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The Naive Set Theory ($\mathsf{NST}$) is an inconsistent, trivial theory, but its positive fragment was shown to be consistent and is basis of e.g. positive set theory $\mathsf{GPK}_{\infty}^{+}$ of Olivier Esser.

In the positive fragment, instances of unrestricted comprehension are restricted to only positive formulas (the smallest class of formulas containing atomic membership and equality formulas and closed under $\land$, $\lor$, $\exists$ and $\forall$):

positive comprehension: $\exists y\forall x(x\in y \leftrightarrow \varphi(x))$ s.t. $y$ is not free in $\varphi$ and $\varphi$ is a positive formula.

My question is the following: Is the sentence $\exists y[\forall x(x\in y\leftrightarrow x\in x)\land y\in y]$ provable in the positive fragment of $\mathsf{NST}$ (or at least in $\mathsf{GPK}_{\infty}^{+}$)?

Just for comparison: the sentence above is not provable in $\mathsf{ZFC}$ (since $\{x:x\in x\}$ is equal to $\emptyset$) nor $\mathsf{NF}$ (since $\{x:x\in x\}$ does not exist). In $\mathsf{ZFC{-}FA}$ it is independent.

  • Isn't it independent? If you say it is, then it is (as it satisfies), so it is (our conclusion). No contradiction. If you say it isn't, then it isn't (as it doesn't satisfy), so it isn't (our conclusion). No contradiction. – Darmani V Oct 14 '24 at 09:44
  • @DarmaniV it could be independent, that is why I added "independence" tag. the statement surely is independent in some fragments of NST, not sure about the positive fragment. – Timotej Šujan Oct 14 '24 at 09:47
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    @DarmaniV In fact, this sentence is not independent - it is provable (see the answer to the linked duplicate). This is closely related to the fact that, as a consequence of Löb's theorem, the sentence of arithmetic that intuitively says "I am provable" is provable. – Alex Kruckman Oct 14 '24 at 13:44

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