Laws of Algebra of Sets Can someone help me to prove that:
$$\left ( A-B \right )-\left ( C-D \right )=\left [ A-\left ( B\cup C \right ) \right ]\cup \left [ \left ( A\cap D \right )-B \right ]$$
What I did was:
$\left ( A-B \right )-\left ( C-D \right )=\left [ A\cap \left ( B\cup C \right ) ^{c}\right ]\cup \left [ \left ( A\cap D \right )\cap B^{c} \right ]$
But I do not know, how to continue. Can anyone help me?
 A: So if $x \in (A-B)-(C-D)$ what does this mean? It means $x \in A-B$ but $x \notin C-D$. So $x \in A$ and $x \notin B$ and $x \notin C-D$, which can happen in two ways: either $x \notin C$ or $x \in D$. In the first case, $x \notin (B \cup C)$ and in the second case $x \in A \cap D$. In either case it will be in the right hand set. The reverse inclusion goes the same way. 
Purely set-algebraically:
$$(A-B)-(C-D)= (A-B)\cap (C-D)^\complement = (A \cap B^\complement) \cap (C \cap D^\complement)^\complement=$$ $$= (A \cap B^\complement) \cap \left(C^\complement \cup D\right)=
(A\cap B^\complement \cap C^\complement) \cup (A \cap B^\complement \cap D)=$$ $$= (A \cap (B \cup C)^\complement) \cup (A \cap D \cap B^\complement) = (A-(B \cup C)) \cup ((A \cap D)- B) $$
using $X-Y= X \cap Y^\complement$ repeatedly, plus distributivity of $\cap$ over $\cup$, and de Morgan twice.
A: \begin{align*}
x \in (A - B) - (C - D) & \iff x \in (A - B) \wedge x \not \in (C - D) \\
& \iff (x \in A \wedge x \not \in B) \wedge x \not \in (C \cap D^c) \\
& \iff (x \in A \cap B^c) \wedge (x \not \in C \lor x \not \in D^c) \\
& \iff (x \in A \cap B^c \wedge x \not \in C) \lor (x \in A \cap B^c \wedge x \not \in D^c) \\
& \iff (x \in (A \cap B^c)-C) \lor (x \in (A \cap B^c)-D^c) \\
& \iff (x \in (A \cap (B^c \cap C^c)) \lor (x \in (A \cap (B^c \cap (D^c)^c)) \\
& \iff (x \in (A \cap (B \cup C)^c) \lor (x \in (A \cap D) \cap B^c) \\
& \iff (x \in A - (B \cup C)) \lor (x \in (A \cap D) -  B)) \\
& \iff x \in (A - (B \cup C))\cup (A \cap D) - B).
\end{align*}
This should do the trick.
