For my univerity studies I had to prove this:
$ Let \,\, n \,\ \in N \,\, and \,\, L1,\, L2 \, \in R(n \times n) \,\, be \,\, both \,\, lower \,\, triangular \,\, matrices.$
$ Show \,\, that \,\, L := L1L2 \,\, is \,\, also \,\, a \,\, lower \,\, triangular \,\, matrix.$
I proved it like this (and I need some verification for the proof):
$L_{ij} := (L1L2)_{ij}$
$Now \,\, just \,\, look \,\, at \,\, (L1L2)_{ij} \,\, where \,\, j > i.$
$(L1L2)_{ij} = \sum_{r=1}^n l1_{ir}l2_{rj} = \sum_{r=j}^i l1_{ir}l2_{rj} = \sum_{r=1}^n l1_{ir}l2_{rj} = \begin{cases}
0, & \text{if $j>i$ is even} \\
a \in R, & \text{else}
\end{cases}
$
$\Rightarrow \text{L is lower triangular}$
$\square$
Best Answer
Let me call the matrices $A$ and $B$ to avoid too many subscripts. Fix $i<j$. Then $$ (AB)_{ij} = \sum_{k=1}^n A_{ik}B_{kj}. $$ But either $i<k$ or $k<j$ holds (because the negation $k\leq i<j\leq k$ is impossible), so $A_{ik}=0$ or $B_{kj}=0$. Therefore all the addends are zero and $(AB)_{ij}=0$.
Analysis of the OP's proof
When you write $$ \sum_{r=1}^n L1_{ir} L2_{rj} = \sum_{r=j}^i L1_{ir} L2_{rj}, $$ what are you trying to say? Notice also that the starting index of the summation is larger than the ending one. What is it that you want to express?
The next step in your computation then is just reverting back to the previous formula. Why is that?
Finally, what does "$j>i$ is even" mean? And what about $a\in R$?
In summary, I would not consider your proof a proof at all.