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After drawing a diagram of this statement, I believe it to be false. However, I'm having trouble approaching how to disprove this. Do I try to prove the negation? Or what else can I do?

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The statement is true (if you're talking about complements)... – David Mitra Sep 25 '12 at 16:03
Ok, yeah, you're right. I just found out that it is true based on Demorgans law. However, I can't use Demorgans law for the proof. So, how do you suggest approaching proving this? – wrd1534 Sep 25 '12 at 16:08
There are two ways I'd suggest. One is given by @Babak below, namely show that $(A\cup B)'\subseteq A'\cap B'$ and that $A'\cap B'\subseteq (A\cup B)'$. Another is to use Venn diagrams. Depending on your comfort, you might find one easier than the other. – Rick Decker Sep 25 '12 at 16:16

$x\in (A'\cap B')\to x\in A'$ and $x\in B'$. If $x\in A'$ then $x\notin A$ and if $x\in B'$ then $x\notin B$. In both cases, $x\notin A\cup B$. This means that $x\in (A\cup B)'$. The other direction is analogous. You can do it.

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Um, how does $x \not\in A$ imply $x \not\in A \cup B$? – Johannes Kloos Sep 25 '12 at 16:16
@JohannesKloos: I fixed it. Thanks for noting me on time. :-) – Babak S. Sep 25 '12 at 16:31
You are getting very close to $20$K! +1 – amWhy Mar 22 '13 at 0:17
Good answer anyway – Adi Dani Mar 22 '13 at 1:02

$x\in(A\cup B)'$ $\Rightarrow$ $(x\notin A\text{ and } x\notin B)$ $\Rightarrow$ $(x\in A'\text{ and } x\in B')$ $\Rightarrow$ $x\in A'\cap B'$ $\Rightarrow (A\cup B)'\subseteq A'\cap B'$

$x\in A'\cap B'$ $\Rightarrow$ $(x\in A'\text{ and } x\in B')$ $\Rightarrow$ $(x\notin A\text{ and }x\notin B)$ $\Rightarrow$ $x\in(A\cup B)'$ $\Rightarrow A'\cap B'\subseteq (A\cup B)'$


$(A\cup B)'\subseteq A'\cap B'$ and $A'\cap B'\subseteq (A\cup B)'$ $\Leftrightarrow$ $(A \cup B)' = A' \cap B'$

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