[joe_the_user]

One can write down a concrete polynomial p ∈ Z[x1,...x9] such that the statement "there are integers m1,...,m9 with p(m1,...,m9)=0" can neither be proven nor disproven in ZFC (assuming ZFC is consistent).

So you were to specify such a thing and the thing was small enough it's value could be determined by a command line program and you found integers m1.... m9 such that when you typed them at the command line, the value returned was 0, would "reality" have determined a "truth" that was not deducible?

Still, I can't see each of the steps wouldn't be easily determined by existing axioms.

Anyway, my head hurts.

[joe_the_user]

OK, the way I'd figure it out is: for such a polynomial, you definitely can't find those integers. There isn't any concrete m1...m9 satisfying the condition. The stumbling block is you can't find a proof for this fact in ZFC.

But this seems to go against the idea that for any proposition independent from an axiom system, there is a model of the axiom system where that proposition is true and another where it is false.

Someone enlighten me (not sarcastic).

[dodobirdlord]

> OK, the way I'd figure it out is: for such a polynomial, you definitely can't find those integers. There isn't any concrete m1...m9 satisfying the condition. The stumbling block is you can't find a proof for this fact in ZFC.

If this were formalizable it would be a proof.

I encourage you to read through the excellent piece on the busy beaver function and computability. It's not entirely related, but it's fun! And it touches on the same theme as your question here, that human intuition about mathematics is weak. The writeup describes and proves the non-computability of a particular function of positive integers with definite (but non-computable!) value.

https://www.scottaaronson.com/writings/bignumbers.html

[Dylan16807]

> If this were formalizable it would be a proof.

A proof, but not a proof that can be expressed inside ZFC.

[dodobirdlord]

Obviously, the original statement has been proved to be independent from ZFC.

[OscarCunningham]

You're right that if ZFC is consistent then it has a model in which this polynomial has an integer root. The issue is that the "integers" in the model are different from the actual integers. So you can't take the root out of the model and use it to prove ZFC inconsistent.

[joe_the_user]

That is certainly how I would understand the situation.

However, it seems intuitively you construct the notion of finite, could engage in the ordinary computation and have a way of distinguishing these "weird roots" from regular roots.

I suppose if you make the position of (something like) "every polynomial whose root cannot be found by finite calculation does not, in fact, have one" an axiom but then you would have discarded the notion of finite axioms (you'd have a second order system).

Still, might actually have a way to specify "truth", something I'd thought was a bit beyond logic at this point.

[OscarCunningham]

You might be interested in https://en.wikipedia.org/wiki/Tarski%27s_undefinability_theo....

[OscarCunningham]

The point is that if ZFC is consistent then there are no such integers. So a computer program searching for them would run forever without returning any output. But since we can't prove that ZFC is consistent you would never know whether your computer program was really running for ever. Even after a million years you wouldn't be able to prove that it wouldn't finally give some output in five minutes time.

[joe_the_user]

Consistency or inconsistency seems like a red herring here. The proposition concerns conjures independent of ZFC if ZFC is consistent.

Assuming is ZFC is consistent, the theory says there are polynomials whose zeros we could forever fruitless search for but which couldn't be proved to have no zeros in ZFC. We could forever fruitlessly search for a proof of the polynomail having no zero, in fact (putting it this way, returns the situation to something marginally understandable, in fact makes something that would follow from the halting problem).

If ZFC is inconsistent, we actually could prove that this polynomial had a zero and we could prove it didn't have a zero, since we could prove anything, at least anything in the vocabulary of ZFC.