Induction on 2 numbers 
Show that for the Fibonacci numbers it holds that
$f_{x+y}=f_{y+1}f_{x}+f_{y}f_{x-1}$ for $x\ge 2, y\ge 1$

The proof goes as follows:
First we show that the claim holds for all $x\ge2$,$y=1$ and that it holds for $x\ge2, y=2$(easy)
In the Induction step we take $y\ge2, x\ge2$. It holds that
$f_{x+(y+1)}=f_{x+y}+f_{x+(y-1)}=f_{y+1}f_x+f_yf_{x-1}+f_yf_x+f_{y-1}f_{x-1}$ by the induction hypothesis
$=f_{y+2}f_x+f_{y+1}f_{x-1}$
Is this really a complete proof? I  know that there are several ways to do an induction on $2$ numbers. But these usually involve an induction step on each variable or an induction on the sum of both.But in this one we never really do anything with $a$.
Could someone please explain/motivate this or just give me some reference for this kind of induction?
 A: Actually, you're not doing induction on two variables here at all. You're only doing induction on $y$, while $x$ is assumed to be arbitrary. But, as such, it is a good proof! 
A: It's subtle but this induction is not doing: Assume true for $x$ and assume true for $y$ then show it is true for $x+1$ and $y+1$.
That would not be valid as that proves only  for $x$ and $y$ where $x - y = constant$.
BUT this is actually doing something different.  It is assuming it is true for $n = x+y$ and showing it is therefore true for $n = (x+y) + 1 = (x+1) + y = x + (y+1)$.
That is valid.
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On second reading, I see Bram28, is correct and this is only an induction on the variable $y$. and not on $x$ at all. 
To see that this proof is complete, you should note the base cases (for $y= 1; x\ge 2$ and $y= 2; x\ge 2$)[1] was for any arbitrary $x$ and not for an $x = 2$. So induction on $x$ is unnecesary.
I'm going to leave my answer up, because I think it illustrates a good point of looking at an induction proof involves carefully evaluating what the variable being inducted upon is; not merely what it looks like.
The poof could have been identically done on $x+y \implies x + y+ 1$ (only the framing of the base case would differ).
[1] Both base cases were omitted which isn't really fair.  
$y=1; f_{x+y}=f_{x+1}=1*f_x + 1*f_{x-1}=f_2f_x + f_1f_{x-1}=f_yf_x + f_{y+1}f_x$.
$y=2; f_{x+y}=f_{x+2}=f_{x+1} + f_x=f_x + f_{x-1} + f_x = 2*f_{x} + 1*f_{x-1} = f_3*f_x + f_2*f_{x-1} = f_{y+1}f_x + f(y)*f_{x-1}$.
