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May 29

Is the proposition $\exists x\,((\emptyset \in x)\land (x\in x))$ decidable in ZF ?

ZF, including the Axiom of Regularity. HOTmag (talk) 18:56, 29 May 2018 (UTC)[reply]

Yes, the axiom of foundation (which some people call "regularity") implies that no set is an element of itself. --Trovatore (talk) 21:20, 29 May 2018 (UTC)[reply]

Thanx. BTW, not only "some people" (as you call them), but also our article Axiom of regularity. HOTmag (talk) 21:48, 29 May 2018 (UTC)[reply]

Unfortunately. --Trovatore (talk) 22:43, 29 May 2018 (UTC)[reply]

It seems like you dislike this name, and I wonder why... HOTmag (talk) 22:46, 29 May 2018 (UTC)[reply]

Oh, I could come up with reasons, I suppose, but it boils down to the fact that "foundation" is the more usual name in my experience, which of course is influenced by the particular set-theoretic subculture I worked in. --Trovatore (talk) 22:48, 29 May 2018 (UTC)[reply]

May 29

Is the proposition ∃ x ( ( ∅ ∈ x ) ∧ ( x ∈ x ) ) {\displaystyle \exists x\,((\emptyset \in x)\land (x\in x))} decidable in ZF ?

Is the proposition $\exists x\,((\emptyset \in x)\land (x\in x))$ decidable in ZF ?