Property for comparing two composition series Suppose we have two composition series
$$
 M = M_0 \unrhd M_1 \unrhd \cdots \unrhd M_r = 1
$$
and
$$
 M = N_0 \unrhd N_1 \unrhd \cdots \unrhd N_s = 1
$$
for a finite group $M$. Then we know that $r = s$. But for given $i$, let $j$ be choosen such that
$$
 M_{i-1} \cap N_j \le M_i, \quad\mbox{but}\quad M_{i-1} \cap N_{j-1} \nleq M_i.
$$

How to show that $N_{j-1} \cap M_i \le N_j$?

As 
\begin{align*}
 N_{j-1} \cap M_{i-1} \le N_j 
 \Leftrightarrow N_{j-1} \cap M_{i-1} & = (N_{j-1} \cap M_{i-1}) \cap N_j \\
 & = M_{i-1} \cap N_j \\
 & \le M_i
\end{align*}
which is excluded, we must have $N_{j-1} \cap M_{i-1} \nleq N_j$. So if we choose $k$ such that
$$
 N_{k-1} \cap M_i \le N_k, \quad\mbox{but}\quad N_k \cap M_{i-1} \nleq N_k
$$
i.e. I guess as both conditions are a maximality requirement for $k$ we then must have $k \le j$. Just for sets these inclusion do not hold, so I guess there must be facts about groups and composition series involved, but I am unable to find the right argument. So I am asking for any help!?

To avoid all those suffixes, we have simple subnormal sections $A/B$ and $C/D$ of $G$ with $A \cap D \le B$ and $A \cap C \not\le B$, and we want to prove $B \cap C \le D$.
By the 2nd Isomorphism Theorem $(A \cap C)/(A \cap D) \cong (A \cap C)D/D \le C/D$, and since this quotient is nontrivial, $C/D$ is simple, and $A$ is subnormal in $G$, we have $(A \cap C)D = C$ and $(A \cap C)/(A \cap D)$ is simple.
Now $A \cap D = B \cap D \unlhd B \cap C \unlhd A \cap C$. But we know that $B \cap C \ne A \cap C$, so $B \cap D = B \cap C$ and the resut follows.
