Is there a way to prove that $A \setminus C = (A \setminus (B \cup C)) \cup ((A \cap B) \setminus C)$? Drawing out the Venn Diagrams I know this is true.
$(A \setminus (B \cup C))$ is just all the values that are solely in A.
$((A \cap B) \setminus C)$ is just all the values in both A and B but not in C.
And taking the union of these two leads us to the values that are either only in A or in only A and B (but not C).
I am unsure how to rigorously prove this, however. Would starting off by letting $x \in (A \setminus (B \cup C))$ be a good start?
Venn Diagram
 A: You pretty much have it.  It is a proof by cases that :


*

*Take any $x$ where $x\in(A\smallsetminus(B\cup C))\cup((A\cap B)\smallsetminus C)$.


*Thus we have that $x\in((A\smallsetminus(B\cup C))$ or $x\in ((A\cap B)\smallsetminus C)$

*

*In the left case: $x\in A$ but $x\notin(B\cup C)$. The latter means $x\notin B$ and $x\notin C$. Thus in this case: $x\in A$ and $x\notin C$, which is to say: $x\in A\smallsetminus C$.


*In the right case: $x\in(A\cap B)$ but $x\notin C$.  The former means $x\in A$ and $x\in B$. Thus in this case: $x\in A$ and $x\notin C$, which is to say: $x\in A\smallsetminus C$.




*So in each case: $x\in A\smallsetminus C$.
Therefore $(A\smallsetminus(B\cup C))\cup((A\cap B)\smallsetminus C)\subseteq A\smallsetminus C$.

However, that is but half or the proof.
Demonstrating the converse uses the law of excluded middle: "Either $x\in B$ or $x\notin B$, so..."
A: To prove sets $S,\,T$ are equal, i.e. $x\in S\iff x\in T$, you should try to prove both directions of the $\iff$ simultaneously by only using $\iff$ steps. To wit:$$\begin{align}x\in A\setminus C&\iff x\in A\land x\notin C\\&\iff(x\in A\land x\notin B\land x\notin C)\lor(x\in A\land x\in B\land x\notin C)\\&\iff(x\in A\land x\notin B\cup C)\lor(x\in A\cap B\land x\notin C)\\&\iff(x\in A\setminus(B\cup C))\lor(x\in(A\cap B)\setminus C)\\&\iff x\in(A\setminus(B\cup C))\cup((A\cap B)\setminus C).\end{align}$$
