# Showing that $ZFC$ can't prove the existence of Strongly Inaccessible Cardinals. Why the extra steps?

It is possible I am missing something exceedingly obvious. We know that $$V_{\kappa} \vDash ZFC$$ when $$\kappa$$ is strongly inaccessible.

Pretty much every proof of $$ZFC \not \vdash "\exists$$ inaccessible cardinals" goes like:

Suppose $$ZFC \vdash "\exists$$ strong inaccessibles" (I'll shorten this from now one as $$\exists I$$). Then pick the least such one, $$\kappa$$. $$V_{\kappa} \vDash "\exists I"$$. By standard absoluteness, what $$V_{\kappa}$$ thinks are strongly inaccessible will indeed be strongly inaccessibles, contradicting the choice of $$\kappa$$ as the smallest such cardinal.

And this argument is fine to me. But I don't see why we have to go through the trouble of estabilishing absoluteness, picking the smallest one and everything. Why can't we simply do:

Suppose $$ZFC \vdash "\exists I"$$. Since we may formalize model theory, the truth predicate for set models, consistency statements, etc, we have that $$ZFC \vdash "\exists I" \rightarrow Con(ZFC)$$ (by exhibiting a set model of $$ZFC$$). Then we have by basic modus ponens that $$ZFC \vdash Con(ZFC)$$. This contradicts Godel's incompleteness theorem and then we're done.

This argument seems fine to me, and more natural too. However, most arguments I can see seem to follow the first argument instead. And the first one isn't hard per se, but it makes me wonder why we're doing all these extra steps.

Am I missing something? Does the second argument not work?

By contrast, in the first approach we give a method for building models of $$\mathsf{ZFC}$$ + "There are no inaccessibles" from models of $$\mathsf{ZFC}$$. Moreover, this method "preserves niceness" in many senses, e.g. if the starting $$\mathsf{ZFC}$$-model is well-founded then the $$\mathsf{ZFC\neg I}$$-model is well-founded as well. Note that this is not something incompleteness can do for us alone - e.g. there is no nice way to take a well-founded model of $$\mathsf{ZFC}$$ and produce a well-founded model of $$\mathsf{ZFC+\neg Con(ZFC)}$$ since there are no well-founded models of $$\mathsf{ZFC+\neg Con(ZFC)}$$ in the first place.
• Why is there no well founded model of $ZFC + \neg Con(ZFC)$? From what little I know, the only thing I can tell about a model of $ZFC + \neg Con(ZFC)$ is that it has non standard natural numbers Jan 19, 2021 at 2:32
• @WhyIsSetTheorySoHard If it's got nonstandard natural numbers, it's not well-founded! (The interesting situation is the converse: ill-founded $\omega$-models of $\mathsf{ZFC}$ exist and are quite fun beasties, but that's highly nontrivial.) Jan 19, 2021 at 2:32
• I don't see how that follows :( I suspect I'm missing something stupid again. As I understand it, there will just be a bunch of $\mathbb{Z}$ chains after $\mathbb{N}$ which code "contradictions". Jan 19, 2021 at 2:34