Prove $(A \cup B)' = A' \cap B'$ I would like some assistance in verifying this proof? 
(I understand the last conjecture about "symmetry" is probably shaky, but I just want to know if the first part is right since going backwards will more than likely be the same as is usually the case in such proofs). 
We seek to prove $(A \cup B)' = A' \cap B'$.

Suppose we have the universal set $U$.
We know $x \in (A \cup B)'$.
$\implies x \in U - (A \cup B)$.
Thus $x \in U \land x \notin (A \cup B)$.
$x \notin A \lor x \notin B \implies x \notin (A \cap B)$.
So $x \in U \land (x \notin (A \cap B))$.
$\implies x \in U \land (x \notin A \land x \notin B) $.
$\implies (x \in U \land x \notin A) \land (x \in U \land x \notin B)$
$\implies x \in (U-A) \land x \in (U-B)$ 
$\implies x \in A' \land x \in B'$
$\implies x \in A' \cap B'$
By the symmetry of the situation we see that $x \in A' \cap B' \implies x \in (A \cup B)' $.
Therefore the sets are subsets of one another, therefore they are equal.

We can prove using the contradiction method.
Let us assume there is an $x$ such that $x\in(A\cup B)'$ and $x\notin (A' \cap B')$.
As $x \in (A\cup B)'$, thus $x\notin (A \cup B)$. This would mean that, $x\notin A$ and $x\notin B$. Thus, $x\in A'$ and $x\in B'$. Which implies that $x\in (A'\cap B')$. But that is a contradiction.
Thus there exists NO element $x$ such that $x\in(A\cup B)'$ and $x\notin (A' \cap B')$. Thus both sets are equal.
