Not a duplicate of
If $a\leq b$ and $-a\leq b$, then $|a|\leq b$.
if $-a\leq b\leq a$, then $|b|\leq a$
Is my proof of $|a| \leq b \iff -b \leq a \leq b$ correct?
Prove that for all real numbers $a$ and $b$, $|a| \leq b$ iff $-b \leq a \leq b$
This is exercise $3.5.12.a$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Prove that for all real numbers $a$ and $b$, $|a|\leq b$ iff $-b\leq a\leq b$.
I am familiar with the routine proof of the above theorem but I was wondering whether we could prove the right-to-left direction of the above theorem in the following simple way:
Suppose $-b\leq a\leq b$. Since $a\leq b$ then $a\leq b$ or $-a\leq b$ and thus by definition $|a|\leq b$. Therefore if $-b\leq a\leq b$ then $|a|\leq b$. $Q.E.D.$
I am suspicious of my proof! Is it correct$?$ If not, then why$?$
Thanks for your attention.