What can be said about $P (A \setminus B) \setminus (P (A) \setminus P (B))$? This is one of the problem I have been solving in Velleman's How to prove book:

Suppose A and B are sets. What can you prove about $P (A \setminus B) \setminus (P (A) \setminus P (B))$ ?

Now, I started solving it like this assuming that $ x \in P (A \setminus B) \setminus (P (A) \setminus P (B))$:
$ (x \in P(A \setminus B)) \land (x \notin (P(A) \setminus P(B))) $ 
$ (x \subseteq A \setminus B) \land (\neg (x \in P(A) \setminus P(B))) $
$ (x \subseteq A \setminus B) \land (\neg (x \in P(A) \land x \notin P(B))) $
$ (x \subseteq A \setminus B) \land ((x \notin P(A) \lor x \in P(B))) $
$ (x \subseteq A \setminus B) \land ((x \notin P(A) \lor x \subseteq B)) $
Now, I see that in the solution they have concluded that $x$ is $\emptyset$. But I cannot figure out how $\emptyset$ comes for $x$ ?
 A: Here is a more 'logical' and 'calculational' version of this proof.  Note how the $\;\emptyset\;$ automatically falls out of the calculation.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$We calculate which $\;x\;$ are in the given set, by expanding the definitions and simplifying using set theory and logic laws, as follows:
$$\calc
x \in P (A \setminus B) \setminus (P (A) \setminus P (B))
\calcop={definition of $\;\setminus\;$, twice}
x \in P (A \setminus B) \;\land\; \lnot(x \in P (A) \land \lnot (x \in P (B)))
\calcop={definition of $\;P(\phantom\square)\;$, three times}
x \subseteq A \setminus B \;\land\; \lnot(x \subseteq A \land \lnot (x \subseteq B))
\calcop={set theory: $\;x \subseteq A \setminus B\;$ implies $\;x \subseteq A\;$}
x \subseteq A \setminus B \;\land\; \lnot(\true \land \lnot (x \subseteq B))
\calcop={logic: simplify}
x \subseteq A \setminus B \;\land\; x \subseteq B
\calcop={set theory: subset of multiple sets iff subset of their intersection}
x \subseteq (A \setminus B) \cap B
\calcop={set theory: simplify}
x \subseteq \emptyset
\calcop={set theory: simplify using $\;\emptyset \subseteq x\;$}
x = \emptyset
\endcalc$$
Therefore, by the definition of singleton sets, this proves $\;P (A \setminus B) \setminus (P (A) \setminus P (B)) \;=\; \{\emptyset\}\;$.
A: If $x\in\wp(A\setminus B)$, then either $x=\varnothing$, or $x\subseteq A$ and $x\setminus B
\ne\varnothing$. In the latter case, $x\in\wp(A)\setminus\wp(B)$ and is therefore not in $\wp(A\setminus B)\setminus\big(\wp(A)\setminus\wp(B)\big)$. Thus, $\wp(A\setminus B)\setminus\big(\wp(A)\setminus\wp(B)\big)=\{\varnothing\}$.
A: Let us do it step by step: With $x \subseteq A \Leftrightarrow x\in PA$, your conclusion is equivalent to:
$$
    (x \subseteq A \setminus B) \wedge (x\not\subseteq A \vee x \subseteq B)
$$
And with $\alpha \to \beta \Leftrightarrow \neg\alpha \vee \beta$ also to:
$$
    (x \subseteq A \setminus B) \wedge (x\subseteq A \rightarrow x \subseteq B)
$$
This can be simplified the following way. You know that
$$
    (x \subseteq A\setminus B) \Rightarrow x\subseteq A
$$
So your conclusion implies:
$$
    (x \subseteq A \setminus B) \wedge (x\subseteq A) \wedge (x\subseteq A \rightarrow  x \subseteq B)
$$
With modus ponens we can follow:
$$
    (x \subseteq A \setminus B) \wedge (x \subseteq B)
$$
As $S \subseteq T$ is defined as $\forall s \in S: s\in T$, we get:
$$
    (\forall a \in x: a \in A \setminus B) \wedge (\forall b\in x: b \in B)
\\ \Leftrightarrow
    (\forall a \in x: a \in A \wedge a\not \in B) \wedge (\forall b\in x: b \in B)
$$
Putting the quantifiers together (note that $\forall$ commutes over $\wedge$), we get:
$$
    (\forall a \in x: a \in A \wedge a \not\in B \wedge a \in B)
$$
Of course, $\alpha \wedge \neg\beta \wedge \beta$ is equivalent to $\bot$ (which denotes the constantly false formula):
$$
    \forall a \in x: \bot
\Leftrightarrow
\forall a: a\in x\rightarrow \bot
\Leftrightarrow
\forall a: \neg(a\in x)
\Leftrightarrow
\not\exists a: a \in x
$$
I.e. there exists no $a$ which is an element of $x$, so $x = \emptyset$.
A: I'll give an alternative answer consisting of just a few lines from where you left off;
$ (x \subseteq A \setminus B) \land ((x \not\subseteq A \lor x \subseteq B)) $
$ ((x \subseteq A \setminus B) \land (x \not\subseteq A)) \lor ((x \subseteq A \setminus B) \land (x \subseteq B))$  
(I applied the distributive property)
The left part of the or can't by satisfied by any x, since any subset of $ A \setminus B $ is also a subset of $ A $;
The right part of the or is satisfied by $ \emptyset $, since any bigger subset of $ B $ would contain elements of B, and therefore would not be a subset of $ A\setminus B $.
So $x=\emptyset$.
