0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.