Disprove $\mathcal{P}(A\Delta B) \setminus \{\emptyset\} = \mathcal{P}(A) \Delta\mathcal{P}(B) $. Is my proof good? Proof that the following equality is wrong:
$$\mathcal{P}(A\Delta B) \setminus \{\emptyset\} = \mathcal{P}(A) \Delta\mathcal{P}(B) $$
I choose elements of sets $A$ and $B$ so that they are separate.

*

*$A = \{1, 2\}$

*$ \mathcal{P}(A) = \{\{1\}, \{2\}, \{1, 2\}, \emptyset\}$

*$B = \{3, 4\}$

*$\mathcal{P}(B) = \{\{3\}, \{4\}, \{3, 4\}, \emptyset\}$

*$A\cup B = \{1, 2, 3, 4\}$

*$A\cap B = \emptyset$
Now I see that:
On the left side:
$\mathcal{P}(A\Delta B) = \mathcal{P}((A\cup B) \setminus (A\cap B)) = \mathcal{P}((A\cup B) $
$ \mathcal{P}((A\cup B) =\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\},\emptyset \} $
$ \mathcal{P}((A\cup B) \setminus \{\emptyset\} =\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}\} $
On the right side:
$\mathcal{P}(A) \Delta\mathcal{P}(B) = (\mathcal{P}(A) \cup \mathcal{P}(B)) \setminus (\mathcal{P}(A) \cap \mathcal{P}(B)) = \{ \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\} \} $
Therefore:
$\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}\} \neq \{ \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\} \} $
i.e. $\mathcal{P}(A \,\Delta\,B) \setminus \{\emptyset\} \neq \mathcal{P}(A) \Delta\mathcal{P}(B)$.
Is my proof valid? Are there any errors? I could take smaller $A$ and $B$ sets but I was scared that it wouldn't be so obvious that way.
 A: Yes, your counterexample to $\mathcal{P}(A \,\triangle\, B) \setminus \{\emptyset\}= \mathcal{P}(A) \,\triangle\, \mathcal{P}(B)$ is correct. And your argument is  correct as well.
By the way, you can find an easier counterexample to  $\mathcal{P}(A \,\triangle\, B) \setminus \{\emptyset\}= \mathcal{P}(A) \,\triangle\, \mathcal{P}(B)$.
Take
\begin{align} A &= \{1\} & B &= \{2\}.
\end{align}
Then, $A \cap B = \emptyset$ and $A \cup B = \{1,2\}$ and so $A \,\triangle\, B = (A \cup B) \setminus (A \cap B) = \{1,2\}$. Therefore, $\mathcal{P}(A \,\triangle\, B) = \{\{1\},\{2\},\{1,2\},\emptyset\}$.
On the other hand, $\mathcal{P}(A) = \{\{1\},\emptyset\}$ and $\mathcal{P}(B) = \{\{2\},\emptyset\}$, hence $\mathcal{P}(A) \cup \mathcal{P}(B) = \{\{1\}, \{2\}, \emptyset\}$ and $\mathcal{P}(A) \cap \mathcal{P}(B) = \{\emptyset\}$.
Thus, $\mathcal{P}(A) \,\triangle\, \mathcal{P}(B) = \big(\mathcal{P}(A) \cup \mathcal{P}(B)\big) \setminus \big(\mathcal{P}(A) \cap \mathcal{P}(B)\big) = \{\{1\}, \{2\}\}$
Summing up,
\begin{align}
\mathcal{P}(A \,\triangle\, B) \setminus \{ \emptyset\} = \{\{1\},\{2\},\{1,2\}\} \neq \{\{1\},\{2\}\} = \mathcal{P}(A) \,\triangle\, \mathcal{P}(B)
\end{align}
