# Conway's Version of the Axiom of Regularity

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?

• "Conway's version" is better known as $\in$-induction. (And I don't think that attributing this to Conway is historically correct either.) Mar 30 '18 at 6:41
• 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. Mar 30 '18 at 6:48
• Look at the title of your question. Mar 30 '18 at 6:48
• That's just an informal title. You are making too much of this. I have nothing more to say on this matter. Mar 30 '18 at 6:50

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 $y\cap A=\varnothing$.