# Provable soundness of finite fragments of ZFC

Fix some reasonable computable enumeration of the axioms of ZFC, and let $$ZFC_n$$ be the theory consisting of the first n axioms.

Is it the case that, for each natural number n, and each sentence $$\phi$$, ZFC proves the sentence

$$(ZFC_n \vdash \phi ) \rightarrow \phi$$?

(Where we formalize provability in ZFC in some reasonable way.)

$$\forall n \forall \phi \ : \ (ZFC_n \vdash \phi ) \rightarrow \phi$$

(Which ZFC trivially does not, if it is consistent.)

• Yes, this is true. The proof is basically a variation of the proof of the reflection theorem, that $\mathsf{ZFC}$ proves the consistency of each of its finite fragments. (I'm adding this as a comment rather than an answer since that variation isn't immediate; if I can find the time later I'll write an answer with the prof.) Amusingly, this argument is internal to $\mathsf{ZFC}$! That is, $\mathsf{ZFC}$ proves "$\mathsf{ZFC}$ proves the soundness of each of its finite fragments." Nov 1 '20 at 20:48
• Of course, this is again different from $\mathsf{ZFC}$ proving "each of $\mathsf{ZFC}$'s finite fragments is sound," which $\mathsf{ZFC}$ can't do unless it's inconsistent. Nov 1 '20 at 20:51
• @Noah thanks for the reply. I would def be interested to see the nontrivial adaptation of the RT proof. I had come to believe that the statement was true, and had tried to prove it without doing "real work" (as a simple corollary of various theorems) but was unable to. Nov 2 '20 at 0:18

Consider any finite list $$\Phi$$ of axioms of ZFC and any other sentence $$\phi$$. By the Lévy-Montague reflection theorem, there is some rank-initial segment $$V_\theta$$ of the universe for which all the sentences in $$\Phi$$ and also $$\phi$$ are absolute between $$V_\theta$$ and $$V$$. Since the sentences of $$\Phi$$ are part of ZFC, they are true in $$V$$ and hence also in $$V_\theta$$. In particular, $$V$$ looks upon $$V_\theta$$ as a model of $$\Phi$$, according to the truth predicate that it can define for this set structure. Therefore, if $$V$$ thinks that $$\Phi\vdash\phi$$, then it will think that $$V_\theta\models\phi$$. Since $$\theta$$ was chosen so that this sentence is absolute, this implies $$\phi$$ holds in $$V$$, as desired. So we've established any instance of the implication.
Addendum. Let me explain that one can also strengthen the conclusion somewhat, by assuming not that the sentences of $$\Phi$$ are part of ZFC, but rather merely that they are true. In other words, I claim that ZFC proves every instance of the scheme: $$(\psi\vdash\phi)\to(\psi\to\phi).$$ If we take $$\psi$$ to be the conjunction of the sentences in $$\Phi$$, this generalizes your scheme. But the same proof works here. By the Lévy-Montague reflection theorem, there is $$V_\theta$$ for which both $$\psi$$ and $$\phi$$ are absolute between $$V_\theta$$ and $$V$$. Now, if $$\psi\vdash\phi$$ and $$\psi$$ is true (in $$V$$), then $$\psi$$ is true in $$V_\theta$$, and so $$\phi$$ also is true there, and so $$\phi$$ is true in $$V$$, as desired.
• Thanks JDH. I had come up with something similar, but got confused because I thought this argument only showed that $\phi$ is true in V, and not that ZFC proves $\phi$ from the assumption $ZFC_n \vdash \phi$. But now I realize that ZFC actually proves $(V_\theta \models \phi) \leftrightarrow \phi$, and also $(ZFC_n \vdash \phi) \rightarrow (V_\theta \models ZFC_n \ \rightarrow \ V_\theta \models \phi)$. Nov 2 '20 at 14:01