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We have the result that $ZFCfin$$\mathsf{ZFCfin}$, the usual $ZFC$$\mathsf{ZFC}$ axioms with the axiom of infinity replaced by its negation, is bi-interpretable with $PA$$\mathsf{PA}$, first order Peano Arithmetic. We also know of a natural way to weaken $PA$$\mathsf{PA}$ into fragments, by restricting the induction axiom schema to forumlae of a specific complexity, so $Q+I\Sigma_3$$\mathsf{Q+I\Sigma_3}$ is the non inductive axioms of $PA$$\mathsf{PA}$ plus induction restricted to formulae of at most $\Sigma_3^0$$\mathsf{\Sigma_3^0}$ complexity in the language of first order arithmetic.

Does weakening the axiom of separation and the axiom of replacement in $ZFCfin$$\mathsf{ZFCfin}$ result in the above outlined fragments of PA$\mathsf{PA}$? For example, does weakening the two axiom schema to formulae of $\Sigma_3$$\mathsf{\Sigma_3}$ complexity in the language of set theory give a set theory bi-interpretable with $Q+I\Sigma_3$$\mathsf{Q+I\Sigma_3}$? Or does the axiom of powerset also need to be dropped and then replacement replaced with collection?*

If restricting separation and replacement is the correct way of getting bi-interpretable theories, does a theory bi-interpretable with $Q$$\mathsf{Q}$, Robinson Arithmetic, result from dropping both axiom schemes entirely?

*The reason I say this is because I know that $KP$$\mathsf{KP}$, Kripke-Platek set theory, does away with powerset and without powerset collection and replacement are not equivalent. I am wondering if that plays a factor and if it isn't too much for one question, if anyone can explain the interplay between getting rid of powerset and the strength of fragments of arithmetic.

We have the result that $ZFCfin$, the usual $ZFC$ axioms with the axiom of infinity replaced by its negation, is bi-interpretable with $PA$, first order Peano Arithmetic. We also know of a natural way to weaken $PA$ into fragments, by restricting the induction axiom schema to forumlae of a specific complexity, so $Q+I\Sigma_3$ is the non inductive axioms of $PA$ plus induction restricted to formulae of at most $\Sigma_3^0$ complexity in the language of first order arithmetic.

Does weakening the axiom of separation and the axiom of replacement in $ZFCfin$ result in the above outlined fragments of PA? For example, does weakening the two axiom schema to formulae of $\Sigma_3$ complexity in the language of set theory give a set theory bi-interpretable with $Q+I\Sigma_3$? Or does the axiom of powerset also need to be dropped and then replacement replaced with collection?*

If restricting separation and replacement is the correct way of getting bi-interpretable theories, does a theory bi-interpretable with $Q$, Robinson Arithmetic, result from dropping both axiom schemes entirely?

*The reason I say this is because I know that $KP$, Kripke-Platek set theory, does away with powerset and without powerset collection and replacement are not equivalent. I am wondering if that plays a factor and if it isn't too much for one question, if anyone can explain the interplay between getting rid of powerset and the strength of fragments of arithmetic.

We have the result that $\mathsf{ZFCfin}$, the usual $\mathsf{ZFC}$ axioms with the axiom of infinity replaced by its negation, is bi-interpretable with $\mathsf{PA}$, first order Peano Arithmetic. We also know of a natural way to weaken $\mathsf{PA}$ into fragments, by restricting the induction axiom schema to forumlae of a specific complexity, so $\mathsf{Q+I\Sigma_3}$ is the non inductive axioms of $\mathsf{PA}$ plus induction restricted to formulae of at most $\mathsf{\Sigma_3^0}$ complexity in the language of first order arithmetic.

Does weakening the axiom of separation and the axiom of replacement in $\mathsf{ZFCfin}$ result in the above outlined fragments of $\mathsf{PA}$? For example, does weakening the two axiom schema to formulae of $\mathsf{\Sigma_3}$ complexity in the language of set theory give a set theory bi-interpretable with $\mathsf{Q+I\Sigma_3}$? Or does the axiom of powerset also need to be dropped and then replacement replaced with collection?*

If restricting separation and replacement is the correct way of getting bi-interpretable theories, does a theory bi-interpretable with $\mathsf{Q}$, Robinson Arithmetic, result from dropping both axiom schemes entirely?

*The reason I say this is because I know that $\mathsf{KP}$, Kripke-Platek set theory, does away with powerset and without powerset collection and replacement are not equivalent. I am wondering if that plays a factor and if it isn't too much for one question, if anyone can explain the interplay between getting rid of powerset and the strength of fragments of arithmetic.

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Is restricting FoundationReplacement and Separation enough to make $Q+I\Sigma_n$ bi-interpretable with Set Theory?

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We have the result that $ZFCfin$, the usual $ZFC$ axioms with the axiom of infinity replaced by its negation, is bi-interpretable with $PA$, first order Peano Arithmetic. We also know of a natural way to weaken $PA$ into fragments, by restricting the induction axiom schema to forumlae of a specific complexity, so $Q+I\Sigma_3$ is the non inductive axioms of $PA$ plus induction restricted to formulae of at most $\Sigma_3^0$ complexity in the language of first order arithmetic.

Does weakening the axiom of separation and the axiom of replacement in $ZFCfin$ result in the above outlined fragments of PA? For example, does weakening the two axiom schema to formulae of $\Sigma_3$ complexity in the language of set theory give a set theory bi-interpretable with $Q+I\Sigma_3$? Or does the axiom of powerset also need to be dropped and then replacement replaced with collection?*

If restricting separation and replacement is the correct way of getting bi-interpretable theories, does a theory bi-interpretable with $Q$, Robinson Arithmetic, result from dropping both axiom schemes entirely?

*The reason I say this is because I know that $KP$, Kripke-Platek set theory, does away with powerset and without powerset collection and replacement are not equivalent. I am wondering if that plays a factor and if it isn't too much for one question, if anyone can explain the interplay between getting rid of powerset and the strength of fragments of arithmetic.

We have the result that $ZFCfin$, the usual $ZFC$ axioms with the axiom of infinity replaced by its negation, is bi-interpretable with $PA$, first order Peano Arithmetic. We also know of a natural way to weaken $PA$ into fragments, by restricting the induction axiom schema to forumlae of a specific complexity, so $Q+I\Sigma_3$ is the non inductive axioms of $PA$ plus induction restricted to formulae of at most $\Sigma_3^0$ complexity in the language of first order arithmetic.

Does weakening the axiom of separation and the axiom of replacement in $ZFCfin$ result in the above outlined fragments of PA? For example, does weakening the two axiom schema to formulae of $\Sigma_3$ complexity in the language of set theory give a set theory bi-interpretable with $Q+I\Sigma_3$? Or does the axiom of powerset also need to be dropped and then replacement replaced with collection?

If restricting separation and replacement is the correct way of getting bi-interpretable theories, does a theory bi-interpretable with $Q$, Robinson Arithmetic, result from dropping both axiom schemes entirely?

We have the result that $ZFCfin$, the usual $ZFC$ axioms with the axiom of infinity replaced by its negation, is bi-interpretable with $PA$, first order Peano Arithmetic. We also know of a natural way to weaken $PA$ into fragments, by restricting the induction axiom schema to forumlae of a specific complexity, so $Q+I\Sigma_3$ is the non inductive axioms of $PA$ plus induction restricted to formulae of at most $\Sigma_3^0$ complexity in the language of first order arithmetic.

Does weakening the axiom of separation and the axiom of replacement in $ZFCfin$ result in the above outlined fragments of PA? For example, does weakening the two axiom schema to formulae of $\Sigma_3$ complexity in the language of set theory give a set theory bi-interpretable with $Q+I\Sigma_3$? Or does the axiom of powerset also need to be dropped and then replacement replaced with collection?*

If restricting separation and replacement is the correct way of getting bi-interpretable theories, does a theory bi-interpretable with $Q$, Robinson Arithmetic, result from dropping both axiom schemes entirely?

*The reason I say this is because I know that $KP$, Kripke-Platek set theory, does away with powerset and without powerset collection and replacement are not equivalent. I am wondering if that plays a factor and if it isn't too much for one question, if anyone can explain the interplay between getting rid of powerset and the strength of fragments of arithmetic.

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