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Let $T$ be a finite collection of axioms of $\mathrm{ZF}$, let $\sigma$ be a sentence in the language of $\mathrm{ZF}$ and consider the statement

$\tau$: “any transitive countable model of $T$ satisfies $\sigma$”.

Then $\mathrm{ZFC}\vdash\tau$ implies $\mathrm{ZFC}\vdash\sigma$, by the classical argument using Reflection, Downward Löwenheim–Skolem and Mostowki’s collapse to get a countable transitive model where finitely many sentences are absolute.

My question is: does $\mathrm{ZF}\vdash\tau$ imply $\mathrm{ZF}\vdash\sigma$? This may look trivial but, without choice, Downward Löwenheim–Skolem can’t be used as above. On the other hand, it could be argued that $\mathrm{ZF}\vdash$ “there is a proof of $\sigma$ from $T$”, but this does not mean that such a proof can be found in the meta-theory.

Thank you for your help with this (possibly trivial) matter.

Emil Jeřábek
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dragoon
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    I don’t know about the rest, but ZF certainly proves reflection in the form “if there is a proof of $\sigma$ in $T$, then $\sigma$”. This holds for any sequential theory that proves induction for all formulas in its language, see https://mathoverflow.net/a/87249 . – Emil Jeřábek Nov 14 '19 at 15:06
  • I'm unclear about your assertion concerning ZFC. I think it's consistent that every transitive model of a large enough fragment $T$ of ZFC is also a model of $\mathrm{Con}(\mathsf{ZFC})$. But as we all know, $\mathrm{Con}(\mathsf{ZFC})$ is not a theorem of ZFC. The point is that you might need non-well-founded (and hence not transitive) models to get a model for $T + \neg \mathrm{Con}(\mathsf{ZFC})$. – Will Brian Nov 14 '19 at 15:19
  • @WillBrian See e.g. Remark 3 after Theorem 12.14 in Jech’s Set Theory (2006 edition, if it matters). – Emil Jeřábek Nov 14 '19 at 15:30
  • @EmilJeřábeksupportsMonica: OK, the remark says that if $\sigma$ is true then there is a transitive model for $T + \sigma$, where $T$ is any finite sub-theory of ZFC. This is the reflection/L-S/collapse argument that the OP hints at. My point is that maybe it's possible to have $\sigma$ be true, but not a theorem of ZFC, while at the same time having every countable transitive model of ZFC be a model for $\sigma$. I don't see anything in this remark that contradicts that possibility. Am I missing something? – Will Brian Nov 14 '19 at 15:37
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    @WillBrian Work inside ZFC, and assume $\neg\sigma$. By Jech’s remark, there exists a countable transitive model of $T+\neg\sigma$. But by assumption, all countable transitive models of $T$ satisfy $\sigma$, a contradiction. Thus, $\sigma$. (By the way, this actually shows the stronger statement $\mathrm{ZFC}\vdash\tau\to\sigma$.) – Emil Jeřábek Nov 14 '19 at 15:53
  • @EmilJeřábeksupportsMonica: Your argument shows that if $\neg \sigma$ is false, then there is a countable transitive model for $T + \neg \sigma$. Or, a little more generally, if $\sigma$ is not a theorem of ZFC, then there is a universe $\tilde V$ such that in that universe, there is a countable transitive model for $T + \neg \sigma$. But this is not how I interpret the OP's claim. What I think he's claiming is that inside some fixed universe $V$ (the real one, if you think that way), if every countable transitive model satisfies $T + \sigma$, then $\sigma$ is a theorem of ZFC . . . – Will Brian Nov 14 '19 at 16:07
  • . . . but it seems plausible to me that there is a statement $\sigma$ (for example, $\mathrm{Con}(\mathsf{ZFC})$) such that $\sigma$ is true in $V$, and in fact true in every countable transitive model of $T$, but at the same time $\sigma$ is not a theorem of ZFC. Again, please correct me if I'm missing something, but I don't see how your previous comment addresses this. – Will Brian Nov 14 '19 at 16:08
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    @WillBrian That’s not what the OP is claiming. The claim is absolutely clear: $\mathrm{ZFC}\vdash\tau$ implies $\mathrm{ZFC}\vdash\sigma$. I have shown that even $\mathrm{ZFC}\vdash(\tau\to\sigma)$. Your interpretation is $\mathrm{ZFC}\vdash(\tau\to\mathrm{Pr}_{\mathrm{ZFC}}(\sigma))$, which may be false. – Emil Jeřábek Nov 14 '19 at 16:16
  • Yes, you're right -- I was misreading the OP's claim. Thanks for being patient and helping me get that sorted out. – Will Brian Nov 14 '19 at 16:19
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    See https://mathoverflow.net/questions/269682/absoluteness-reflection-to-ctms-and-choice-in-outer-models – Elliot Glazer Nov 14 '19 at 21:34
  • Thanks for the many useful comments (btw, what is OP?).

    @EmilJeřábeksupportsMonica thanks for the observation that $\mathrm{ZFC}\vdash(\tau\Rightarrow\mathrm{Pr}_{\mathrm{ZFC}}(\sigma))$ may be false. I somewhat was considering it be to be true, so that $\sigma$ could be obtained as mentioned in your first comment, but now I see this is not the way.

     – dragoon Nov 14 '19 at 23:34
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    @ElliotGlazer The answer to the post you suggested indicates a way to answer my question affirmatively. Thank you. – dragoon Nov 14 '19 at 23:34
  • @dragoon, OP = original post or original poster, i.e. you. –  Nov 15 '19 at 05:03

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