Comment by wat10000
8 months ago
My argument has nothing to do with the universe. My argument is that there is a single definition of the BB function and its definition does not allow for different values in different circumstances.
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?
> Or can I say that ZFC’’ says integers only go up to 100 and therefore BB(4) is 100 in that model?
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'.)
> An axiomatic system can be consistent, but wrong.
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.
Am I understanding you correctly that there’s is one specific finite integer which equals BB(748), but that some models of ZFC will say it’s a different one, and it’s just not correct?
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?
In all models where BB(748) = F and F is actually finite, then F will be the same in all such models. There can't be two models that disagree about the value of F for some actual natural number. It's only in models where BB(748) = Q where Q != F then Q is necessarily not actually finite and hence not an actual natural number.
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.
17 replies →