Axiomatic Independence and Truth

Question

So let the sentence $s$ be independent from an axiomatic system $A$ (like PA or ZFC), i.e. neither $s$ nor $\lnot s$ is provable there. Now I read that $s$ is true in case $s$ = $\forall x Px$ and $s$ is false in case $s$ = $\exists x Px$.

1. My understanding is this: If for example $s$ = $\forall x Px$ and it's independent from $A$ then the truth of $s$ hangs in the air. Our only way is to go to a more powerful (and sound) system than $A$ - let's call it $A'$ - and prove there that indeed there is no x in the domain that fulfills $\lnot P$, so basically one proves $s$ in $A'$ and since $A'$ is sound $s$ is true and that means $s$ was already true in $A$ but there we couldn't prove it due to $A$'s limitations. So far so good, but $A'$ could also prove $s$ false, by proving that there's indeed some object $o$ with $\lnot P(o)$ which was overseen by $A$, couldn't it? So that would make it false to say with my book that if $s$ is independant from some $A$ but has the form $\forall x Px$ it must be true, right? It depends. Am I correct or my book?

2. If what I read in the book was true one could basically make an axiom that says: if a statement and its negation is not provable with the usual tools (axioms, theorems via inference rules), but has the form "$\forall x Px$", it's considered to be proven, else the negation is proved. Then Goedel's incompleteness would not be a problem anymore, wouldn't it?

Answer 1

There are multiple questions on this site already about independence implying truth/falsity for appropriate formulas/systems; however, as far as I can tell this is not an exact duplicate, primarily due to the idea behind (1) in the OP.

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Let's work in PA (= first-order Peano arithmetic) for simplicity.

You haven't precisely stated the relevant principle: what is true, and PA proves this, is the following:

If $s$ has the form $\exists xP(x)$ for a "simple" formula $P$ and $s$ is independent of PA, then $s$ is false.

If $s$ has the form $\forall xP(x)$ for a "simple" formula $P$ and $s$ is independent of $PA$, then $s$ is true.

Here "simple" has a technical meaning, namely using only bounded quantifiers. More snappily, using the jargon of the arithmetical hierarchy we have:

Every $\Sigma^0_1$ sentence independent of PA is false, and every $\Pi^0_1$ sentence independent of PA is true.

And to reiterate, the above statement is provable in PA. The simplicity requirement here is crucial: in general, we can't deduce $s$ or $\neg s$ from PA + "$s$ is independent of PA."

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So why is that true?

Well, it's much easier to think about when rephrased as follows:

PA proves every true $\Sigma^0_1$ sentence.

That is, PA is $\Sigma^0_1$-complete. The idea is that a $\Sigma^0_1$-sentence is true iff there is some natural number serving as a "witness" to it in a way which can be checked by basic calculations; every natural number is given by a term in PA, so PA can produce a "guess-and-check" proof of any specific true $\Sigma^0_1$-sentence. (For the fact that PA proves that PA is $\Sigma^0_1$-complete, see the discussion here.)

The $\Sigma^0_1$-completeness of PA immediately proves the above "truth/falsity-from-independence" results:

• If $s$ is $\Sigma^0_1$ and PA-unprovable, then since PA is $\Sigma^0_1$-complete $s$ must be false.

• If $s$ is $\Pi^0_1$ and PA-un-disprovable, then just think about $\neg s$.

Hopefully this demystifies the situation.

At this point I ought to address your point (1). I think the issue comes down to the requirement that the new system $A'$ be sound: we can't have one sound system decide a sentence one way and another sound system decide it a different way! Soundness means not proving any false statements, so two sound systems can never disagree - although of course one might prove something the other can't. So the "divergent sound extensions" phenomenon you're describing can't actually happen.

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Now what about your point (2)?

Well, the issue is that someone still needs to prove the relevant independence fact! For example, the Godel sentence $G_{PA}$ is $\Pi^0_1$, hence true if PA-independent. Moreover, PA proves that - that is, PA proves "If $G_{PA}$ is PA-independent, then $G_{PA}$." However, PA doesn't quite prove "$G_{PA}$ is PA-independent." All PA can do is prove "If PA is consistent then $G_{PA}$ is PA-independent," but that's not quite enough.

So we're stuck with the same issue as always: what system are we using to prove things in? Once we fix an appropriate system $T$ (say, PA), there will be relevant independence phenomena which $T$ can't prove. In particular, $T$ won't be able to prove any independence phenomenon over itself (if it did, it would prove its own consistency: nothing is independent over an inconsistent theory!).