# What things does ZFC not know if it knows?

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?)

• Independent does not mean "unprovable" but "neither provable nor disprovable". – Asaf Karagila Nov 10 '17 at 21:00
• @AsafKaragila I'm aware. Does something in my question suggest that I meant unprovable? – PyRulez Nov 10 '17 at 21:02
• Read the first sentence of the second paragraph. It literally says that you mean "unprovable". – Asaf Karagila Nov 10 '17 at 21:02
• 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. – Joel David Hamkins Nov 10 '17 at 23:21
• @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. – PyRulez Nov 10 '17 at 23:25

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.

• @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". – Asaf Karagila Nov 10 '17 at 21:05
• @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. – Andrés E. Caicedo Nov 10 '17 at 21:07
• @AsafKaragila Of course. – Andrés E. Caicedo Nov 10 '17 at 21:11
• @Andrés: Okay, what about Con(ZF+there is a super-Berkeley cardinal)? – Asaf Karagila Nov 10 '17 at 21:12
• @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. – Carl Mummert Nov 11 '17 at 18:40