26

First, some context. Ever since I was a high schooler, I have been fascinated with large numbers. As I have grown in mathematical maturity, I have become both disappointed and fascinated to see that the process of naming larger and larger numbers requires a sort of philosophical tradeoff.

For example, an ultrafinitist might reject $10^{10^{100}}$ as being a valid representation of a number. A finitist might reject numbers defined using ordinal collapsing functions with large cardinals. A person who believes only in the constructible universe might reject the same ordinal collapsing functions if cardinals incompatible with $V=L$ are used. In general, the more you are willing to accept philosophically, the "larger" a number you can name concisely.

The interesting thing about these tradeoffs is that some of these philosophical stances have concrete impacts on the "real" world. For example, there is a number $n$ for which new axioms in addition to those of ZFC would be needed to prove the value of $BB(n)$. So in a way, this "real world" value changes depending on our philosophical stance.

My question then is: what sorts of axioms might we accept to strengthen as much as possible the values of $BB(n)$ which we can prove? And more generally, is there a way we can evaluate whether certain statements independent of ZFC are "true" based on their implications for the "real world" value of $BB(n)$ (if there is even a "true" value for this number).

EDIT: useful reading for those not familiar with the phenomenon of independence of $BB(n)$ from ZFC:

  • Adam Yedidia, Scott Aaronson, A Relatively Small Turing Machine Whose Behavior Is Independent of Set Theory, Complex Systems 25(4) (2016), journal, arXiv:1605.04343

EDIT: As requested, an example of a "real world" consequence of a set theoretical axiom would be how Con(ZFC) would prove the Turing machine in the above paper doesn’t halt, whereas the negation of Con(ZFC) proves it halts in finite time.

David Roberts
  • 33,851
exfret
  • 479
  • 3
    "For example, there is a number $n$ for which new axioms in addition to those of ZFC would be needed to prove the value of $\operatorname{BB}(n)$": is this a suspicion, or a theorem? (And what does it mean to "prove the value" of something?) – LSpice Jan 11 '21 at 21:08
  • 2
    @LSpice I asked myself the same question, and wikipedia led me to this article: https://arxiv.org/abs/1605.04343 – user2520938 Jan 11 '21 at 21:10
  • 4
  • 2
    The premise of this question seems very dubious to me. It would be good if you could give an example where making extra set-theoretic assumptions allowed one to find the answer to some numerical question of a discrete/finitistic nature. One reason to be suspicious: why wouldn't that just give you the answer in ZFC itself? – Sam Hopkins Jan 11 '21 at 21:11
  • 3
    @Sam Hopkins For example, if we assume Con(ZFC) then the Turing machine referenced in Yedidia and Aaronson doesn’t halt, and assuming the negation of Con(ZFC) gives that this Turing machine does halt. – exfret Jan 11 '21 at 21:14
  • To add another thought, it is true that the halting of a Turing machine isn’t quite finitistic, but as you pointed out, any truly finitistic statement can be proven simply by exhaustion of cases in ZFC. However, whether a Turing machine halts is still thought of as a “real world” testable thing by many of my peers, at the very least. Perhaps people who have thought about this longer than me think otherwise, though. – exfret Jan 11 '21 at 21:33
  • 7
    You could always add an axiom of the sort: $BB(5)=193894749339392$. This one is almost surely inconsistent (because I made up that large integer). With such an axiom, the value of $BB(5)$ is very easy to find. – Pace Nielsen Jan 11 '21 at 21:57
  • When you say "a constructivist" I guess you mean someone who wants to work in the constructible universe, rather than someone who does not posit LEM. – David Roberts Jan 11 '21 at 22:53
  • @David Roberts Yes, my philosophy of math knowledge is not very extensive. I was thinking I might have been using that word wrongly. – exfret Jan 12 '21 at 00:00
  • @exfret no problems. Probably the best description is "someone who believes only in the constructible universe", there being no real school of thought that seriously accepts this (to my knowledge), but it's no biggie, it's clear what you meant. – David Roberts Jan 12 '21 at 01:03
  • I took the liberty of making the edit while make another more substantial edit, @exfret, I hope you don't mind. – David Roberts Jan 12 '21 at 04:39
  • @exfret. Indeed it is a common error to think that a Turing Machine is "real world." Its use of an infinite tape means it is not "real world." – abo Jan 12 '21 at 09:27
  • 1
    @abo Each local step of a turing machine is physically realizable, though (by expanding the tape as needed). In fact, I would argue it is not the infinite tape, but the infinite time that makes the "real-worldness" of turing machines iffy (and in fact, this iffy-ness is why I use the word "real world" in quotes). It is only because many of my peers (who are often much smarter than I) see them as real-world that I posited that this is a "real-world" phenomenon. One has even gone so far to say that a mathematical question is "real" exactly when it can be tested by a turing machine. – exfret Jan 12 '21 at 10:30
  • @DavidRoberts I am very grateful for your clarifying edits (and I see you made my hastily added link into a proper citation). Thank you! – exfret Jan 12 '21 at 10:32
  • @exfret. "Expanding the tape as needed" is not "real world." A physical "real world" computer cannot do so, since it has a finite tape. – abo Jan 12 '21 at 13:44
  • @PaceNielsen "With such an axiom, the value of $BB(5)$ becomes very easy to find" - I laughed out loud, but now I see this as a real moneymaking opportunity. I'm officially adding the Riemann Hypothesis to ZFC as an axiom, in order to force the Clay Mathematics Institute to award me one million dollars. What are they going to say--"No"? Can't argue with an axiom. – Rivers McForge Jan 13 '21 at 05:27
  • @abo https://en.wikipedia.org/wiki/Self-replicating_spacecraft – user76284 Jan 17 '21 at 19:20
  • @SamHopkins Consider a machine M that searches for a contradiction in ZFC and halts it if finds one. Gödel’s second incompleteness theorem implies that ZFC cannot prove M never halts. – user76284 Jan 17 '21 at 19:23
  • @PaceNielsen Isn’t that easily falsifiable by ZFC (assuming it’s incorrect)? Take all 5-state machines and run them for that number of steps. – user76284 Jan 17 '21 at 19:38
  • @user76284 Easily? If so, add a googleplex more digits. – Pace Nielsen Jan 18 '21 at 00:24

3 Answers3

22

There's a certain confusion underlying your question, which Andreas Blass's answer is trying to point out. Let me see if I can explain it in different words.

You say, “the negation of Con(ZFC) proves it halts in finite time” and you are trying to use this fact to argue about which axioms beyond ZFC to accept. The best sense I can make out of your comment is that ZFC + $\neg$Con(ZFC) has no $\omega$-model (i.e., no model in which the natural numbers are standard) and you treat this fact as an argument against accepting $\neg$Con(ZFC) as an axiom. So your question sounds like, do any of the standard strengthenings of ZFC have the property that they have no $\omega$-model? As Andreas Blass explained, the answer is no. So under this interpretation of your question, the answer is that none of the standard large cardinal axioms can be rejected on such grounds.

But I suspect that this is not your real question. I am guessing that when you refer to the “real world” you mean the standard integers, not the integers in some model of ZFC. Let me see if I can state what I think is your intended question.

Suppose for the sake of argument that BB(3) = 17 but that this fact is unprovable in ZFC. (Of course, this is absurd; but to write down realistic numbers in place of 3 and 17 would be, shall we say, inconvenient, and it does not matter for what I am going to say what the actual numbers are.) What is it about this statement that is unprovable? Well, BB(3) $\ge$ 17 is actually provable in ZFC, and in fact in a much weaker system, because you can just write down a specific size 3 Turing machine and verify that it halts in 17 steps. The troublesome part is proving that all those other size 3 Turing machines that take at least 18 steps never halt at all. For some of these machines, ZFC is unable to prove that they do not halt.

So what about stronger axioms? It certainly can be the case that adding a large cardinal axiom to ZFC will allow us to prove that some of those non-halting size 3 Turing machines do not halt. For example, adding large cardinal axioms yields new consistency statements, and consistency statements can be cast as statements about certain Turing machines (those that search for contradictions) not halting. So if this is what you mean, then the answer is yes, in principle, we can partially evaluate large cardinal axioms on the basis of their “real-world” consequences. However, I do not think that the BB function is particularly helpful in this regard. The trouble is that we do not know of any “reasonable” axiom that would let us prove that BB(3) = 17. And even if we did, how would that help us evaluate the axiom? Presumably, we would try to accumulate empirical evidence by enumerating size 3 Turing machines that run for at least 18 steps, and watching to see if they ever halt. The longer they run, the more confident we become that they will indeed never halt, so the happier we are that the predictions of our new axiom are being validated. The trouble is that 18 is so ridiculously large that we will never reach it, let alone sit around waiting even longer to see if the machine halts. So we have no practical way of accumulating the relevant empirical evidence.

Timothy Chow
  • 78,129
  • 7
    "Of course, this is absurd" It's true for sufficiently large values of $3$ and $17$! – Noah Schweber Jan 12 '21 at 05:17
  • 1
    Maybe this is getting too far away from the OP's question, but is there any work in the philosophy of math on treating Turing machines as physical objects which we can investigate empirically (in order to determine mathematical truths)? That seems to be a viewpoint certain "realistically" minded people want to take (see also this other MO question https://mathoverflow.net/questions/34710/succinctly-naming-big-numbers-zfc-versus-busy-beaver) – Sam Hopkins Jan 12 '21 at 05:35
  • An easier question (perhaps?) .... let the bounding number of PA be the smallest value $n$ such that there exists some machine of size $n$ for which PA can't prove its halting. Now bounding number of PA is $\leq$ to set theory. For some standard agreed upon computation model (such as in question), is it known that bounding number of PA is strictly smaller than set theory? In fact while we are at it, what is known about relation between bounding numbers of HA and PA (for example) or other theories? – SSequence Jan 12 '21 at 08:07
  • I meant "can't prove its non-halting" in last comment. – SSequence Jan 12 '21 at 08:53
  • Thank you for your answer. I think I had all these thoughts brewing in my head, but hearing them out loud really clarifies things. I have only taken two graduate courses in logic, and that was quite some time ago. But I still like to think about such things, so sometimes I need to ask outside sources for clarification. – exfret Jan 12 '21 at 10:22
  • One parting thought: Perhaps we can take it as granted that if ZFC is consistent, then any turing machine whose halting-ness is unprovable in it should actually run forever (otherwise ZFC should prove that it halts, since this is something that can be verified in finite time). Is it conceivable that we could evaluate an axiom on this basis, or would this just be equivalent to showing whether an axiom is consistent with ZFC? I am thinking the latter, but often I get myself confused on these matters, as you can see. – exfret Jan 12 '21 at 10:26
  • @exfret My logic knowledge is also quite limited, but I "think" that's one of the issues with difference between "consistency" and "arithmetic soundness" for a sufficiently strong theory $T$. If $T$ is consistent (and strong enough) then if a program actually halts then its halting will always be provable. If a program doesn't halt then its non-halting/looping may or may not be provable (in a sufficiently strong theory). However, if the theory is not "arithmetically sound" (and just "consistent") it may end up "proving wrongly" the halting of some program which actually loops forever. – SSequence Jan 12 '21 at 11:40
  • Now "arithmetic soundness" should entail being correct about all arithmetic statements. But since statements about halting/non-halting are subset of arithmetic statements, being correct on halting/non-halting should be necessary for the soundness condition to hold. [the thoughts on previous and this comment are my own .... I think probably other people will agree (or tend to agree) but I don't know for sure] – SSequence Jan 12 '21 at 11:49
  • 4
    @exfret : SSequence is correct. The technical way to say "any Turing machine whose non-haltingness is provable in ZFC does not halt" is "ZFC is $\Sigma_1$-sound" or "ZFC is 1-consistent." This is a stronger assumption than "ZFC is consistent." (It is also weaker than "ZFC is arithmetically sound" which would mean that any arithmetic statement provable in ZFC is true of the standard integers). So you could demand that any proposed new axiom A have the property that not only is ZFC + A consistent, but it is also $\Sigma_1$-sound. Almost everyone would agree with you, I think. – Timothy Chow Jan 12 '21 at 13:34
  • @Noah: While we can't muck around with $3$ all that much, $17$ is very easy to make large. If you want $17$ to be equal to some very large $n$, simply work in base $n-7$. – Asaf Karagila Jan 12 '21 at 13:43
18

$\DeclareMathOperator\BB{BB}$Philosophical issues, like acceptance (or non-acceptance) of large cardinals, won't affect $\BB(n)$, because the busy beaver function is defined arithmetically and so depends only on the natural numbers. Specifically, suppose I believe in some large cardinal, say supercompact. Then within my set-theoretic world, there is the submodel of Gödel's constructible sets, and within that there's an even smaller model containing only sets of rank below the first inaccessible cardinal. In this tiny sub-universe, there are no large cardinals, not even inaccessible ones, let alone measurable or bigger (like supercompact). But $\BB$ is exactly the same as it was in my original, rich universe. I could even cut down further to the minimal transitive model of ZF (or even of weaker theories) without changing $\BB$.

To change $\BB$, you need to move to non-standard models of arithmetic. So there will be infinite natural numbers in that new world. Then of course the domain of $BB$ will change (since it'll include all the integers) but, more importantly, the value $\BB(n)$ could change even for some standard $n$. Such a change could, however, only be of one sort: The new value of $\BB(n)$ is a non-standard integer. (Some Turing machine computation that never halted in the standard world now halts after a non-standard number of steps.)

LSpice
  • 11,423
Andreas Blass
  • 72,290
  • $\DeclareMathOperator\BB{BB}$I don't really understand this answer. Why would the definition of $\BB$ in such a model allow the Turing machine to run for a non-standard number of steps? – user2520938 Jan 11 '21 at 21:55
  • 4
    @user2520938 The definition of BB allows Turing machines to run for any natural number of steps. In worlds with nonstandard natural numbers, they're included in "any natural number". – Andreas Blass Jan 11 '21 at 22:05
  • Hm I see. I'm not sure that your answer answers the question though. The question is not about changing the values of BB, but about making it possible to prove the exact values of BB. – user2520938 Jan 11 '21 at 22:08
  • @user2520938 True, but the question also mentions that "in a way, this “real world” value changes depending on our philosophical stance." – Andreas Blass Jan 11 '21 at 22:13
  • Yes, but doesn’t this rely on a platonist philosophical stance that BB(n) is well-defined? At the very least, an ultrafinitist might object to the well-definedness of BB(5), for example. – exfret Jan 12 '21 at 00:03
  • In any case, it is true that this is not at the core of my question. In fact, I am asking that if the BB numbers are fixed and well-defined (even if I do make the comment that certain people could object to this), can we use this to justify or refute certain set theoretical axioms? For example, say we have some type of evidence that the Turing machine in the article I linked to doesn’t halt. Then this would also be evidence for the “truth” of Con(ZFC). My question is whether this sort of translation from knowledge of BB can translate into a definite way to determine the actual truth of... – exfret Jan 12 '21 at 00:08
  • ...statements independent of ZFC. – exfret Jan 12 '21 at 00:08
  • One small thing is that if I believe that inaccessible cardinals are inconsistent, then from your point of view I must be working in a non-standard universe, since in your world supercompact cardinals exists so the consistency of ZFC+inaccessible cardinals is a true statement. That has a slew of other philosophical problems, though. It's not solving or significantly subverting the issue in your answer, of course. – Asaf Karagila Jan 12 '21 at 13:40
  • @AsafKaragila Note that I wrote "Suppose I believe in some large cardinal, say supercompact." That "suppose" is important. In fact, I am not at all confident about the existence of supercompact cardinals (though I think they have a reasonable chance to be consistent). My actual beliefs about existence (not just consistency) of large cardinals goes up through various sorts of indescribables but has not reached subtle cardinals. Nevertheless, your conclusion remains correct: If you think inaccessibles are inconsistent [continued in next comment] – Andreas Blass Jan 12 '21 at 19:12
  • then, from my point of view, you are working in a nonstandard universe. Specifically, any proof of that inconsistency must, from my point of view, be nonstandardly long. Therefore, if you manage to show me a standard proof that inaccessibles don't exist, then I'll have to change my point of view (but that's likely to be the least of my problems). – Andreas Blass Jan 12 '21 at 19:14
  • 1
    @Andreas: I think that other than the heartbreak, most of the problems belong to the people of my generation and those that follow. Especially those without a permanent job... :-P – Asaf Karagila Jan 12 '21 at 19:16
8

Knowing the values of the Busy Beaver function is the same as knowing the truth values of $\Pi^0_1$-statements (ie statements of the form $\forall n \in \mathbb{N} \ P(n)$ for decidable properties $P$). Knowing a particular $\mathrm{BB}(m)$ means knowing all $\Pi^0_1$-statements of complexity $m$ (for a suitable complexity measure).

For any c.e. theory $T$, $\mathrm{Con}(T)$ is a $\Pi^0_1$-statement. Moreover, if $T$ is of the form "$\mathrm{ZFC}$ + some short axiom", the complexity of $\mathrm{Con}(T)$ (in the sense above) will not be much more than the complexity of $\mathrm{Con}(\mathrm{ZFC})$. Hence if we want to settle values of the Busy-Beaver function on "medium-sized" inputs, we would need to add axioms that settle a lot of consistency statements.

Proving that adding an axiom actually increases the range of provable Busy-Beaver values will be non-trivial. It does not suffice to show that it settles more $\Pi^0_1$-statements, but one needs to show that actually increases the least complexity of an undetermined $\Pi^0_1$-statement. Unfortunately, the complexity notion here is essentially Kolmogorov complexity -- which means that we cannot compute the complexity of a logical formula from that formula.

Arno
  • 4,471