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? 

NO : $x∉A∪B$ means that $x∉A$ and $x∉B$. 
To be element of the union of two sets means to belong at least to one of them. 
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 $B$.
