$x \ge |a| \leftrightarrow x \ge a \land x \ge -a $?

Question

$x \ge |a| \leftrightarrow x \ge a \land x \ge -a $ ?

WTS $x \ge |a| \rightarrow x \ge a \land x \ge -a $
    Since $|a| > -a$ then we have $x \ge -a$
    Since $|a| > a$ then we have $x \ge a$
Then $x \ge |a| \rightarrow x \ge a \land x \ge -a $
WTS $x \ge a \land x \ge -a\rightarrow x \ge |a| $
    Case 1: $a \ge 0$
        Since $a = |a|$ then we have $x \ge |a|$
        Since $|a| > -a$ we still have $x \ge -a$ by $x \ge |a| > -a$
    Case 2: $a < 0$
        Since $-a = |a|$ then we have $x \ge |a|$
        Since $|a| > a$ we still have $x \ge a$ by $x \ge |a| > a $
    In both cases we have $x \ge |a|$
Then $x \ge a \land x \ge -a\rightarrow x \ge |a| $
Then $x \ge |a| \leftrightarrow x \ge a \land x \ge -a $

Is my proof correct? Are there any unnecessary components? Is there a way to do this without cases? Also is it possible to derive this equivalence from either

$|a| \le b \leftrightarrow -b \le a \le b$

or

$|a| \ge b \leftrightarrow a \le -b \lor b \le a$

?

Answer 1

Your proof seems basically correct if somewhat verbose. There are, however, two places where you say "since $|a|>a$" and two where you say "since $|a|>-a$"; all of these should be $\ge$, not $>$.

An easier way to do this, with fewer cases, is as follows: first suppose $a\ge 0$. Then $x\ge |a|$ is the same as $x\ge a$, and since $-a\le 0$, we have $-a\le a\le x$, so that $x\ge -a$ as well. If on the other hand $a\le 0$, then $x\ge |a| = |-a|$ implies (by the first part) that $x\ge -a$ and $x\ge -(-a) = a$, so again the result holds.

As you point out, you can also derive this from the fact that $|a| \le b \leftrightarrow -b \le a \le b$: $x\ge |a| \leftrightarrow -x \le a \le x$, so that $-x\le a$ and $a\le x$; multiplying the first inequality by $-1$ gives $x\ge -a$ and $x\ge a$.