I often run into some version of the statement
$$\forall x \; (x \notin x)$$
...but it is not entirely clear to me how exactly one derives it formally.
Is this statement part of the definition of $\in$, somehow?
Alternatively, can it be deduced from the ZFC axioms?
I am aware of the fact that the RHS of the definition below
$$Q := \{x\;:\; x \notin x\}$$
plays a central role in Russell's paradox, but from this I can conclude only that $Q$ (whatever it may be) cannot be a set; I don't know exactly how to use this result to prove the universally-quantified expression given at the beginning of this post.
EDIT: Here's my clumsy/noob-like attempt to flesh out @MauroALLEGRANZA's first comment.
It follows from the Axiom of Pairing, that for every set $x$, the set $\{x, x\} = \{x\}$ exists.
The Axiom of Regularity can be expressed formally, as follows: $$ \forall \;\mathrm{sets}\; v \;(v \neq \varnothing \Rightarrow (\exists w \in v)\; v \, \cap \, w = \varnothing) $$
This means, in particular, that for every set $x$, since $\{x\} \neq \varnothing$ (by the Axiom of Extensionality), there must exist a set $w \in \{x\}$ such that $\{x\} \, \cap \, w = \varnothing$. Since, by definition, $x$ is the only set $y$ such that $y \in \{x\}$, $x$ must be the set $w$ whose existence the Axiom of Regularity guarantees. Therefore, we must have that $\forall x \; (\{x\} \, \cap x = \varnothing)$.
This assertion implies that $\forall x\,\forall z (z \in \{x\} \Rightarrow z \notin x)$. In particular, $$\forall x (x \in \{x\}) \Rightarrow \forall x (x \notin x)$$.
Once I write this proof all out in tedious detail, I realize that the Axiom of Regularity, at least as stated above, depends on the existence of the set $\varnothing$.
The one proof I know of the existence of $\varnothing$ uses the Axiom Schema of Separation as justification for the existence of the set $\{u\;:\; u \in X \;\wedge\; u \neq u\}=:\varnothing$, where $X$ is the set whose existence is guaranteed by the Axiom of Infinity.
In order for this definition of $\varnothing$ to conform to the usual sematics of $\varnothing$, we must have that $$\neg \, (\exists x \; (x \neq x))$$
...or, equivalently,
$$\forall x \; (x = x)$$
This is not a ZFC axiom, as far as I can tell, so I figure this must be either an axiom of logic or part of the definition of $=$ . (Sheesh, the search for "rock bottom" sure feels like it's "turtles all the way down"...)
