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I have a basic question about propositional truncation $||$-$||$ and double negation $\neg\neg$.

According to the recursion rule of $||$-$||$, $A\rightarrow B=||A||\rightarrow B$ as long as $B$ is a mere proposition (i.e., proof-irrelevant). Now let $B=\neg\neg A$, since $\neg\neg A$ is a mere proposition, and $A\rightarrow\neg\neg A$ is a tautology, we naturally conclude that

$$||A||\rightarrow\neg\neg A.\quad\quad(1)$$

Since $||A||\rightarrow\neg\neg A$ is a tautology, then $\neg\neg(||A||\rightarrow\neg\neg A)$ is also a tautology. Since $\neg\neg$ distributes over $\rightarrow$, we get the following

$$\neg\neg||A||\rightarrow\neg\neg A.\quad\quad(2)$$

Therefore, the following is also true:

$$\neg\neg(||A||\rightarrow A).\quad\quad(3)$$

But (3) looks quite crazy because it's almost the inverse of $A\rightarrow||A||$, though under $\neg\neg$. I don't know if these are all correct. I would appreciate if someone can tell if it's correct or if there is anything wrong with my derivations.

More: Since there is a mapping for $A\rightarrow||A||$, its double negation $\neg\neg(A\rightarrow||A||)$ is also a tautology. Together with (3), we have $\neg\neg(A\leftrightarrow||A||)$. Is there any place wrong with my derivation?

Emini Jask
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  • What do you mean by "$\neg\neg$ distributes over $\to$"? – Mike Shulman Jul 11 '20 at 17:14
  • @MikeShulman Thanks for the reply. It's $\neg\neg(\phi\rightarrow\psi)\leftrightarrow(\neg\neg\phi\rightarrow\neg\neg\psi)$. – Emini Jask Jul 11 '20 at 17:26
  • That's not immediately obvious to me; why is it true? – Mike Shulman Jul 11 '20 at 18:15
  • @MikeShulman Hi Mike, you can take a look at Lemma 6.2.2 in van Dalen's textbook Logic and Structure (Click the link. It's an electronic version). The proof for this equivalence in intuitionistic logic is given on pages 159-160. – Emini Jask Jul 11 '20 at 19:22
  • @MikeShulman, This is not the first time I see people ask this question, so I wrote up three proofs here. – Z. A. K. Jul 11 '20 at 20:50
  • @Z.A.K.Thanks for the proofs! Do you have any advice for my question in the post? – Emini Jask Jul 11 '20 at 20:55
  • @EminiJask Not yet, but I am thinking about it. I don't have much intuition about whether $\neg\neg(||A|| \rightarrow A)$ is crazy or not: I'm predisposed to thinking that truncations are weird anyway. – Z. A. K. Jul 11 '20 at 21:01
  • @Z.A.K. Thanks! I read that post before and also similar analyses in a paper by the same author Nicolai Kraus. I feel that it's crazy because it shows that $||A||\rightarrow A$ is always inhabited although the proof object is anonymous (due to $\neg\neg$)... – Emini Jask Jul 11 '20 at 21:33
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    Regarding craziness, even having a function $\Vert A \Vert \to A$ for all $A$ would not be an inverse of $A\to \Vert A \Vert$, only a section of it. That's essentially a strong axiom of choice, "every inhabited type has a specified element" -- strong enough to contradict univalence, but consistent to assume under UIP. Note that the weaker statement that $\Vert (\Vert A \Vert \to A)\Vert$ holds for all $A$ is not contradictory, and indeed follows from LEM, but is still not constructively provable. – Mike Shulman Jul 12 '20 at 23:03

2 Answers2

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Your arguments are correct as stated. It is indeed the case that the ($\Pi$-closures of the) types $\neg\neg (||A|| \rightarrow A)$ and $\neg\neg (||A|| \leftrightarrow A)$ are inhabited, and for the reasons outlined above.

The second part of your question far subtler: it asks about the "craziness" of the fact that these types are inhabited. I'm not sure how to answer this (although I suspect that a good answer is possible, and will eventually be penned by someone else). In the meantime, I offer a few remarks.

It's not surprising that we have functions $|| \bot || \rightarrow \bot$ or $||\mathbb{N}|| \rightarrow \mathbb{N}$ for fixed types $\bot, \mathbb{N}$. However, we certainly would not expect any term to inhabit $\Pi A:\mathbf{U}.||A|| \rightarrow A$. However, actually proving that there is no term of that type seems like a task that would involve fairly elaborate semantic reasoning (assuming that this is indeed the case; I might ask a followup question about this later).

The fact that we can construct an inhabitant of the type $\Pi A:\mathbf{U}.\neg\neg (||A|| \rightarrow A)$ is a pleasant surprise, but it does not say much about the inhabitedness of the former, as we still can't prove $\neg\neg \Pi A:\mathbf{U}.(||A|| \rightarrow A)$. In fact, there is a fairly direct proof of $\Pi A:\mathbf{U}.\neg\neg(||A|| \rightarrow A)$ that does not invoke the distributivity of $\neg\neg$, and might illuminate what's going on.

We can prove $\Pi A:\mathbf{U}.\neg(||A|| \rightarrow A) \rightarrow \neg A$ simply by taking $f : \neg(||A|| \rightarrow A)$ and $a : A$, and constructing $f(\lambda x.a) : \bot$. Similarly, we can prove $\Pi A:\mathbf{U}. \neg A \rightarrow ||A|| \rightarrow A$ since given $a : || A ||$ and $f: A \rightarrow \bot$ we can get $\mathrm{squashrec}(f): || A || \rightarrow \bot$, and then $\mathrm{squashrec}(f)(a) : \bot$. Putting these together we have a term of type $\Pi A:\mathbf{U}.\neg(||A|| \rightarrow A) \rightarrow || A || \rightarrow A$, from which $\Pi A:\mathbf{U}.\neg\neg(||A|| \rightarrow A)$ readily follows.

Z. A. K.
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    "I suspect that a good answer is possible, and will eventually be penned by someone else" -- indeed, it looks like one might have appeared while I was writing this answer. – Z. A. K. Jul 11 '20 at 22:40
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    One can prove internally that $\Pi A:\mathbf{U}. \Vert A\Vert\to A$ contradicts the univalence axiom; this is Exercise 3.11 in the HoTT Book. Thus, since the univalence axiom is consistent (which might be considered "elaborate semantic reasoning"), that type is uninhabitable. – Mike Shulman Jul 12 '20 at 22:58
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Everything you’ve said works, and there’s a simple intuition for it all: $\neg\neg\cdot$ is a modality, and so is $\Vert\cdot\Vert$, and the latter is a “loosening” of the former. Specifically, they’re equivalent under the former (that is, $\prod_{A : \mathcal{U}} \neg\neg (\Vert A\Vert \simeq \neg\neg A)$), because we can prove $\mathrm{LEM}_{\neg\neg} :\equiv \prod_{A : \mathcal{U}} \neg\neg (A + \neg A)$. So if we assume $\mathrm{LEM}_{-1} :\equiv \prod_{A : \mathcal{U}} \mathrm{isProp}(A) \to \left\Vert A + \neg A\right\Vert$, then the two modalities are purely equivalent too.

The way I like to think about this is that, in a classical setting, propositional truncation is just the same thing as double negation, which is already well known to produce a classical environment in an otherwise constructive setting. So in the general constructive-by-default setting of MLTT and its descendants, propositional truncation isn’t automatically equivalent to double negation, but you also can’t prove that it’s disequivalent without an explicit anticlassicality principle—and they’ll always be equivalent in the “forced classical” environment underneath double negation.

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