Theorems of ZF through countable transitive models

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.

• 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 mathoverflow.net/a/87249 . – Emil Jeřábek Nov 14 '19 at 15:06
• @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
• @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
• – Elliot Glazer Nov 14 '19 at 21:34
• @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