Prove a first-order theory is complete if given $\vdash P\lor Q$ we have $\vdash P$ or $\vdash Q$ In my textbook, completeness is defined to be:

A theory is complete if, given any sentence $P$, either $P$ or NOT–$P$ is a theorem.

And now it seems to be a simpler way to see that if a theory is complete or not. 
It says:

a first-order theory is complete if given $\vdash P\lor Q$ we have $\vdash P$ or $\vdash Q$.

I cannot find an explicit connection between the later statement and the orginal definiton. So may I ask for some explaination? And I would really appreciate if someone can give a completely formal proof. Or some reference ae also helpful.
 A: First of all, you should read the 'if's in these statements as 'if and only if's.
Second, the thing to remember is that theories are closed under logical consequence, i.e. It Includes all of its logical consequences.
Now:
Given the first statement (the definition), we can derive the second: $T$ is complete if and only if for any $P$ and $Q$: If $T\vdash P \lor Q$, then either $T \vdash P$ or $T \vdash Q$.
We'll first do left to right:
Suppose theory $T$ is complete.  Now suppose $T \vdash P \lor Q$. By the definition we know either $T \vdash P $ or $T \vdash \neg P$, and the same is true for $Q$. Now, we want to show that either $T \vdash P$ or $T \vdash Q$, so if either $T \vdash P$ or $T \vdash Q$, we're there. So, assume that neither is the case. Then we have $T \vdash \neg P$ and $T \vdash \neg Q$. But given $T \vdash P \lor Q$, that means $T \vdash \bot$ ($\bot$ is a logical consequence of $P \lor Q$, $\neg P$, and $\neg Q$). But if $T \vdash \bot$, then $T \vdash P$ (anything follows from a contradiction), and so we're good: we always have $T \vdash P$ or $T \vdash Q$, as desired.
Now we go from right to left:
Assume that for any $P$ and $Q$: if $T \vdash P \lor Q$, then either $T \vdash P $ or $T \vdash Q$. Now take any statement $P$. We want to show that either $T \vdash P$ or $T \vdash \neg P$. Well, we know that $T \vdash P \lor \neg P$ since $P \lor \neg P$ is a tautology. So, we have that either $T \vdash P$ or $T \vdash \neg P$, as desired. So, $T$ is complete.
