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I'm attempting to understand the answer to the question, Is there a statement whose undecidability is undecidable (as in independent, not a decision problem)? The answer appears to be "Yes". However, in both top-voted answers, the answer depends on this:

Specifically, it's impossible for ZFC to prove that it can't prove something, because such a statement is tantamount to the consistency of ZFC; if ZFC were inconsistent then it could prove everything, so proving that there's something that can't be proved is equivalent to proving the consistency of ZFC, and of course this is forbidden by the second incompleteness theorem.

At first I didn't understand this, but now I think I do: If ZFC is inconsistent, then ZFC could prove anything. Thus, proving that it can't prove something proves that ZFC is consistent, since if it were inconsistent then it could prove everything. And if ZFC can prove that it is consistent, then ZFC is inconsistent, as per Godel's second incompleteness theorem.

However, there are statements known to be undecidable in ZFC. If a statement $S$ is known to be undecidable, is this not a proof that $(ZFC \not\vdash S) \land (ZFC \not\vdash \neg S)$ - meaning ZFC shows it can't prove something (two things, to be exact)?

Or is it that the proof of undecidability is not within ZFC and so ZFC is safe?

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  • $\begingroup$ On the Wikipedia page you link to, it says these statements are undecidable assuming ZFC is consistent. They become both provable and disprovable if ZFC is inconsistent $\endgroup$ – Ross Millikan Jan 30 '14 at 1:16
  • $\begingroup$ @RossMillikan: Right, but if ZFC is consistent, and ZFC proves that a statement is undecidable, then that's proof of ZFC's consistency, which means ZFC is inconsistent. So I'm assuming the answer is that the proofs the statements are undecidable don't lie within ZFC? Put another way: How do we know those statements are undecidable assuming ZFC is consistent? $\endgroup$ – Claudiu Jan 30 '14 at 1:21
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    $\begingroup$ "the proofs the statements are undecidable don't lie within ZFC" ... Correct: they lie within ZFC+Con(ZFC). $\endgroup$ – GEdgar Jan 30 '14 at 1:49
  • $\begingroup$ @GEdgar: What is Con(ZFC)? I am new at this and couldn't google it easily. $\endgroup$ – Claudiu Jan 30 '14 at 2:05
  • $\begingroup$ Con(ZFC) denotes statement of consistency of ZFC. Assuming that ZFC is consistent, Con(ZFC) is unprovable in ZFC, and thus ZFC+Con(ZFC) is a stronger theory -- sufficiently strong to prove some assertions the ZFC itself could not (Con(ZFC) is one such). $\endgroup$ – Peter Košinár Jan 30 '14 at 2:58
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Or is it that the proof of undecidability is not within ZFC and so ZFC is safe?

Exactly. Proofs of undecidability in ZFC always assume a stronger axiom such as Con(ZFC) that is not provable in ZFC. This means that the overall undecidability result is not provable in ZFC, it is only provable in the stronger system.

As you saw, if ZFC is inconsistent then it proves every statement in its language. So any proof that demonstrates an undecidable proposition in ZFC must assume ZFC is consistent - this axiom is exactly Con(ZFC).

One way of assuming Con(ZFC) is assuming that ZFC has a model (that is, a set that is a model of ZFC). This is equivalent to ZFC being consistent, by the completeness theorem for first-order logic, and so it is not provable in ZFC itself. Thus any undecidability proof that proceeds by constructing a model must assume Con(ZFC).

Most set theorists view Con(ZFC) as a relatively innocuous assumption (particularly because it is required in order to prove any undecidability result). They avoid assuming extra axioms beyond ZFC if possible, but not because they doubt Con(ZFC). The benefit of doing as much as possible in ZFC itself (or in weaker systems) is that - when possible - it leads to results that are more generally applicable. But when it is not possible to prove a certain result without going beyond ZFC, set theorists routinely look for additional axioms to help with the proof.

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