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In the Appendix to Part Zero of On Numbers and Games, John Horton Conway gives what he says is an equivalent version of the axiom of regularity in ZF.

If $P$ is a proposition that holds of a set $x$ whenever it holds for all members of $x$, then $P$ holds for every set.

This inductive principle is certainly more intuitive than what Conway calls "the peculiarly opaque form" of the axiom of regularity:

Every nonempty Class [sic] $X$ has some member disjoint from X.

I should mention that, although Conway speaks of ZF, he allows proper classes. In fact, the capitalization of "Class" in the statement of the axiom of regularity indicates that $X$ may be a proper class.

I've been trying to show the equivalence.

One direction is easy. Suppose that the axiom of regularity holds, and let $P$ be a proposition with the property above. Let $Y=\{x|\neg P(x)\}$ By the axiom of regularity, $\exists y\in Y(y\cap Y=\emptyset)$ But then all the members of $y$ satisfy $P$ and $y$ does not satisfy $P$, contradiction.

For the other direction, assume that Conway's version holds, and let $$ P(x)=x=\emptyset \vee \exists a\in x(a\cap x = \emptyset)$$ I need to show that if $P(y)$ is true for all $y\in x$ then $P(x)$ is true. So, $\forall y\in x(\exists z\in x\cap y),$ and we have $z \in y \in x$ However, $z\in x$ so we can apply the same argument repeatedly to get an infinite descending sequence $\dots \in z \in y \in x$ which I know is prohibited by the axiom of regularity, but I don't see how to use in this situation.

Can you help me to complete the proof, or show me a line that succeeds if this approach is hopeless?

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    $\begingroup$ "Conway's version" is better known as $\in$-induction. (And I don't think that attributing this to Conway is historically correct either.) $\endgroup$
    – Asaf Karagila
    Mar 30 '18 at 6:41
  • $\begingroup$ I didn't attribute it to Conway. I said he gives it in the Appendix to Part Zero of On Numbers and Games Look it up if you don't believe me. $\endgroup$
    – saulspatz
    Mar 30 '18 at 6:48
  • $\begingroup$ Look at the title of your question. $\endgroup$
    – Asaf Karagila
    Mar 30 '18 at 6:48
  • $\begingroup$ That's just an informal title. You are making too much of this. I have nothing more to say on this matter. $\endgroup$
    – saulspatz
    Mar 30 '18 at 6:50
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Apply Conway's version with $P(x)$ being the property "every Class $A$ that has $x$ as an element also has an element $y$ such that $y\cap A=\varnothing$.

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  • $\begingroup$ Thank you. That seems so simple, but I doubt that I would ever have thought of it myself. $\endgroup$
    – saulspatz
    Mar 30 '18 at 1:47

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