Proof Fibonacci derivation I was wondering how to prove that
$$f(n+m+2) = f(n+1)f(m+1) + f(n)f(m)$$
where $f$ is the fibonacci sequence and n, m are positive integers.
Can be this done with induction?
I'm lost with this method of proof, because there are two variables.
Any idea or suggestion is welcome.
 A: 
Can be this done with induction?

It can. More specifically, it can be done with strong induction on two variables. First I suggest looking at https://math.stackexchange.com/a/7665/146030 and thinking of why, in both cases, the first three statements implies the fourth.
We will prove the claim that
$$f(n+m+2)=f(n+1)f(m+1)+f(n)f(m).$$
To begin we define the fibonacci sequence as
\begin{align}
f(0)&=0 \\
f(1)&=1 \\
f(n)&=f(n-1)+f(n-2), \text{for } n\ge2.
\end{align}
When $n=0$ and $m=0$ then
\begin{align}
f(n+m+2) &=
f(2) \\
&= 1 \\
&= 1 \cdot1 + 0\cdot0 \\
&= f(1)f(1)+f(0)f(0) \\
&= f(n+1)f(m+1)+f(n)f(m)
\end{align}
and so the statement is true when $n=m=0$.
To prove the statement true for all nonnegative $n,m$, we first induct on $n=k$ for a fixed $m$. Assume the statement true for all $0\leq k\leq n$. We now prove the statement for $k+1$.
\begin{align}
f((k+1)+m+2) &=
f(k+m+3) \\
&= f(k+m+2) + f(k+m+1) \\
&= f(k+m+2) + f((k-1)+m+2) \\
&= \big[f(k+1)f(m+1)+f(k)f(m)\big] + \big[f(k)f(m+1)+f(k-1)f(m)\big] \\
&= \big[f(k+1)f(m+1)+f(k)f(m+1)\big] + \big[f(k)f(m)+f(k-1)f(m)\big] \\
&= \big[f(k+1)+f(k)\big]f(m+1) + \big[f(k)+f(k-1)\big]f(m) \\
&= f(k+2)f(m+1) + f(k+1)f(m) \\
&= f((k+1)+1)f(m+1) + f(k+1)f(m)
\end{align}
And so by mathematical induction the statement is true for all $n$ and that fixed $m$. We can see that a similar inductive proof works for a fixed $n$ and $m=k$. Thus we can conclude the statement is true.
A: I'm afraid this statement is impossible to prove as it is wrong. Indeed this can be seen with $n=1, m=0$: 
$$f(1+0+2)=f(3)=f(2)+f(1)=2$$
whereas for your proposition 
$$f(1+0+2)=f(1+1)f(0+1)+f(0)f(1)=f(2)f(1)+f(0)f(1)=1 \cdot 1+0 \cdot 1=1$$
As for Trevor Fancher's induction, it is completely right except that his basic step is incomplete. 
(I shall refer to the statement as $P(n,m)$).
Indeed what he does is prove that for a fixed p, $P(n,p)$ and $P(n+1,p)$ true $\Rightarrow P(n+2,p)$ true $\forall n \in \mathbb{N}$. By symetry he also proves that for a fixed p, $P(p,m)$ and  $P(p,m+1) \Rightarrow P(p,m+2)$ $\forall m \in \mathbb{N}$.
Then to conclude the proof what we do is: 


*

*$P(0,0)$ true + P(1,0) true + Induction on n $ \Rightarrow P(n,0)$ true
$\forall n$

*$P(0,1)$ true + P(1,1) true + Induction on n $ \Rightarrow P(n,1)$ true $\forall n$

*$P(n,0)$ true + P(n,1) true + Induction on m $ \Rightarrow P(n,m)$ true $\forall n, m$
But in this proof we only see that $P(0,0)$ is true, not $P(0,1)$,$P(1,0)$ nor $P(1,1)$ hence this proof is incorrect as they are necessary to conclude the proof (and thus for the induction to be valid).
I believe the true similar statement you were looking to prove is 
$$f(n+m+1)=f(n+1)f(m+1)+f(n)f(m)$$
which is proven is a quasi-identical manner.
A: Your proposition is not true (as it can be seen for $P(0,1) : f_{0+1+2} ≠ f_{0+1}f_{1+1} + f_{0}f_{1}$). Some elements are missing to build such a proof in a rigourous way : 
You use both ranks $−1$ and $$. Then your inductive step is based on two consecutive ranks, means we want to use this for our induction : $P(k,n) \land P(k+1,n)\implies P(k+2,n)$. It comes that you have to prove $P(0,n)\implies P(1,n)$: which is false. One should have proved that the proposition holds for $P(0,0), P(0,1), P(1, 0), P(1,1)$ to be able to use 2 consecutive ranks in the induction.
The correct proposition is rather $f_{m+n+1}$ = $f_{m+1}f_{n+1} + f_mf_n$
$(Analysis II  w/ Anna) > All$
