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The statement "ZFC $\vdash 0=1$" is independent of ZFC due to Goedel's second incompleteness theorem. That got me wondering, for what other statements $\phi$ is "ZFC $\vdash \phi$" independent of ZFC?

Now of course, for any statement $\phi$ for which "ZFC $\nvdash \phi$" is true, then "ZFC $\vdash \phi$" is independent of ZFC. So instead, I'll ask which "ZFC $\vdash \phi$" are independent of ZFC + Con(ZFC), and stronger theories?

(Of course, easy examples are Con(T) for T=ZFC+Con(ZFC) or stronger theory, but are there others?)

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  • $\begingroup$ Independent does not mean "unprovable" but "neither provable nor disprovable". $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 10, 2017 at 21:00
  • $\begingroup$ @AsafKaragila I'm aware. Does something in my question suggest that I meant unprovable? $\endgroup$
    – Christopher King
    Commented Nov 10, 2017 at 21:02
  • $\begingroup$ Read the first sentence of the second paragraph. It literally says that you mean "unprovable". $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 10, 2017 at 21:02
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    $\begingroup$ And also, now you are asking for independence over $\text{ZFC}+\text{Con}(\text{ZFC})$, rather than merely independence over $\text{ZFC}$. These are not the same thing, and that change affects everything I have said so far. $\endgroup$
    – Joel David Hamkins
    Commented Nov 10, 2017 at 23:21
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    $\begingroup$ @JoelDavidHamkins It shouldn't be interpreted as"statements independent of (ZFC + Con(ZFC) and stronger theories) but as "statements independent of (ZFC + Con(ZFC)) and statements independent of (stronger theories)". Each theory gives a separate question. $\endgroup$
    – Christopher King
    Commented Nov 10, 2017 at 23:25

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You seem to use "independent of $T$" to mean "unprovable in $T$", so I'll interpret the question that way (not as "neither provable nor refutable in $T$).

If $ZF\vdash\phi$ is true, then it can be proved in ZF and in fact in much weaker systems, just by taking a proof of $\phi$ in ZF and verifying that it is indeed a proof (and ends with $\phi$). Conversely, if $ZF\vdash\phi$ is false, then it can't be proved in ZF or in ZF + Con(ZF) or in any arithmetically sound theory. (My Platonism is showing here, as I take it for granted that ZF + Con(ZF) is arithmetically sound because it's true.) So $ZF\vdash\phi$ is unprovable in ZF + Con(ZF) iff it is false, i.e., if $\phi$ is unprovable in ZF.

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    $\begingroup$ @PyRulez: How do you what is true in true arithmetic? Is Con(ZFC) true? Is Con(ZFC+Con(ZFC)) true? is Con(ZFC+there is a measurable cardinal) true? Is Con(ZF+DC+AD) true? I can continue for more or less forever with these questions (and for each answer I would probably ask a follow up of "Why? How do you know that?"), at some point you're going to have to give me a better answer than just "true in true arithmetic". $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 10, 2017 at 21:05
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    $\begingroup$ @AsafKaragila I do not see why anything more is needed. An arithmetic statement is true if it holds in the standard model, that is the standard interpretation of the term. Maybe I do not understand what you are asking. $\endgroup$
    – Andrés E. Caicedo
    Commented Nov 10, 2017 at 21:07
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    $\begingroup$ @AsafKaragila Of course. $\endgroup$
    – Andrés E. Caicedo
    Commented Nov 10, 2017 at 21:11
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    $\begingroup$ I agree with @NoahSchweber's last comment about Platonism, and I'd add that, when I say that an arithmetical statement is true, I mean exactly what most mathematicians who are not logicians mean by that. $\endgroup$
    – Andreas Blass
    Commented Nov 11, 2017 at 2:10
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    $\begingroup$ @Qfwfq: I would say it is not. Most non-logician mathematicians have an even firmer sense of the natural numbers "as a structure", and when they say "true" they mean "true in that structure". At the same time few know much of anything about formal logic. // Separately, phrases such as "true in PA" are somewhat painful to read, even if I know what is meant, because they mix truth with provability in a way that the two should not mix. $\endgroup$
    – Carl Mummert
    Commented Nov 11, 2017 at 18:40
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I think you're asking what happens if you iterate adding consistency statements. There's a whole book about that, "Inexhaustibility" by Torkel Franzen, though it starts with arithmetic rather than ZFC. This article is also good:

https://xorshammer.com/2009/03/23/what-happens-when-you-iterate-godels-theorem/

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