Some hypergeometric transformation In Concrete Mathematics, the identity
$$\sum_{k\geq 0}\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\=\sum_{k\geq 0}\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^{k}$$
is proven, where $n\geq 0$ and $m$ are integers, and $x,y,r$ are complex numbers.
From this identity, the hypergeometric equality
$$F(a,-n;c;z)=\frac{(a-c)^{\underline n}}{(-c)^\underline n}F(a,-n;1-n+a-c;1-z)$$
can then "clearly" be derived, where $n\geq 0$ is an integer, and $a,b,c$ are complex numbers.
I don't understand how this hypergeometric equality can be proven using the first identity. Any suggestions?

We represent the binomial identity
\begin{align*}
\sum_{k\geq 0}&\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\
&=\sum_{k\geq 0}\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^k\tag{1}
\end{align*}
using hypergeometric series by following the presentation in Concrete Mathematics.
LHS
We start with the LHS of (1) and calculate the quotient $\frac{t_{k+1}}{t_{k}}$ of consecutive terms of the sum. We obtain
\begin{align*}
t_k&=\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\
&=\frac{(m+r)!(n+k)!}{(m-n-k)!(r+n+k)!n!k!}x^{m-n-k}y^k\\
\\
\frac{t_{k+1}}{t_k}&=\frac{(n+k+1)(m-n-k)}{(r+n+k+1)(k+1)}\,\frac{y}{x}\\
&=\frac{(k+\color{blue}{n+1})(k+\color{blue}{n-m})}{(k+\color{blue}{n+r+1})(k+1)}\left(\color{blue}{-\frac{y}{x}}\right)
\end{align*}
The first summand $k=0$ of the LHS of (1) is $\binom{m+r}{m-n}x^{m-n}$ and we conclude
\begin{align*}
\sum_{k\geq 0}&\binom{m+r}{m-n-k}\binom{n+k}{n}x^{m-n-k}y^k\\
&=\binom{m+r}{m-n}x^{m-n}\color{blue}{F\left(n+1,n-m;n+r+1;-\frac{y}{x}\right)}\tag{2.1}
\end{align*}
RHS
We proceed in the same way with the RHS and obtain
\begin{align*}
u_k&=\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^k\\
&=\frac{(-r)!(n+k)!}{(m-n-k)!(-r-m+n+k)!n!k!}\left(-x\right)^{m-n-k}(x+y)^k\\
\\
\frac{u_{k+1}}{u_k}&=\frac{(n+k+1)(m-n-k)(-1)}{(-r-m+n+k+1)(k+1)}\,\frac{x+y}{x}\\
&=\frac{(k+\color{blue}{n+1})(k+\color{blue}{n-m})}{(k+\color{blue}{n-m-r+1})(k+1)}\,\color{blue}{\frac{x+y}{x}}
\end{align*}
The first summand $k=0$ of the RHS of (1) is $\binom{-r}{m-n}(-x)^{m-n}$ and we conclude
\begin{align*}
\sum_{k\geq 0}&\binom{-r}{m-n-k}\binom{n+k}{n}(-x)^{m-n-k}(x+y)^{k}\\
&=\binom{-r}{m-n}(-x)^{m-n}\color{blue}{F\left(n+1,n-m;n-m-r+1;\frac{x+y}{x}\right)}\tag{2.2}
\end{align*}

Replacing $n$ with $q$ in (2.1) and (2.2) we obtain from (1)
\begin{align*}
&\color{blue}{F\left(q+1,q-m;q+r+1;-\frac{y}{x}\right)=\binom{-r}{m-q}\binom{m+r}{m-q}^{-1}(-1)^{m-q}}\\
&\qquad \color{blue}{\cdot F\left(q+1,q-m;q-m-r+1;\frac{x+y}{x}\right)}\tag{2.3}
\end{align*}

Substitutions
Finally we show the identity (2.3) can be written as
\begin{align*}
\color{blue}{F(a,-n;c;z)=\frac{(a-c)^{\underline n}}{(-c)^{\underline n}}F(a,-n;1-n+a-c;1-z)}\tag{3.1}
\end{align*}
Comparing the LHS of (2.3) and (3.1) we use the substitutions
\begin{align*}
a&:=q+1\\
n&:=-q+m\\
c&:=q+r+1\tag{3.2}\\
z&:=-\frac{y}{x}
\end{align*}
With these substitutions the LHS of (2.3) and (3.1) coincide. The upper parameters $q+1$ and $q-m$ are the same at both sides of (2.3). We now use the substitutions to derive the argument, the lower parameter and the factor in front of the hypergeometric series.

*

*From $z=-\frac{y}{x}$ we see
\begin{align*}
1-z=1+\frac{y}{x}=\frac{x+y}{x}
\end{align*}
and the arguments coincide.


*Looking at the lower parameter $n-m-r+1$ in (2.3) we obtain with (3.2)
\begin{align*}
q-m-r+1&=(-n)-r+1\\
&=(-n)-(c-q-1)+1\\
&=(-n)-(c-a)+1\\
&=1-n+a-c
\end{align*}
in accordance with the lower parameter of the RHS in (3.1).


*We obtain
\begin{align*}
\binom{-r}{m-q}&=\binom{a-c}{n}=(a-c)^{\underline{n}}\\
\binom{m+r}{m-q}(-1)^{m-q}&=\binom{-q-r-1}{m-q}(-1)^{m-q}=\binom{-c}{n}\\
&=(-c)^{\underline{n}}
\end{align*}
and the claim (3.1) follows.
