Proving double inequality by cases I am puzzled by an exercise in Vellemen's How To Prove It, here is the exercise: 
Prove that for all real numbers $a$ and $b$, $|a| \le b$ iff $-b \le a \le b $.
There is actually no official solution for this exercise, but one solution I found online (recommended by Peter Smith from Cambridge, so I assume much of it is correct) suggests a proof by case approach. For the left to right direction, he considers case 1: $a\ge0$ and case 2: $a\lt0$, so that we can assume $|a|=a$ or $|a|=-a$ for the respective cases.
How he completed each case is pretty easy, what I don't understand is how he used the conclusion from each case to reach the conclusion. The conclusion for case 1 and 2, are $a\le b$ and $-b \le a$ respectively.
Then he concluded, 'So either $a\le b$ or $-b \le a$, therefore $-b \le a \le b $.'
I don't understand how is this valid, i.e. how you can join the conclusion from the two cases together with a disjunction to form the conclusion? Because to me, double inequality should be a conjunction, i.e. b is greater or equal to a, AND a is greater or equal to -b. 
Furthermore, shouldn't he have concluded $-b \le a \le b $ from EACH case, and then say that since the cases exhaust all logical possibilities, thus the conclusion is true? How come he can get away with only proving one side of the double inequality with each case?
Thank you very much in advance!
 A: I don't know Peter Smith's proof, but the cases play out as follows:
1) ($0 \le a$) Then $|a|\le b$ iff $-b\le a \le b$ because $-b\le 0$ and $a = |a|$.
2) ($a < 0$) Then $|a| = -a$. By 1), $|-a|\le b$ iff $-b \le -a \le b$. Because $x\le y$ iff $-y\le -x$, and $|a| = |-a|$, we get $|a|\le b$ iff $-b \le -a = |a| \le --b = b$.
A: First note that $|a|\le b \Rightarrow b\ge 0$ since $|a| \ge 0$ for all $a$ in $\mathbb{R}$
Now from:
$$
\left( a \ge 0 \Rightarrow |a|=a\right) \land \left( a < 0 \Rightarrow |a|=-a\right)
$$
we have:
$$
\left( |a| \le b \right) \iff \left[ \left( a \ge 0\right) \land \left( a \le b\right) \right] \lor \ \left[ \left( a < 0\right) \land \left( -a \le b\right) \right]
$$
and, since $\left( -a \le b\right) \iff \left(-b \le a\right)$ we have:
$$
\left( |a| \le b \right) \iff \left[ \left( a \ge 0\right) \land \left( a \le b\right) \right] \lor \ \left[ \left( a < 0\right) \land \left( -b \le a\right) \right]
$$
Now, since $b \ge 0$ we have $-b \le b$ so $-b\le a \le b$
