Let (A1) be the axiom of extensionality: $\forall x,y ( x = y \longleftrightarrow \forall z \in x \leftrightarrow z \in y))$ and let (A2) be the empty set axiom $\exists x \forall y (y \notin x)$.
Then my book asks me the following:
(a) Show that $\langle \omega , \in \rangle \models (A1) \land (A2)$.
(b) Show that $\langle \{ \varnothing , \{\{\varnothing\}\}, \{\varnothing, \{\varnothing\}\}\}, \in \rangle \models \lnot (A1) \land (A2)$.
The exercise is classified as "difficult" but my attempt at an answer is easy and hence suspicious and I must be missing something:
(a) $\varnothing \in \omega \implies (A2)$.
If $x,y \in \omega$ and $x=y$ then $z \in x \iff z \in y$ (though this doesn't quite look like a proof, I can't think of anything else to write). Similarly, for the other direction: If $z \in x \iff z \in y$ then $y = x$.
(b) Let $M = \{ \varnothing , \{\{\varnothing\}\}, \{\varnothing, \{\varnothing\}\}\}$. Then $\varnothing \in M \implies (A2)$.
We have $\varnothing \neq \{\{\varnothing\}\}$ but $\forall z ( z \in \varnothing \iff z \in \{\{\varnothing\}\}$ hence $\lnot (A1)$.
What am I missing? Thanks for your help.