Comment by ChadNauseam
8 months ago
The number itself is not independent of ZFC. (Every integer can be expressed in ZFC.) What's independent of ZFC is the process of computing BB(748).
8 months ago
The number itself is not independent of ZFC. (Every integer can be expressed in ZFC.) What's independent of ZFC is the process of computing BB(748).
I think the more correct statement is that there are different models of ZFC in which BB(748) are different numbers. People find that weird because they don't think about non-standard models, as arguably they shouldn't.
How is that possible? That implies there’s at least one specific program whose execution changes based on the ZFC model. The rules of program execution are so simple, it doesn’t make sense that they’d change based on anything like that.
Because what it means to "halt in finite time" has different meanings in different models, because time is measured with different numbers.
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Isn't that incompatible with the models being consistent?
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.
The issue is that X and Y are not actual natural numbers. They are mathematical objects that satisfy all the ZFC axioms and Peano arithmetic but are infinitely large. The issue is that ZFC underspecifies natural numbers.
Sure, if someone just gives you the number, ZFC can represent it. But ZFC cannot prove that the value is correct, so how do you know you have the right number? Use a stronger proof system? Go a bit bigger and same issue.
Not an expert, but I've read about this a bit because it bothered me also and I think this is the answer:
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?)
And also, if BB were computable, then it could be used to solve the halting problem: run the Turing machine of size n for BB(n) steps, and if it hasn't halted yet, it never will. So the BB function is clearly not computable.
But to me as a layman that seems true regardless of formal axioms chosen, but I guess I need to read that linked thesis.
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I am also not an expert, but this does not sound right to me. Godel's incompleteness theorem shows that there are certain things that cannot be proven. Being independent of ZFC means that something is such a case. So BB(643) being independent of ZFC means that we cannot prove or disprove that a certain number is BB(643). Aka we don't have the math to know for certain.
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> 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.
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.
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It likely comes from the smallest machine that someone has been able to construct that can diagonalize over all proofs in ZFC, or something similar.
It has to come from a finite value (specifically, the amount of complexity that can be enocoded in 745 pieces of information https://turingmachinesimulator.com/shared/vgimygpuwi), because the finite size 745 with infinite tape leads to uncomputability, but the size 5 does not.
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.
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We need to distinguish between a computer that's equivalent to BB(n), and a computer big enough to compute the value of the number that is BB(n). By (terrible) analogy: a 4004 can be made to write a finite loop that describes how many FLOPs the number 1 supercomputer can compute without, itself, being able to usefully perform the computations of that supercomputer. (The 4004 will run out of memory/addressable disk space.) Similarly, we can no longer build decidable programs in ZFC that can compute the number BB(748). Scott is saying that they now think this "disassociation" might occur at BB(7)!
Nobody can give you that number, because it's way bigger than what can be represented in the universe.
To try and help people digging into this, the following helped me.
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
Looking at [3], they seem to argue that the system isn’t complete for the usual Gödel reasons, which, sure, it isn’t, but then they call the claim that the system fails to decide, which is a statement about probability, a “scientific fact”. This seems to me like a mistake?
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.
It looks like the authors of [3] misunderstood Chaitin. What Chaitin said about the limits of provability is that no statements of the form "K(x) > c_F" can be proven in formal system F where c_F is some constant depending on F.
I will admit that I added that cite mostly because of the very real barriers to even learning RUSS.
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?
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