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My question is related to this question (Meaning of the Axiom of regularity (foundation)).

Why wouldn't $p\neq q$ be ruled out without axiom of foundation? And how does Axiom of foundation helps in rolling it out?
Also I appreciate if someone tells me why the first program of @Asaf's answer in mentions question hold which is.
"Such sets $(x=\{ x\} )$are sometimes called Quine Atoms. And indeed if x,y are two different Quine atoms, then $x\neq y$ because $x\notin y $ and $y \in y$ (and vice versa) "?

I am fairly new to this topic, therefore I am sorry if my question sounds naive.

K. Smith
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1 Answers1

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It's not that $p\neq q$ is ruled out by the axiom of regularity. It is the existence of a set of the form $x=\{x\}$, or a Quine atom, altogether which is ruled out.

The problem is that we often think about sets as "defined by a formula" or "a property". So the property $x=\{x\}$ is something that if two sets satisfy, we cannot distinguish them "as easily", and we need to more or less require them to be distinct axiomatically.

Asaf Karagila
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  • Thank for your answer. A couple of things I want to clear out for myself: 1. Am I right that thinking about sets having "a property" runs us into the problem of mixing sets with proper classes? 2. Can we even have $x={ x}$ as set? A set containing itself! – K. Smith Sep 22 '17 at 16:28
  • To the proposer: None of the axioms of ZFC (Zermelo-Frankel- Choice) actually mentions "sets". And none of the axioms defines "$\in$". It is just a binary-relation symbol. Other binary relations can be reflexive, e.g. $ x\leq x.$ – DanielWainfleet Sep 23 '17 at 01:37