Help showing that $ (D \setminus A) \setminus (C \setminus B) = D \setminus (A \cup (B \cup C)) $ I wanted to show that
$$ (D \setminus A) \setminus (C \setminus B) = D \setminus (A \cup (B \cup C)) $$
Since I did not know how to approach this, I let wolfram alpha do the work and got the following results: https://www.wolframalpha.com/input/?i=(D%5CA)%5C(C%5CB)%3DD%5C(A+union+B+union+C)
It says: undetermined (can be true or false).
So, it can be true or false. But on what does it depend whether the solution is true or false?
If I take for example the sets:
$$ A=\{1,2,3,4,5\} $$
$$ B=\{2,1,3,4,6\} $$
$$ C=\{3,1,2,4,7\} $$
$$ D=\{4,1,2,3,8\} $$
then $$ (D \setminus A)\setminus (C\setminus B) = \{8\} $$
$$ D \setminus (A \cup B \cup C) = \{8\} $$
But what would be cases, in which this relation does not hold?
 A: Well you found one example where it is true.
And here is an example where it is false:
$$
A =\{0\}, B =\{1\}, C =\{1\}, D =\{1\}
$$
Then $(D \setminus A) = \{1\} \setminus \{0\} = \{1\}$ and $(C \setminus B) = \{1\} \setminus \{1\} = \emptyset$.
Hence $(D \setminus A) \setminus (C \setminus B) = \{1\} \setminus \emptyset = \{1\}$.
However $A \cup (B \cup C) = \{0\} \cup (\{1\} \cup \{1\}) = \{0, 1\}$.
Hence $D \setminus (A \cup (B \cup C)) = \{1\} \setminus \{0, 1\} = \emptyset$.
So $(D \setminus A) \setminus (C \setminus B) = \{1\} \neq \emptyset = D \setminus (A \cup (B \cup C))$ in this particular case.
However it is true that at least one part of the equality namely $\boxed{D \setminus (A \cup (B \cup C)) \subseteq (D \setminus A) \setminus (C \setminus B)}$ does always hold.
Proof: For let $x \in D \setminus (A \cup (B \cup C)$ be an arbitrary element.
Then $x \in D$ but $x \notin A$ and $x \notin B$ and $x \notin C$.
So firstly as $x \in D$ and $x \notin A$, we have $x \in (D \setminus A)$.
And secondly as $x \notin C$ we have in particular that $x \notin (C \setminus B)$ also.
Thus combining the results we get $x \in (D \setminus A)$ and $x \notin (C \setminus B)$ are both true or to say the same thing $x \in (D \setminus A) \setminus (C \setminus B)$.
Thus $D \setminus (A \cup (B \cup C)) \subseteq (D \setminus A) \setminus (C \setminus B)$ indeed.
A: $A = C = \emptyset$, $B = D = \{1\}$ is a counterexample.
A: You're not necessarily removing the elements of B from D in the LHS, but are in the RHS.
Sample: 
A = {1}
B = {2}
C = {3}
D = {1, 2, 3, 4, 5}
$C\setminus B=\{3\}$, so $(D\setminus A)\setminus(C\setminus B)=\{2,4,5\}$  but $D\setminus(A\cup B\cup C)=\{4,5\}$
