Comment by dwohnitmok
1 month ago
Your comment to JoJoModding in Math Stackexchange is incorrect.
"That's because there are nonstandard models of PA which contain infinite natural numbers. So PA may be able to prove that it produces a proof, but can't prove that the proof is of finite length. And therefore it might not be a valid proof."
If PA proves "PA proves X" then PA can prove X. This is because the key observation is not that there are nonstandard models, but rather that the standard natural numbers model PA.
Therefore if PA proves "PA proves X", then there is in fact a standard, finite natural number that corresponds to the encoded proof of "PA proves X". That finite natural number can be used then to construct a proof of X in PA.
I haven't had the time to go through your argument in more detail, but it's important to note (because the natural language version of what you've presented is ambiguous) that you haven't shown "PA proves 'Provable(forall n, G(n))'" in which case it would be the case that in fact "PA proves 'forall n, G(n)'", but rather "PA proves 'forall n, Provable(G(n))'".
My logic is very rusty at this point, but if someone could give me an argument that you cannot move the 'Provable' outside the 'forall', I would really appreciate that, without making reference to Goodstein sequences. In other words, that in general for propositions 'P' it is not true that proving "forall n, Provable(P(n))" implies you can prove "Provable(forall n, P(n))".
> If PA proves "PA proves X" then PA can prove X.
Not true.
From PA we can construct a function that can search all possible proofs that can be constructed in PA. In fact I outlined one way to do this at the end of my answer.
With this function, we can construct a function will-return that analyzes whether a given function, with a given input, will return. This is kind of like an attempted solution to the Halting Problem. We know that it doesn't always work. But we also know that it works a lot of the time.
From will-return we can create a function opposite-return that tries to return if a given function with a given input would not, and doesn't return if that function would. This construction is identical to the one in the standard proof of the Halting Problem.
Now we consider (opposite-return opposite-return opposite-return). (Actually you need a step to expand the argument into this recursive form. I've left that out, but that is identical to the one in the standard proof of the Halting problem.)
PA can prove the following:
- PA proves that if PA can prove that opposite-return returns, then it doesn't. - PA proves that if PA can prove that opposite-return doesn't return, then it does. - PA proves that if it can prove everything that it proves that it can prove, then PA must have a proof of one of the two previous statements. - Therefore PA proves that if it can prove everything that it proves that it can prove, then PA is inconsistent.
This is a form of Gödel's second incompleteness theorem.
And, therefore, there must be a distinction to be made between "PA proves" and "PA proves that it proves".
I am not sure I follow and it is possible I may not be on the same wavelength, but here is another thought.
For any sentence S, if first-order PA (or any first-order theory) proves S, then that means that S holds true in every model of that theory, via the completeness (not incompleteness) theorem.
The statement "PA proves x" is equivalent to saying "there exists a natural number N which is a Gödel number encoding a proof of x." The letter "x", here, is a natural number that is assumed to encode some sentence we want to prove, that is, x = #S for some sentence S.
The above is a predicate with one existential quantifier: Provable(x) = there exists N such that IsProof(N, x) holds true, where IsProof says that N is the Gödel number of a valid proof ending in x.
If PA proves "Provable(x)", that means that the predicate "Provable(x)" holds in every model of PA. This means every model of PA would have some natural number N that satisfies IsProof(N, x). Of course, this number can vary from model to model.
However: the standard model, which has only standard natural numbers, is also another model of PA. So if PA proves "Provable(x)," and Provable(x) thus holds in every model, it must also hold in the standard model of PA. This means that there must be a regular, standard, finite natural number N_st that encodes a proof of x.
Since the standard model is an initial segment of every other model of PA, then every other model includes the standard model and all the standard natural numbers. Thus, if PA proves Provable(x), then the standard number N_st must exist in all models.
So we cannot have a situation where PA proves Provable(x) but all proofs of x must be nonstandard and infinite in length. This would mean no such proof exists in the standard model of PA - but then, via completeness, PA would not be able to prove "Provable(x)".
For every n, the function will return a function, that PA proves has a proof in PA.
Suppose that we're in a nonstandard model. For all standard n, there really will be a proof. For nonstandard n, there may or may not actually be a proof.
And so PA cannot prove from the fact that it proved the existence of a proof, that PA actually proves it.
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> > If PA proves "PA proves X" then PA can prove X.
If we assume PA to be sound, then indeed everything it proves is true.
> Not true.
Now you're saying PA is unsound.
But your article wasn't about PA proves "PA proves X", it was about "forall n : PA proves G(n)".
For PA not to prove "forall n: G(n)", there is no soundness issue, only a ω-consistency issue.
I think we are saying the same thing.
If PA proves that it proves a statement, PA cannot conclude from that fact that it proves that statement.
If PA proves that it proves a statement, and then fails to prove it, PA is unsound.
There exist collections of statements such that PA proves that it proves each statement, and PA does prove each statement, but PA does not prove the collection of statements.
Our understanding of the last is helped by understanding that "PA proves that PA proves S" is logically not the same statement as, "PA proves S". Even though they always have the same truth value.
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> PA proves that if it can prove everything that it proves that it can prove, then PA must have a proof of one of the two previous statements.
I don't believe this is true. I don't know what result you're using here, but I think you're mixing up "provable" and "true".
In particular your line of reasoning violates Lob's Theorem, which is a corollary of the second incompleteness theorem.
The standard model is a model of PA only if PA is consistent, which it cannot prove unless it is inconsistent (Godel's theorem). So your proposed proof doesn't work in PA itself, which is the point of that comment, I believe.
I believe my proof actually works in systems even weaker than PA (but the metatheory of how exactly to do the encoding in a first-order theory is a bit of a headache for me at the moment).
In particular I'm not relying on the consistency of PA. If PA is inconsistent, "If PA proves 'PA proves X' then PA can prove X" also holds (trivially), because then PA can prove anything.
Hmm, earlier you wrote:
> This is because the key observation is not that there are nonstandard models, but rather that the standard natural numbers model PA.
This fact (that the standard natural numbers model PA) implies that PA is consistent, so your argument is implicitly assuming that from the outset, right? If PA is consistent then this is not something that PA can prove, so I don't see how you could run your proof in PA itself. And certainly not how you could do it in a weaker theory. Logic is subtle though... so what am I missing?
As far as I know what you're describing (namely that "Provable(X)" implies "X") is called the uniform reflection principle and added to PA is much stronger than PA alone.
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