Proof of De-Morgan's law. The law states : $(A \cup B)' = A' \cap B' $
Although I can do it using Venn diagrams, I tried it using this method:
Let $x$ be any arbitrary element which exists in the set $(A \cup B)'$
$x \in (A \cup B)'$
$\Longrightarrow  x \notin (A\cup B)$
$\Longrightarrow  x \notin A $ and $x \notin B $
$\Longrightarrow  x \in A' $ or $x \in B' $
But now I'm getting or so that means union , so the final proof which I'm getting is :
$(A \cup B)' = (A' \cup B') $
Please tell where I am going wrong.
 A: Your last implication:  Why are you switching from "and" to "or"?  You shouldn't.
Your proof only leaves the job half done.  You have only proved $(A \cup B)' \subseteq A' \cap B'$.  To establish equality, you also need to prove the other containment:  if $x \in A' \cap B'$, then $x \in (A \cup B)'$.
A: I think you've gone wrong with your last statement.
I.E.
$\Longrightarrow  x \epsilon A' $ or  $ x \epsilon B'$
Note that $A'$includes $(A \cup B)'$ and $B-A$
And, $B'$includes $(A \cup B)'$ and $A-B$
So, according to your last statement, x can belong to $(A \cup B)'$, as well as $B-A$ or $A-B$. And that is not what you mean by your second last statement, i.e. $\Longrightarrow  x \notin A $ and  $ x \not in B'$
Thus, you shouldn't switch from AND to OR. You'll get the required relation, then.
A: Welcome to Math Stack Exchange. You correctly had $x\notin A$ and $x\notin B$, which is precisely what it means to say that $x\in A'$ $\textbf{and}$ $x\in B'$. So $x\in A'\cap B'$. If $x\in A'\cap B'$, then $x\notin A$ and $x\notin B$, so $x$ is not in A or B, i.e., $x\notin A\cup B$. So $x\in (A\cup B)'$.
