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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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  • $\begingroup$ Technically doesn't one of the seven axioms of $\mathsf{Q}$, $y = 0 \lor \exists x (Sx = y)$, not belong to the finite and non-inductive fragment of $\mathsf{PA}$? Not that it makes any difference for the question given that it appears to be already be a consequence of $\mathsf{I\Sigma_1}$. Just commenting on this for other readers like me who became confused when they conflated the non-inductive fragment of $\mathsf{PA}$ with $\mathsf{Q}$ -- they aren't quite the same. $\endgroup$ Jan 28, 2023 at 15:06

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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.

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  • $\begingroup$ This is a very dumb question which is somewhat besides the point of this wonderful and thoughtful answer, but does the definition of $\mathsf{ZFfin}$ = $\mathsf{ZFCfin}$ also require the empty set axiom in order to exclude the empty model? (Because normally existence of at least one set can be seen to follow from the axiom of infinity, which is negated here, and then empty set follows by applying separation.) Or does our semantics / metatheory preclude ever using empty models, so we're fine? $\endgroup$ Jan 28, 2023 at 4:33
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    $\begingroup$ @Chill2Macht It entirely depends on the version of first-order logic one uses. Usually, as I did in my answer, people use first-order logic that doesn't allow empty models and hence empty set axiom is not needed, since existence of some set is guaranteed by the underlying logic and next the empty set could be constructed by comprehension. However, in principle, one could have a version of first-order logic without this restriction en.wikipedia.org/wiki/Free_logic and then of course empty set axiom had to be added (or some other axiom implying the existence of at least one set). $\endgroup$ Jan 28, 2023 at 16:12

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