背景：
前期已论证关于三个二项式系数乘积的等式，即：
$$
\require{AMSmath}
\begin{cases}
\sum_{k} \binom{m-r+s}{k} \binom{n+r-s}{n-k} \binom{r+k}{m+n} = \binom{r}{m}\binom{s}{n} \\
\sum_{k} (-1)^k \binom{m+l}{l+k}\binom{m+n}{m+k}\binom{n+l}{n+k} = \frac{(l+m+n)!}{l!m!n!}
\end{cases} \\
\Downarrow \\
\sum_{j,k}(-1)^k \binom{l+m}{l+k}\binom{l+k}{j}\binom{m-k}{l-k-j}\binom{m+n+j}{m+l}
$$
问题：
基于上述背景知识，尝试着证明下列等式关系的成立性：
$$
\sum_{j,k} (-1)^{j+k} \binom{j+k}{k+l} \binom{r}{j}\binom{n}{k}\binom{s+n-j-k}{m-j} =
(-1)^l \binom{n+r}{n+l} \binom{s-r}{m-n-l}
$$
解构：
令\(j=n+l, k= r-l\)，则上式变换为：
$$
\begin{align}
(-1)^{j+k} \binom{j+k}{k+l} \binom{r}{j}\binom{n}{k}\binom{s+n-j-k}{m-j}
&=
(-1)^{n+r} \binom{n+r}{r}\binom{r}{n+l}\binom{n}{r-l} \binom{s-r}{m-n-l} \\
&=
(-1)^{n+r}\frac{(n+r)!}{r!n!}\frac{r!}{(n+l)!(r-n-l)}\frac{n!}{(r-l)!(n-r+l)!}\binom{s-r}{m-n-l} \\
&=
(-1)^{n+r}\binom{n+r}{n+l} \binom{0}{r-n-l}\binom{s-r}{m-n-l}\\
&= (-1)^{l}\binom{n+r}{n+l}\binom{s-r}{m-n-l}\\
\end{align}
$$
因此，现在的工作重心转移至论证：求和变量j,k仅在\(j=n+l,k=r-l\)的情况下有效
$$
\begin{align}
\sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\binom{n}{k}\binom{s+n-j-k}{m-j}
&= \sum_{\alpha,\beta}(-1)^{\alpha} \binom{\alpha}{\beta+l}\binom{r}{\alpha-\beta} \binom{n}{\beta} \binom{s+n-\alpha}{m-\alpha+\beta} \\
&= \sum_{k} (-1)^k \binom{n}{k} \left( \sum_{j} (-1)^j \binom{j+k}{k+l}\binom{r}{j}\binom{s+n-k-j}{m-j} \right) \\
&= \sum_{k} (-1)^k \binom{n}{k} \left( \sum_{j} (-1)^j \binom{r}{j} \binom{j+k}{k+l}\binom{s+n-k-j}{s+n-m-k} \right) \\
\end{align}
$$
下面对求和\(\sum_{j}(-1)^{j+k} \binom{r}{j} \binom{j+k}{k+l}\binom{s+n-k-j}{s+n-m-k}\)展开解构：
以简单的形式作为切入口，即讨论\(n=0\)时的情况
$$
\begin{align}
\sum_{j}(-1)^{j+k} \binom{r}{j} \binom{j+k}{k+l}\binom{s+n-k-j}{s+n-m-k}
&= \sum_{j}(-1)^j \binom{r}{j} \binom{j}{l}\binom{s-j}{s-m} \\
&= \sum_{j}(-1)^j \frac{r!}{j!(r-j)!}\frac{j!}{l!(j-l)!} \binom{s-j}{s-m} \\
&= \sum_{j} (-1)^j \binom{r-l}{r-j}\binom{r}{l}\binom{s-j}{s-m}\\
&= \binom{r}{l} \sum_{j} (-1)^j \binom{r-l}{j-l}\binom{s-j}{s-m} \\
&= \binom{r}{l} (-1)^l \sum_{j} (-1)^j \binom{r-l}{j}\binom{s-l-j}{s-m} \\
&= \binom{r}{l} (-1)^l \binom{s-r}{m-l}
\end{align}
$$
现在，正式展开原等式的解构过程：
假设上述等式在\(n-1\)的情况下成立，即满足关系
$$
\sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\binom{n-1}{k}\binom{s+n-1-j-k}{m-j}
= (-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n+1-l}
$$
现基于上述假设关系，证明等式在n的情况下成立：
$$
\begin{align}
\sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\binom{n}{k}\binom{s+n-j-k}{m-j}
&= \sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\left( \binom{n-1}{k}+\binom{n-1}{k-1}\right) \left(\binom{s+n-1-j-k}{m-j} + \binom{s+n-1-j-k}{m-1-j} \right) \\
&= (-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n+1-l} + (-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n-l} +\sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\binom{n-1}{k-1}\binom{s+n-1-j-k}{m-j} + \sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\binom{n-1}{k-1}\binom{s+n-1-j-k}{m-1-j} \\
&= (-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n+1-l} + (-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n-l} -\sum_{j,k}(-1)^{j+k}\binom{j+1+k}{k+l+1}\binom{r}{j}\binom{n-1}{k}\binom{s+n-1-j-k-1}{m-j} – \sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l+1}\binom{r}{j}\binom{n-1}{k}\binom{s+n-1-j-k-1}{m-1-j} \\
&= (-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n+1-l} + (-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n-l} + (-1)^l \binom{n-1+r}{n+l}\binom{s-1-r}{m-n-l} – (-1)^l \binom{n-1+r}{n-1+l}\binom{s-1-r}{m-n+1-l} +(-1)^l \binom{n-1+r}{n+l}\binom{s-1-r}{m-1-n-l} – (-1)^l \binom{n-1+r}{n-1+l}\binom{s-1-r}{m-n-l}\\
&= (-1)^l \binom{n-1+r}{n-1+l}\binom{s-1-r}{m-n-l} + (-1)^l \binom{n-1+r}{n-1+l}\binom{s-1-r}{m-1-n-l} + (-1)^l \binom{n-1+r}{n+l}\binom{s-r}{m-n-l} \\
&=(-1)^l \binom{n-1+r}{n-1+l}\binom{s-r}{m-n-l} + (-1)^l \binom{n-1+r}{n+l}\binom{s-r}{m-n-l} \\
&=(-1)^l \binom{n+r}{n+l}\binom{s-r}{m-n-l}\\
&=
\end{align}
$$
$$
\sum_{j} (-1)^{r-j} \binom{r}{j}\binom{\sigma+j}{\phi} = \binom{\sigma}{\phi-r} \\
\sum_{k} (-1)^{k} \binom{n}{k}\binom{\delta-k}{\Phi} = \binom{\delta-n}{\delta-\Phi} \\
\Downarrow \\
\sum_{j} (-1)^{r-j} \binom{r}{j}\binom{k+j}{k+l} = \binom{k}{k+l-r} \\
\sum_{k} (-1)^{k} \binom{n}{k}\binom{s+n-j-k}{m-j} = \binom{s-j}{s+n-m} \\
\Downarrow \\
\left( \sum_{j} (-1)^{r-j} \binom{r}{j}\binom{k+j}{k+l} \right) \left(\sum_{k} (-1)^{k} \binom{n}{k}\binom{s+n-j-k}{m-j} \right) = \sum_{j,k}\binom{k}{k+l-r} \binom{s-j}{s+n-m} \\
\sum_{j,k} (-1)^{k+j}\binom{r}{j}\binom{k+j}{k+l}\binom{n}{k}\binom{s+n-j-k}{m-j} = (-1)^l \sum_{j,k}\binom{r-l-k-1}{r-l} \binom{s-j}{s+n-m} \\
\sum_{j,k} (-1)^{k+j}\binom{r}{j}\binom{k+j}{k+l}\binom{n}{k}\binom{s+n-j-k}{m-j} = (-1)^l \sum_{j,k}\binom{r-l-k-1}{r-l} \binom{s-j}{s+n-m} \\
$$
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