Set Theory question I have some exercises to prove different laws of set theory but my study guide does not provide any answers for the exercise. I have completed one of the exercises and just want to make sure I am getting the correct answers.
$$(A\cap B)^c=A^c\cup B^c$$
I have:
Let $P(x) : x\in A$; $Q(y) : y\in B$  
(I did this step because this is how the book started each example.)
$$\begin{align*}
(A\cap B)^c&\iff\neg\big(P(x)\land Q(y)\big)\\
&\iff\neg P(x)\lor\neg Q(y)\\
&\iff A^c\cup B^c
\end{align*}$$
Thanks for any help.
Tony
 A: Your first and last $\iff$’s are meaningless, I’m afraid. The statement $P\iff Q$ means that either $P$ and $Q$ are both true, or $P$ and $Q$ are both false. In particular, $P$ and $Q$ have to be statements, things that can be true or false. $(A\cap B)^c$, however, is a set, not a statement: it can no more be true or false than a symphony can be pink.
The statement $\neg\big(P(x)\land Q(y)\big)$ actually says that it is not the case that $x\in A$ and $y\in B$. What are $x$ and $y$? They’ve come out of nowhere. What you want, I think, is to say that an object belongs to $(A\cap B)^c$ if and only if it is not the case that it belongs to $A$ and to $B$:
$$x\in(A\cap B)^c\iff\neg\big(P(x)\land Q(x)\big)\;.$$
Now you can proceed much as you did before, applying one of the logical De Morgan’s laws:
$$\neg\big(P(x)\land Q(x)\big)\iff\neg P(x)\lor\neg Q(x)\;.$$
That step was fine, except that you were working with an expression that wasn’t quite right to begin with. In the last step, however, you have the same problem that you had in the first step: $A^c\cup B^c$ is not a statement, so it makes no sense to connect it to something with $\iff$. What you want here is
$$\neg P(x)\lor\neg Q(x)\iff x\notin A\lor x\notin B\iff x\in A^c\lor x\in B^c\iff x\in A^c\cup B^c\;.$$
And since you have $\iff$ ‘if and only if’ at each step, this argument shows that the elements of $(A\cap B)^c$ are exactly the same as those of $A^c\cup B^c$ and hence that $(A\cap B)^c=A^c\cup B^c$: you’re done at that point.
A: It seems you have the right idea. You can think it about it this way:
If $x\in (A\cap B)^c$, then $x\notin A\cap B$. How can $x$ fail to be in $A\cap B$? If $x\notin A$ or $x\notin B$, which means $x\in A^c\cup B^c$.
Can you do the other direction?
This is simply a restatement of De Morgan laws:
$$\neg (a\wedge b)\equiv \neg a\vee \neg b$$
$$\neg (a\vee b)\equiv \neg a\wedge \neg b$$
A: Well, because this is your first exercise in set theory, I think it's useful to give you a general hint for such questions that ask you to prove two sets are equal.
In most cases, when you're asked to prove that $A=B$, you'll have to prove $A \subseteq B$ and then $B \subseteq A$. 
Logically, $A \subseteq B$ is defined as 
$\forall x: x \in A \implies x \in B$
and $B \subseteq A$ is defined the same way just by exchanging $A$ and $B$. So only the direction reverses and it becomes:
$\forall x: x \in B \implies x \in A$
If you prove both of these statements to be true, then their intersection must be true, which is logically equivalent to $\forall x: x \in A \iff x \in B$ which is the definition of equality for sets.
You can also use logically equivalent statements to arrive at $\forall x: x \in A \iff x \in B$ directly like you did in your post.
So, that proof is correct, even though you've missed to write $\forall x: x \in \cdots$. You need to write that down to make it meaningful. Then you replace logically equivalent statements to arrive at the desired statement. But it really depends on how you define such things. Some books on elementary(naive) set theory, prefer to introduce sets first and then study logic with the help of associating solution sets to predicates. Some books, like yours I guess, do it the opposite way. They first introduce propositions, truth values, propositional functions(predicates), ... and then they prove set theoretic results.
In case that your book is doing the later one, first you should check with truth tables that De Morgan's laws hold in propositional calculus and then you case use it to prove De Morgan's laws for sets.
