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Nov 17, 2022 at 5:38 vote accept Dylan Pizzo
Nov 17, 2022 at 5:38 comment added Dylan Pizzo Nice! My intuition of the answer is that we could define the "same elements" relation $$x \equiv y \leftrightarrow \forall t (t\in x \leftrightarrow t\in y)$$ Notice that this can can be flexibly stratified with any assignment of values for $x$ and $y$. Thus, the flexibly stratified formula $\exists y (y \equiv x \wedge y \not\in x)$ is a form of Russell's Paradox. In fact, with extensionality, it's exactly the same, but it looks like the paradox goes through even without extensionality, which is really cool!
Nov 17, 2022 at 3:48 comment added bof Very nice. It might be slightly simpler to say that $x\in y$ is logically equivalent to $$\exists z[x\in z\land\forall w(w\in z\to w\in y)],$$ whence $x\notin x$ is logically equivalent to the "flexibly stratified" formula $$\forall z[x\in z\to\exists w(w\in z\land w\notin x)].$$
Nov 16, 2022 at 19:38 history edited Emil Jeřábek CC BY-SA 4.0
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Nov 16, 2022 at 19:06 history answered Greg Kirmayer CC BY-SA 4.0