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Let T be a finite collection of axioms of ZF, let σ be a sentence in the language of ZF and consider the statement

τ: “any transitive countable model of T satisfies σ”.

Then ZFCτ implies ZFCσ, 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 ZFτ imply ZFσ? 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 ZF “there is a proof of σ 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.

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 σ in T, then σ”. 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 Con(ZFC). But as we all know, Con(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+¬Con(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 σ is true then there is a transitive model for T+σ, 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 σ be true, but not a theorem of ZFC, while at the same time having every countable transitive model of ZFC be a model for σ. 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 ¬σ. By Jech’s remark, there exists a countable transitive model of T+¬σ. But by assumption, all countable transitive models of T satisfy σ, a contradiction. Thus, σ. (By the way, this actually shows the stronger statement ZFCτσ.) – Emil Jeřábek Nov 14 '19 at 15:53
  • @EmilJeřábeksupportsMonica: Your argument shows that if ¬σ is false, then there is a countable transitive model for T+¬σ. Or, a little more generally, if σ is not a theorem of ZFC, then there is a universe ˜V such that in that universe, there is a countable transitive model for T+¬σ. 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+σ, then σ is a theorem of ZFC . . . – Will Brian Nov 14 '19 at 16:07
  • . . . but it seems plausible to me that there is a statement σ (for example, Con(ZFC)) such that σ is true in V, and in fact true in every countable transitive model of T, but at the same time σ 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: ZFCτ implies ZFCσ. I have shown that even ZFC(τσ). Your interpretation is ZFC(τPrZFC(σ)), 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 ZFC(τPrZFC(σ)) may be false. I somewhat was considering it be to be true, so that σ 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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