proof of 1) $|a| = 0 \iff a = 0$ and 2) $|a| \ge 0$ On the internet there are many proofs but very summary

Definition
$|a| = $
$a$ if $a \ge 0$
$-a$ if $a &lt; 0$

Proposition 1 $|a| = 0 \Leftarrow \Rightarrow a = 0$

$\Rightarrow$)

Suppose that $|a| = 0$
This implies that $a\ge 0$ xor $a&lt;0$ 
Case $a&lt;0$
$|a| = -a$ by hypothesis $-a= 0$
As $a&lt;0$
$a + (-a)&lt;0 + 0$
$0&lt;0$ (!contradiction)
Case $a \ge 0$
$|a| = a$ by hypothesis $a = 0$


$\Leftarrow$)
Suppose that $a= 0$
$|a| = a$ by hypothesis $|a| = 0$

proposition 2
$|a| \ge 0$
case $a \ge 0$
$|a| = a$ by hypothesis $|a| \ge 0$
case $a &lt; 0$
$|a| = -a$
As $a&lt;0$
$a+(-a)&lt;0+(-a)$
$0&lt;-a$
$0&lt;|a| \Rightarrow 0 \le |a|$

are valid my proofs?
I think you do a little bit complicate. $$|a|=0\iff (a=0\quad \text{or}\quad  -a=0)\iff a=0.$$

Moreover, you don't need that fact that $a\geq 0$.
By the definition,

$$|a|=0\implies ( 0=a\land a\ge0)\lor(0=-a\land a&lt;0)\iff ( 0=a\land a\ge0)\iff 0=a.$$

And obviously,

$$a=0\implies |a|=0.$$

For the second proposition,

$$a\ge0\implies|a|=a\ge0,\\
a&lt;0\implies |a|=-a&gt;0$$
so that $a\ge0$ holds as all $a\in\mathbb R$ have been covered.