# Confused about $\mathsf{ZFC} \nvdash \mathrm{Con}(\mathsf{ZFC}) \to \mathrm{Con}(\mathsf{ZFC} + \mathsf{I})$

I am studying with Jech's book. He claims that

The existence of inaccessible cardinals is not provable in $$\mathsf{ZFC}$$. Moreover, it cannot be shown that the existence of inaccessible cardinals is consistent with $$\mathsf{ZFC}$$.

And he gives the proof for the second part:

To prove the second part, assume that it can be shown that the existence of inaccessible cardinals is consistent with $$\mathsf{ZFC}$$; in other words, we assume if $$\mathsf{ZFC}$$ is consistent, then so is $$\mathsf{ZFC} + \mathsf{I}$$ where $$\mathsf{I}$$ is the statement “there is an inaccessible cardinal.” We naturally assume that $$\mathsf{ZFC}$$ is consistent. Since $$\mathsf{I}$$ is consistent with $$\mathsf{ZFC}$$, we conclude that $$\mathsf{ZFC} + \mathsf{I}$$ is consistent. It is provable in $$\mathsf{ZFC} + \mathsf{I}$$ that there is a model of $$\mathsf{ZFC}$$. Thus the sentence “$$\mathsf{ZFC}$$ is consistent” is provable in $$\mathsf{ZFC} + \mathsf{I}$$. However, we have assumed that “$$\mathsf{I}$$ is consistent with $$\mathsf{ZFC}$$” is provable, and so “$$\mathsf{ZFC} + \mathsf{I}$$ is consistent” is provable in $$\mathsf{ZFC} + \mathsf{I}$$. This contradicts Gödel’s Second Incompleteness Theorem.

“it cannot be shown” means: It cannot be shown by methods formalizable in $$\mathsf{ZFC}$$.

So his proof claims that if we assume that $$\mathsf{ZFC}$$ is consistent, $$\mathsf{ZFC} \nvdash \mathrm{Con}(\mathsf{ZFC}) \to \mathrm{Con}(\mathsf{ZFC} + \mathsf{I})$$, right? I am confused here. Can we claim that $$\mathsf{ZFC} + \mathsf{I}$$ is consistent in the metatheoretical sense provided that a consistent theory($$\mathsf{ZFC}$$) proves that if itself is consistent, then $$\mathsf{ZFC} + \mathsf{I}$$ is consistent? If $$\mathsf{ZFC} \vdash \neg\mathrm{Con}(\mathsf{ZFC})$$ (then $$\mathsf{ZFC}$$ is not 1-consistent) still $$\mathsf{ZFC} \vdash \mathrm{Con}(\mathsf{ZFC}) \to \mathrm{Con}(\mathsf{ZFC} + \mathsf{I})$$ holds, but then since $$\mathsf{ZFC} + \mathsf{I} \vdash \mathrm{Con}(\mathsf{ZFC})$$, $$\mathsf{ZFC} + \mathsf{I}$$ is inconsistent.

• You raise a good point. If ZFC proves itself inconsistent, then it does prove the relative consistency of ZFC and ZFC + I... namely, it proves they're both inconsistent. (But then again, we can't trust it on these matters, since it's arithmetically unsound, if not inconsistent.) But I'm not sure I understand the question you're asking, or what point you're trying to make by showing (perfectly correctly) that if ZFC proves itself inconsistent, then ZFC+I is inconsistent. Apr 15, 2020 at 14:47
• @spaceisdarkgreen I am confused while reading his proof. To use the Gödel’s Second Incompleteness Theorem to make a contradiction, we should know that $\mathsf{ZFC} + \mathsf{I}$ is consistent metatheoretically.
– Ris
Apr 15, 2020 at 14:58
• Of course we don't know that, though. But what if it is inconsistent? What can we conclude then? Apr 15, 2020 at 15:03
• @spaceisdarkgreen If $\mathsf{ZFC} + \mathsf{I}$ is inconsistent, then $\mathsf{ZFC} + \mathsf{I} \vdash \mathrm{Con}(\mathsf{ZFC} + \mathsf{I})$ does not lead to any contradiction, so I think his proof is wrong.
– Ris
Apr 15, 2020 at 15:04
• No, but only in the corner case you mentioned where ZFC proves itself inconsistent. If ZFC+I is inconsistent, then ZFC can prove that, so the only way it can prove Con(ZFC)-> Con(ZFC+I) is to prove not Con(ZFC). Apr 15, 2020 at 15:28

After writing this answer, I realized that spaceisdarkgreen already explained this in the comment thread above; if they leave an answer, I'll delete this one.

Yes, there's an issue here. What we really have is the following:

"In $$\mathsf{ZFC}$$ (or indeed much less$$^1$$), we can prove that the following are equivalent:

1. $$\mathsf{ZFC}\not\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$$.

2. $$\mathsf{ZFC}\not\vdash \neg Con(\mathsf{ZFC})$$.

Note that the latter is intermediate between $$Con(\mathsf{ZFC})$$ and $$\Sigma_1$$-$$Sound(\mathsf{ZFC})$$ (the latter of which in turn is a very weak fragment of arithmetical soundness).

The $$\neg 2\rightarrow \neg 1$$ direction is exactly what you've observed: if $$\mathsf{ZFC}\vdash \neg Con(\mathsf{ZFC})$$, then $$\mathsf{ZFC}\vdash Con(\mathsf{ZFC})\rightarrow\varphi$$ for every sentence $$\varphi$$.

Now we want to show $$\neg1\rightarrow\neg 2$$. This basically parallels Jech's argument. There are three steps, each of which is provable in $$\mathsf{ZFC}$$ (or indeed much less):

• Monotonicity. Suppose $$\mathsf{ZFC}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$$. Then a fortiori we have $$\mathsf{ZFC+I}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$$, and so $$\mathsf{ZFC+I}\vdash Con(\mathsf{ZFC+I})$$.

• Godel's second incompleteness theorem. From this and the previous bulletpoint we get $$\neg Con(\mathsf{ZFC+I})$$.

• Note - addressing one of your comments - that no additional assumption here is necessary: "if $$\mathsf{ZFC+I}$$ is consistent then GSIT applies and so $$\mathsf{ZFC+I}$$ is inconsistent" is already a deduction of $$\neg Con(\mathsf{ZFC+I})$$.
• $$\Sigma_1$$-completeness. The previous bulletpoint implies $$\mathsf{ZFC}\vdash\neg Con(\mathsf{ZFC+I})$$. But now combining this with our original hypothesis $$\neg 1$$, we get $$\mathsf{ZFC}\vdash \neg Con(\mathsf{ZFC+I})\wedge[Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})],$$ which in turn yields $$\mathsf{ZFC}\vdash\neg Con(\mathsf{ZFC})$$ as desired.

$$^1$$Mathematical limbo - how low can we go?

As the argument above shows, we really just need our metatheory to prove three things:

• Monotonicity of $$\vdash$$.

• Godel's second incompleteness theorem.

• The $$\Sigma_1$$-completeness of $$\mathsf{ZFC}$$.

The first is basically trivial (e.g. even Robinson arithmetic does that), while this fascinating paper of Visser mentions $$\mathsf{EA}$$ as an upper bound for the third ($$\mathsf{EA}$$ is incredibly weak, as that same paper demonstrates). Meanwhile, I believe - but don't have a source for the claim - that $$\mathsf{EA}$$ also proves GSIT, which would make $$\mathsf{EA}$$ in fact a sufficient metatheory!

However, going all the way down to $$\mathsf{EA}$$ - if we even can - is really just showing off. For almost all purposes it's enough to observe that $$I\Sigma_1$$ (a weak fragment of $$\mathsf{PA}$$) is enough. $$I\Sigma_1$$ has a number of nice properties which in my opinion do make it a better stopping point than the more-famous $$\mathsf{PA}$$: basically, it's the weakest "natural" theory capable of "naturally" developing basic computability theory (for example, the provably total functions of $$I\Sigma_1$$ are exactly the primitive recursive functions). It's also finitely axiomatizable, which is sometimes quite useful. And finally, it's the first-order part of $$\mathsf{RCA_0}$$, meaning that a reduction to $$I\Sigma_1$$ fits quite nicely in the program of reverse mathematics.