4 ms·

> AFAIK, there's no known hard, theoretical reason why the above laid out future history cannot take place. Probably the biggest issue is that they'd have no m

> AFAIK, there's no known hard, theoretical reason why the above laid out future history cannot take place. Probably the biggest issue is that they'd have no method to establish that Aleph* is consistent. To continue this BB chain indefinitely, you must invent further and further first-order theories, each of which might not be consistent, let alone Σ₁-sound. And with an Σ₁-unsound theory, any halting proof might not hold up in the standard model of the integers. You'd effectively have to postulate an indefinite amount of oracular knowledge. Also, another physical issue: you can show that within any consistent, recursively axiomatizable theory, the length of the shortest proof of "the longest-running n-state machine is M" must grow at an uncomputable rate in terms of n. Ditto for the shortest proof of "machine M halts", where M is factually the longest-running n-state machine. Otherwise, a machine could use a computable bound on the proof length to solve the halting problem. Therefore, the proof should very quickly become too large to fit within our light cone. In any case, the BB-related evidence for that position rested on BB(5) being determinable by extending the techniques used for BB(4). But in fact, it turns out that similar extensions don't even get you to BB(6). So there isn't anything to support the position, other than the pure speculation that anything is physically achievable given enough time.

srcreigh 2y ago
Thanks for sharing interesting info! How do we know that there would be consistency issues or Σ₁-soundness issues? Your claim about proof size categorizing n-state machine halting status is new to me. Do you have any links to read more about this? The argument doesn't make sense to me. Rather it seems like more of a consequence of BB being incomputable in the first place. The proof sizes for each BB(n) aren't expected to be computable at all. There is necessarily a different theory for each n (or intervals of n where each theory applies with limits on each). > So there isn't anything to support the position, other than the pure speculation that anything is physically achievable given enough time. Something something burden of proof something. It would be extremely fascinating to have a conclusive argument that large BB numbers cannot be solved.
- LegionMammal978 2y ago
  > How do we know that there would be consistency issues or Σ₁-soundness issues? From Gödel's second incompleteness theorem, no consistent first-order recursively-axiomatizable theory (i.e., a theory that can have its proofs validated by a Turing machine) can prove its own consistency. Thus, to prove that your current theory (e.g., ZFC) is consistent, you must move to a stronger one (e.g., Aleph*). But then you can't prove the consistency of that without an even stronger theory, and so on. Thus, you end up with an infinite regression, and you can't ultimately prove the consistency of any of these theories. > Your claim about proof size categorizing n-state machine halting status is new to me. Do you have any links to read more about this? Not really, other than some of my own ramblings on the bbchallenge Discord server. But it's not that long: Suppose that the longest-running n-state machine M can always be proven to halt using a proof of under f(n) symbols, where f(n) is some fixed computable function. Then, you could construct a TM that, given n, enumerates every valid proof of length less than f(n) symbols, and checks whether it shows that a particular n-state machine halts. The TM then simulates all of the proven halters to see how long they run. By assumption, the longest-running machine M is included in this list. So this TM can compute BB(n), which is an impossibility. Therefore, the function f(n) cannot exist. As a corollary, since it's "uncomputably difficult" to prove that the longest-running machine halts at all, it's no less difficult to fully establish the value of BB(n).
  - srcreigh 2y ago
    > Thus, you end up with an infinite regression, and you can't ultimately prove the consistency of any of these theories. There is similar issue with even ZFC and PA. It’s not really a dealbreaker imo. > Suppose that the longest-running n-state machine M can always be proven to halt using a proof of under f(n) symbols, where f(n) is some fixed computable function. Then, you could construct a TM that, given n, enumerates every valid proof of length less than f(n) symbols The issue with that argument is that the TM which enumerates every valid proof can’t exist in the first place. If you fix an axiomatic theory, it’s already known that the theory has a limit.[1] If every theory has a limit, you need countably infinitely many axiomatic theories together to prove BB(n) for all n. So there’s no TM which can even enumerate all the proofs, since a TM must have finite states, and thus can’t enumerate infinitely many proof systems. (In fact for similar reasons I believe a Halt program, which has infinite states but which works for all TMs with finite states, platonically exists. It’s an emulator and an infinitely long busy beaver number lookup table. The diagonalization argument doesn’t apply, since the infinite Halt doesn’t accept itself as input. This Halt would have countably many states since each busy beaver number is finitely calculated and there’s only countably many of them.) So it’s not clear that f(n) is uncomputable. If f(n) is the symbol count and not the binary encoded length of the symbols, it even seems that it’s trivially bounded by some constant for all n. The proof could be one symbol the meaning of which is encoded in the theory. It is a fascinating question though. I’m sure there is some function of axiomatic theory proof checker TM size and binary encoded proof length which does grow with n. It’s unclear if it would be uncomputable though. The consequence of it being uncomputable is that the universe doesn’t have the resources to even encode the theory and/or represent it’s proofs. In fact I suppose even as long as it grows at all, there would be a limit to BB(n) which can be possibly determined. Very fascinating [1]: page 5 https://www.scottaaronson.com/papers/bb.pdf https://www.scottaaronson.com/papers/bb.pdf

srcreigh 2y ago

Thanks for sharing interesting info! How do we know that there would be consistency issues or Σ₁-soundness issues? Your claim about proof size categorizing n-state machine halting status is new to me. Do you have any links to read more about this? The argument doesn't make sense to me. Rather it seems like more of a consequence of BB being incomputable in the first place. The proof sizes for each BB(n) aren't expected to be computable at all. There is necessarily a different theory for each n (or intervals of n where each theory applies with limits on each). > So there isn't anything to support the position, other than the pure speculation that anything is physically achievable given enough time. Something something burden of proof something. It would be extremely fascinating to have a conclusive argument that large BB numbers cannot be solved.

LegionMammal978 2y ago
> How do we know that there would be consistency issues or Σ₁-soundness issues? From Gödel's second incompleteness theorem, no consistent first-order recursively-axiomatizable theory (i.e., a theory that can have its proofs validated by a Turing machine) can prove its own consistency. Thus, to prove that your current theory (e.g., ZFC) is consistent, you must move to a stronger one (e.g., Aleph*). But then you can't prove the consistency of that without an even stronger theory, and so on. Thus, you end up with an infinite regression, and you can't ultimately prove the consistency of any of these theories. > Your claim about proof size categorizing n-state machine halting status is new to me. Do you have any links to read more about this? Not really, other than some of my own ramblings on the bbchallenge Discord server. But it's not that long: Suppose that the longest-running n-state machine M can always be proven to halt using a proof of under f(n) symbols, where f(n) is some fixed computable function. Then, you could construct a TM that, given n, enumerates every valid proof of length less than f(n) symbols, and checks whether it shows that a particular n-state machine halts. The TM then simulates all of the proven halters to see how long they run. By assumption, the longest-running machine M is included in this list. So this TM can compute BB(n), which is an impossibility. Therefore, the function f(n) cannot exist. As a corollary, since it's "uncomputably difficult" to prove that the longest-running machine halts at all, it's no less difficult to fully establish the value of BB(n).
- srcreigh 2y ago
  > Thus, you end up with an infinite regression, and you can't ultimately prove the consistency of any of these theories. There is similar issue with even ZFC and PA. It’s not really a dealbreaker imo. > Suppose that the longest-running n-state machine M can always be proven to halt using a proof of under f(n) symbols, where f(n) is some fixed computable function. Then, you could construct a TM that, given n, enumerates every valid proof of length less than f(n) symbols The issue with that argument is that the TM which enumerates every valid proof can’t exist in the first place. If you fix an axiomatic theory, it’s already known that the theory has a limit.[1] If every theory has a limit, you need countably infinitely many axiomatic theories together to prove BB(n) for all n. So there’s no TM which can even enumerate all the proofs, since a TM must have finite states, and thus can’t enumerate infinitely many proof systems. (In fact for similar reasons I believe a Halt program, which has infinite states but which works for all TMs with finite states, platonically exists. It’s an emulator and an infinitely long busy beaver number lookup table. The diagonalization argument doesn’t apply, since the infinite Halt doesn’t accept itself as input. This Halt would have countably many states since each busy beaver number is finitely calculated and there’s only countably many of them.) So it’s not clear that f(n) is uncomputable. If f(n) is the symbol count and not the binary encoded length of the symbols, it even seems that it’s trivially bounded by some constant for all n. The proof could be one symbol the meaning of which is encoded in the theory. It is a fascinating question though. I’m sure there is some function of axiomatic theory proof checker TM size and binary encoded proof length which does grow with n. It’s unclear if it would be uncomputable though. The consequence of it being uncomputable is that the universe doesn’t have the resources to even encode the theory and/or represent it’s proofs. In fact I suppose even as long as it grows at all, there would be a limit to BB(n) which can be possibly determined. Very fascinating [1]: page 5 https://www.scottaaronson.com/papers/bb.pdf https://www.scottaaronson.com/papers/bb.pdf