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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.

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    “Similarly, for the other direction: If $z\in x \Leftrightarrow z\in y$ then $y=x$” needs more justification. This is the only non-trivial part of the question.2012-11-24
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    It looks like you're probably supposed to write down exactly what A1 and A2 are when _relativized to_ $\omega$ respectively $M$. Since you're getting the right result in (b), you're probably _understanding_ it correctly, but writing it down explicitly will help with clarity.2012-11-24
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    Dear @HenningMakholm, thank you. What does "relatively to $\omega$ respectively $M$" mean?2012-11-24
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    @Matt: Another way to express it would be, say, "unfold the definition of $\vDash$ in $\langle M,{\in}\rangle\vDash\text{(A1)}$".2012-11-24
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    By now I think I have to translate it into a statement that quantifies over the elements in $\omega$.2012-11-24
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    (or $M$, respectively)2012-11-24

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