I have found the following Fibonacci Identity (and proved it).


If $F\_n$ denotes the nth Fibonacci Number, we have the following identity
\begin{equation}
 F\_{n-r+h}F\_{n+k+g+1} - F\_{n-r+g}F\_{n+k+h+1} = (-1)^{n+r+h+1} F\_{g-h}F\_{k+r+1}
\end{equation}
where $F\_1 = F\_2 = 1$, $r \leq n$, $h \leq g$, and $n, g, k \in \mathbb{N}$. 


It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.


I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!



Is this a new Fibonacci Identity?
Here is an expanded comment of user44191. The basic observation is that one can extend $F\_n$ to all $n\in {\mathbb Z}$ by requiring $F\_{-n}=(-1)^{n+1}F\_n$. Then by [Vajda's formula](https://en.m.wikipedia.org/wiki/Cassini_and_Catalan_identities), one has $$F\_{n'+a'}F\_{n'+b'}-F\_{n'}F\_{n'+a'+b'}=(-1)^{n'}F\_{a'}F\_{b'}=(-1)^{n'+a'+1}F\_{-a'}F\_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.


