Equinumerous sets, power set 
(ZF) Prove that for an arbitrary set A the following holds:

*

*$ \bar{A} = \bar{A ∪  [{A}]} ⇒ \bar{P(A)} =\bar{(P(A) ∪ [P(A)])}$ ;


*$ \bar{A} =  \bar{(A ∪ [A])} ⇒  \bar{P(P(A))} =  \bar{P(P(A) × P(P(A))} $.
What I have made so far is:

*

*By def, A and B are equinumerous if there is a bijection f: A->B.

Let  $ x  \subset A =>x \subset \bar{A ∪ {A}} $.
Then $ Dom(f) \subseteq A$   and $Rng(f) \subseteq \bar{A ∪  [{A}]} $. Obviously this is injection since $x\subset A$ or $x \equiv sing A $.
But because $ x \in A \Rightarrow x\in P(A).$ But $f(x)\subseteq P(A)$ and $f(x)\subseteq A\Rightarrow f(x) \in P(A)\cup [A].$ Hence there is a function $g$ for which $g(f(x))$ is $P(A)\mapsto (P(A)∪[P(A)])¯  $;
And hence they are equinumerous.


*We know that $ (X)^(B∪C)=X^B×X^C $. But afterwards I do not know what to do.

With [x] I denote the singleton of the set x.
I am sorry for the syntax, but I am a beginner in Math Text Editors.
I will be really thankful to anyone who share the ideas he/she has about the problem since I am stuck on it.
I tried to explain what I think that will be useful with the solution even though I am not quite sure what to do about 2.
Thanks in advance!
 A: I’m afraid that your argument for $(1)$ simply doesn’t make much sense. You start by saying that if $x\subseteq A$, then $x\subseteq\overline{\overline{A\cup\{A\}}}$. However, this makes no sense, because $\overline{\overline{A\cup\{A\}}}$ has not been given an independent meaning as a set: it’s just part of the notation that you’re using to say when there is a bijection between two sets. What is true, and what I suspect you meant, is that if $x\subseteq A$, then $x\subseteq A\cup\{A\}$.
You then conclude from this that $\operatorname{dom}f\subseteq A$ and $\operatorname{ran}f\subseteq\overline{\overline{A\cup\{A\}}}$. There are multiple problems with this, starting with the fact that you have not defined $f$. We can guess that you mean the embedding $f:A\to A\cup\{A\}$, though the fact that you were originally looking at subsets $x$ of $A$ rather than elements of $A$ actually suggests that $f$ is a map from $\wp(A)$ to $\wp(A\cup\{A\})$, in which case the domain of $f$ is not a subset of $A$ but rather of $\wp(A)$. And in any case you are once again treating $\overline{\overline{A\cup\{A\}}}$ as if it were a set.
Then you say that $x\in A$ implies that $x\in\wp(A)$; this is simply false. And I can make no sense of the last sentence (about $g$).
You want to show that if there is a bijection between $A$ and $A\cup\{A\}$, then there is a bijection between $\wp(A)$ and $\wp(A)\cup\{\wp(A)\}$. The natural way to start is:

Let $f:A\to A\cup\{A\}$ be a bijection.

The rest is completely revised and corrected.
To get a bijection between $\wp(A)$ and $\wp(A)\cup\{\wp(A)\}$ we can use the fact that there is an obvious bijection $s$ from $A$ to $\mathscr{S}=\big\{\{a\}:a\in A\big\}$: let
$$s:A\to\mathscr{S}:a\mapsto\{a\}\,.$$
The map $f$ is a bijection from $A$ to a set, $A\cup\{A\}$, that is $A$ together with one extra element, so we should be able to use it and $s$ to get a bijection from $\mathscr{S}$ to the set $\mathscr{S}\cup\{\wp(A)\}$, which is $\mathscr{S}$ together with one extra element. Specifically, let
$$\hat f=f\circ s^{-1}:\mathscr{S}\to A\cup\{A\}\,,$$
and let
$$g:A\cup\{A\}\to\mathscr{S}\cup\{\wp(A)\}:x\mapsto\begin{cases}
s(x),&\text{if }x\in A\\
\wp(A),&\text{if }x=A\,;
\end{cases}$$
$\hat f$, being a composition of bijections, is a bijection, $g$ is easily seen to be a bijection as well.
Now let
$$\begin{align*}
&h:\wp(A)\to\wp(A)\cup\{\wp(A)\}:\\
&\quad X\mapsto\begin{cases}
g\left(\hat f(X)\right),&\text{if }X\in\mathscr{S}\\
X,&\text{if }X\in\wp(A)\setminus\mathscr{S}\,;
\end{cases}
\end{align*}$$
it’s not hard to check that $h$ is also a bijection.
