It boggles my mind that a number (an uncomputable number, granted) like BB(748) can be "independent of ZFC". It feels like a category error or something.
replies(12):
Most of these 'uncomputable' problems are uncomputable in the sense of the halting problem: you can write down an algorithm that should compute them, but it might never halt. That's the sense in which BB(x) is uncomputable: you won't know if you're done ever, because you can't distinguish a machine that never halts from one that just hasn't halted yet (since it has an infinite number of states, you can't just wait for a loop).
So presumably the independence of a number from ZFC is like that also: you can't prove it's the value of BB(745) because you won't know if you've proved it; the only way to prove it is essentially to run those Turing machines until they stop and you'll never know if you're done.
I'm guessing that for the very small Turing machines there is not enough structure possible to encode whatever infinitely complex states end up being impossible to deduce halting from, so they end up being Collatz-like and then you can go prove things about them using math. As you add states the possible iteration steps go wild and eventually do stuff that is beyond ZFC to analyze.
So the finite value 745 isn't really where the infinity/uncomputability comes from-it comes from the infinite tape that can produce arbitrarily complex functions. (I wonder if over a certain number of states it becomes possible to encoding a larger Turing machine in the tape somehow, causing a sort of divergence to infinite complexity?)
In a very real sense, a deep kind of infinite complexity can be generated from 745 objects of certain kind, but not from 5 objects of that kind..
Turing machines have infinite tape, not infinite state. The entire set of all halting machines of a given size collectively only use finite tape. Totally finite. Only (some of) the non-halting machines use infinite tape.
The problem is that we don't know in advance how large the (definitely finite) upper bound on the amount of tape all the size-N halting machines use, until after enough of them (one per known equivalence class) halt. And we don't know (in general) how to run all the halting ones until they halt, without also running a non-halting program for an unbounded amount of time.
TL:DR: unbounded is not infinite, but big enough to be a problem.
But the overall 'complexity' (at a timestep, say) is going to be due to the states and the tape together. The BB(5) example that was analyzed, iirc, was a Collatz-like problem (Aaronson describes it here: https://scottaaronson.blog/?p=8088 ). My interpretation of this is that:
1. collatz-like functions have a lot of complexity just due to math alone 2. 5 states turned out to be enough to "reach" that one that 3. more states means you're going to reach more possible Collatz-like functions (they don't have to be Collatz-like; it's just easier to think about them like that) 4. eventually you reach ones that ZFC cannot show to halt, because there is effectively no way to prove it other than running them, and then you would have to solve the halting problem.
The part that was helpful for me to be less unsettle by BB(745) being independent of the ZFC was the notion that it eventually boils down to a halting problem, and asking ZFC to "solve" it... which is more agreeable than the idea that "ZFC cannot compute a function that seems to be solvable by brute force".
But to me as a layman that seems true regardless of formal axioms chosen, but I guess I need to read that linked thesis.
Incidentally, Gödel's theorem eventually comes down to a halting-like argument as well (well, a diagonal argument). There is a presentation of it that is in like less than one page in terms of the halting problem---all of the Gödel-numbering stuff is essentially an antiquated proof. I remember seeing this in a great paper which I can't find now, but it's also mentioned as an aside in this blog post: https://scottaaronson.blog/?p=710
wait jk I found it: https://arxiv.org/abs/1909.04569
Two lenses for trying to understand this are potentially Chastain's limits on output of a lisp program being more complex than the program itself [1] or Markov's proof that you can't classify manifolds in d>= 4.
If you try the latter and need/want to figure out how the Russian school is so different this is helpful [2]
IMHO the former gives an intuition why, and the latter explains why IMHO.
In ZFC, C actually ends up implying PEM, which is why using constructionism as a form of reverse math helped it click for me .
This is because in the presence of excluded middle, every sequentially complete metric space is a complete space, and we tend to care about useful things, but for me just how huge the search space grows was hidden due to the typical (and useful) a priori assumption of PEM.
If you have a (in my view) dislike for the constrictive approach or don't want/have to invest in learning an obscure school of it, This recent paper[3] on the limits for finding a quantum theory of everything is another lens.
Yet another path is through Type 2 TMs and the Borel hierarchy, where while you can have a uncomputable number on the input tape you algorithms themselves cannot use them, while you can produce uncomputable numbers by randomly selecting and/or changing an infinite sequence.
Really it is the difference between expressability and algorithms working within what you can express.
Hopefully someone else can provide more accessible resources. I think a partial understanding of the limits of algorithms and computation will become more important in this new era.
[1] https://arxiv.org/abs/chao-dyn/9407003 [2] https://arxiv.org/abs/1804.05495 [3] https://arxiv.org/abs/2505.11773
Usually, "what happens" is that the machines become large enough to represent a form of induction too strong for the axioms to 'reason' about. It's a function of the axioms of your theory, and you can add more axioms to stave it off, but of course you can't prove that your new axioms are consistent without even more axioms.
> There is a presentation of it that is in like less than one page in terms of the halting problem---all of the Gödel-numbering stuff is essentially an antiquated proof.
Only insofar as you can put faith into the Church–Turing thesis to sort out all the technicalities of enumerating and verifying proofs. There still must be an encoding, just not the usual Gödel numbering.
> There is a presentation of it that is in like less than one page in terms of the halting problem
Those are two very different ideas. Your second sentence says that Gödel's theorem is easy to prove if you have results about the halting problem. Your first one says that in order to prove Gödel's theorem, you need to establish results about the halting problem.
Like, a TOE is not expected to decide all statements expressible in the theory, only to predict particular future states from past states, with as much specificity as such past states actually determine the future states. It should not be expected to answer “given a physical setup where a Turing machine has been built, is there a time at which it halts?” but rather to answer “after N seconds, what state is the machine (as part of the physical system) in?” (for any particular choice of N).
Whether a particular statement expressed in the language of the theory is provable in the theory, is not a claim about the finite-time behavior of a physical system, unless your model of physics involves like, oracle machines or something like that.
Edit: it later says: “ Chaitin’s theorem states that there exists a constant K_{ℱ_{QG}} , determined by the axioms of ℱ_{QG} , such that no statement S with Kolmogorov complexity K(S) > K_{ℱ_{QG}} can be proven within ℱ_{QG} .”
But this, unless I’m badly misinterpreting it, seems very wrong? Most formal systems of interest have infinitely many distinct theorems. Given an infinite set of strings, there is no finite universal upper bound on the Kolmogorov complexity of the strings in that set.
Maybe this was just a typo or something?
They do then mention something about the Bekenstein bound, which I haven’t considered carefully yet but seems somewhat more promising than the parts of the article that preceded it.
By the typos etc.. you. can probably also tell I was doing this on mobile, unfortunately as a passenger in a car.
To quote Chaitin’s explanation here:
> In contrast I would like to measure the power of a set of axioms and rules of inference. I would like to be able to say that if one has ten pounds of axioms and a twenty-pound theorem, then that theorem cannot be derived from those axioms.
This paper's notation does seem to be confusing, but I still think it is essentially complete with the above.
"K_{ℱ_{QG}}" would probably most commonly be L in most descriptions, a natural number that is the upper bound of complexity for provable statements in a formal system S
L is not a limit on complexity, it means that there is no formal proof for S that its Kolmogorov complexity exceeds L, for any string.
You can still prove that there are strings far more complex than L with S, and in fact there will often be far more of those strings than the ones equal to or less than L.
It is a limit on what you can prove about those strings with a greaterKolmogorov complexity in S.
Or to rewrite the above:
"There exists a natural number L such that we can't prove the Kolmogorov complexity of any specific string of bits is more than L."
Does that help or did I miss the mark on your objection?
ZF & ZFC are as important as they are because they're the weakest set theories capable of working as the foundations of mathematics that we've found. We can always add axioms, but taking axioms away & still having a usable theory on which to base mathematics is much more difficult.
Suppose model A proves BB(748) = X and model B proves BB(748) = Y > X. But presumably the models can interpret running all size 748 Turing machines for Y steps. Either one of the machines halts at step Y (forming a proof within A that BB(748) >= Y contradicting the assumed proof within A that BB(748) = X < Y) or none of the machines halts at step Y (forming a proof within B that BB(748) != Y contradicting the assumed proof within B that BB(748) = Y).
I'm guessing the only way this could ever work would be some kind of nastiness like X and Y aren't nailed down integers, so you can't tell if you've reached them or not, and somehow also there's a proof they aren't equal.
> So presumably the independence of a number from ZFC is like that also: you can't prove it's the value of BB(745) because you won't know if you've proved it; the only way to prove it is essentially to run those Turing machines until they stop and you'll never know if you're done.
These aren't similar ideas. You can't know if a machine that hasn't halted yet will ever halt. But you can easily know if a machine that has already halted was going to halt.
Independence is the second case. For the value of BB(x) to be independent of ZFC, one of two things must hold:
(1) ZFC is inconsistent, and therefore all statements are independent of it.
(2) ZFC is consistent with two different statements, "BB(x) = a" and "BB(x) = b" for two different a, b. This means that a disproof of either statement cannot exist.
This, in turn, means that there is no observation you could ever make that would distinguish between the values a and b (for the identity of BB(x)). No matter what you believe the value of BB(x) might secretly be, there are no consequences; nothing anywhere could ever change if the value turned out to be different. Because, if there were an observable consequence of the value being different, the hypothetical observation of that consequence would be a disproof of the value that didn't cause it, and no such disproof can exist.
Neither value, a or b, can be more true than the other as the answer to the question "what is BB(x)?". It doesn't make sense to consider that question to have an answer at all.
Running for fewer steps is extremely well defined and I don’t imagine that enters into this.
That means there’s issue is “never stop”? That also seems pretty well defined to me. For BB(748) to vary based on your model, if the machines that run for fewer steps don’t change, then that means one of the machines that never stops in one model will stop in another. Or the BB winner for our model will never stop in another model.
How can changing your model make it so a specific Turing machine goes from stopping after 10,000 steps to never stopping, or from never stopping to stopping after 11,000 steps?
So in one model a Turing Machine called R never stops. In another model R stops after Q steps. But here's the issue... Q isn't an actual natural number, what it is is some mathematical object that satisfies all of the properties of a natural number in ZFC, but is not an actual natural number. What it actually is is some infinitely large object that satisfies all of the Peano axioms of what a natural number is as well as satisfies the following set of rules:
Q > 0
Q > 1
Q > 2
Q > 3
...
So within this model, the Turing machine R halts after Q steps, and since from within the model Q is finite then from within this model BB(748) is at least equal to Q.
If BB(748) is actually 10,000, then we can add this as an axiom to ZFC to get a new formal theory ZFC + "BB(748) = 10000".
In this new theory the previous structure that contained Q as an element will not satisfy the definition of a natural number, so we don't have to worry about Q anymore... however, there will exist some number T > 748 where BB(T) is independent of our new theory. For BB(T), there will exist some other model that has its own Q* which satisfies all of our axioms including the axiom that BB(748) = 10000, but also that
Q* > 0
Q* > 1
Q* > 2
Q* > 3
...
If there’s another model where this machine doesn’t stop, then that means that at some point during this process, you reach a particular machine state and tape contents and transition to a different state than you did in the first model. That has to happen, because otherwise the execution follows the same process as before, and halts at Q steps. But the mechanics of the machine don’t depend on your theory. They’re just state transitions and tape operations.
Q isn't a natural number because natural numbers must be finite, but Q is infinitely large.
>If you had unlimited time and paper, you could sit down and run the machine by hand, counting each step, until it reaches the halting state. You will have counted Q steps.
What if the machine never stops? How many steps will you run before you decide that the machine never halts?
>There’s no such thing as a machine that stops after a number of steps defined by an infinitely large construct.
There's no such thing as an actual machine that stops after an infinite number of steps, but that's not the issue. The issue is that ZFC has different models with conflicting definitions of what infinite is. In one model there is an object called Q that satisfies all of the properties in ZFC of being a natural number, but is infinitely large. In this model the Turing Machine halts after Q steps. But there is another model, called the standard model, and in this model there is no Q, all elements of this model are actually finite, and in this model the Turing machine never halts.
ZFC doesn't know which of these two models is the "real" model of natural numbers. From within ZFC both of these models satisfy all properties of natural numbers. It's only from outside of ZFC that one of these models is wrong, namely the model that contains Q as an element.
You can add more axioms to ZFC to get rid of the model that has Q as an element, but if the resulting theory containing your new axiom is consistent, then it necessarily follows that there is some other model that will contain some element Q* which is also infinitely large but from within the theory satisfies all of the new/stronger properties of being a natural number.
That doesn’t make any sense. A Turing machine can’t halt after a infinite number of steps. It either halts after a finite number of steps, or it never halts.
I’m sure there are models of hypercomputation and corresponding “what’s the largest number of steps they can run?” functions that would admit infinities, but those would not be Turing machines and the function would not be the Busy Beaver.
What the commenter above you said doesn't make sense in our daily life, but it makes perfect sense when in comes to non-standard models.
You got confused because you're thinking natural numbers as something we can count in real physical world, which is a perfectly sane mental model, and that is why there was a comment above said:
> People find that weird because they don't think about non-standard models, as arguably they shouldn't.
Q is not a number you can actually count, so it doesn't fit into our intuition of natural number. The point is not that Q exists in some physical sense in real life, like "3" in "3 apples" (it doesn't). The point is that ZF itself isn't strong enough to prevent you from defining random shit like Q as a natural number.
Ultrafinitism? If you'd run the Turing machine that performs BB(748) steps in a physical universe that admits it, you'd get a physical representation of BB(748). If you have a competing theory about which Turing machine computes BB(748), you can run them both alongside in this universe and see with your own eyes which one finishes first.
I guess from ultrafinitist's point of view such universe has different mathematics, but isn't it a fringe viewpoint in mathematical circles?
Sure, you can come up with a set of axioms where the natural numbers include infinities. You may be able to use it to prove interesting things. But all that does here it make it so that the set of numbers describing how many steps a Turing machine runs before it stops is no longer the “natural numbers.”
I'm not sure what flavor of ultrafinitism you're referring here. If it's the "very big numbers, like TREE(3), are not natural numbers as they are far bigger than the number of atoms in this universe..." kind, then it has nothing to do with what this is about.
> physical representation
> your own eyes
Non standard models of ZFC have nothing to do with our physical world. That's why no physicist or engineer cares about them (or cares about axiom systems at all). So we need to be very careful when connecting the idea of physical, running "stuff" to the discussion of ZFC.
Anyway, back to
> you can run them both alongside in this universe and see which one finishes first
There are two Turing Machines, Foo and Bar. We build and run them in our physical universe. Foo halts at the standard BB(748) steps. Bar just keeps running and running. That's what we will see with our own eyes.
The issue is that when we try to reason out whether Bar will ultimately halts, ZFC doesn't prevent us from defining a non-standard model where Bar halts after a non-standard number of steps. Note that the physical Bar will not halt in our universe. The "non-standard number of steps" is as nonsense as it sounds. It's just that ZFC doesn't prevent us from defining such a nonsense. The point of ZFC is it's compatible with almost all the useful, sane math. It's not necessarily incompatible with bullshit and insane math.
That is it. The fact that Bar is still keeping running in our universe is completely irrelevant.
But it does prevent you from defining a non-standard model where Bar halts after a finite number of steps. Since BB is finite by definition, the non-standard number of steps after which Bar halts cannot be BB(748).
I’m pretty sure you and the other commenter have this mixed up. The fact that BB(748) is independent of ZFC doesn’t mean there are different models that have different values of BB(748). It means that ZFC is insufficient to determine the value of BB(748). That value is still some finite integer, you just can’t prove which one it is. Equivalently, there is some 748-state Turing machine which never halts but ZFC cannot prove never halts.
And no, you can’t change your model such that this Turing machine halts in some non-standard number of steps. Or rather, you can, but that doesn’t actually change anything. The machine still doesn’t halt for the purposes of defining BB(748).
> This, in turn, means that there is no observation you could ever make that would distinguish between the values a and b (for the identity of BB(x)). No matter what you believe the value of BB(x) might secretly be, there are no consequences; nothing anywhere could ever change if the value turned out to be different. Because, if there were an observable consequence of the value being different, the hypothetical observation of that consequence would be a disproof of the value that didn't cause it, and no such disproof can exist.
There's one part of this I don't understand. "BB(x) = n" means "there is at least one x-state Turing machine that halts after exactly n steps, and there are no x-state Turing machines that halt after more than n steps", right? Then why wouldn't this approach work (other than the numbers being way too big to actually do in this universe)? WLOG, assume a < b. Run all possible x-state Turing machines for b steps. If any halted on step b, then you've disproved "BB(x) = a". If not, then you've disproved "BB(x) = b".
We really don't.
> that BB(748) is independent of ZFC
> there are different models that have different values of BB(748)
> ZFC is insufficient to determine the value of BB(748)
These three statements are equivalent.
f(n)=X is independent of ZFC means there are different models of ZFC that have different values of f(n). It's a very trivial theorem[0]. If you don't like it, I can't convince you otherwise.
> that doesn’t actually change anything
Changing the model will not change how any machine works in our physical, mechanical universe. However, it does change the value of BB(748).
I understand your line of thinking: There is only one mechanical universe, which is the one where we exist. We can build Turing machines in this universe. BB(n) depends on Turning machines. Since there is only one single universe, there is only one single value of BB(n).
It's a perfectly fine mental model for most cases. This was exactly how I thought when the first time I heard about BB(n). But it's not the kind of math than Scott Aaronson et al. are doing.
Bar keeps running in our mechanical universe. But it can also halt in some non-standard number of steps. This weird, absurd-sounding proposition works because non-standard numbers simply don't map to anything in mechanical universe. They're purely abstract objects living in ZFC+~Con(ZFC).
[0]: Given f(n)=X is independent of ZFC. Which means f(n)=X and ~(f(n)=X) are both consistent relative to ZFC. Therefore, if there is any model of ZFC, there is a model M1 that entails ZFC+(f(n)=X), and a model M2 that entails ZFC+~(f(n)=X). The value of f(n) cannot be the same in M1 and M2.
That is we can add a nonsensical axiom and get a consistent nonsensical theory that has nothing to do with actually running Turing machines (no matter in which physical or abstract universe they run). Er, OK, fine I guess.
A universally inapplicable theory.
No. I can't wrap my head around it. Successors for the tape state are defined for the initial segment of a non-standard natural numbers. How the proof of termination would even look like? Something non-constructive that doesn't allow to choose the machine among a finite number of the machines?
What is “a model” here? Can I say that there’s a model ZFC’ which is the same as ZFC except that 107 is considered to be equivalent to 200, and therefore BB(4) in ZFC’ is actually 200? Or can I say that ZFC’’ says integers only go up to 100 and therefore BB(4) is 100 in that model? Or is it something more restricted than that?
You'd be defining a new axiomatic system here, not just a model of ZFC. I don't know how we're going to formalize Turning machine in this system, but if we managed to do it, the value of BB(4) is likely to be indeed 100, at least for some models of this new system.
Roughly speaking, a model of ZFC is a set and a binary relationship over the set, whose members all satisfy every axiom of ZFC. Obviously this super simplified definition does a crazy amount of handwaving.
But we don't need to accept or understand the idea of model. What we need to accept is this simple idea:
An axiomatic system can be consistent, but wrong.
For example, if ZFC is consistent, then T = ZFC+~Con(ZFC) would be consistent as well. But this T is wrong, as it believes ZFC is inconsistent.
Similarly, if ZFC is indeed consistent, then T is wrong about which Turing machines halt. Therefore it would have a wrong value of BB(748) (and many other BB(n)).
However, since ZFC can't prove its own consistency, it can't prove that value is wrong. That's why there are different values of BB(748). Those values are not necessarily equally correct, it's just that ZFC isn't strong enough to prove which one is wrong.
Models, nonstandard natural numbers, etc... are more or less technical details (so mathematicians can avoid scary terms like 'wrong'.)
And since we can find a four-state Turing machine that runs for more than 100 steps before halting, ZFC’’ is just not correct when it says that BB(4) = 100, but we still say that 100 is the value in that model?
From within those models Q satisfies all the properties of being a natural number but it's not actually a natural number. Q is some successor of 0, you can add 1 to Q to get another distinct mathematical object, there is some predecessor to Q called P so that P + 1 = Q, etc etc... Q satisfies all the properties within ZFC of being a natural number but it isn't an actual natural number.
Furthermore if ZFC is consistent then it's impossible for any model of ZFC to have BB(4) = 100. ZFC is sufficiently powerful to prove that BB(4) != 100, it is not sufficiently powerful enough to prove that BB(748) = F for some actual natural number F.
Yes, between you and me we know that BB(n) needs to be a natural number, but we have no way to formally and uniquely define what natural numbers are. The best we can do is come up with a formal definition of natural numbers that includes the actual natural numbers but will also include other number systems that contain mathematical objects that are infinitely big and hence are not actual natural numbers. Hence our formal definition of natural numbers will not uniquely define a single set of numbers {0, 1, 2, 3, ...}, there will be other sets of numbers such as {0, 1, 2, 3, ..., Q - 1, Q, Q + 1, ...} for some infinitely large object Q that satisfy the formal definition of natural numbers. Between you and me, we both know Q isn't an actual natural number, but what we don't know is what formal rule we need to add to our formal definition of natural numbers in order to get rid of Q. In fact, it's worse than that; we know that even if we add a rule that gets rid of Q, there will always be some other number system {0, 1, 2, 3, ..., Q* - 1, Q*, Q* + 1, ...} to take its place. No formal definition can ever uniquely define the natural numbers (unless that formal system is inconsistent).
It's also true that a model where BB(748) = Q is not the actual BB, it's some other function. The problem is that this model satisfies all of the rules of ZFC and all of the properties that ZFC says natural numbers must satisfy and hence it satisfies the formal definition of BB even though it isn't the actual BB. Remember it is impossible for the formal definition of BB to include the property that BB(n) = an actual natural number, because there is no formal definition that uniquely singles out what the actual natural numbers are. Since there exists one such model that satisfies all of the rules about BB but isn't actually the real BB then we can't use ZFC to formally prove what the value of BB(748) actually is.
What we can do is add new rules to ZFC get rid of this model, but that will only push the issue down to BB(749) or BB(750) or maybe if pick a really powerful rule we push the issue down to BB(800)... but the point stands that adding new rules only pushes the problem further down the road, it never eliminates the problem entirely.
The issue is that what they said seemingly was not
"There exists a natural number L such that we can't prove the Kolmogorov complexity of any specific string of bits is more than L."
But
"There exists a natural number L such that we can't prove (in ℱ_{QG}) any statement S whose complexity is more than L.",
which is wrong.
They later go on to say “These strings cannot be generated by programs of length <= n, and hence cannot correspond to provable statements in ℱ_{QG}.” which follows from the previous wrong statement but doesn’t follow from the accurate statement you gave, which seems to suggest that they really did mean the inaccurate statement that they wrote, not the correct one you wrote.
How could such a definition give rise to such a set with Q in it?
What it means specifically? ZFC+~(BB(748)=N) allows to extend definition of the Turing machine to a non-standard number of steps?
Can "BB(748) is undefined" be a provable theorem in ZFC+~(BB(748)=N) instead?
While we know that ZFC+~(BB(748)=N) is consistent, we don't know whether \exist Q!=N where ZFC+(BB(748)=Q) is consistent.
Intuitively I see it as: by adding axiom ~(BB(748)=N), we codify that this formal system isn't powerful enough to describe a Turing machine with 748 states (and probably all the machines with number of states greater than 748).
What you described doesn't rule out {0,1,2... Q-1,Q,Q+1...}, because you only defined how to yield new natural number, but not exclude things that are not yielded that way from N (the set of all natural numbers).
Now, our intuition is to add this missing part into our axioms, right?
So instead:
> 0 is a natural number, and if n is a natural number then so is S(n)
We say:
> For any X⊆N, if 0 ∈ X, and for every n ∈ X, S(n) ∈ X, then X=N.
This is a perfect valid axiom. And it does rule out the nonstandard shit: for a set N' that looks like {0,1,2... Q-1,Q,Q+1...}, we can get X = {0,1,2...}, which is a subset of N'. According to this axiom, if N'=N then X=N', but it clearly doesn't because Q∈X while ~(Q∈N'). Therefore, N' isn't N.
However, this axiom is not included in the commonly accepted Peano Arithmetic! The reason is that this uses second-order logic, and Peano Arithmetic is a first-order theory.
The above axiom effectively defines a predicate, f(X), which accepts a set as input and returns whether the set is N. This is second-order logic.
Peano Arithmetic, being first-order logic, doesn't have such predicate. This is why we can't rule out these nonstandard {0,1,2... Q-1,Q,Q+1...}.
When it comes to ZFC, it's more complicate as in ZFC, 'natural numbers' are ordinals of sets. But ZFC is written in first-order logic as well, and it's known that an axiomatic system written in first-order logic will have nonstandard models. Even if you can rule out {0,1,2... Q-1,Q,Q+1...} by defining PA in ZFC in some unusual way or adding new axioms to ZFC, as long as it's still a first-order theory, it will have 0 (if inconsistent) or multiple (if consistent) models[0].
[0]: https://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_...
The interesting bit is they were able to construct a machine that halts if ZFC is consistent. Since a consistent axiomatic system can never prove its own consistency (another famous proof) ZFC can't prove that this machine halts. And ZFC can't prove that it never halts without running it for infinite steps.
That ZFC-consistency-proving machine has 643 states, so BB(643) either halts after the ZFC-consistency-proving machine or the ZFC-consistency-proving machine never halts. If BB(643) halts after the ZFC-consistency-proving machine, then ZFC is consistent and ZFC can't prove BB(643) halts since ZFC can't prove the ZFC-consistency-proving machine halts.
But the other commenter said:
"Yes, between you and me we know that BB(n) needs to be a natural number, but we have no way to formally and uniquely define what natural numbers are. The best we can do is come up with a formal definition of natural numbers that includes the actual natural numbers but will also include other number systems that contain mathematical objects that are infinitely big and hence are not actual natural numbers."
Is there some subtlety that allows both of these statements to be true, or is this just a contradiction? Was the other commenter implicitly assuming "unless you involve second-order logic"?
In first order logic it's impossible to uniquely define any property that would also uniquely define the natural numbers.
What second order logic does let you do is deal with only one single model, called the categorical model. So instead of having a theory that has a whole bunch of different models including the actual natural numbers along with undesirable models that contain infinitely large values... you can force your theory to have one single model, no more "It's true in this model, but it's false in that other nonstandard model that's getting in the way." So yes in a particular theory of second order logic BB(748) has one single value because there is only one single model.
So problem solved right? Not even close... because that one single categorical model being used to represent the natural numbers may not be the actual natural numbers, the intended natural numbers where every number is actually finite. Having one single categorical model does not imply working with the actual model you intended to work with. Depending on your choice of second order theory you may be operating within a theory where the single categorical model is indeed unbeknownst to you {0, 1, 2, 3, ..., Q - 1, Q, Q + 1, ...} and hence BB(748) is equal to Q and you'll have no way of knowing this before hand since you lack an effective proof system.
>For any X⊆N, if 0 ∈ X, and for every n ∈ X, S(n) ∈ X, then X=N.
You're right that this is a perfectly valid axiom in SOL, but it's not true that it rules out non-standard "shit". What this axiom does is it forces your theory to have a single model, the categorical model. But this axiom does not in anyway pick out the standard model as the categorical model, it doesn't force the categorical model to be the standard model. It's possible that N = {0, 1, 2, ..., Q - 1, Q, Q + 1, ...} in which case the categorical model ends up being nonstandard. This axiom has no way to force N to be intended standard model.
But then its unsound, isn't it? Isn't our background assumption that ZFC is consistent and sound? It can't prove its own consistency, but we are assuming that under standard models, it is sound.
> For example, if ZFC is consistent, then T = ZFC+~Con(ZFC) would be consistent as well.
It would be consistent if ZFC didn't also prove ZFC+Con(ZFC), but then it would indeed be unsound.
> Similarly, if ZFC is indeed consistent, then T is wrong about which Turing machines halt. Therefore it would have a wrong value of BB(748) (and many other BB(n)).
No, if it's sound, it just doesn't have a proof of the form "BB(748)=K" for any K.
> However, since ZFC can't prove its own consistency, it can't prove that value is wrong. That's why there are different values of BB(748). Those values are not necessarily equally correct, it's just that ZFC isn't strong enough to prove which one is wrong.
No, ZFC is just not strong enough to prove any of these.
We can come up with some function BB' that admits that, but it's just a different function.
It seems we can't even define a function with standard domain and non-standard codomain while not using literals for non-standard numbers in its definition.
That is ~(BB(748)=ActualValueOfBB748) is false even if it can't be proven in ZFC. In a sense, busy beaver creates its own mathematical reality.
>I think the more correct statement is that there are different models of ZFC in which BB(748) are different numbers.
You asked how this was possible and that's the specific question that I am addressing. As I mentioned elsewhere, to fully appreciate this answer requires parsing some very subtle and nuanced details that simply can not be glossed over or dismissed, if you genuinely want to know how it's possible that ZFC can be consistent even though different models give different values of BB(748).
If you want to argue something else, about what model is correct or what model is incorrect, that's a perfectly fine argument to have but it's more in the realm of philosophy than it is in the realm of formal mathematics.