Simplification of set formulas: $[ (A - B)^c - (B - A)^c ]\cap A = \varnothing$ Using the set properties, I have to demonstrate that
$[ (A - B)^\mathsf{c}  -  (B - A)^\mathsf{c}] \cap A = \emptyset. $
So far I've seen some logic properties, but never applied to sets. Could you guys help me?
 A: $$\begin{align}&x\in((A\setminus B)^{\mathsf c}\,\setminus\, (B\setminus A)^{\mathsf c})\cap A\\\implies&x\in(A\setminus B)^{\mathsf c}\land x\notin(B\setminus A)^{\mathsf c}\land x\in A\\
\implies&x\notin(A\setminus B)\land x\in (B\setminus A)\land x\in A\\
\implies&x\in (B\setminus A)\land x\in A\\
\implies&x\in B\land x\notin A\land x\in A\\
\implies& x\notin A\land x\in A\\
\implies &\perp
\end{align}$$
A: The definition of $A - B$ is $\{x \mid x \in A \text{ and } x \notin B\}$, which can be rewritten as $A \cap B^c$. This fact allows us to write
$$
\begin{align*}
[(A - B)^\mathsf{c}  -  (B - A)^\mathsf{c}] \cap A &= [(A \cap B^\mathsf{c})^\mathsf{c} - (B \cap A^\mathsf{c})^\mathsf{c}] \cap A\\
&= [(A \cap B^\mathsf{c})^\mathsf{c} \cap (B \cap A^\mathsf{c})] \cap A\\
&\subseteq A^\mathsf{c} \cap A\\
&= \emptyset.
\end{align*}
$$
A: The key fact with such set equations is that they can usually be translated into corresponding logical formulae.
For example:


*

*$x \in A \cap B$ if and only if $x \in A \wedge x \in B$.

*$x \in A^\mathsf{c}$ if and only if $\neg (x \in A)$

*$x \in A - B$ if and only if $x\in A \wedge x \not\in B$.


Applying these rules repeatedly ought to transform your statement about sets into a logical statement. Membership of the left hand side corresponds to a statement which can't possibly be true, so the set has no members. Two sets are equal exactly when they have the same members, so that means it is the empty set.
A: In my notation $ A^c = \{ x : x \notin A\} $. Then we must prove that
\begin{equation}
[ (A \cap B^c )^c \cap (B \cap A^c)^c ] \cap A = \emptyset.
\end{equation}
Notice that
\begin{equation}
(A \cap B^c)^c = (A^c \cup B)
\end{equation}
Then
\begin{eqnarray}
[(A^c \cup B) \cap (B^c \cup A)] \cap A & =&\left \{ [(A^c \cup B) \cap B^c] \cup [(A^c \cup B) \cap A]\right \} \cap A \\
&=& \left \{ [A^c \cap B^c] \cup [(A^c \cup B)] \right \} \cap A\\
&=& \left \{ [A^c \cap B^c] \cap A \right \} \cup \left \{ [(A^c \cup B)] \cap A \right \}\\
&=& \emptyset \cup  \emptyset \\
&=& \emptyset.
\end{eqnarray}
