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

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!

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.

• Thank you, Asaf! This is much clearer for me now. I had my own doubts when I wrote $(\spadesuit)$ down. When $(\spadesuit)$ is formalized within $\mathsf{ZFC}$, it cannot be proven if $\mathsf{ZFC}$ is consistent. It was meant to be a sentence in the meta-theory, and it definitely cannot be proven if we assume that the external $\mathsf{ZFC}$ on which the meta-theory is based is consistent. In order to discuss the consistency of the external $\mathsf{ZFC}$, we need a meta-meta-theory. – Berrick Caleb Fillmore Apr 17 '15 at 16:16
• I was thinking that if we stay within the meta-theory, then although we know in our guts that $(\spadesuit)$ cannot be proven, strictly speaking, we have to pass over it in silence, like how Wittgenstein would put it. – Berrick Caleb Fillmore Apr 17 '15 at 16:16
• Also, if you could, would you provide me with a reference to these strange mathematical universes in which $\mathsf{ZFC}$ actually proves $\neg \text{Con}(\mathsf{ZFC})$? Thanks! – Berrick Caleb Fillmore Apr 17 '15 at 16:20
• One thing to note is that in the meta-theory we assume $\sf\operatorname{Con}(ZFC)$, otherwise the entire thing is moot. And the proofs that various things like inaccessible cardinals imply the consistency of $\sf ZFC$ can be done in very weak meta-theories. So the choice of meta-theory is moot here. As for your second question, take a model of $\sf ZFC$ in which there are no models for $\sf ZFC+\operatorname{Con}(ZFC)$, and internally to that model (i.e. the model is now the meta-theory) it will be true that $\sf ZFC$ is consistent, but proves its own inconsistency. – Asaf Karagila Apr 17 '15 at 16:31
• @AsafKaragila I've always understood the metatheory in set theory as implicitly finitist, in the sense that the goal is to establish (p.r.) algorithms that convert proofs, such as you hand me a proof of $\text{ZFC} + \mathfrak{c} = \aleph_7 \vdash \varphi$ for some arithmetical $\varphi$ and I find some algorithm that provably turns it into a proof of $\text{ZFC} \vdash \varphi$. I also never thought of the metatheory to have an associated model. (cont.) – Jori Feb 1 at 0:18

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$(ZFC)$.

• This is not much of an open problem, just provably unprovable. Not to mention that in the case that $\sf ZFC\vdash\lnot\operatorname{Con}(ZFC)$, the model and the meta-theory disagree on the integers, which gives a good philosophical justification, in my opinion, why this is not really the case. – Asaf Karagila Apr 17 '15 at 16:48
• @AsafKaragila I take it you don't believe in a consistent but arithmetically unsound $ZFC$. – Sam Apr 17 '15 at 17:14
• I don't believe in anything. I just find the idea of using an unsound foundational theory a bit... strange. – Asaf Karagila Apr 17 '15 at 17:58
• Thanks, Sam. You’ve put it very nicely. I prefer saying “It’s an open problem” or “I don’t know” to contending with phrases like ‘provably unprovable’. – Berrick Caleb Fillmore Apr 17 '15 at 20:29
• @AsafKaragila What do you mean exactly when you say "The solution is that it depends on the meta-theory"? I'm asking because any meta-theory strong enough to prove the claim "$\mathsf{ZFC}\nvdash\neg\mathsf{RH}$" will itself be strong enough to actually establish $\mathsf{RH}$. So in this situation, the solution to $\mathsf{RH}$ will (as you put it) "depend on the meta-theory". – user52534 Apr 17 '15 at 23:28