Problem with two binomial identities from Concrete mathematics book Concrete mathematics (second edition) claims that:

$$\sum_{k}\binom{l}{m+k}\binom{s+k}{n}\left(-1\right)^{k}=\left(-1\right)^{l+m}\binom{s-m}{n-l}\tag{I}$$

Which is valid for $l\ge0$ and $m,n \in \mathbb Z$

$$\sum_{k \le l}\binom{l-k}{m}\binom{s}{k-n}\left(-1\right)^{k}=\left(-1\right)^{l+m}\binom{s-m-1}{l-m-n}\tag{II}$$

Which is valid for $l,m,n \in \mathbb Z$ such that $l,m,n \ge 0$

I tried to prove them:
$(\text{I})$
$$\sum_{k}\binom{l}{m+k}\binom{s+k}{n}\left(-1\right)^{k}$$$$=\sum_{k}\binom{l}{m+k}\binom{s+k}{s+k-n}\left(-1\right)^{k}\tag{Pascal's rule}$$$$=\left(-1\right)^{s-n}\sum_{k}\binom{l}{m+k}\binom{-n-1}{s+k-n}$$$$=\left(-1\right)^{s-n}\sum_{k}\binom{l}{m+k}\binom{-n-1}{-s-k-1}\tag{Pascal's rule}$$
Setting $m+k \mapsto k$ follows:
$$=\left(-1\right)^{s-n}\sum_{k}\binom{l}{k}\binom{-n-1}{-s-1+m-k}$$$$=\left(-1\right)^{s-n}\binom{l-n-1}{-s-1+m}\tag{ Vandermonde's identity}$$$$=\left(-1\right)^{s-n}\binom{l-n-1}{l-n-m+s}\tag{Pascal's rule}$$$$=\left(-1\right)^{l-m}\binom{s-m}{s-m+l-n}$$$$=\left(-1\right)^{\color{red}{l-m}}\binom{s-m}{n-l}\tag{Pascal's rule}$$

$(\text{II})$
$$\sum_{k \le l}\binom{l-k}{m}\binom{s}{k-n}\left(-1\right)^{k}$$
$$=\sum_{k \le l}\binom{l-k}{l-k-m}\binom{s}{k-n}\left(-1\right)^{k}\tag{Pascal's rule}$$
$$=\sum_{k \le l}\binom{-m-1}{l-k-m}\binom{s}{k-n}\left(-1\right)^{l-m}$$
Setting $k-n \mapsto k$ follows:
$$=\left(-1\right)^{l-m}\sum_{k \le l}\binom{-m-1}{l-m-n-k}\binom{s}{k}$$$$=\left(-1\right)^{\color{red}{l-m}}\binom{s-m-1}{l-m-n}\tag{ Vandermonde's identity}$$
In both of the identities the problem is that my power is $\color{red}{l-m}$, while it should be $l+m$.
Also I really want to know how exactly we can determine over what conditions such identities with more that one variable hold (I'm asking that since it's not easy at all for me and I think maybe there exist something that I'm not aware of)
 A: Here is a different type of proof, for variety's sake. We seek
to evaluate
$$\sum_k {l\choose m+k} {s+k\choose n} (-1)^k.$$
where $l\gt 0$ and $m,n$ integers. 
Now the first binomial coefficient is definitely zero when $k\lt -m.$
Furthermore with $l$ non-negative we also certainly have
$l^\underline{m+k} = 0$ when $l \lt m+k$ or $k \gt l-m.$ So we get
$$\sum_{k=-m}^{l-m} {l\choose m+k} {s+k\choose n} (-1)^k
= \sum_{k=0}^{l} {l\choose k} {s-m+k\choose n} (-1)^{k-m}
\\ = (-1)^m \sum_{k=0}^{l} {l\choose k} (-1)^k [z^n] (1+z)^{s-m+k}
\\ = (-1)^m [z^n] (1+z)^{s-m}
\sum_{k=0}^{l} {l\choose k} (-1)^k (1+z)^{k}
\\ = (-1)^m [z^n] (1+z)^{s-m} (1-(1+z))^l
\\ = (-1)^{m+l} [z^{n-l}] (1+z)^{s-m}
= (-1)^{l+m} {s-m\choose n-l}$$
as claimed. We use $\;\underset{w}{\mathrm{res}} \frac{1}{w^{m+k+1}} (1+w)^l$ for ${l\choose m+k}$ because the residue vanishes when $m+k\lt 0.$ The same residue yields zero with $l$ positive when $l\lt m+k$ and $m+k \ge 0$ because in that case we have a coefficient extractor $[w^{m+k}] (1+w)^l.$
