Using the method of "searching for a falsifying truth assignment" prove the De Morgan's law

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DeMorgan's Laws : 1. $(A\cap B)'=A'\cup B'$

2. $(A\cup B)'=A'\cap B'$

Let us assume (2) is false, and $x\in (A\cup B)' \implies x \notin (A\cup B)$

$\implies x\notin A and x\notin B \implies x\in A' and x\in B'$

$x\in A'\cap B'$ .

As x was a general element; $(A\cup B)'=A'\cap B'$ , which is a contradiction.

Hence, our assumption was false and (2) is true.

Similarly, (1) can be proved as well.