Logical equivalence + Quantifiers (Universal and Existential) I'm taking a Discrete Mathematics course and we're using Rosen's book (which I hate because it seems like it makes difficult material to understand even more incomprehensible). Trouble is, I am lost when it comes to combining what I learned about logical equivalency and quantifiers.

Sample problem from homework:

Show that ∃x( P(x) v Q(x) ) and ∃xP(x) v ∃xQ(x) are logically equivalent.

What is the step by step process to answering this question?

Edit: Does anyone have a recommendation for a better textbook on Discrete Math?
Hint: You are over complicating this problem. 

Think about what both expressions mean. The "proof" that 
$$
\exists x(P(x)\lor Q(x)) \Longleftrightarrow \exists xP(x)\lor \exists xQ(x)
$$
is remarkably simple. 

Proof. Both expressions are true precisely when at least one of $P(x)$ and $Q(x)$ is true for at least one value of $x$ in the universe of discourse. $\Box$



Of course, if you want to prove it in a different manner, then you are welcome to try, but that explanation is really all you need. 



Here's a simple example with trivial constructions, but it may help you see why the two expressions are equivalent. Assume we are working with integers as the domain of discourse.

Start by constructing statements $P(x)$ and $Q(x)$.


$P(x) : x\neq x$
$Q(x) : x=1$


Now think about what both expressions mean.

Expression 1: $\exists x(P(x)\lor Q(x))$ means that there exists an $x\in\mathbb{Z}$ where at least one of $P(x)$ and $Q(x)$ is true. Well, $x\neq x$ is never true for any integer $x\in\mathbb{Z}$. Thus, we know $P(x)$ is not true. However, $Q(x)$ is true. Certainly there exists an $x\in\mathbb{Z}$ such that $x=1$, namely $x=1$. Okay cool. So expression 1 is true for our sample constructions of $P$ and $Q$. 

Expression 2: $\exists xP(x)\lor \exists xQ(x)$ means that (1) there exists an $x\in\mathbb{Z}$ such that $P(x)$ is true, OR (2) there exists an $x\in\mathbb{Z}$ such that $Q(x)$ is true. Well, we just saw that $x\neq x$ is never true in $\mathbb{Z}$; thus, we can throw out $\exists xP(x)$, but what about $\exists xQ(x)$? Again, certainly there exists an $x\in\mathbb{Z}$ such that $x=1$, namely $x=1$. Thus, $\exists xQ(x)$ is true, thereby making $\exists xP(x)\lor \exists xQ(x)$ a true statement. 



Notice that Expression 1 and Expression 2 are really saying the same thing; that is, they are actually logically equivalent.
$\exists{x}(P(x)\lor{Q(x)})\equiv\neg \forall{x}\neg(P(x)\lor{Q(x)})\equiv\neg \forall{x}(\neg P(x)\land\neg{Q(x)})\equiv \neg\forall{x}\neg P(x) \land \neg\forall{x}\neg{Q(x)} \equiv \exists{x}P(x)\land \exists{x}Q(x) $