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Is there a way of saying in second order arithmetic that the number of sets $X$ such that $\phi$ equals the number of sets $X$ such that $\psi$, where $\phi$ and $\psi$ are formulas with $X$ free, and where we don't care about the distinction between different infinite cardinalities (i.e., number is something in $\omega$ or $+\infty$)?

We can define the concept of there being finitely many $X$ such that $\phi$ in second order logic (or more generally monadic second order logic with an infinity predicate for sets) using the trick in the proof of Proposition 7 of Bárány, Kaiser, and Rabinovich - Expressing cardinality quantifiers in monadic second-order logic over chains, so the only interesting case is where $\phi$ and $\psi$ both have finite numbers of satisfiers. (The same paper's Corollary 18 shows that we can define the uncountability of the number of sets satisfying a formula in MSO on $(\mathbb N,<)$ so given the Continuum Hypothesis, we don't need the restriction that "number" doesn't care about which infinite cardinality we have, and the finite case is still the only case to consider.)

Is there a way of saying in second order arithmetic that the number of sets $X$ such that $\phi$ equals the number of sets $X$ such that $\psi$, where $\phi$ and $\psi$ are formulas with $X$ free, and where we don't care about the distinction between different infinite cardinalities (i.e., number is something in $\omega$ or $+\infty$)?

We can define the concept of there being finitely many $X$ such that $\phi$ in second order logic (or more generally monadic second order logic with an infinity predicate for sets) using the trick in the proof of Proposition 7 of Bárány, Kaiser, and Rabinovich - Expressing cardinality quantifiers in monadic second-order logic over chains, so the only interesting case is where $\phi$ and $\psi$ both have finite numbers of satisfiers.

Is there a way of saying in second order arithmetic that the number of sets $X$ such that $\phi$ equals the number of sets $X$ such that $\psi$, where $\phi$ and $\psi$ are formulas with $X$ free, and where we don't care about the distinction between different infinite cardinalities (i.e., number is something in $\omega$ or $+\infty$)?

We can define the concept of there being finitely many $X$ such that $\phi$ in second order logic (or more generally monadic second order logic with an infinity predicate for sets) using the trick in the proof of Proposition 7 of Bárány, Kaiser, and Rabinovich - Expressing cardinality quantifiers in monadic second-order logic over chains, so the only interesting case is where $\phi$ and $\psi$ both have finite numbers of satisfiers.

Is there a way of saying in second order arithmetic that the number of sets $X$ such that $\phi$ equals the number of sets $X$ such that $\psi$, where $\phi$ and $\psi$ are formulas with $X$ free, and where we don't care about the distinction between different infinite cardinalities (i.e., number is something in $\omega$ or $+\infty$)?

We can define the concept of there being finitely many $X$ such that $\phi$ in second order logic (or more generally monadic second order logic with an infinity predicate for sets) using the trick in the proof of Proposition 7 of Bárány, Kaiser, and Rabinovich - Expressing cardinality quantifiers in monadic second-order logic over chains, so the only interesting case is where $\phi$ and $\psi$ both have finite numbers of satisfiers. (The same paper's Corollary 18 shows that we can define the uncountability of the number of sets satisfying a formula in MSO on $(\mathbb N,<)$ so given the Continuum Hypothesis, we don't need the restriction that "number" doesn't care about which infinite cardinality we have, and the finite case is still the only case to consider.)

Is there a way of saying in second order arithmetic that the number of sets $X$ such that $\phi$ equals the number of sets $X$ such that $\psi$, where $\phi$ and $\psi$ are formulas with $X$ free, and where we don't care about the distinction between different infinite cardinalities (i.e., number is something in $\omega$ or $+\infty$)?

We can define the concept of there being finitely many $X$ such that $\phi$ in second order logic (or more generally monadic second order logic with an infinity predicate for sets) using the trick in the proof of Proposition 7 here, so the only interesting case is where $\phi$ and $\psi$ both have finite numbers of satisfiers.

Is there a way of saying in second order arithmetic that the number of sets $X$ such that $\phi$ equals the number of sets $X$ such that $\psi$, where $\phi$ and $\psi$ are formulas with $X$ free, and where we don't care about the distinction between different infinite cardinalities (i.e., number is something in $\omega$ or $+\infty$)?

We can define the concept of there being finitely many $X$ such that $\phi$ in second order logic (or more generally monadic second order logic with an infinity predicate for sets) using the trick in the proof of Proposition 7 of Bárány, Kaiser, and Rabinovich - Expressing cardinality quantifiers in monadic second-order logic over chains, so the only interesting case is where $\phi$ and $\psi$ both have finite numbers of satisfiers.