Analyzing logical forms clarification This example is from Velleman's "How To Prove It":

Example 2.3.6. Analyze the logical forms of the following statements.

  
*$x \in \bigcup \{ \mathscr{P}(A)| A \in F \}$

On the next page, the solution is written as $\exists A \in F(x \in \mathscr{P}(A))$. This makes sense to me, but it does not seem to follow the expansion rules. I worked it out as
\begin{align*}
    x & \in \bigcup \{ \mathscr{P}(A)| A \in F \} \\
    \exists B & \in \{ \mathscr{P}(A)| A \in F \} (x\in B) & \text{(definition of union)} \\
    \exists B & (B \in \{ \mathscr{P}(A)| A \in F \} \wedge x \in B) \\
    \exists B & (\exists A \in F(B =\mathscr{P}(A) ) \wedge x \in B)
\end{align*}
How does one go from $\exists B (\exists A \in F(B =\mathscr{P}(A) ) \wedge x \in B)$ to $\exists A \in F(x \in \mathscr{P}(A))$? Should it just be inferred, or can we transform one side into the other using basic rules?
 A: I guess you are missing that the last step of your derivation
$$\exists B  (\exists A \in F(B =\mathscr{P}(A) ) \wedge x \in B)$$
is equivalent to (for a "formal" proof, see $(*)$ below) 
$$\exists B  (\exists A \in F(B =\mathscr{P}(A) \wedge x \in B))$$
which is equivalent to (since the order of two quantifiers of the same kind can be inverted)
$$\exists A \in F(\exists B (B =\mathscr{P}(A) \wedge x \in B))$$
Now, it is clear that $\exists B (B =\mathscr{P}(A) \wedge x \in B)$ is equivalent to $x \in \mathscr{P}(A)$.

$(*)$ Indeed, $$\exists A \in F(B =\mathscr{P}(A)) \wedge x \in B$$ means that 
$$\exists A (A \in F \land B =\mathscr{P}(A)) \wedge x \in B,$$ which is equivalent to 
$$\exists A (A \in F \land (B =\mathscr{P}(A) \wedge x \in B)),$$ which is equivalent to $$\exists A \in F (B =\mathscr{P}(A) \wedge x \in B).$$
A: $\begin{align}
&x\in\bigcup\{\mathcal P(A)\mid A\in F\}\tag 1
\\& \exists B~\in\{\mathcal P(A)\mid A\in F\}~.(x\in B)\tag 2
\\&\exists B~.\exists A\in F~.(B=\mathcal P(A)\wedge x\in B)\tag 3
\\&\exists B~.\exists A\in F~.(B=\mathcal P(A)\wedge x\in\mathcal P(A)) \tag 4
\\&\exists A\in F~.\exists B~.(B=\mathcal P(A)\wedge x\in\mathcal P(A)) \tag 5
\\&\exists A\in F~.(\exists B~.(B=\mathcal P(A))\wedge x\in\mathcal P(A)) \tag 6
\\&\exists A\in F~.(\top\wedge x\in\mathcal P(A)) \tag 7
\\&\exists A\in F~.(x\in\mathcal P(A)) \tag8
\end{align}$
