$A \in M_n$ is reducible if and only if there is a permutation $i_1, ... , i_n$ of $1,... , n$ Let $A \in M_n$. $A$ is called reducible if $A$ is permutation similar to a matrix of the form

$$\begin{pmatrix}
B &amp; 0\\
C &amp;D
\end{pmatrix}
$$
where $B$ and $D$ are square matrices. 

Reading a book I got stuck in the following lines:

Obviously, $A \in M_n$ is reducible if and only if there is a permutation $i_1, ... , i_n$ of $1,... , n$ and an integer $s$ with $1 \leq s \leq n-1$ such that
$A [i_1,. . . , i_s | i_{s+1} , . .. , i_n] = 0$.

I am confused with... $A [i_1,. . . , i_s | i_{s+1} , . .. , i_n] = 0$...what does this means?
It means that the submatrix of $A$ given by
$$
\begin{bmatrix}
a_{i_1,i_{s+1}} &amp; \cdots &amp; a_{i_1,i_n}\\
\vdots &amp; &amp; \vdots\\
a_{i_s,i_{s+1}} &amp; \cdots &amp; a_{i_s,i_n}
\end{bmatrix}
$$
is equal to the zero matrix.
It means that there are $s$ rows with indexes $i_1$,\dots, $i_s$ and $n-s$ columns with indexes $i_{s+1}$,\dots, $i_n$ such that the submatrix of these rows and columns is a zero matrix.