Do we know if the consistency of $ \mathsf{ZFC} $ is sufficient to prove that $ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \to \mathsf{SM} $?

Question

This is basically a sequel to this earlier post. There, I asked:

If $ \mathsf{ZFC} $ is consistent, then is $ \mathsf{ZFC} \nvdash \neg \text{Con}(\mathsf{ZFC}) $, i.e., $ \neg \text{Con}(\mathsf{ZFC}) $ is unprovable in $ \mathsf{ZFC} $?

If $ \mathsf{I} $ denotes the sentence ⟪There exists an inaccessible cardinal⟫, then the enlightening response provided by Asaf to my question says:

Asaf’s response. If $$ (\spadesuit) \quad \mathsf{ZFC} ~ \text{is consistent} ~ \Longrightarrow ~ \mathsf{ZFC} + \mathsf{I} ~ \text{is consistent}, $$ then $ \mathsf{ZFC} \nvdash \neg \text{Con}(\mathsf{ZFC}) $ when $ \mathsf{ZFC} $ is consistent.

As is well-known, $ (\spadesuit) $ is an open question in set theory.

The motivation behind my question is the following theorem:

Theorem. If $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is consistent, then $ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \to \mathsf{SM} $, where $ \mathsf{SM} $ is the statement in the language of set theory saying that there exists a standard model of $ \mathsf{ZFC} $.

For the sake of convenience, let me provide a proof.

Proof

Suppose that $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is consistent. As the sentence $ \text{Con}(\mathsf{ZFC}) $ is absolute, we have $$ \mathsf{ZFC} \vdash \mathsf{SM} \to \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})). $$ Assume for the sake of contradiction that $ \mathsf{ZFC} \vdash \text{Con}(\mathsf{ZFC}) \to \mathsf{SM} $. Then modus ponens yields $$ \mathsf{ZFC} \vdash \text{Con}(\mathsf{ZFC}) \to \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})). $$ By the Resolution Theorem of first-order logic, we obtain $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \vdash \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})), $$ which contradicts Gödel’s Second Incompleteness Theorem as $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ was assumed to be a consistent theory. $ \quad \blacksquare $

My question is:

Question. Do we know if we can relax ‘$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is consistent’ to simply ‘$ \mathsf{ZFC} $ is consistent’ in order for the theorem to hold still?

If the consistency of $ \mathsf{ZFC} $ implies the consistency of $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $, then we are done by the theorem above. This is the reason for my earlier post. If $ \text{Con}(\mathsf{ZFC}) $ is independent of $ \mathsf{ZFC} $ when $ \mathsf{ZFC} $ is consistent, then $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is consistent when $ \mathsf{ZFC} $ is consistent and the theorem follows.

Also in my earlier post, Zhen Lin mentioned that if $ \mathsf{ZFC} $ is $ \omega $-consistent, then $ \neg \text{Con}(\mathsf{ZFC}) $ is not provable in $ \mathsf{ZFC} $. user52534 also suggested that if $ \mathsf{ZFC} $ is $ \Sigma^{0}_{1} $-sound, then we get the same thing. Hence, we can replace ‘$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is consistent’ by ‘$ \mathsf{ZFC} $ is $ \omega $-consistent’ or even better, ‘$ \mathsf{ZFC} $ is $ \Sigma^{0}_{1} $-sound’.

I would therefore appreciate an answer to my question in one of the following three forms:

• ‘Yes’ with a careful proof sketch (for a beginner like myself) or a reference.
• ‘No’ with a careful proof sketch or a reference.
• ‘This is an open problem’ with a reference.

Thanks!

Answer 1

Let me first clarify the issue with an inaccessible, since it seems to me that my words were somewhat misunderstood.

1. I only mentioned inaccessible cardinals, since it is a plausible axiom, and if you believe it to be consistent, then of course there will be a (standard) model of $\sf ZFC$. And in that case we cannot prove $\sf\lnot\operatorname{Con}(ZFC)$.

   What I didn't say is that "If the consistency of $\sf ZFC$ implies the consistency of $\sf ZFC+I$ then ..." because we can easily show that this cannot be the case, unless $\sf ZFC$ was inconsistent to begin with.

2. It is not an open question whether or not inaccessible cardinals are consistent. We know that this statement cannot be proved. Whether or not inaccessible cardinals are inconsistent is an open question, but I don't think any set theorist today believes that.

The point is that in some mathematical universes, $\sf ZFC$ is consistent, but $\sf ZFC+\operatorname{Con}(ZFC)$ is not consistent. In those universes, $\sf ZFC$ proves its own inconsistency, which sounds weird, and indeed those universes will necessarily disagree with the meta-theory about the integers, so as far as models of set theory go, they will be non-standard models.

But in such mathematical universe, $\sf ZFC\vdash\lnot\operatorname{Con}(ZFC)$. And therefore, for any statement $\varphi$, $\sf ZFC\vdash\operatorname{Con}(ZFC)\rightarrow\varphi$. Simply because a false premise implies everything.

Finally, we can relax the assumption on $\sf\operatorname{Con}(ZFC)$ and essentially require that $\sf\operatorname{Con}(ZFC)$ is not refutable. This follows from both $\omega$-consistency and $\Sigma^0_1$-soundness.

Answer 2

To summarize, and to answer the question, it seems that we have an unknown problem.

The reason being: the OP has established "If $ZFC\nvdash$ $\neg$Con$(ZFC)$, then $ZFC\nvdash$ Con$(ZFC)$ $\implies SM$." in the statement of his question. Also, Asaf has established (the almost trivial) "If $ZFC\vdash$ $\neg$Con$(ZFC)$, then $ZFC\vdash$ Con$(ZFC)$ $\implies SM$."

Putting these two together, we have established "$ZFC\vdash$ $\neg$Con$(ZFC)$ if and only if $ZFC\vdash$ Con$(ZFC)$ $\implies SM$.

(*) So since it is not known whether $ZFC\vdash$ $\neg$Con$(ZFC)$ or not (even under the extra assumption that $ZFC$ is consistent), then it follows from the above paragraph that is not know whether $ZFC\vdash$ Con$(ZFC)$ $\implies SM$ or not (even under the extra assumption that $ZFC$ is consistent).

Regarding (*): It is a big assumption for me to say that something is not known, so the best I can do is to direct you to the discussion about the arithmetic soundness of $ZFC$ over at FOM, where the posters dabble in the possibility of a consistent $ZFC$ such that $ZFC\vdash$ $\neg$Con$(ZFC)$.