Prove $(A \cup B) \setminus C \subseteq A \cup (B\setminus C)$ I have to prove that $$(A \cup B) \setminus C \subseteq A \cup (B \setminus C) $$
I am not sure how to do this, but one of my ideas is to show that there is an $x$ that belongs to $(A \cup B) \setminus C$, and it is arbitrary. And then to show that $x$ belongs to $A \cup (B \setminus C)$  as well.
Do you think that it is a good way to do it or you have other ideas? 
 A: HINT: 
Prove that 
$$
(A \cup B)\setminus C=(A\setminus C) \cup(B\setminus C) \tag{1}
$$

I will show how to prove $(1)$.
Assume $x\in (A \cup B)\setminus C$, then $x\in A$ or $x \in B$, and $x \notin C$. That means $x \in A \setminus C$ or $x \in B \setminus C$. Then $x \in (A \setminus C) \cup (B \setminus C)$. So that $(A \cup B)\setminus C\subseteq(A\setminus C) \cup(B\setminus C) $.
From the other side, assume $x\in (A\setminus C) \cup(B\setminus C)$, then $x \in A\setminus C $ or $x \in B\setminus C$, which means $x \notin C$ and $x $ belongs to either $A$ or $B$. So $x \in (A \cup B) \setminus C$. Thus
$$
(A \cup B)\setminus C=(A\setminus C) \cup(B\setminus C) \subseteq A\cup(B\setminus C)
$$
A: Let $x\in (A\cup B)\setminus C$. Then, by definition, $x\in A\cup B$ but $x\notin C$. If $x\in A$, then so be it; the fact that $x\notin C$ is irrelevant. If $x\in B$, though, of course, we still have $x\notin C$. Hence $x\in A\cup(B\setminus C)$. Therefore, $ (A\cup B)\setminus C\subseteq A\cup(B\setminus C)$.
