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Consider structures $(A,f)$ encoding a Boolean algebra $A$ endowed with an automorphism $f$. There is an obvious notion of isomorphism between such structures.

Consider the endomorphism $\hat{\Phi}$ of the Boolean algebra $2^\omega$ of subsets of $\omega$ given by $A\mapsto \{a\in\omega:a+1\in A\}$. It induces an automorphism $\Phi$ of the quotient Boolean algebra $2^\omega/\mathrm{fin}$, where $\mathrm{fin}$ is the ideal of finite subsets. (Under Stone duality, this corresponds to the self-homeomorphism of the Stone-Čech remainder of $\omega$ induced by $n\mapsto n+1$.)

Whether $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^{-1})$ are isomorphic is essentially unknown (see my previous related question for more details).

My question is

Are $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^{-1})$ elementary equivalent?

That is, do they satisfy the same first-order sentences (in the language of Boolean algebras endowed with an automorphism)?

Note that $\Phi^n$ has exactly $2^{|n|}$ fixed points for $n\neq 0$. In particular, $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^n)$ are not not equivalent for $|n|\ge 2$.

If the question has a positive answer, it is tempting to ask whether one can characterize simply those $(A,f)$ having the same first-order theory as $(2^\omega/\mathrm{fin},\Phi)$.

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    $\begingroup$ If they are elementary equivalent, then a back and forth argument shows that they are isomorphic under CH. (And whether they are isomorphic under CH is unknown.) So I think this is open. $\endgroup$ Feb 19, 2020 at 15:17
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    $\begingroup$ @PaulMcKenney this was indeed my hope to pass from EE to isomorphism, but at the same time if they're not EE, it should be for some "concrete"reason. So I hope this approach leads to something. $\endgroup$
    – YCor
    Feb 19, 2020 at 15:30
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    $\begingroup$ @PaulMcKenney how does the back-and-forth argument work? is $(2^\omega/\mathrm{fin},\Phi)$ $\omega_1$-saturated? since it seems to reduce to my previous question, I think this would be worth an answer. $\endgroup$
    – YCor
    Feb 19, 2020 at 16:44
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    $\begingroup$ That's what I was thinking, but it's no longer clear to me that $(2^\omega / \mathrm{fin}, \Phi)$ is saturated. $\endgroup$ Feb 19, 2020 at 22:56
  • $\begingroup$ @PaulMcKenney: I was thinking a bit more about this question today and, for what it's worth, I can show that $(2^\omega/\mathrm{fin},\Phi)$ is not $\omega_1$-saturated. I have a somewhat tedious proof of this assertion, but this comment is too short to contain it . . . but feel free to email me if you'd like some details. $\endgroup$
    – Will Brian
    Aug 11, 2020 at 14:39

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Yes, these two structures are elementarily equivalent.

This is proved as a corollary to another theorem, which states

Theorem: CH implies that $\Phi$ and $\Phi^{-1}$ are conjugate to each other in the automorphism group of $\mathcal P(\omega) / \mathrm{fin}$.

You can find a proof of this on the arXiv (here), and a short overview of the proof in my answer to this MO question. Let me add a few words explaining why the theorem quoted above implies that the structures $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$ are elementarily equivalent (even without assuming CH).

The short version is: we can force CH without changing the theory of $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$, so if they're isomorphic in the forcing extension, and their theories have not changed, they must have been elementarily equivalent to begin with.

In more detail, let $\mathbb P$ denote the poset of countable partial functions $\omega_1 \times \omega \to \omega$, ordered by extension, i.e., the usual poset for forcing CH with countable conditions. Let $G$ be a $V$-generic filter on $\mathbb P$. Because $\mathbb P$ is a countably closed notion of forcing, no new subsets of $\omega$ are added by $\mathbb P$, which means that $V \cap \mathcal P(\omega) / \mathrm{fin} = V[G] \cap \mathcal P(\omega) / \mathrm{fin}$. Furthermore, because $\Phi$ is definable by the simple formula $\Phi([A]) = [A+1]$, the action of $\Phi$ is the same in $V$ and in $V[G]$. Thus the structures $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle^V$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle^{V[G]}$ are one in the same. It follows (by the ``absoluteness of the satisfaction relation'') the theory of $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ is the same in $V$ and in $V[G]$. The same argument applies to $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$, so the theory of $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$ is also the same in $V$ and in $V[G]$. Because $V[G] \models$ CH, the theorem quoted above implies $\langle \mathcal P(\omega) / \mathrm{fin},\Phi \rangle$ and $\langle \mathcal P(\omega) / \mathrm{fin},\Phi^{-1} \rangle$ are conjugate in $V[G]$. In particular, they are elementarily equivalent in $V[G]$. Because the theories of these two structures is the same in $V$ and in $V[G]$, this means they have the same theory in $V$ as well.

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