How to show one De Morgan's Law implies the other (Set Theory)

Question

$1.)\space\space (A\space \cap B)' = A' \cup B'$

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

How can I show that $1 \rightarrow 2$? I understand how to get from the left side of each law to the right side, but not how to show that the first implies the second.

Answer 1

Applying $1$ to $A'$ and $B',$ $$(A'\cap B')'=A''\cup B''=A\cup B$$ Thus we have, $$(A'\cap B')'=A\cup B.$$

Using the fact that two sets are equal if and only if their complements are equal we get,

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

Now, for all sets $P,$ we have $P''=P.$ Therefore,

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

Answer 2

I'll introduce related boolean variables for clarity.

Let $A' = C,\ B' = D$. Note that $C' = A,\ D' = B$.

From the first one, taking the complement of both sides and reversing the sides, we get:

$(C \ \cup D)' = A \ \cap B = C' \ \cap D'$

So the first implies the second. The implication is reversible, and it's easy to show this by working on the second and going to the first.