$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Observe that $x\ge 0$ iff $x+(-x)\ge 0+(-x)$ iff $0\ge -x$ as needed.

Since $0$ is an identity element with respect to $+$ then $x+0=x$, specifically for $x=0$ by definition this means that $-0=0$ as needed.

Observe that $x\ge y$ is the same as $x-y\ge 0$ is the same as $0\ge y-x$ is the same as $-x\le -y$ as needed.

Let $x\in ℝ$ if $x>;0$, then $x=\left|x\right|$ by definition, if $x<;0$, then $\left|x\right|=-x$, or rather $x=-\left|x\right|$

Recall the definition of $\max (x,-x)$ which is that it's value is $x$ if $x\ge -x$ and $-x$ if $x\le -x$. Suppose that $x\ge 0$ then $|x|=x$ and since $x\ge -x$ then we know that $\max (x,-x)=x$ and so in this case we see that $|x|=\max \left(x\right)$.

Now in the case that $x<0$ then $-x>0$ and therefore $-x>x$ therefore $\max (x,-x)=-x$ and since $x<0$ then $|x|=-x$.

Therefore it's been verified in any case so that $|x|=\max (x,-x)$.

$⟹$ we want to prove that $|a|\le x$. Case one, if $a\ge 0$ then $|a|=a$ and since we already know that $a\le x$ then we obtain that $|a|\le x$ as needed. In the second case if $a<0$ then $|a|=-a$ so we want to show that $-a\le x$ which is the same as $a\ge -x$ which we know by assumption as well.

$⟸$ Suppose that $|a|\le x$ if $a\ge 0$ then $|a|=a$ and so we're really assuming that $a\le x$ therefore $-a\ge -x$ since $a\ge -a$ then by chaining inequalities we get $x\ge a\ge -a\ge -x$ which implies $x\ge a\ge -x$ as needed. Now for the case of $a\le 0$ then we've assumed that $-a\le x$ now $-a$ is positive and by symmetrically following our previous case with $-a$ in place of $a$ we see that $x\ge -a\ge a\ge -x$ and so $x\ge a\ge -x$ again, as needed.

Let's recall that $c\le |c|\le \max (|c|,|a|)$, and on the other hand that $a\ge -|a|\ge -\max (|a|,|c|)$ combining inequalities, that says $$-\max (|a|,|c|)\le b\le \max (|a|,|c|)$$ therefore $$|b|\le |\max (|a|,|c|)|=\max (|a|,|c|)$$ as needed.

Since $1<\left|c\right|$ then $1\cdot \left|x\right|\le \left|c\right|\cdot \left|x\right|=\left|cx\right|$

TODO

Suppose that $a=b$ for contradiction, then $-\left|a\right|<a<\left|a\right|$ , but either $a=\left|a\right|$ or $a=-\left|a\right|$, so this yields a contradiction.

We prove it true by induction based on the size of $X$. Base case $\left|X\right|=2$ is covered by the regular triangle inequality. Now suppose that it holds true for $\left|Y\right|=k$ we'll show it holds true for $\left|Z\right|=k+1$, let $z\in Z$, then we have $|\sum (Z\setminus \left\{z\right\})|\le {\sum }_{a\in Z\setminus \left\{z\right\}}\left|a\right|$ since $|Z\setminus \left\{z\right\}|=k$. Now by the regular triangle inequality we obtain that $$\begin{array}{cccc}& |\sum Z|& =|\sum (Z\setminus \left\{z\right\})+z|& \text{<span class="tml-eqn"></span>}\\ & & \le |\sum (Z\setminus \left\{z\right\})|+\left|z\right|& \text{<span class="tml-eqn"></span>}\\ & & \le \sum _{a\in Z\setminus \left\{z\right\}}\left|a\right|+\left|z\right|& \text{<span class="tml-eqn"></span>}\\ & & =\sum _{z\in Z}\left|z\right|& \text{<span class="tml-eqn"></span>}\end{array}$$

We have the following: $$\begin{array}{ccc}& |a-b|\le c& \text{<span class="tml-eqn"></span>}\\ & \Updownarrow & \text{<span class="tml-eqn"></span>}\\ & |a-b|+|b|\le c+|b|& \text{<span class="tml-eqn"></span>}\\ & \Updownarrow & \text{<span class="tml-eqn"></span>}\\ & |a-b+b|\le |a-b|+|b|\le c+|b|& \text{<span class="tml-eqn"></span>}\\ & \Updownarrow & \text{<span class="tml-eqn"></span>}\\ & |a|\le c+|b|& \text{<span class="tml-eqn"></span>}\end{array}$$ Where in the second last line we used the triangle inequality