Rewrite set theory formal I'm having trouble with rewriting this expression only with the use of:
Brackets, variables, $A$, $B$, $\neg$,$\exists$,$\lor$,$\in$

I tried various solution but the only one that convinces me a little bit is the following one:
$$\forall a \in A \to \neg \exists b \in B \space (a \in B) \equiv \neg(\forall a \in A) \lor \neg \exists b \in B\space (a\in B) \equiv\exists a \in A \lor \neg \exists b \in B (a \in B)$$
Is this right? Could you otherwise give me a hint on how to translate those expressions only with the use of symbols?
•B=$\emptyset$
•$\bigcup A$
•If A and B are sets than exists $C=\{A,B\}$
 A: Basically, make sure you know the definitions of all the involved symbols, expand them, and see where you end up.
Here is the expanding part:
\begin{align}
& A \not\subseteq \mathcal P(B) \\
\equiv & \qquad \text{"definition of $\;\not\ \;$"} \\
& \lnot (A \subseteq \mathcal P(B)) \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& \lnot \langle \forall V : V \in A : V \in \mathcal P(B) \rangle \\
\equiv & \qquad \text{"definition of $\;\mathcal P\;$"} \\
& \lnot \langle \forall V : V \in A : V \subseteq B \rangle \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& \lnot \langle \forall V : V \in A : \langle \forall x : x \in V : x \in B \rangle \rangle \\
\equiv & \qquad \text{"..."} \\
\end{align}
Now get rid of the $\;\forall\;$ using the rules of predicate logic, and you are done.
A: Hint: 
We want to say that there is an $x$ such that $x\in A$ and $x$ is not in the powerset of $B$.
Recall that $P\land Q$ is equivalent to $\lnot(\lnot P\lor \lnot Q)$.
So the "and" can be eliminated. 
Now we want to say that $x$ is not in the powerset of $B$. This says that there is a $t$ such that $t\in x$ and $t$ is not in $B$. Eliminate "and" as before, and put the pieces together.  
A: As answer on your first question: the following statements are equivalent
$A\nsubseteq\wp\left(B\right)$
$\exists a\left[a\in A\wedge\neg a\in\wp\left(B\right)\right]$
$\exists a\left[a\in A\wedge\exists x\left[x\in a\wedge\neg x\in B\right]\right]$
$\exists a\neg\left[\neg a\in A\vee\neg\exists x\left[x\in a\wedge\neg x\in B\right]\right]$
$\exists a\neg\left[\neg a\in A\vee\neg\exists x\neg\left[\neg x\in a\vee x\in B\right]\right]$
