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I'm familiar with a few independence proofs e.g. Continuum Hypothesis, but I was wondering whether there's a vague since in which we got "lucky" with these, and some statements which are undecidable in ZFC will also have their undecidability be undecidable.

My knee-jerk reaction was no, because by Godel's Completeness Theorem if we have some statement $S$ and $D$:="$S$ is independent of ZFC", then if $D$ is independent of ZFC, there exists a witness in some models to a proof or a disproof of $S$, and since proofs are syntactic a statement can't be (dis)provable in only some models. But this is very hand-wavy, and moreover seems like could equally well be used to show that $CON(PA)$ is not independent of $PA$.

I'm guessing the error is that, even though there are witnesses in some models which the model "thinks" encode a (dis)proof of $S$, they're an analog of nonstandard numbers in $PA$, and can't actually be decoded into a finite proof

So I guess my question is two-fold: 1. Is the above reasoning valid (if a bit hand-wavy), 2. Can the independence of a statement itself indeed be independent? Further, can this "recurse" infinitely, such that for some statements not only can we never decide it, we can't decide that we can't decide it, and we can't decide that we can't decide that...?

dahak27
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    The answer to your title question is "yes, and always". ZFC can't prove ZFC is consistent, so can't prove that anything is independent of ZFC. – spaceisdarkgreen Dec 23 '24 at 16:08
  • @spaceisdarkgreen Doesn't it kind of depend what is meant by that statement? E.g. CH being independent of ZFC means that ZFC has models in which CH and ones in which it's false (assuming ZFC has models at all). However, I'm not sure how you would describe the question in the title with models. – QuantumWiz Dec 24 '24 at 11:35
  • @QuantumWiz Well, yes, the meaning of the question does affect the correct answer :). I don't think there's much of a difference between framing things in terms of model existence and consistency, since those are equivalent. In terms of models, I would interpret independence of $\phi$ from $\sf ZFC$ to mean that there is a model of $\sf ZFC + \phi$ and a model of $\sf ZFC + \lnot\phi$. The real distinction is between independence relative to consistency of $\sf ZFC$ (i.e. what David Gao's answer is talking about and when you add "assuming ZFC has models at all") and outright independence. – spaceisdarkgreen Dec 24 '24 at 20:31
  • Yes, I interpreted independence in the same way. What was confusing me is that "independence of an independence statement" (as per the title) would seem to ask whether $ZFC$ has models such that $ZFC+\phi$ and $ZFC+\lnot\phi$ both have models and that sounds kind of nonsensical. (Disclaimer: I have to admit that I'm not super familiar with mathematical logic so I might be misunderstanding something here.) – QuantumWiz Dec 25 '24 at 03:27

2 Answers2

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If there is any statement $\phi$ independent of theory $T$, then $T$ is consistent (as inconsistent theory can prove anything). Thus, if there is any statement independent of ZFC, then ZFC is consistent. And this is simple enough to be proved in ZFC, i.e. for any $\phi$, $\text{ZFC} \vdash \neg\Box \ulcorner\phi\urcorner\rightarrow \neg \Box \ulcorner\bot\urcorner$.

As $\text{ZFC}\not\vdash \neg \Box \ulcorner\bot\urcorner$, there is no statement $\phi$ s.t. ZFC can prove $\neg \Box \ulcorner\phi\urcorner$ - in particular, there is no statement that ZFC can prove to be independent of itself.

mihaild
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A more interesting question is to interpret “independence” to mean a statement and its negation are both relatively consistent, i.e., in our case,

$$\text{ZFC} \vdash \text{Con(ZFC)} \to (\text{Con(ZFC} + \phi) \wedge \text{Con(ZFC} + \neg\phi))$$

This is what is meant when we say, for example, ZFC can show CH is independent. Under this interpretation, a standard example of a statement that is (assumed to be) independent of ZFC but this independence cannot be proved in ZFC is the existence of a large cardinal (basically every large cardinal property can work, except the few that are known to be inconsistent with ZFC). Such a statement always has its negation being consistent with ZFC (and this fact can be proved in ZFC, assuming $\text{Con(ZFC)}$), but the relative consistency of such a statement cannot be proved in ZFC (but this fact can actually be proved in ZFC, again assuming $\text{Con(ZFC)}$).

David Gao
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    A more direct example is $\text{Con(ZFC)}$ itself, whose negation is relatively consistent and this relative consistency can be proved in ZFC. But ZFC cannot prove $\text{Con(ZFC)}$ is relatively consistent (but this fact can be proved in ZFC, assuming $\text{Con(ZFC)}$). – David Gao Dec 23 '24 at 16:47