This also follows from the Chu-Vandermonde identity ${s+t \choose n}=\sum_{k=0}^n {s \choose k}{t \choose n-k}$ and the upper negation rule for binomial coefficients $\binom{r}{k} = (-1)^k \binom{k-r-1}{k}$.

$$\sum_{\substack{i+j=s \atop i\geq 0, j \geq 0}}\binom{A-n+j}{j}\binom{n-j}{i} = \sum_{j=0}^s \binom{A-n+j}{j}\binom{n-j}{s-j}.$$ Then apply upper negation to get $$\sum_{j=0}^s (-1)^j \binom{-A+n-1}{j} (-1)^{s-j}\binom{s-n-1}{s-j}. $$ Chu-Vandermonde followed by upper negation again yields $$= (-1)^s \binom{-A+s-2}{s} = \binom{A+1}{s}.$$

This also follows from the Chu-Vandermonde identity ${s+t \choose n}=\sum_{k=0}^n {s \choose k}{t \choose n-k}$ and the upper negation rule for binomial coefficients $\binom{r}{k} = (-1)^k \binom{k-r-1}{k}$.

$$\sum_{\substack{i+j=s \atop i\geq 0, j \geq 0}}\binom{A-n+j}{j}\binom{n-j}{i} = \sum_{j=0}^s \binom{A-n+j}{j}\binom{n-j}{s-j}.$$ Then apply upper negation to get $$\sum_{j=0}^s (-1)^j \binom{-A+n-1}{j} (-1)^{s-j}\binom{s-n-1}{s-j}. $$ Chu-Vandermonde followed by upper negation again yields $$= (-1)^s \binom{-A+s-2}{s} = \binom{A+1}{s}.$$