Other elementary question about image and preimage of relations.

Other elementary question about image and preimage of relations.

Hi everyone I continue working with set theory. I'm really stuck with some
exercise. I thought to have the answer but I tried to give an example and
is different than I proved.
The problem is to prove the next two statements and give some
counterexample for what the equality doesn't hold.
(1) $R^{-1}[\,R[\,A\,]\,] \supseteq A \cap \mathscr{D}(R)\,$ ( where
$\mathscr{D}(R)$ is the domain of the relation ).
Proof:
($\,\Rightarrow\,$) Suppose $x\in A \cap \mathscr{D}(R).$ So, $x\in
\mathscr{D}(R)$ iff there is a $y$ such that $\langle x,y \rangle \in R.$
Then there is an $x\in A$ such that $\langle x,y \rangle \in R,$ therefore
$y\in R[\,A\,].$ Symmetrically, there is an $y\in R[\,A\,]$ such that
$\langle x,y \rangle \in R,$ therefore $x \in R^{-1}[\,R[\,A\,]\,],$ as
desired.
($\,\Leftarrow\,$) Now suppose $x\in R^{-1}[\,R[\,A\,]\,].$ Then there
exist some $y\in R[\,A\,]$ such that $\langle x,y \rangle \in R.$ On the
other hand, $y\in R[\,A\,] $ if there exist an $x'\in A$ such that
$\langle x',y \rangle \in R.$
Assume the property: $\,x\,R\,y \text{ and } x'\,R\,y$ implies $x=x'.$
Therefore $x\in A$ and also, as $x\,R\,y$ it follows that $x\in \mathscr
{D} (R)$. Hence, $x \in A \cap \mathscr {D} (R).$
So, as I proved (assuming that is correct) the equality holds using an
extra-property. But if a give the next example where the property does not
hold:
$A: = \left\{\, x, x^{*}, x^{**}\, \right\} \text { and } R:= \left\{
\langle x,y \rangle, \langle x^{*},y \rangle \right\}.$
$A \cap \mathscr{D}(R) = \left\{\, x, x^{*} \,\right\} \text{ and }
R^{-1}[\,R[\,A\,]\,] = R^{-1} [\left\{\,y \, \right\}] = \left\{\, x,
x^{*} \,\right\} $
So, I am not assuming the necessary property but the equality holds. My
question is: where is wrong my reasoning?
As always thanks in advanced.



Sorry I forgot to put the statement 2.