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I'm looking to validate (or invalidate) whether this intuition about what non-standard integers can mean is correct.

In $ZFC$, we can define a Turing machine $M$ that enumerates all proofs of $ZFC$ and only stops if it finds one of $0=1$.

Because of Gödel's Second Incompleteness Theorem, if $ZFC$ is consistent then $Con(ZFC)$ is independent from it and so is $\neg Con(ZFC)$. As such, $ZFC$ cannot prove whether $M$ halts or not, so a proposition $H$ saying $M$ halts is also independent of $ZFC$.

If $ZFC$ is consistent, then $ZFC + H$ is consistent. In models of this theory, $M$ halts.

We can define $m$ as the number of steps it takes for $M$ to halt.

From my understanding, $m$ would have to be a non-standard integer. Is that correct? Because if it was some standard integer $n$, we could run $M$ step by step all the way up to $n$ in $ZFC$ to formulate a proof of $\neg Con(ZFC)$.

And if $m$ is non-standard, is the following correct? $ZFC + H$ proves "there is an integer $m$ such that $M$ halts at the $m$th step", but does not prove any of the following statements: $m = 0$, $m = S(0)$, $m = S(S(0))$, $m = S(S(S(0)))$, $m = S(S(S(S(0))))$, etc. So even though $M$ halts, we can't "really" reach $m$, it won't be a number we can actually write down as the result of applying actually finitely many times the successor operation on $0$, even though $H$ says $M$ halts and thus we should be able to?

This feels correct to me, but as an outsider I'm hesitant to be 100% confident.

Uretki
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  • You might benefit from reading about general recursive functions, primitive recursive functions and how the Ackermann function, which only produces honest and true integers, is not primitive. – Eric Towers Jul 01 '24 at 08:33
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    @EricTowers sorry, what does it mean? It's very easy to build a TM that computes Ackermann function (and full PA can prove that this machine always halts). – mihaild Jul 01 '24 at 08:35
  • @mihaild : And yet, understanding why the given problem is in the set of general recursive functions and whether that level of computability entails "magical integers" is equivalent to understanding that the Ackermann function does not require "magical integers". – Eric Towers Jul 01 '24 at 08:37
  • @EricTowers I'm not seeing the relation to the Ackermann function, could you be more explicit? – Uretki Jul 01 '24 at 08:52
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    If ZFC is consistent (which is hopefully the case) , then this machine does not halt. If ZFC is not consistent , it will probably take very long to find a proof of $0=1$ , but the number of steps , no matter how large , is just an integer , not a non-standard integer. If the machine does not halt , it does not make sense to count the number of steps. – Peter Jul 01 '24 at 08:56
  • Hence I would say that this is not correct. Either the machine halts , then the number of steps is any integer and can be reached , or it does not halt , then it neither halts after a "nonstandard number of steps" , whatever this should mean. – Peter Jul 01 '24 at 08:59
  • @Peter would you say that in models of $ZFC + H$ the machine doesn't halt, but the model (incorrectly) believes that it does after some number of step $m$ which has to be non-standard, since $M$ doesn't actually stop? I'm assuming this is what being unsound is, correct? Proving statements ("$M$ halts") that aren't actually true, without necessarily being able to prove contradictions. – Uretki Jul 01 '24 at 09:21
  • I must admit that I am not an expert in this topic , but for me it makes no sense that a turing machine halts , but in a "nonstandard number of steps". Maybe , you mean that ZFC cannot prove that this turing machine does not halt. This is in fact true , since then ZFC would have proven that it is consistent which is impossible (Goedel). If the machine halts , then ZFC is inconsistent and we have no contradiction because that a machine halts can be proven anyway. – Peter Jul 01 '24 at 09:50
  • @Peter Yes, and since $H$ is independent of $ZFC$ (you stated the reason why), then there should be models for $ZFC + H$. I think it should follows from Gödel's completeness theorem, since if $\neg H$ was true in all models of $ZFC$, then $ZFC$ would be able to prove $\neg H$, which it can't, so $ZFC$ must have models with $H$. But I'm no expert, and maybe I'm misunderstanding something. From Qiaochu Yuan's answer, it seems to be incorrect to say that $M$ stops in those models, but surely $m$ is still definable, and thus would have to be non-standard because $M$ doesn't actually stop? – Uretki Jul 01 '24 at 10:07
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    This blog post (and the post it discusses) have some interesting thoughts on non-standard halting times if you haven't seen them before: https://johncarlosbaez.wordpress.com/2016/04/02/computing-the-uncomputable/ – cobbal Jul 01 '24 at 17:22
  • The difficulty may be due to a conflation of metalanguage natural numbers and the object-language (formal) natural numbers. Talk about "m=0, m=S(0), m=S(S(0)), m=S(S(S(0))), m=S(S(S(S(0)))), etc." obviously refers to the meta-integers. Meanwhile, once a Turing machine is formalized, one can obviously ask questions about how it behaves at any formal integer (including ones viewed as "nonstandard"). A related recent discussion is at https://math.stackexchange.com/questions/4936355/why-does-forall-n-in-mathbbn-vdash-pn-not-imply-forall-n-in-mathbb – Mikhail Katz Jul 02 '24 at 10:46

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Yes, this is basically right, although I would quibble a bit with this:

In models of this theory, $M$ halts.

I think it would be more accurate to say that models of this theory contain a nonstandard number which is "supposed to be" the halting time of $M$. You may find this MO answer of mine a helpful read.

Qiaochu Yuan
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  • Do I understand correctly then that despite $H$ saying that $M$ halts, it doesn't, the theory just "thinks" it does. Are we still able to define $m$ as what the theory will think is $M$'s "halting" time? (Also, I indeed found your link helpful.) – Uretki Jul 01 '24 at 09:17
  • Clarifying, by "it doesn't" I meant "$M$ doesn't". – Uretki Jul 01 '24 at 09:29
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    @Uretki Yes, ZFC + H proves the sentence "M halts" and any models of this theory satisfy the sentence "M Halts". Yes, "the number of steps it takes $M$ to halt" is definable in ZFC + H, hence definable without parameters in any models of ZFC +H. Yes, assuming Con(ZFC), $M$ doesn't actually halt (so ZFC + H and its models are wrong), and any model of ZFC (necessarily not of ZFC + H) whose natural numbers are isomorphic to the actual natural numbers will satisfy "M does not halt" (more generally any model sound with respect to $\Pi_1^0$ arithmetical statements). – spaceisdarkgreen Jul 01 '24 at 15:21
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    @Uretki I think you nail it in your second-to-last paragraph of the question that if we externally tried to "run" the "M" of a given model of ZFC + H, we'd just see the same behavior as "the real M" (which I guess is also what Qiaochu means when he quibbles with saying "M halts in models of ZFC+H"). But of course the model has more things it thinks are natural numbers after the standard cut (so any recursion internal to the model extends through them), and there is an $m$ out there where the model satisfies "$M$ halts at step $m$". – spaceisdarkgreen Jul 01 '24 at 15:47
  • @Uretki: so the reason I quibbled here is that nonstandard models don't provide any kind of actual semantics for what it "really means" to run a Turing machine for a "nonstandard number of steps" (I was confused about this point for awhile and I admit I may still be confused about it now). The "actual behavior" of a Turing machine consists of how it behaves after a standard number of steps. – Qiaochu Yuan Jul 01 '24 at 20:54
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    @QiaochuYuan "The tape" is nonstandardly long, but you can still ask what's on the tape, which cell the read-head is at, and which internal state the TM is in at (possibly nonstandard) time $m$ and there's an answer (abstractly speaking)... and then the next transition just depends on the (entirely standard) local information of what's on the cell and what state it's in and looks like normal. But the global "running" is odd, since from a nonstandard time $m$, history infinitely regresses through $m-1$, $m-2$, etc and similarly there's no "next step" after a given infinite sequence. – spaceisdarkgreen Jul 02 '24 at 01:19
  • @spaceisdarkgreen, it is not odd. We know that a nonstandard model is not well-ordered, but obviously the language in question does not know that, since $\mathbb N$ is well-ordered. See also my comment below the question. – Mikhail Katz Jul 02 '24 at 15:12
  • @MikhailKatz I think it’s pretty odd… I think nonstandard models are a little odd, but one gets used to these things of course. But yes, the reason the recursive definition still produces a unique “time evolution” is because the model satisfies first order induction (or more simply because the well-definedness can be proved in ZFC). – spaceisdarkgreen Jul 02 '24 at 15:22
  • @MikhailKatz e.g., it’s “odd” that a non-well-founded thing can satisfy first-order induction. – spaceisdarkgreen Jul 02 '24 at 15:30
  • @spaceisdarkgreen, you are basically saying that you find nonstandard models of arithmetic "odd". But they have been around since Skolem's 1933 paper and are no longer considered mysterious. – Mikhail Katz Jul 03 '24 at 06:34
  • @MikhailKatz Yes, I said that, and yes, them being "odd" in my opinion doesn't mean we don't understand them. I think we're just arguing over the appropriate level of 'mystification' of structures you would presumably admit have at least some (initially) counterintuitive attributes and a less-than-fully constructive nature (and I'd easily admit that they're hardly unique in this respect). Bear in mind my comment was pushing back against the notion that we can't understand the goings on of a Turing machine inside a non-$\omega$ model. – spaceisdarkgreen Jul 03 '24 at 20:42
  • @spaceisdarkgreen, OK good. An additional remark is that from certain perspectives, such as Hamkins' multiverse, there is no standard model of $\mathbb N$. Every model is nonstandard with respect to a smaller model (in a different instance of a set-theoretic universe inside his multiverse). This makes such "external non-well-ordering" even more natural. Of course, the metalanguage naturals are present in all of these and can be thought of as a sorites-type subcollection (not a set). – Mikhail Katz Jul 04 '24 at 10:01