalf of
>> shu...@sandiego.edu> wrote:
>>
>>>3. He includes the axiom of choice (p12) formulated in terms of his
>>>(proof-irrelevant) propositions, as well as what seems to be a Hilbert
>>>choice operator (though it's not clear to me whether this applies in
>>>open contexts or not).  Since he has powerclasses with propositional
>>>extensionality, I think this means that Diaconescu's argument proves
>>>LEM, which he obviously wouldn't want.  It's harder for me to guess
>>>how this should be fixed, since without some kind of AC, setoids don't
>>>satisfy the principle of unique choice.
>>
>> Why not? If we identify propositions with setoids that are internally
>> propositions (all elements are equal) and identify propositions upto
>> logical equality we get unique choice.
>>
>> What do I miss here?
>> Thorsten
>>
>>
>>>
>>
>>
>>
>>
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