For any two sets , the power sets : $\mathcal P(A\setminus B) = \mathcal P(A) \setminus \mathcal P(B)$? For any two sets , the power sets : $\mathcal P(A\setminus B) = \mathcal P(A) \setminus \mathcal P(B)$?
Is this proof correct? 
$$\begin{split}x \in \mathcal P(A\setminus B) &\iff x \subseteq A\setminus B  \\ &\iff x \subseteq A \land x \nsubseteq B \\ & \iff x \in\mathcal P(A) \land x \notin \mathcal P(B) \\ & \iff x \in \mathcal P(A) \setminus \mathcal P(B)\end{split}$$
 A: It is not the case that:
$$
x \subseteq A \setminus B \iff x \subseteq A \land x \not\subseteq B
$$
Indeed, the statement you're trying to prove is actually false. For a counterexample, consider $A = \{1, 2\}$ and $B = \{2, 3\}$. Notice that:
$$
P(A \setminus B) = P(\{1\}) = \{\varnothing, \{1\}\}
$$
while on the other hand:
$$
P(A) \setminus P(B) = \{\varnothing, \{1\}, \{2\}, \{1, 2\}\} \setminus \{\varnothing, \{2\}, \{3\}, \{2, 3\}\} = \{\{1\}, \{1, 2\}\}
$$
A: This proof can't be right, because, say, if $A = \{1\}$ and $B = \{1, 2\}$, then $P(A - B) = P(\varnothing) = \{\varnothing\}$. Meanwhile, $P(A) - P(B) = \{\varnothing, \{1\}\} - \{\varnothing, \{1\}, \{2\}, \{1, 2\}\} = \varnothing$.
A: Inclusion does not hold in either direction.
$\mathcal P(A\setminus B)\subseteq\mathcal P(A)\setminus\mathcal P(B)$ is always false, because $\emptyset\in\mathcal P(A\setminus B)$ while $\emptyset\notin\mathcal P(A)\setminus\mathcal P(B).$
$\mathcal P(A)\setminus\mathcal P(B)\subseteq\mathcal P(
A\setminus B)$ is sometimes false: if $x\in A\cap B$ and $y\in A\setminus B,$ then $\{x,y\}\in\mathcal P(A)\setminus\mathcal P(B)$ while $\{x,y\}\notin\mathcal P(A\setminus B);$
in fact, $\mathcal P(A)\setminus\mathcal P(B)\subseteq\mathcal P(
A\setminus B)$ holds only when $A\subseteq B$ or $A\cap B=\emptyset.$
