This is HW, and I have read Sum of two polyhedra is a polyhedron but I don't understand the solution posted (is $M$ a polyhedron? Why do we take the projection? Why is the projection of $M$ a polyhedron?).
Thus I ask the question again, with first the definition of polyhedron.
Definition 2.4. A subset $\mathscr{P}$ in $\mathbb{R}^{n}$ is called a polyhedron if it can be described in the following form: $$ \mathscr{P}=\left\{x \in \mathbb{R}^{n} \mid A x \leq b\right\} $$ where $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^{m}$ are given.
Below is the statement to prove:
If $P$ and $Q$ are polyhedral in $\mathbb{R}^{n},$ then the direct sum $P+Q=\{z=x+y \mid x \in P, y \in Q\}$ is also a polyhedron.
To be honest I feel that I lack tools to attack this, so any help will be appreciated.
$M$ is a polyhedron since we can view
$$M=\{ (x,y,z): -Ax \le -a, -By \le -b, -x-y+z \le 0, x+y-z \le 0 \}.$$
We take the projection because we know the result that projection of a polyhedron gives us another polyhedron. Here is a link that has a proof of that.
Hence, the idea is that first construct the polyhedron $M$, after which they use the result about polyhedron to show that the projection obtained by dropping $x$ and $y$ from $M$ would give us the desired polyhedron.