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Has an equivalent of the set-theoretic hierarchies (arithmetical, hyperarithmetical, Levy etc.) that uses the uniqueness quantification, $\exists !$ (and its dual, $\neg\exists!\neg$) been studied somewhere? That is a hierarchy such that $\exists! x \, \phi(x)$ is $\widetilde\Sigma_{n+1}$ when $\phi$ is $\widetilde\Pi_n$ and $\neg\exists!x \, \neg\phi(x)$ is $\widetilde\Pi_{n+1}$ when $\phi$ is $\widetilde\Sigma_{n}$.

The uniqueness quantification would be defined as $\exists !x \varphi(x) := \exists x \varphi(x) \wedge \forall y (\varphi(y) \rightarrow y=x)$. So we can establish, for example in the Levy hierarchy, that $\widetilde\Sigma_{n} \subseteq \Delta_{n+1}$. From there we obtain that $\Delta_n \subseteq\widetilde\Sigma_{n+1} \subseteq \Delta_{n+2}$ and a few questions arise that are non-trivial to me: Are those inequalities strict? What is a predicate in $\Delta_{n+2} \backslash \widetilde\Sigma_{n+1}$? In $\widetilde\Sigma_{n+1}\backslash\Delta_{n}$? etc.

I'm interested in works with any such defined set-theoretic hierarchy or any results that show that this hierarchy is not so interesting, is somehow equivalent to the classical hierarchies etc. I'm also interested in such results in constructive models such as $L$, $L_{\omega_1}$, ….

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    $\begingroup$ Where did you get the idea that $\tilde\Sigma_n\subseteq\Delta_n$? Your formula (which is misbracketed) introduces two quantifier alternations, though the inner one will merge with the one from the previous level when $n>1$, but anyway, it only establishes $\tilde\Sigma_n\subseteq\Sigma_{n+1}$. In fact, if you expand $\exists!x\,\phi(x)$ as $\exists x\,\phi(x)\land\forall x,y\,(\phi(x)\land\phi(y)\to x=y)$ instead, you can show $\tilde\Sigma_n\subseteq\Delta_{n+1}$ (more precisely, each $\tilde\Sigma_n$ formula is equivalent to a conjunction of a $\Sigma_n$ and a $\Pi_n$ formula). ... $\endgroup$ Commented Feb 2, 2022 at 16:39
  • $\begingroup$ ... That’s still one level higher than what you claim. Your claim is demonstrably wrong, since it is easy to see that in the arithmetical hierarchy, $\tilde\Sigma_n=\Sigma_n$. $\endgroup$ Commented Feb 2, 2022 at 16:42
  • $\begingroup$ The end of my last comment should read $\tilde\Sigma_n\supseteq\Sigma_n$. $\endgroup$ Commented Feb 2, 2022 at 16:55
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    $\begingroup$ One important ingredient in the study of the usual hierarchies is that several consecutive $\exists$ quantifiers can be replaced by a single $\exists$ (and similarly for $\forall$). But I see no way to collapse $\exists!x\,\exists!y$ into a single $\exists!$. Also, $\exists!$ behaves quite badly with respect to propositional connectives. And with respect to itself, in that $\exists!x\,\exists!y$ isn't equivalent to $\exists!y\,\exists!x$. In short, I'd expect a theory based on $\exists!$ to be a mess. $\endgroup$ Commented Feb 2, 2022 at 17:09
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    $\begingroup$ For arithmetical hierarchy, the upper bound I gave above is tight: a set is $\tilde\Sigma_n$ iff it is definable by a conjunction of a $\Sigma_n$ formula and a $\Pi_n$ formula, i.e., iff it can be written as a difference of two $\Sigma_n$ sets. $\endgroup$ Commented Feb 2, 2022 at 17:41

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