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In this paper, author Nik Weaver warns that there could be questions of $\Sigma_1$-validity of ${\mathrm{ZFC}}$ set theory. As I understand it, he suggests that the axioms of a set theory might be strong enough to decide PA-undecidable statements about Turing machines, and moreover decide them wrongly. Wrongly in the sense that e.g. ${\mathrm{ZFC}}$ might prove for a particular machine $t$ there to "exist" a $n\in {\mathbb N}$ such that $t$ halts at step $n$ (i.e. ${\mathrm{ZFC}}$ proves $\exists n.T(n)$ for some statement $T$ about a particular coding up of a machine $t$), but "in reality" the machine actually can't do that. This wouldn't be a problem w.r.t. consistency, since it affects a statement where Peano arithmetic (${\mathrm{PA}}$) axioms were too weak to make a call on it.

See here for some relevant definitions and this related SE question. The relation between ${\mathrm{PA}}$ and ${\mathrm{ZFC}}$ is subtle but e.g. from reading the answer on the mentioned questions, one would think ${\mathrm{PA}}$ got us covered for any given Turing machine w.r.t. $\sigma_1$. So I'm trying to understand where a loophole is that Weaver points out. To clarify, I'm not interested in an inconsistent ${\mathrm{PA}}$ situation or an inconsistent ${\mathrm{ZFC}}$ situation, but I won't presume any soundness properties of the systems. And for the question, I'm also only interested in machines that could (edit: assuming we could wait arbitrary long and provide the memory capacity) be put on the quest to search for a natural and fail to ever halt - despite ${\mathrm{ZFC}}$ claims.

I know that there are conservativity results of ${\mathrm{PA}}$ over ${\mathrm{HA}}$ on, I think, that level of the hierarchy, and similarly with second order-${\mathrm{PA}}$ v. second order-${\mathrm{HA}}$, as well as ${\mathrm{ZF}}$ over ${\mathrm{IZF}}$. I.e. we know that for $\exists n.(f(n)=0)$-statements of ${\mathrm{HA}}$, won't see anything new in ${\mathrm{PA}}$ (${\mathrm{PA}}$ can't fail the world of HA on that level, so to speak). But do we know anything more about ${\mathrm{ZFC}}$ speaking about PA statements? I'm not sure if results I just listed do in fact clarify the $\Sigma_1$-validity issue on their own. Further (since I read someone bring it up in this context) I wonder whether or how ${\mathrm{ZF}}$ models of itself or of ${\mathrm{PA}}$ could clarify what's possible and what's inconsistent. Is this issue discussed somewhere on that level of the hierarchy, where validity issues in connection with "the real world" would still be "hands on" in this sense (i.e. not stuff about Turing jumps somewhere removed from recursive machines)? Assuming ${\mathrm{PA}}$ is right about the world, can ${\mathrm{ZFC}}$ be wrong about the world in this "wrong promise/wait forever" kind of way? I do understand that "wrong promise" might be orthogonal to a hard "wrong" notion in proof theory.

Nikolaj-K
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  • I've heard that iteration of a process which at step $n$ writes an integer in "recursive base $2n$" (where the exponents also get translated into recursive base $2n$), change all bases to $2n+1$, and then subtracts one, always terminating is sufficient to show consistency of PA, and therefore can't be proved by PA. But ZFC does prove termination of this calculation with any starting input, essentially as a result of the fact that $\varepsilon_0 = \omega^{\omega^{\cdots}}$ is well-ordered. Though I guess that's probably $\Pi_2$ not $\Sigma_1$. – Daniel Schepler Sep 19 '20 at 00:03
  • @DanielSchepler Sounds interesting. It's not a bad apple case, though, is it? Weaver, in chapter 2 starting at around page 6, 7 in the above text, speaks about (quote) "if ZFC gives us bad information about which Turing machines halt", where, for this to be persuasive, I suppose there needs to be a means of judging "bad". If your ordinal-type consistency statement can't be dismissed by other means (the machine would indeed just halt), then this would just give another fair theorem and not a false promise of sorts. Or is a real-world-halting that PA doesn't predict automatically bad? – Nikolaj-K Sep 19 '20 at 00:14
  • @Nikolaj-K "is a real-world-halting that PA doesn't predict automatically bad?" You have to distinguish between single halting instances and universally halting programs. The former are pretty boring here: $\mathsf{PA}$ is $\Sigma^0_1$-complete (indeed this is gigantic overill; see e.g. the discussion here), so there are no individual instances of halting computations which $\mathsf{PA}$ or $\mathsf{ZFC}$ miss. However, the statement that such-and-such program halts on all inputs is significantly more complicated - it is $\Pi^0_2$ in general. – Noah Schweber Sep 19 '20 at 02:04
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    (As Daniel said above.) By Godel, no "reasonable" theory (namely: no consistent computably axiomatizable theory interpreting $\mathsf{Q}$) is $\Pi^0_1$-complete, so a fortiori there will be true $\Pi^0_2$ statements which (say) $\mathsf{ZFC}$ misses. As to the question you ask, it is indeed possible that $\mathsf{PA}$ is fully sound but $\mathsf{ZFC}$ is inconsistent, or that $\mathsf{PA}$ is fully sound and $\mathsf{ZFC}$ is consistent but that $\mathsf{ZFC}$ is not $\Sigma_1$-sound, or so on. – Noah Schweber Sep 19 '20 at 02:06
  • @Nikolaj-K I find your question as written a bit unclear, hence my comments as opposed to an answer. If my comments addressed your question I'll be happy to turn them into an answer, let me know - if not, in what way do they not? – Noah Schweber Sep 19 '20 at 02:10
  • @NoahSchweber: Okay, thanks. I clarified it in the question and added links. In short, I'm not interested in an inconsistent PA- or a ZFC situations, and also only interested in machines that could (assuming arbitrary large fixed memory but possibly infinite time) be put on the quest to search for a natural and fail to ever halt - despite ZFC claims. PS I found a question from some years ago that goes in the same direction as mine, except it's more about consistency and existing theorems. – Nikolaj-K Sep 19 '20 at 12:21
  • @Nikolaj-K Reading your updated question, much is still unclear to me. Why would one think "PA got us covered for any given Turing machine wrt $\Sigma_1$"? And why would that cast doubt on Weaver's "loophole" that e.g. ZFC might prove a false $\Sigma_1$ sentence? – spaceisdarkgreen Sep 19 '20 at 17:44
  • From reading the other question, it sounds like ZF statements would be backed up by PA if for any machine $t$ someone could come up with start (the starting process meaning the input is fixed) and for which ZF proved it halted, PA would prove that (for the given input) too. – Nikolaj-K Sep 19 '20 at 22:24
  • @Nikolaj-K I'm not sure where you get that from the other question. Only thing I can see is Carl mentions PA proves every true $\Sigma_1$ sentence. That's true, but you're saying that if ZF proves a false $\Sigma_1$ sentence, then PA must prove it also, which is very different. There's no good reason for thinking that that I'm aware of, apart from the typical reasons for thinking that ZF doesn't prove any false $\Sigma_1$ statements in the first place. – spaceisdarkgreen Sep 20 '20 at 01:11
  • @spaceisdarkgreen I don't think I've written anything about PA proving false statements? Again, I'm wondering about existence statements that PA neither proves nor disproves (the statement is undecidable in PA), and such that ZF can prove them positively (ZF proves existence of an integer), and such that when you'd ran the machine that the statement is about, the claimed existence result would not actually be confirmed (I'm asking whether it could be that ZF made a false promise in that sense). If that doesn't make sense, then I'd just like to understand the argument brought forward better. – Nikolaj-K Sep 20 '20 at 17:33
  • @Nikolaj-K Well, you said that if ZF proved a computation terminated, then PA would. And since ZF and PA both prove all the true cases, the only potentially interesting thing is the false cases that one or both of them might prove. Hence I took your claim to be effectively that if ZF proves that a computation terminates, when in fact it doesn't, then PA must also. – spaceisdarkgreen Sep 20 '20 at 18:09
  • @Nikolaj-K I haven't looked into Weaver's argument, but ZF is much stronger than PA, so it seems reasonable to me for a skeptic of one variety or another to believe that PA doesn't prove any false instances and yet ZF might. If ZF proves a false instance but PA doesn't, then either ZF is inconsistent or the instance is undecidable in PA, since if PA refuted (correctly) the false instance, ZF would too. So yes, the case Weaver is talking about here is what you describe: a statement that a computation halts that is undecidable in PA, provable in ZF, and false. But what is your question? – spaceisdarkgreen Sep 20 '20 at 18:26
  • Do you really mean "assuming arbitrary large but fixed memory but possibly infinite waiting time"? The halting problem for machines with a fixed amount of memory is decidable since without unlimited memory a non-halting computation has to repeat itself. What Weaver is actually pointing out is that (working in a reasonable "base theory") it's conceivable that $\mathsf{ZFC}$ proves some false $\Sigma_1$ sentences even if $\mathsf{PA}$ is $\Sigma_1$-sound (or more). Your "$\mathsf{PA}$ has you covered" reflects the fact that - assuming consistency - neither proves any false $\Pi_1$ sentences. – Noah Schweber Sep 21 '20 at 18:15
  • That is, what $\mathsf{PA}$ (indeed much less) is strong enough to do is prove every true $\Sigma_1$ sentence; so if we have a consistent theory at least as strong as $\mathsf{PA}$ (and again that's overkill), that theory can't prove any false $\Pi_1$ sentences. Put another way, consistency is equivalent to $\Pi_1$-soundness but $\Sigma_1$-soundness is stronger. – Noah Schweber Sep 21 '20 at 18:18
  • Right, the fixed memory didn't make sense. Thanks for the answer you posted, looks like you straight out confirm that it's indeed possible. I'll give it a day or two. – Nikolaj-K Sep 21 '20 at 20:46

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The question and comment thread seem a bit all over the place; let me collect the various facts which I think together summarize the situation appropriately.

Any "appropriate" theory $T$ - like $\mathsf{PA}$ or $\mathsf{ZFC}$ - is strong enough to verify all true halting facts. That is, such theories are $\Sigma_1$-complete. Conversely, per Godel no such theory is $\Pi_1$-complete. In particular, if $T$ is "appropriate" in the relevant sense then so is $T+\neg\mathsf{Con}(T)$; since this latter theory is $\Sigma_1$-complete, consistent, and proves a false $\Sigma_1$-sentence, it (and a fortiori $T$ itself) must not be $\Pi_1$-complete.

So a consistent theory of the type we're looking at cannot afford to make any false $\Pi_1$ assertions, but it could conceivably make a false $\Sigma_1$ assertion. This distinction between $\Pi_1$ and $\Sigma_1$ explains why there is no tension between Weaver's observation and the "sufficiency" of $\mathsf{PA}$ (and indeed much less) for $\Sigma_1$ propositions.

Finally, $\mathsf{ZFC}$ could indeed be consistent but $\Sigma_1$-unsound even if $\mathsf{PA}$ is fully sound - at least, working within a reasonably weak background theory. Specifically, assuming $\mathsf{ZFC+Con(ZFC)}$ is consistent in the first place then so is $\mathsf{ZFC+Con(ZFC)+\neg Sound_{\Sigma_1}(ZFC)+Sound(PA)}$. (The "$+\mathsf{Sound(PA)}$"-bit is redundant since $\mathsf{ZFC}$ already proves that $\mathsf{PA}$ is fully sound, I'm just including it for explicitness.) So basically we cannot use the correctness of $\mathsf{PA}$ to justify even the $\Sigma_1$-correctness of $\mathsf{ZFC}$ without baking the latter assumption into our base theory at the outset. However, this happens long before we hit $\mathsf{ZFC}$: the same holds for $\Pi^1_1$-$\mathsf{CA_0}$, if memory serves, which is a tiny fragment of $\mathsf{Z_2}$ which is itself a tiny tiny tiny fragment of $\mathsf{ZFC}$. So leaping all the way to $\mathsf{ZFC}$ here is massive overkill, and I think makes this phenomenon appear more mysterious than it actually is.

Noah Schweber
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