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It turns out that one cannot have two fundamentally different parallel set membership relations $\in$ and $\in^*$, if they both satisfy ZF with respect to the common language, for in this case they must in fact be isomorphic. In particular, in this situation, they cannot satisfy different theories.

Theorem. If $\in$ and $\in^*$ are membership relations each satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ is isomorphic to $\langle V,\in^*\rangle$.

Proof. One can see that $\in^*$ must be well-founded, since for every set $x$ the $\in^*$-class $\{y\mid y\in x\}$ has an $\in^*$-minimal element, by the $\in^*$-foundation axiom, and such an element is an $\in^*$-minimal element of $x$. Similarly, $\in$ is well-founded with respect to $\in^*$.

If one also knows that the relations $\in^*$ and $\in$ are set-like with respect to each other, then we may consider the Mostowski collapse $$\pi(x)=\{\pi(y)\mid y\in^* x\},$$ which is an isomorphism of $\langle V,\in^*\rangle$ with some transitive class $\langle M,\in\rangle$. But in fact, we must have $M=V$ since if every element of a set $z$ is $\pi(y)$ for some $y$, then let $x$ be the $\in^*$ with elements $y$ whenever $\pi(y)\in z$. This set exists by replacement, using that $\in$ is set-like with respect to $\in^*$, and it follows that $\pi(x)=z$. So by $\in$-induction, every set is in $M$, and we have the desired isomorphism.

But for the general case, one doesn't know at first that the relations are set-like and so more care is needed. Consider the situation of $$\pi:\langle V_\alpha,\in\rangle\cong\langle V^*_{\alpha^*},\in^*\rangle,$$ where $\alpha$ is an $\in$-ordinal and $\alpha^*$ is an $\in^*$-ordinal and $V_\alpha$ and $V^*_{\alpha^*}$ are the rank-initial segments of the universe as constructed with respect to the two membership relations. Since transitive sets are rigid, these isomorphisms are unique when they exist. We can always extend to one more level by considering the power sets, which map across by their pointwise action. And we can take unions at limit stages. It cannot be that one side runs out of ordinals before the other, for in this case we would have a bijection of the whole universe to a set, either in the $\in$-universe or in the $\in^*$ universe. So this provides the desired isomorphism, as well as a proof that the relations are in fact set-like. $\Box$

Since isomorphic models have the same theory, it follows that:

Corollary. If $\in$ and $\in^*$ are membership relations satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ has the same theory as $\langle V,\in^*\rangle$.

I think this theorem is classically known. Similar ideas are used in my paper:

Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016). DOI:10.12775/LLP.2016.007, ZBL1369.03047.

It turns out that one cannot have two fundamentally different parallel set membership relations $\in$ and $\in^*$, if they both satisfy ZF with respect to the common language, for in this case they must in fact be isomorphic. In particular, in this situation, they cannot satisfy different theories.

Theorem. If $\in$ and $\in^*$ are membership relations each satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ is isomorphic to $\langle V,\in^*\rangle$.

Proof. One can see that $\in^*$ must be well-founded, since for every set $x$ the $\in^*$-class $\{y\mid y\in x\}$ has an $\in^*$-minimal element, by the $\in^*$-foundation axiom, and such an element is an $\in^*$-minimal element of $x$. Similarly, $\in$ is well-founded with respect to $\in^*$.

If one also knows that the relations $\in^*$ and $\in$ are set-like with respect to each other, then we may consider the Mostowski collapse $$\pi(x)=\{\pi(y)\mid y\in^* x\},$$ which is an isomorphism of $\langle V,\in^*\rangle$ with some transitive class $\langle M,\in\rangle$. But in fact, we must have $M=V$ since if every element of a set $z$ is $\pi(y)$ for some $y$, then let $x$ be the set with $\in^*$-elements $y$ whenever $\pi(y)\in z$. This set exists by replacement, using that $\in$ is set-like with respect to $\in^*$, and it follows that $\pi(x)=z$. So by $\in$-induction, every set is in $M$, and we have the desired isomorphism.

But for the general case, one doesn't know at first that the relations are set-like and so more care is needed. Consider the situation of $$\pi:\langle V_\alpha,\in\rangle\cong\langle V^*_{\alpha^*},\in^*\rangle,$$ where $\alpha$ is an $\in$-ordinal and $\alpha^*$ is an $\in^*$-ordinal and $V_\alpha$ and $V^*_{\alpha^*}$ are the rank-initial segments of the universe as constructed with respect to the two membership relations. Since transitive sets are rigid, these isomorphisms are unique when they exist. We can always extend to one more level by considering the power sets, which map across by their pointwise action. And we can take unions at limit stages. It cannot be that one side runs out of ordinals before the other, for in this case we would have a bijection of the whole universe to a set, either in the $\in$-universe or in the $\in^*$ universe. So this provides the desired isomorphism, as well as a proof that the relations are in fact set-like. $\Box$

Since isomorphic models have the same theory, it follows that:

Corollary. If $\in$ and $\in^*$ are membership relations satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ has the same theory as $\langle V,\in^*\rangle$.

I think this theorem is classically known. Similar ideas are used in my paper:

Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016). DOI:10.12775/LLP.2016.007, ZBL1369.03047.

If one has two parallel set-theoretic membership relations $\in$ and $\in^*$, but they are aware of one another in the sense that one has ZF for each of them in the combined language (so each relation is available as a class with respect to the other relation), then I claim that it follows that they are isomorphic. In particular, in this situation, they cannot satisfy different theories.

Proof. Define the isomorphism $\pi:\langle V,\in^*\rangle\to \langle V,\in\rangle$ by $\in^*$-recursion as follows: $$\pi(x)=\{ \pi(y)\mid y\in^* x\}.$$ What I mean is that the elements of $\pi(x)$ with respect to the $\in$-relation are the sets $\pi(y)$ for which $y\in^*x$. It follows immediately that $y\in^*x$ if and only if $\pi(y)\in \pi(x)$, and from this it follows that $\pi$ is an isomorphism of $\langle V,\in^*\rangle$ with its range under $\in$. It is now easy to see that $\pi$ is onto, because if every element of a set $z$ is of the form $\pi(y)$ for some $y$, then let $z^*$ be the set for which $y\in^*z^*$ just in case $\pi(y)\in z$. This set exists by the $\in^*$-replacement axiom using ZFC in the combined language. It follows easily that $z=\pi(z^*)$, and so there can be no $\in$-minimal element not in the range of $\pi$, and so $\pi$ is onto. So it is an isomorphism. $\Box$

Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016).  DOI:10.12775/LLP.2016.007,  ZBL1369.03047.

It turns out that one cannot have two fundamentally different parallel set membership relations $\in$ and $\in^*$, if they both satisfy ZF with respect to the common language, for in this case they must in fact be isomorphic. In particular, in this situation, they cannot satisfy different theories.

Proof. One can see that $\in^*$ must be well-founded, since for every set $x$ the $\in^*$-class $\{y\mid y\in x\}$ has an $\in^*$-minimal element, by the $\in^*$-foundation axiom, and such an element is an $\in^*$-minimal element of $x$. Similarly, $\in$ is well-founded with respect to $\in^*$.

If one also knows that the relations $\in^*$ and $\in$ are set-like with respect to each other, then we may consider the Mostowski collapse $$\pi(x)=\{\pi(y)\mid y\in^* x\},$$ which is an isomorphism of $\langle V,\in^*\rangle$ with some transitive class $\langle M,\in\rangle$. But in fact, we must have $M=V$ since if every element of a set $z$ is $\pi(y)$ for some $y$, then let $x$ be the $\in^*$ with elements $y$ whenever $\pi(y)\in z$. This set exists by replacement, using that $\in$ is set-like with respect to $\in^*$, and it follows that $\pi(x)=z$. So by $\in$-induction, every set is in $M$, and we have the desired isomorphism.

But for the general case, one doesn't know at first that the relations are set-like and so more care is needed. Consider the situation of $$\pi:\langle V_\alpha,\in\rangle\cong\langle V^*_{\alpha^*},\in^*\rangle,$$ where $\alpha$ is an $\in$-ordinal and $\alpha^*$ is an $\in^*$-ordinal and $V_\alpha$ and $V^*_{\alpha^*}$ are the rank-initial segments of the universe as constructed with respect to the two membership relations. Since transitive sets are rigid, these isomorphisms are unique when they exist. We can always extend to one more level by considering the power sets, which map across by their pointwise action. And we can take unions at limit stages. It cannot be that one side runs out of ordinals before the other, for in this case we would have a bijection of the whole universe to a set, either in the $\in$-universe or in the $\in^*$ universe. So this provides the desired isomorphism, as well as a proof that the relations are in fact set-like. $\Box$

Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016).  DOI:10.12775/LLP.2016.007,  ZBL1369.03047.

If one has two parallel set-theoretic membership relations $\in$ and $\in^*$, but they are aware of one another in the sense that one has ZF for each of them in the combined language (so each relation is available as a class with respect to the other relation), then I claim that it follows that they are isomorphic. In particular, in this situation, they cannot satisfy different theories.

Theorem. If $\in$ and $\in^*$ are membership relations each satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ is isomorphic to $\langle V,\in^*\rangle$.

Proof. Define the isomorphism $\pi:\langle V,\in^*\rangle\to \langle V,\in\rangle$ by $\in^*$-recursion as follows: $$\pi(x)=\{ \pi(y)\mid y\in^* x\}.$$ What I mean is that the elements of $\pi(x)$ with respect to the $\in$-relation are the sets $\pi(y)$ for which $y\in^*x$. It follows immediately that $y\in^*x$ if and only if $\pi(y)\in \pi(x)$, and from this it follows that $\pi$ is an isomorphism of $\langle V,\in^*>$ with its range under $\in$. It is now easy to see that $\pi$ is onto, because if every element of a set $z$ is of the form $\pi(y)$ for some $y$, then let $z^*$ be the set for which $y\in^*z^*$ just in case $\pi(y)\in z$. This set exists by the $\in^*$-replacement axiom using ZFC in the combined language. It follows easily that $z=\pi(z^*)$, and so there can be no $\in$-minimal element not in the range of $\pi$, and so $\pi$ is onto. So it is an isomorphism. $\Box$

Since isomorphic models have the same theory, it follows that:

Corollary. If $\in$ and $\in^*$ are membership relations satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ has the same theory as $\langle V,\in^*\rangle$.

I think this theorem is classically known. Similar ideas are used in my paper:

Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016). DOI:10.12775/LLP.2016.007, ZBL1369.03047.

If one has two parallel set-theoretic membership relations $\in$ and $\in^*$, but they are aware of one another in the sense that one has ZF for each of them in the combined language (so each relation is available as a class with respect to the other relation), then I claim that it follows that they are isomorphic. In particular, in this situation, they cannot satisfy different theories.

Theorem. If $\in$ and $\in^*$ are membership relations each satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ is isomorphic to $\langle V,\in^*\rangle$.

Proof. Define the isomorphism $\pi:\langle V,\in^*\rangle\to \langle V,\in\rangle$ by $\in^*$-recursion as follows: $$\pi(x)=\{ \pi(y)\mid y\in^* x\}.$$ What I mean is that the elements of $\pi(x)$ with respect to the $\in$-relation are the sets $\pi(y)$ for which $y\in^*x$. It follows immediately that $y\in^*x$ if and only if $\pi(y)\in \pi(x)$, and from this it follows that $\pi$ is an isomorphism of $\langle V,\in^*\rangle$ with its range under $\in$. It is now easy to see that $\pi$ is onto, because if every element of a set $z$ is of the form $\pi(y)$ for some $y$, then let $z^*$ be the set for which $y\in^*z^*$ just in case $\pi(y)\in z$. This set exists by the $\in^*$-replacement axiom using ZFC in the combined language. It follows easily that $z=\pi(z^*)$, and so there can be no $\in$-minimal element not in the range of $\pi$, and so $\pi$ is onto. So it is an isomorphism. $\Box$

Since isomorphic models have the same theory, it follows that:

Corollary. If $\in$ and $\in^*$ are membership relations satisfying the ZF axioms in the combined language, then $\langle V,\in\rangle$ has the same theory as $\langle V,\in^*\rangle$.

I think this theorem is classically known. Similar ideas are used in my paper:

Hamkins, Joel David; Kikuchi, Makoto, Set-theoretic mereology, Log. Log. Philos. 25, No. 3, 285-308 (2016). DOI:10.12775/LLP.2016.007, ZBL1369.03047.