Consider the set in $\Bbb{R}^3$ given by: $P1 = \{(x,y,z) \in \Bbb{R}^3 : |x| \le 1\, |y| \le 1\, |z| \le 1\}$. Show that the set is a polyhedron

Question

I am encountering trouble with this mathematical proof question for an optimisation context.

Question : Consider the set in $\Bbb{R}^3$ given by: $P1 = \{(x,y,z) \in \Bbb{R}^3 : |x| \le 1\, |y| \le 1\, |z| \le 1\}$. Show that the set is a polyhedron

A polyhedron $P ⊆ \Bbb{R}^n$ is the set of all points $x ∈ \Bbb{R}^n$ that satisfy a finite set of linear inequalities. Mathematically, the definition of a polyhedron would be (definition - a) Polyhedron = $\{x | Ax \ge b\}$. For some matrix $A ∈ \Bbb{R}^{m\times n}$ and a vector $b ∈ \Bbb{R}^m$.

More interestingly, the definition of a polyhedron can also equal to Polyhedron = $\{x | Ax \le b\}$. (Definition - b)

Although there are $2$ definitions and I would like to prove using definition A, I feel that presently, it would be easier to prove using definition b first, and then use it as a springboard to prove using definition A. This is because I feel as the question uses ${\le}$ for the absolute values of x,y and z.

I have separated the absolute value equations into 6 equations such as $|x| {\le 1}$ into $+x {\le 1}$ and $-x {\le 1}$ and continuing for $y$ and $z$. I know the $x$ matrix in the polyhedron corresponds to the $x,y$ and $z$ variables in the set.

1. However, I am confused about the next step of translating these 6 sets of equations, namely $+x {\le 1}$ and $-x {\le 1}$, $+y {\le 1}$ and $-y {\le 1}$, $+z {\le 1}$ and $-z {\le 1}$ into the polyhedron definition of Polyhedron = $\{x | Ax \le b\}$ for the proof. Would it simply suffice to substitute all the coefficient of the equations into $A$ such that it is a $6 \times 3$ matrix and showing $b$ as a column vector of $1$ for the final step in the proof, or would further steps need to be taken? Any method would also be appreciated

2. Next it is still the same question problem in the title. As shown in point $1$, I was trying to prove the set to be a polyhedron using definition $2$. However, I would still like to prove using definition 1 and my intent is to use the point 1 as a springboard. Assuming I am using the method in point 1 such that $\{x | Ax \le b\}$ with $A$ as a $6 \times 3$ matrix using the coefficients of the $6$ equations earlier,I am trying to prove using definition $1$ as an end-goal, that is $\{x | Ax \ge b\}$. For this $\{Ax \le b\}$ part, I have a negative sign forming $\{-Ax \ge -b\}$ to change the negative sign. However, I was wondering how to convert this to a Polyhedron = $\{x | Ax \ge b\}$ as there are now negative signs.

Again, any method would be appreciated if there is a flaw in my approach.

Please help me with this question and thank you guys!