Being not "close" to ZFC and not "far away" from it at the same time

Question

For each formula $\varphi$ we can have a look at the following two properties: $$1) ZFC \not \vdash Con (ZFC) \rightarrow Con (ZFC + \varphi)$$ $$2) ZFC + \varphi \not \vdash Con (ZFC)$$
Intuitively, 1) implies that $\varphi$ is not "close" to ZFC, whereas 2) says that $\varphi$ is not "far away" from ZFC.
If we assume that ZFC + $\varphi$ is consistent, then it's impossible that both 1) and 2) doesn't hold (otherwise we had $ZFC + \varphi \vdash Con(ZFC + \varphi)$, which would be a contradiction to Gödel's second incompleteness theorem). So, if ZFC + $\varphi$ is consistent, at least 1) or 2) holds. Assuming that ZFC is consistent, it's possible to find a suitable $\varphi$ such that exactly one of them holds: If $\varphi$ is the formula "There is an inaccessible cardinal", then 1) holds and 2) doesn't hold. By setting $\varphi = CH$, we have that 2) holds and 1) doesn't hold. Is it possible that both 1) and 2) hold?

Answer 1

Yes, although no natural examples are known to the best of my knowledge, and in fact the easiest proof I know is nonconstructive:

Suppose there were no such $\varphi$ - that is, for any $\varphi$ exactly one of the two possibilities holds. Each possibility is universal - the set of $\varphi$ without property $1$ is computably enumerable, and same for the set of $\varphi$ without property $2$. Since, under our assumption, property $1$ and property $2$ are the negations of each other, this means that each of these sets is in fact computable.

But it's easy to show that the set of sentences satisfying $2$ is non-computable. For suppose otherwise. Then we can build a computable consistent complete theory extending ZFC, contradicting Godel's theorem, as follows:

• Fix a computable listing $\{\theta_e:e\in\mathbb{N}\}$ of the first-order sentences in the language of ZFC.

• We now define a new sequence of sentences $\chi_e$, for $e\in\omega$, as follows:

  • $\chi_0=\theta_0$ if ZFC+$\theta_0\not\vdash$ Con(ZFC), and $\chi_0=\neg\theta_0$ otherwise.

  • Having defined $\chi_0, ..., \chi_n$, we let $\chi_{n+1}$ be $\theta_{n+1}$ if ZFC+$(\chi_0\wedge...\wedge\chi_n)\wedge\theta_{n+1}\not\vdash$Con(ZFC), and $\neg\theta_{n+1}$ otherwise.

Let $T=$ ZFC $\cup$ $\{\chi_i:i\in\mathbb{N}\}$. $T$ is complete and consistent (in particular by induction it doesn't prove Con(ZFC)) - and since by assumption we can computably check if a sentence, together with ZFC, proves Con(ZFC) we know that $T$ is computable. Contradiction.

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What about property $1$? This seems much harder:

• First, we need an additional mild-but-nontrivial assumption about ZFC to make property $1$ nontrivial: conceivably ZFC could be consistent yet prove its own inconsistency, in which case every sentence has property $1$. So we need (some weak form of) the assumption that ZFC is sound.

• Even with that assumption, though, the inductive idea above doesn't seem to apply here easily: while it is true that (under our soundness assumption) $\varphi$ has property $1$ if it's inconsistent with ZFC, it's possible that neither $\varphi$ nor $\neg\varphi$ have property $1$, so we can't use property $1$ to detect inconsistency.

So while I suspect that the set of sentences satisfying property $1$ is non-computable, I don't immediately see how to prove it, and it definitely seems conceptually harder.