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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.)

Please note that this is distinct from asking whether ZFC proves

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

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

online_banana
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  • 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." – Noah Schweber Nov 01 '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. – Noah Schweber Nov 01 '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. – online_banana Nov 02 '20 at 00:18

1 Answers1

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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.

As you noted in the question, we get this implication only as a scheme, a separate statement for each instance, because we have the reflection theorem also only as a scheme.

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.

JDH
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  • 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 $(ZFC_n \vdash \phi) \rightarrow (V_\theta \models ZFC_n \ \rightarrow \ V_\theta \models \phi)$. – online_banana Nov 02 '20 at 14:01
  • I watched your great YouTube courses recently and asked a question regarding first-order logic: https://math.stackexchange.com/q/3983971. I would greatly appreciate you could take a look at that. –  Jan 13 '21 at 20:17