TITLE: Is restricting Replacement and Separation enough to make $Q+I\Sigma_n$ bi-interpretable with Set Theory?
QUESTION [8 upvotes]: 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.

REPLY [9 votes]: First let me note that one should be careful with formulation of $\mathsf{ZFCfin}$, for it to be bi-interpretable with $\mathsf{PA}$ (see the paper "On interpretations of arithmetic and set theory" by Richard Kaye and Tin Lok Wong and the paper "$\omega$-models of finite set theory" by Ali Enayat, James H. Schmerl, and Albert Visser). Basically the issue is that for this bi-interpretation to work fine one either need to explicitly add to $\mathsf{ZFCfin}$ the axiom that every set is contained in a transitive set ($\mathsf{TC}$), or alternatively start with the axiomatization of $\mathsf{ZFC}$ where we have scheme of foundation instead of the axiom of regularity. 
For fragments the situation is roughly speaking that the scheme $\Sigma_n\mbox{-}\mathsf{Sep}$ in set theory corresponds to the scheme $\Sigma_n\mbox{-}\mathsf{Ind}$ in arithmetic and the scheme $\Sigma_n\mbox{-}\mathsf{Rep}$ in set theory corresponds to the scheme $\mathsf{B}\Sigma_n$ in arithmetic. More formally, let us choose the following base system of set theory
 $$\mathsf{ZFfin}_1=\mathsf{Ext}+\mathsf{Pair}+\mathsf{Union}+\mathsf{TC}+\mathsf{Reg}+\Sigma_1\mbox{-}\mathsf{Sep}+\Sigma_1\mbox{-}\mathsf{Rep}+\lnot\mathsf{Inf}.$$
Note that although I haven't included powerset axiom in $\mathsf{ZFfin}_1$, it is provable there.
This system is interpretable in $\mathsf{I}\Sigma_1$ by the Ackermann membership $\in_{\mathsf{Ack}}$ that is defined as follows: $$n\in_{\mathsf{Ack}} m \mbox{ iff the $n$-th bit in the binary expansion of $m$ is equal to $1$}.$$
In the other direction $\mathsf{ZFfin}_1$ interpretes $\mathsf{I}\Sigma_1$ by the ordinal arithmetics. With some efforts one could show that this two interpretations form a bi-interpretation. Further, for $n\ge 1$,  it is easy to show that the theory $\mathsf{I}\Sigma_n=\mathsf{I}\Sigma_1+\mathsf{I}\Sigma_n$  proves that $\Sigma_n\mbox{-}\mathsf{Sep}$ holds in Ackermann interpretation and that the theory $\mathsf{ZFfin}_1+\Sigma_n\mbox{-}\mathsf{Sep}$  proves that $\Sigma_n\mbox{-}\mathsf{Sep}$ holds in ordinal arithmetic. This verifies the fact that $\mathsf{I}\Sigma_n$ and $\mathsf{ZFfin}_1+\Sigma_n\mbox{-}\mathsf{Sep}$  are bi-interpretable. By the same kind of argument $\mathsf{I}\Sigma_1+\mathsf{B}\Sigma_n$ and $\mathsf{ZFfin}_1+\Sigma_n\mbox{-}\mathsf{Rep}$  are bi-interpretable. 
Note that the connection that I have outlined above really required using relatively strong theories: we need totality of exponentiation in arithmetic to prove even very basic facts about $\in_{\mathsf{Ack}}$ and due to this the approach wouldn't work if our base set theory wouldn't be able to prove totality of exponentiation in ordinal arithmetic. With a more refined approach it is possible to show bi-interpretability $\mathsf{I}\Delta_0+\mathsf{Exp}$ ($\mathsf{Exp}$ states totality of binary exponentiation function) and certain set theory that includes powerset axiom (see R. Pettigrew "On Interpretations of Bounded Arithmetic and Bounded Set Theory").  Although, strictly speaking, I don't know whether $\mathsf{Q}$ is bi-interpretable with $\mathsf{ZFCfin}-\mathsf{Sep}-\mathsf{Rep}$, it would be very strange for it to be the case.