What are the facts used in each step of this proof ? Suppose that $A\in F^{nm}$ and $B\in F^{ml}$
$$\begin{align}rank A + rank B &= rank\begin{bmatrix}0 & A\\B & 0\\ \end{bmatrix}\\ &\le rank\begin{bmatrix}0 & A\\B & I\\ \end{bmatrix}\\ &= rank\begin{bmatrix}I & A\\0 & I\\ \end{bmatrix}\begin{bmatrix}-AB & 0\\B & I\\ \end{bmatrix}\\&\le rank\begin{bmatrix}-AB & 0\\B & I\\ \end{bmatrix}\\&\le rank\begin{bmatrix}-AB&0\\ \end{bmatrix}+rank\begin{bmatrix}B&I\\ \end{bmatrix}\\&=rankAB+m\end{align}$$ I especially din't understand why we can write this? $$\begin{align}rank\begin{bmatrix}-AB & 0\\B & I\\ \end{bmatrix}\le rank\begin{bmatrix}-AB&0\\ \end{bmatrix}+rank\begin{bmatrix}B&I\\ \end{bmatrix}\end{align}$$ and why $$\begin{align}rank\begin{bmatrix}-AB&0\\ \end{bmatrix}+rank\begin{bmatrix}B&I\\ \end{bmatrix}=rankAB+m\end{align}$$ also I cannot understand the part $$\begin{align}rank A + rank B &= rank\begin{bmatrix}0 & A\\B & 0\\ \end{bmatrix}\\ &\le rank\begin{bmatrix}0 & A\\B & I\\ \end{bmatrix}\end{align}$$
Pick rank-many independent rows from $\begin{pmatrix}-AB&0\\B&1\end{pmatrix}$. Some of these rows live in $\begin{pmatrix}-AB&0\end{pmatrix}$ and some in $\begin{pmatrix}B&1\end{pmatrix}$, where they are still linearly independent hence certainly not more than $\operatorname{rank}\begin{pmatrix}-AB&0\end{pmatrix}$ and $\operatorname{rank}\begin{pmatrix}B&1\end{pmatrix}$, respectively.