Explanation of $\mathrm{ZFC} + \lnot \mathrm{Con(ZFC)}$

Question

I read in JDH's answer to this mathoverflow question that $\mathrm{ZFC}$ is equiconsistent with $\mathrm{ZFC} + \lnot \mathrm{Con(ZFC)}$. But this second statement is a bit weird. What does $\mathrm{ZFC} + \lnot \mathrm{Con(ZFC)}$ mean within set theory such that it is equiconsistent with $\mathrm{ZFC}$ and not just inconsistent?

Answer 1

It means that there are models of $\sf ZFC$ if and only if there are models of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$.

Of course one implication is immediate, if there are models for any theory extending $\sf ZFC$, then they also satisfy $\sf ZFC$ itself. In the other direction we can use the completeness and incompleteness theorems, namely if $\operatorname{Con}\sf (ZFC)$ was true in all the models of $\sf ZFC$, then it would prove its own consistency in contradiction to the second incompleteness theorem. So if $\sf ZFC$ is consistent it must be the case that $\sf ZFC+\lnot\operatorname{Con}(ZFC)$ is consistent as well.

It is worth adding that if $(M,E)$ is a model satisfying $\sf ZFC+\lnot\operatorname{Con}(ZFC)$ then the natural numbers in $M$ must include non-standard integers. Otherwise we would have that $M$ and $V$ (the universe, or the meta-theory, if you prefer to think about this finitistically) agree on basic arithmetical truths like what axioms $\sf ZFC$ have, what are the inference rules and what sort of proofs you can write from $\sf ZFC$, and so in $V$ thinks that $\sf ZFC$ is consistent (because it knows about a model for $\sf ZFC$), then $M$ must think so as well.

So in particular a model of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$ must be ill-founded.