Can $\sf ST$ interpret $\sf GST$?

Question

General set theory ($\sf{GST}$) is the theory defined by the axioms of Extensionality, Adjunction, and Specification (and an existence axiom, if necessary). The theory $\sf{ST}$ is just $\sf{GST}$, but with Specification replaced by Empty Set. My question is whether $\sf{ST}$ can interpret the full strength of $\sf{GST}$. If not, where can I find a proof?

I have determined that $\sf{ST}$ can interpret any finite subset of the $\sf{GST}$ axioms. Given this, if any finite fragment of $\sf{GST}$ can interpret the full $\sf{GST}$, then likewise $\sf{ST}$ can interpret $\sf{GST}$. At the very least, any theory which can at once prove all these interpretations will show that the two are equiconsistent, but this doesn't necessarily mean that they are mutually interpretable.

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EDIT: As pointed out (by me) in the comments, the following work does not actually give the interpretation I wanted, or at least, not in any way that I can see. I'm leaving it here for posterity, but keep in mind that the claims about interpretations and class models are invalid.

For the curious, my proof that $\sf{ST}$ can interpret any finite fragment of $\sf{GST}$ does so by constructing a definable class which models that fragment. In particular, for each formula $\phi(x,v_1,\cdots,v_n)$, we define a class $\mathcal{G}_\phi$ within which Specification by $\phi$ is permitted. Intersecting finitely many of these classes gives us any desired finite fragment of $\sf{GST}$. The definition of $\mathcal{G}_\phi$ is as follows. $$\mathcal{S}_\cup\equiv\{X : \forall Y, (X\cup Y) \text{ is a set}\}$$ $$\mathcal{S}_\cap\equiv\{X : \forall Y, (X\cap Y)\text{ is a set}\}$$ $$\mathcal{S}_\phi\equiv\{X : \forall(\overline{v}), \{x\in X : \phi(x,\overline{v})\}\text{ is a set}\}$$ $$\mathcal{O}_{\phi}\equiv\{X : \forall(Z\subseteq X), Z\in\mathcal{S}_\cup\cap \mathcal{S}_\cap\cap \mathcal{S}_\phi\}$$ $$\mathcal{G}_\phi\equiv\{X : \exists(T\supseteq X), T\in\mathcal{O}_\phi\land\forall(t\in T), t\subseteq T\}$$

To clarify, we identify sets with classes whenever they are extensionally equal, so an expression like "the class C is a set" just means $\exists S, \forall x, x\in S\iff x\in C$. Now, $\mathcal{S}_\cup$ is the class of all sets for which their binary union with any other set will produce another set, and $\mathcal{S}_\cap$ is defined analogously for intersection. The class $\mathcal{S}_\phi$ consists of those sets for which specification by $\phi$ is permitted, and $\mathcal{O}_\phi$ is a subclass of all three, but in such a way that $\mathcal{O}_\phi$ is downwards closed under the subset relation. Finally, the class $\mathcal{G}_\phi$ just consists of all those $X$ admitting a transitive superset within $\mathcal{O}_\phi$.

Using transitive sets guarantees that $\mathcal{G}_\phi$ is a transitive class, so the axiom of Extensionality is preserved. Clearly specification by $\phi$ will also hold in $\mathcal{G}_\phi$, and it's not hard to prove that $\emptyset\in\mathcal{G}_\phi$. Proving that $\mathcal{G}_\phi$ is closed under adjunctions is the hardest part of this, but it is possible. Once we prove all these facts, we see that $\mathcal{G}_\phi$ lets us extend $\sf{ST}$ with the singular Specification instance corresponding to $\phi$. More importantly $\mathcal{G}_\phi$ is downward closed under inclusion, so intersecting finitely many of these classes (built from finitely many different $\phi$) will give us a class which models all those instances of Specification. This is how we interpret any finite fragment of $\sf{GST}$.

Answer 1

The answer is a definite no. The reason is that $\sf{GST}$ is mutually interpretable with $\sf{PA}$, and $\sf{ST}$ is interpretable from a finite fragment of $\sf{PA}$, and $\sf{PA}$ proves that fragment is consistent. If a finite fragment of $\sf{PA}$ could interpret the full $\sf{PA}$, then $\sf{PA}$ would detect the interpretation in such a way that $\sf{PA}$ would prove itself consistent. This is impossible due to Godel incompleteness, so no finite fragment of $\sf{PA}$ could interpret $\sf{PA}$, and consequently $\sf{ST}$ cannot interpret $\sf{GST}$.

That $\sf{PA}$ is mutually interpretable with $\sf{GST}$ is a bit esoteric, but it's a well known result. Using Ackermann's bijection we prove $\sf{PA}$ is bi-interpretable with $\sf{ZF}^{-\infty}$, which is $\sf{ZF}$ with infinity replaced by its negation, and $\sf{ZF}^{-\infty}$ interprets $\sf{GST}$ trivially. The reverse direction follows by defining the class ordinal $\omega$ from within $\sf{GST}$, and then using the well-foundedness of $\omega$ to verify all the $\sf{PA}$ axioms.

That $\sf{ST}$ is interpretable from a finite fragment of $\sf{PA}$ follows from the fact that $\sf{PA}$ interprets $\sf{GST}$, and since $\sf{ST}$ is a finite fragment of $\sf{GST}$, then a finite fragment of $\sf{PA}$ is sufficient to interpret $\sf{ST}$. We'll denote this fragment by $\sf{P}$.

In pursuit of contradiction, suppose $\sf{ST}$ interprets $\sf{GST}$, then $\sf{P}$ interprets $\sf{ST}$ which interprets $\sf{GST}$ which then interprets $\sf{PA}$, so in summary $\sf{P}$ transitively interprets $\sf{PA}$. According to theorem 6.4 of the paper Arithmetization of mathematics in a general setting, there must be an enumeration $\alpha$ of $\sf{P}$ and an enumeration $\beta$ of $\sf{PA}$ such that $\sf{PA}$ proves $\operatorname{Con}_\alpha\implies \operatorname{Con}_\beta$. Since the theory $\sf{P}$ is finite, it's not hard to understand $\operatorname{Con}(\sf{P})\iff\operatorname{Con}_\alpha$, therefore $\sf{PA}$ proves $\operatorname{Con}(\sf{P})\implies\operatorname{Con}_\beta$. Moreover, theorem 5.6 of the same paper shows that $\sf{PA}$ can't prove $\operatorname{Con}_\beta$, so it follows that $\sf{PA}$ also can't prove $\operatorname{Con}(\sf{P})$. This is a contradiction however, since $\sf{PA}$ proves the consistency of all its finite subtheories. We therefore conclude that $\sf{ST}$ cannot interpret $\sf{GST}$.