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I am currently reading the HoTT book and came across exercise 3.14:

Show that assuming $\mathrm{LEM}$, the double negation $\neg \neg A$ has the same universal property as the propositional truncation $\| A \|$, and is therefore equivalent to it. Thus, under $\mathrm{LEM}$, the propositional truncation can be defined rather than taken as a separate type former.

The rules governing the truncation are:

Formation

  • For every type A, $\|A\|$ is a type

Introduction

  • For any $a:A$ we have $ |a| : \| A \|$.
  • For any $x,y:\| A\|$, we have $x=y$.

Elimination resp. Recursion

  • If $B$ is a mere proposition and we have $f:A\to B$, then there is an induced $g:\| A\| \to B$ such that $g(|a|) \equiv f(a)$ for all $a:A$.

My Problem

To proof the recursion principle for $\neg \neg A$ we first fix some $f:A\to B$ and then need to define $g:\neg\neg A\to B$. I don't know how to do that.

There are ways to use $\mathrm{lem:LEM}$, but I have no idea how to convert this terms (to assert judgemental equality), as $\mathrm{lem}$ is a primitive constant.

There is a solution in the Github folder HoTT/HoTT/contrib/HoTTBookExercises.v, that is programmed in Coq. Although I understood the argumentation I still don't see why it should be sufficient to proof $\neg\neg A \simeq \|A\|$ as we would end up with a propositional equality again.

I would really appreciate any help. Thanks in advance.

Best regards, Philipp

EDIT: I assume that $$\mathrm{LEM}=\mathrm{LEM}_{-1}=\prod_{A:\mathcal{U}}\mathrm{isprop}(A)\to A+\neg A. $$ just like the book does.

Philipp
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2 Answers2

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If what you're asking is whether $\neg\neg A$ has the same judgmental computation rule as $\Vert A \Vert$, then I think you're right that the answer is no.

To be precise, the exercise doesn't say to prove that $\neg\neg A$ has the same judgmental computation rule, or even the same induction principle, but the same universal property. The universal property of $\Vert A \Vert$ is unfortunately not stated in this section (it should be), but it appears as the $n=-1$ case of Lemma 7.3.3: for any mere proposition $B$, precomposition with $|-| : A \to \Vert A \Vert$ induces an equivalence

$$ (\Vert A\Vert \to B) \xrightarrow{\simeq} (A\to B).$$

This doesn't require a judgmental computation rule, and to say that $\neg \neg A$ has this same property is indeed the same as saying $\neg\neg A \simeq \Vert A \Vert$ (at least, assuming that $\Vert A \Vert$ exists).

However, it's clearly unreasonable to expect the reader to make this distinction for themselves, especially without having mentioned this universal property, and also especially because Exercise 3.15 does have the judgmental computation rule (at least for the recursion principle). I will submit a fix to the phrasing of the exercise, thanks for pointing this out.

For what it's worth, the judgmental computation rule of propositional equality is not very useful, because the target is always a mere proposition so that all its points are typally (i.e. "propositionally") equal anyway. The main thing I know of that you can do with the judgmental computation rule that you can't do without it is Kraus's magic trick, which might not be regarded as a good thing.

Mike Shulman
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  • Thank you, this answers my question. Do you know why the truncation is defined with judgemental equality then? – Philipp Mar 07 '18 at 06:58
  • Because all other computation rules (for points) are judgmental, and because that's what you get when you define it as a HIT (as long as HITs in general have such computation rules). – Mike Shulman Mar 07 '18 at 10:59
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Since $B$ is a mere proposition, you can use $\mathrm{lem}$ on $B$. If $B$ is true, then you are done, and if $\neg B$, then $\neg A$ using $f$, so $\neg\neg A\to B$ by the principle of explosion.

Mario Carneiro
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  • Hi Mario, thanks for the answer. I guess with $\mathrm{LEM}$ they mean $\mathrm{LEM}_{-1}$, so A would have to be a proposition to argue like that? – Philipp Mar 06 '18 at 15:24
  • @Philipp I don't know your notation, but perhaps it means you can only eliminate when the target is a mere proposition? Okay, so eliminate when the goal is B then. Edited. – Mario Carneiro Mar 06 '18 at 15:27
  • I edited it. You can only use $\mathrm{LEM}$ on mere propositions. – Philipp Mar 06 '18 at 15:33
  • The elimination is only defined for mere propositions anyways... So this has nothing to do with $\mathrm{LEM}$. So as A is not a mere proposition in general I think this proof does not work. – Philipp Mar 06 '18 at 15:49
  • @Philipp Ah, I see. In this case, we have to case on $B$ instead, but it still works out. – Mario Carneiro Mar 06 '18 at 18:55
  • Yes, but the equality would only hold propositionally. The difficulty in my opinion is the judgemental equality. – Philipp Mar 06 '18 at 19:21
  • @Philipp You won't get judgmental equality with this construction, since $\mathrm{lem}$ will block computation. The book is not clear on this point, I think, but all that is being claimed is $\neg\neg A\simeq|A|$, which does not require any definitional equality. – Mario Carneiro Mar 06 '18 at 19:51
  • This is exactly what i think, but then $\neg\neg A$ can't serve as truncation, as the judgemental equality is required, and it makes a difference wether $f(a)\equiv g(|a|)$ is judgemental or propositional. – Philipp Mar 06 '18 at 20:49
  • @Philipp Since the equality $f(a)=g(|a|)$ is contractible, it doesn't make that much difference as far as provable things go. Generally, when we talk about defining some inductive type or HIT by other means, we will only get the computation rule up to provable equality, which is why in practice you still see these HITs added as axiomatic despite their supposed definability, for convenience' sake. – Mario Carneiro Mar 07 '18 at 04:52
  • I see, this answers my question as well. Thank you very much vor the help – Philipp Mar 07 '18 at 07:03