How to prove the two identities in number theory are equivalent? Let 
\begin{align}
Li_2(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}. 
\end{align}
There are two different forms of Abel identities for polylogarithms:
1.
\begin{align}
& Li_2(-x) + \log x \log y \\
& +  Li_2(-y) + \log ( \frac{1+y}{x} ) \log y \\
& +  Li_2(-\frac{1+y}{x}) + \log ( \frac{1+y}{x} ) \log (\frac{1+x+y}{xy}) \\
& +  Li_2(-\frac{1+x+y}{xy}) + \log ( \frac{1+x}{y} ) \log (\frac{1+x+y}{xy}) \\
& +  Li_2(-\frac{1+x}{y}) + \log ( \frac{1+x}{y} ) \log x \\
& = - \frac{\pi^2}{2}. 
\end{align}
2.
\begin{align}
& Li_2(x) + Li_2(y) - Li_2(xy) \\
& = Li_2(\frac{x(1-y)}{1-xy}) + Li_2(\frac{y(1-x)}{1-xy})  + \log(\frac{1-x}{1-xy}) \log(\frac{1-y}{1-xy}).
\end{align}
Are there some references of proving that these two identities are equivalent? Thank you very much.

My comment is a bit too long therefore I write it as an answer.
Thanks for the link, it makes the question much clearer. With $(6.63)$ follows directly $2.$ by the substitution $x\to\frac{x}{1-y}$ and $y\to\frac{y}{1-x}$ . The script uses the symbol-technique $S$ to transform the terms of $(6.63)$ into other terms getting an equation which is somehow equivalent to $1.$ . This equation now has to be retransformed to $1.$ and the constant must be still determined. So it depends on the definition of "equivalent" if $1.$ und $2.$ are equivalent. It's the same question, if e.g. $\frac{1}{(1-x)^2}=\sum\limits_{n=1}^\infty nx^{n-1}$ and $\frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n$ are equivalent. I would say: Because of the difference with the constant it's an incomplete Yes (that $1.$ and $2.$ are equivalent). I hope you will find a better expression than me. :-)
