What does this deduction involving provability imply?

Question

I recently asked this question asking why my reasoning for ZF being inconsistent was wrong. I didn't realize that we have to account for models with non-standard integers. However, I'm left with the question of what proof techniques (not theorems) can be represented within the deductive calculus. For example, say we have the self-referential sentence $\sigma$ such that $$T \vdash (\sigma \leftrightarrow \text{Prb}_T ( \neg \text{Prb}_T \sigma))$$ where $T$ is any sufficiently powerful theory and $\text{Prb}_T \sigma$ represents $T \vdash \sigma$.

Now, assuming throughout that $T$ is consistent, if we have $T \vdash \sigma$ then by the definition of $\sigma$ we have $T \vdash\text{Prb}_T ( \neg \text{Prb}_T \sigma)$. This implies $T \vdash \neg \text{Prb}_T \sigma$. But by the reflection principle for our hypothesis we have $T \vdash \text{Prb}_T \sigma$, a contradiction. So we just proved (meta-mathematically) that assuming $T$ is consistent we must have $T \nvdash \sigma$. This means by the definition of $\sigma$ that $T \nvdash \text{Prb}_T (\neg \text{Prb}_T \sigma)$ implying $T \nvdash \neg \text{Prb}_T \sigma$. This means that there is no proof within $T$ of $T \nvdash \sigma$. We did though provide a meta-mathematical proof that $T \nvdash \sigma$.

I realize that there is no contradiction here: From Godel's incompleteness theorem we know there are many sentences which cannot be proven or disproven in $T$. However, that doesn't negate the fact that we do have a meta-mathematical proof of $T \nvdash \sigma$. So I guess my question is how do we classify this proof if it cannot be represented in $T$. Are there actually models in which $T \models \sigma$ and if so what are they? Or am I confusing two different concepts?

score 2 · Answer 1 · answered Jan 19 '16 at 19:25

2

Yes, there will be models in which $T$ proves $\sigma$. Specifically, in such models there will be a nonstandard natural number which the model thinks is a proof of $\sigma$ from $T$. From the standard point of view, the "proof" coded by such an $n$ may be of nonstandard length, or (if $T$ is an infinite theory) may use nonstandard axioms of $T$.

Describing these models is hard - by Tennenbaum's theorem, there are no recursive such models, and as far as I know no natural nonstandard models of reasonably strong arithmetic have been discovered - but they do exist, as you correctly observe.

answered Jan 19 '16 at 19:25

Noah Schweber

245,398

They might also use non-standard inference rules! – Asaf Karagila Jan 19 '16 at 19:27
@AsafKaragila In the proof systems I'm familiar with, there are only finitely many inference rules. Which one are you thinking of? – Noah Schweber Jan 19 '16 at 19:36
You need at least some sort of schema for quantifiers, no? – Asaf Karagila Jan 19 '16 at 19:37
Oh, I see. Yeah, to me that feels like a standard inference rule being used on a nonstandard term, but that is different from a nonstandard axiom. Good point. – Noah Schweber Jan 19 '16 at 19:42

Thomas Andrews · Accepted Answer · 2016-01-20T16:07:53.790

Godel's theorem is often stated for non-math people as "There is a true statement which is not provable." In reality, what "true" means is that our intuition about the model says it should be true, and undecidability means the axioms do not model our intuition completely. That is, the statement $P$ which is undecidable has an intuitive truth value of "true." In a model where it is false, these are not the natural numbers we were looking for.

Consider a more down-to-earth case, the Goldbach conjecture:

Every even number $n\geq 4$ can be expressed as the sum of two primes.

What would that mean for this to be undecidable? Well, our intuition is that we should add it as an axiom, because our intuition about the natural numbers is that, if Goldbach is false, someone would be able to give us an example $n$ where it failed, and a finite proof that the example fails.

What you are saying is roughly like this, but rather, your intuition is about the way that the model's natural numbers can be used to describe proofs and statements. But if the model of the integers is wilder, then your assumptions about the validity of the integers modeling finite proofs is wrong.

If Goldbach is undecidable, there is some model for Peano or ZF or whatever theory you are working in where it is false. What the meta-mathematics tells us is, this is not the model that we wanted.

(From what I recall, it is not possible for Goldbach to be provably undecidable, however. I think you can convert any proof of undecidability to a proof a Goldbach, but I'm not 100% certain of that. I'm mostly just using it as an example where undecidability would imply, meta-mathematically, truth. But I believe that if Goldbach is undecidable, we'll never know it.)

Asaf be praised.

$$\Huge\stackrel{\stackrel{\frown}{\odot}\ \stackrel{\smile}{\odot}}{\overline{\phantom{\circ\ \circ}}}$$ — Asaf Karagila, Jan 19 '16 at 19:49
Sorry, just a little joke when I give one of my intuitive deeply flawed answers to a question I know you could answer better. Like Catholics who cross themselves when they pass a graveyard. :) @AsafKaragila — Thomas Andrews, Jan 19 '16 at 20:48
I was told that I look a bit "Jesus-y" before, but this is not what I was going for when I first joined the site! ;-) — Asaf Karagila, Jan 19 '16 at 20:48

What does this deduction involving provability imply?

2 Answers2