I hope this question is not a duplicate of the following question (however it is related):
Consider the following axiom:
$\forall x\forall y(\forall z \forall w (z \in x \Leftrightarrow z \in y \wedge(x \in w \Leftrightarrow y \in z)) \Rightarrow x=y ) \quad (\overline{Extensionality})$
Does ZFC-Extenionality+$(\overline{Extensionality})$ prove Extensionality?
I really don't understand the intuition behind this statement.
Nevertheless, $\overline{Extensionality}$ is actually provable from ZFC - Extensionality.
Suppose we have $x, y$. Suppose that for all $z$ and $w$, $z \in x$ if and only if both $z \in y$ and also $(x \in w \iff y \in z)$.
Suppose we have some $z \in x$. By the axiom of pairing, take $w$ such that $x \in w$.
Now because $z \in x$, we see that $z \in y$ and also $(x \in w \iff y \in z)$. Because we picked $w$ such that $x \in w$, we have $y \in z$.
So $z \in y$ and also $y \in z$. Again citing pairing, take $a$ such that $y \in a$ and also $z \in a$. By separation, take $a'$ such that for all $b$, $(b \in a' \iff (b \in a \land (b = z \lor b = y)))$.
By the axiom of regularity, take $c$ such that $c \in a'$ and for all $d$, $\neg (d \in c \land d \in a')$. Since $c \in a'$, either $c = z$ or $c = y$. If $c = z$, then $y \in c$ and $y \in a'$; contradiction. If $c = y$, then $z \in c$ and $z \in a'$; contradiction. So we've reached a contradiction.
From here, it follows that $x = y$.