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Let $M\equiv N$ means that $(M,\in_M)$ and $(N,\in_N)$ satisfy the same sentences of the language of set theory, with $\in_M$ and $\in_N$ being the standard membership relation restricted to $M\times M$ and $N\times N$, respectively.

My question is about what level of Lévy hierarchy the formula $\Psi(\alpha,\beta):= (V_\alpha \equiv V_\beta) $ is in. At first, I thought that was in $\Pi_1$, but after some reflection, I see it is wrong.

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This statement has complexity $\Delta_2$, because it is a locally verifiable feature, meaning one that can be decided yes-or-no inside any sufficiently large $V_\theta$. Any $V_\theta$ above $\alpha$ and $\beta$ will correctly determine whether $V_\alpha\equiv V_\beta$.

I wrote a blog post about local properties in set theory, which explains further details about the complexity calculation.

Let me add that most of the complexity of $V_\alpha\equiv V_\beta$, however, as a relation on $\alpha$ and $\beta$, is just knowing that you have the right $V_\alpha$ and $V_\beta$. In contrast, the relation $M\equiv N$, as a relation on $M$ and $N$, is $\Delta_1$, since you just have to say that the unique satisfaction relations for these structures fulfill the same sentences. Because the satisfaction relations are unique in fulfilling a certain $\Delta_0$ expressible recursion, we can express this in a $\Sigma_1$ way or a $\Pi_1$ way, making them $\Delta_1$. Consequently, your $\Pi_1$ case would be fine if we view it as a relation on pairs $(V_\alpha,V_\beta)$, rather than as a relation on $(\alpha,\beta)$.

One can see that $V_\alpha\equiv V_\beta$ is not complexity $\Pi_1$ in $(\alpha,\beta)$ in all models of set theory by observing that it is not always downwards absolute. For example, perform an Easton-support product to form an extension $V[G]$ by forcing a violation of GCH at every regular cardinal. Now by cardinality considerations there will be ordinals $\alpha\neq \beta$ such that $V[G]_\alpha$ and $V[G]_\beta$ have the same theory, where $\alpha=\kappa+3$ for some regular cardinal $\kappa$. If $V[G^-]\subseteq V[G]$ is the submodel that omits the forcing at coordinate $\kappa$, then $V[G^-]_\alpha$ is not elementarily equivalent to $V[G^-]_\beta$, since the former knows that the missing coordinate is very near the top, but the latter doesn't.

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    $\begingroup$ Answome answer professor. This argument implies that $ V_\alpha\prec V_\beta $ is $\ \Delta_2 $ too? I wil love if you can indicate material to I learn more about these topics $\endgroup$ Apr 14 at 21:16
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    $\begingroup$ Yes, that's right. For this particular kind of calculation, it seems to be mainly folklore, and that is part of why I had written that blog post. So I think that is the best place to start. $\endgroup$ Apr 14 at 21:22

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