# Prove De Morgan's Laws (Set Theory)

Statement: Let $A$ and $B$ be subsets of $R$. Show that $(A \cup B)'=A'\cap B'$.

My attempt: To show that two sets are equal to one another we have to show that $(A \cup B)' \subset A'\cap B'$ and $A'\cap B'\subset (A \cup B)'$. Beginning with the former, suppose that $x \in (A \cup B)'$ then $x\notin A \cup B$. This implies that $x\notin A$ or $x\notin B$, which in turn should imply that $x\in A'$ or $x\in B'$. However the author of my textbook states that: $$x\notin A \text{ or } x\notin B \Rightarrow x\in A' \text{ and } x\in B'$$ Whereas, I think that the following should be true: $$x\notin A \text{ or } x\notin B \Rightarrow x\in A' \text{ or } x\in B'$$

Why did the or change into an and?

• Alright, I got it. I guess I was just cconfused about the words...Please right this as an answer so that I can accept it. – model_checker Sep 29 '16 at 7:48
• If the refrigerator is not in the apartment composed of kitchen and bathroom it means that you cannot find it neither in the kitchen nor in the bathroom. – Mauro ALLEGRANZA Sep 29 '16 at 7:49

NO : $x∉A∪B$ means that $x∉A$ and $x∉B$.
Thus, if $x$ does not belong to the union of $A$ and $B$, it means that $x$ cannot be element neither of $A$ nor of $B$.
$(A \cup B)'$ can be written as $\neg [(x \in A) \vee (x \in B)] \implies \neg (x \in A) \wedge \neg(x \in B) = A' \cap B'$