Prove: $C-(A\cup B)=(C-A)\cap (C-B)$ Please help review this part of the proof.
Suppose that $x\in C-(A\cup B)$. Therefore, $x\in C$ and $x\notin (A \cup B)$. Then, $x\in C \wedge (x\notin A \wedge x\notin B)$. By morgan laws we got that $x\in C \wedge x\notin A$ and $x\in C \wedge x\notin B$. By definition of set subtraction $x\in C-A \wedge x\in C-B$ By intersection definition is clear that $x\in (C-A) \cap (C-B)$. Therefore, $C-(A\cup B)\subset (C-A)\cap (C-B)$.
The other part $ (C-A)\cap (C-B) \subset C-(A\cup B)$ is pretty much similar.
 A: The gist of your argument is correct, though the wording needs polishing.
You have proceeded by using (1) the definition of set difference, (2) de Morgan's Law on complement of a union, (3) distribution, (4) the definition of set difference, then finally (5) the definition of union.   Since each step is biconditional, that is all you need to do.

$${\def\getsto{\mathop{\leftrightarrow}}\begin{align}\because~x\in C\smallsetminus (A\cup B)~&\getsto~(x\in C~\wedge~x\notin(A\cup B)) \tag 1 \\&\getsto~ (x\in C~\wedge~(x\notin A~\wedge~x\notin B)) \tag 2 \\&\getsto~((x\in C\wedge x\notin A)\wedge(x\in C\wedge x\notin B)) \tag 3\\ & \getsto~(x\in (C\smallsetminus A)~\wedge~x\in (C\smallsetminus B)) \tag 4\\ &\getsto~x\in (C\smallsetminus A)\cap (C\smallsetminus B) \tag 5  \\\hline\therefore\quad C\smallsetminus (A\cup B)~&=~~(C\smallsetminus A)\cap(C\smallsetminus B)\end{align}}$$

So just polish up your wordsmithing.


An arbitrary element will be an element of the set $C\smallsetminus (A\cup B)$ exactly when it is an element of $C$ but not of either $A$ nor $B$.   So any such element of our set is exactly an element of $C$, not an element of $A$, and not an element of $B$.   Thus $C\smallsetminus (A\cup B)$ is equal to $(C\smallsetminus A)\cap(C\smallsetminus B)$.

A: $$C\setminus (A\cup B)= {C\cap (A\cup B)^c}_{\text{intersection with a set complement when U defined}}=C\cap(A^c\cap B^c)=$$$$=C\cap (A^c\cap B^c)\cap C=(C\cap A^c)\cap (C\cap B^c)=(C\setminus A)\cap (C\setminus B)$$
$$Q.E.D.$$
