# Proving distributivity of Complement over union (set theory)

I am trying to prove the following identity:

$$M \setminus (N \cup L) = (M \setminus N) \cap (M \setminus L)$$

I thought about saying that $x \in (N \cup L)$ which means that $x$ is in either $N$ or $L$ and is not in $M$ but I'm stuck here. I understand that $M \setminus N$ means that $\{x \in M; \ x \notin N\}$ but I'm confused on how you can prove this identity.

Note that $$M\setminus(N\cup L)= M\cap (N\cup L)^c\tag{defn of setminus}$$ $$= M\cap(N^c\cap L^c) \tag{by DeMorgan}$$ $$= (M\cap N^c) \cap (M\cap L^c)\tag{since M=M\cap M}$$ $$= (M\setminus N)\cap(M\setminus L)\tag{defn of setminus}$$

Let $x \in M\backslash (N \ \cup L)$

$<=>x \in M$ and $x\notin (N \ \cup L)$

$<=>x \in M$ and ($x \notin N$ and $x \notin L$)... De Morgan's law

$<=>(x \in M$ and $x \in M)$ and ($x \notin N$ and $x \notin L$)...$M=M \ \cap M$

$<=>(x\in M$ and $x\notin N)$ and $(x \in M$ and $x \notin L)$... Commutative law and Associative law

$<=>x \in (M\backslash N \ \cap M\backslash L)$