Is this axiom self-contradicting? I was on physics stackexchange and came across an unusual answer where it was stated that the axiom,
$$\forall x ((x \in x) \land (x \notin x))$$
Creates an axiom system where "nothing" exists in it. I kind of misquoted here, here's an exact,
"From that single axiom we can conclude that nothing exists, that the universe of discourse of this axiom system is exactly the empty universe of discourse."
Needless to say, I got into a heated debate on the self-consistency of the axiom. My position is that it can't create a universe or universe of discourse because there are no entities within it. See wikipedia's definition
In English, I interpret the axiom to say,
For all X, X is an element of itself and X is not an element of itself.
This seems self contradicting, akin to Russell's paradox, but the Op insists that it isn't.
I'm open to either take on the issue, but a simple yes/no response will not cut it. A definition review might also help quantify the dialogue.
If it helps, from a semi philosophical-mathematical I consider nothing to be the null set.
 A: This is an example of how mathematical conventions are often divided on the status of degenerate cases.
Some more modern approaches to logic do indeed allow empty universes of discourse. For example, you'd likely see this in any source that works with first-order logic from the point of view of category theory.
However, more traditional approaches reject empty universes of discourse. In the corresponding formulation of first-order logic, your formula would indeed be a contradiction.
A: Working in classical logic, the rules of logic dictate that $x\in x\land x\notin x$ is a false statement.
Therefore the only structure satisfying $\forall x(x\in x\land x\notin x)$ is the empty structure. However due to superficial reasons, we choose not to accept the empty structure as a valid interpretation of a first-order language, so it is not a model of this axiom.
In particular it means that any theory including this axiom is inconsistent. If you want to argue in non-classical logic, perhaps paraconsistent logics, that is a whole other things which should be explicitly mentioned.
(I took a quick look at the Phy.SE answer, and I'd probably stay away from that answer and as a rule of thumb largely ignore any claims about mathematical logic made on Phy.SE without references.)
A: A formula of the form $\phi \land \lnot\phi$ has no models (other than in the model theory of some rather obscure substructural logics). This is unconnected with Russell's paradox and not at all deep. There is no such thing as an "empty model" for a non-empty language, e.g., a language including he membership symbol $\not\in$.
[I should point out that this answer does not conflict with Asaf Karagilia's answer: if you allow models that have empty universes, then $\forall x(\psi(x)\land\lnot\psi(x))$ does not have the form $\phi\land\lnot\phi$ and may hold in a model with an empty universe. However, allowing models with empty universes puts you in the far from classical realm of free logic.]
