Question

Bar of a union b equals to a bar intersection b bar

Answer 1

To prove: $(A\cup B)^{c}=A^{c}\cap B^{c}$

Proof:

Let us take $x\in (A\cup B)^{c}$

Then, $x\in (A\cup B)^{c}$

$\iff x\in S\:but\:x\notin (A\cup B)$

$\iff (x\in S\:but\:x\notin A)\:and\:(x\in S\:but\:\notin B)$

$\iff (x\in A^{c})\:and\:(x\in B^{c})$

$\iff x\in (A^{c}\:and\:B^{c})$

$\iff x\in (A^{c}\cap B^{c})$

Thus we have:

• $(A\cup B)^{c}\subseteq (A^{c}\cap B^{c})$
• $(A^{c}\cap B^{c})\subseteq (A\cup B)^{c}$

Combining them, we get

$\quad (A\cup B)^{c}=A^{c}\cap B^{c}$

Thus proved.

Another method:

• $\therefore (A\cup B)^{c}$
• $=S-(A\cup B)$
• $=S\cap (A\cup B)^{c}$
• $=S\cap (B^{c}\cap A^{c})$
• $=B^{c}\cap A^{c}$
• $=A^{c}\cap B^{c}$
• Hence, proved.