# Reference request for the independence of $\text{Con}(\mathsf{ZFC})$.

Gödel’s Second Incompleteness Theorem says that if $\mathsf{ZFC}$ is consistent, then $\mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC})$, i.e., $\text{Con}(\mathsf{ZFC})$ is not provable in $\mathsf{ZFC}$.

Does anyone know of an authoritative reference that contains the claim that $\neg \text{Con}(\mathsf{ZFC})$ is also not provable in $\mathsf{ZFC}$ if $\mathsf{ZFC}$ is consistent?

Thanks!

• If ZFC is $\omega$-consistent, then $\lnot \mathrm{Con}(\mathrm{ZFC})$ is not provable. (If it were, then by $\omega$-consistency, ZFC is inconsistent – a contradiction!) – Zhen Lin Apr 16 '15 at 8:27
• There is nothing special about ZFC in this regard. The assumptions needed will be the same as for any other theory, and the authoritative reference would just be any reference on the incompleteness theorems. If you are looking for a good book on those, I recommend the one by Peter Smith. – Carl Mummert Apr 16 '15 at 11:14
• @ZhenLin: Hi Zhen Lin. Is there any way we can relax $\omega$-consistency to just consistency? If not, then it seems that the second paragraph of Section 5.2 of this Stanford Encyclopedia article is incorrect. – Berrick Caleb Fillmore Apr 16 '15 at 14:31
• As Asaf's answer alludes, consistency alone is not enough, because ZFC + $\lnot \text{Con}(\text{ZFC})$ is consistent and proves its own inconsistency. – Carl Mummert Apr 16 '15 at 17:21
• I was referring to the question in the post along with Zhen Lin's comment. Zhen Lin said: "If $ZFC$ is $\omega$-consistent, then $\neg$Con$(ZFC)$ is not provable (from $ZFC$)". I'm saying that we also have: "If $ZFC$ is $\Sigma^0_1$-sound, then $\neg$Con$(ZFC)$ is not provable (from $ZFC$)". The significance being that $\Sigma^0_1$-soundness is a weaker condition than $\omega$-consistency. – user52534 Apr 16 '15 at 22:17

Suppose that $\sf ZFC+\operatorname{Con}(ZFC)$ is just inconsistent, and the additional assumption is the one causing the inconsistency. What does that mean? It means that in every model of $\sf ZFC$ it holds that $\lnot\operatorname{Con}\sf (ZFC)$, and therefore it is provable from $\sf ZFC$ that $\lnot\operatorname{Con}\sf (ZFC)$.
Of course if you believe that inaccessible cardinals, for example, are not inconsistent with $\sf ZFC$, then you have all the reason to believe that $\sf ZFC$ does not prove that $\lnot\operatorname{Con}\sf (ZFC)$. Why? Because $\sf ZFC+\exists\kappa\text{ inaccessible}\vdash\operatorname{Con}(ZFC)$.
• Hi Asaf. Thank you for your answer. Hence, if I want to prove that $$\mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \to \mathsf{SM},$$ then I have to assume in the first place that $\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})$ is consistent because the consistency of $\mathsf{ZFC}$ alone is not enough? My comments made below my post will explain to you the motivation behind my question. – Berrick Caleb Fillmore Apr 16 '15 at 18:56
• No, working with $\sf ZFC$ is just fine. All you have to do is note that $\sf ZFC+SM\vdash\operatorname{Con}(ZFC+\operatorname{Con}(ZFC))$. Therefore by the second incompleteness theorem, if $\sf ZFC$ would have proved that $\sf\operatorname{Con}(ZFC)\rightarrow SM$, then $\sf ZFC+\operatorname{Con}(ZFC)$ would have proved its own consistency. – Asaf Karagila Apr 16 '15 at 19:03
• Hmm... There’s still something that I’m not getting. As you’ve said, $$\mathsf{ZFC} + \mathsf{SM} \vdash \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})),$$ or equivalently, $$\mathsf{ZFC} \vdash \mathsf{SM} \to \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})).$$ Hence, if $\mathsf{ZFC} \vdash \text{Con}(\mathsf{ZFC}) \to \mathsf{SM}$, then $$\mathsf{ZFC} \vdash \text{Con}(\mathsf{ZFC}) \to \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})),$$ or equivalently, $$\mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \vdash \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})).$$ – Berrick Caleb Fillmore Apr 16 '15 at 19:57
• By Gödel’s Second Incompleteness Theorem, $\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})$ is inconsistent. This isn’t yet a contradiction as we’ve only assumed that $\mathsf{ZFC}$ is consistent. It now follows that $\mathsf{ZFC} \vdash \neg \text{Con}(\mathsf{ZFC})$, like in your post. Hence, how does this lead to the contradiction that we want in order to say that $$\mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \to \mathsf{SM}$$ without appealing to faith in the existence of inaccessible cardinals? Forgive me if I’m slow to grasp your points! – Berrick Caleb Fillmore Apr 16 '15 at 19:57
• Well, it's quite obvious that $\sf ZFC+SM\vdash\operatorname{Con}(ZFC)$. If $\sf ZFC$ has any model then it is consistent, let alone a standard model. – Asaf Karagila Apr 16 '15 at 19:58