For every real number $x$, $ \lvert 2x – 6 \rvert \gt x \iff \lvert x-4 \rvert \gt 2 $

Question

I was hoping to get a bit of feedback on a proof I've done involving absolute values. This problem is taken from D. Velleman's How to Prove it (#3.5.11). I've only written on one side of the biconditional, but the I believe the other conditional can be structured similarly.

Problem Statement:
Prove that for every real number $x$, $\lvert 2x – 6 \rvert \gt x $ iff $\lvert x – 4 \rvert \gt 2 $.

My question:
Do the statements below constitute a valid proof or do they require supplemental explanation?

---

($\rightarrow$) suppose $ 2x – 6 \gt 0$. Per the definition of $ \lvert 2x – 6 \rvert $ we proceed by cases.

Case 1: $ 2x – 6 \geq 0 $.
Since $ 2x – 6 \geq 0 $, $ \lvert 2x – 6 \rvert = 2x – 6.$
Then
$$ \lvert 2x – 6 \rvert \gt x $$
$$= 2x – 6 \gt x $$
$$= -6 \gt -x $$
$$= 6 \lt x $$
$$= 4+2 \lt x $$
$$= 2 \lt x-4 $$
$$= 2 \lt \lvert x – 4 \rvert $$
$$= \lvert x-4 \rvert \gt 2 $$

Case 2: $ 2x – 6 \lt 0 $. Since $ 2x – 6 \lt 0 $, $ \lvert 2x – 6 \rvert = 6 – 2x $.

Then
$$ \lvert 2x – 6 \rvert \gt x $$
$$= 6 – 2x \gt x $$
$$= 6 \gt 3x $$
$$= 2 \gt x $$
$$= 4 > x + 2 $$
$$= 4 – x \gt 2 $$
$$= \lvert x – 4 \rvert \gt 2 $$

EDIT
I removed the word 'equivalence' from the problem because it might not be the most appropriate word to use. I'm mostly just asking whether the series of statements I've given provide a valid proof of the conclusion.

Answer 1

Looks pretty good to me. But there is an important point you should change:

• As pointed out in the comments, all the "=" preceding the line by line inequalities in your proof should be replaced by an equivalence sign "<=>". Except at the lines where the proof only goes in one direction (when you add the absolute values again to the inequality for instance); at that moment you only put an implies sign "=>"

Concerning the proof line by line (if you want to dot every "i"):

• Case 1: the last line is obvious, no need for it.
• Case 2: you could also add a short explanation about why the transition from the second to last line to the last line is valid, it's not immediately obvious. But math textbooks often skip more steps than that from one line to the next, so I'm not shocked in any way.