Attempted proof of the Contracted Bianchi Identity

[Vanilla Gorilla]

TL;DR

I constructed an attempted proof of the Contracted Bianchi Identity, and cannot identify any errors myself. However, obviously, this does not mean that the proof is watertight. For any errors present, I am hoping to get details and explanations on where I went wrong, how to avoid the mistake in the future, etc.
Criticism welcome!

My Attempted Proof​

$R^{mn}_{;n} - \frac {1} {2} g^{mn} R_{;n} = 0$
$R^{mn}_{;n} = \frac {1} {2} g^{mn} R_{;n}$
So, we want $2 R^{mn}_{;n} = g^{mn} R_{;n}$

Start w/ 2nd Bianchi Identity $R_{abmn;l} + R_{ablm;n} + R_{abnl;m} = 0$
Sum w/ inverse metric tensor twice $g^{bn} g^{am} (R_{abmn;l} + R_{ablm;n} + R_{abnl;m}) = g^{bn} g^{am} (0)$
Distributing $g^{bn} (g^{am} R_{abmn;l} + g^{am} R_{ablm;n} + g^{am} R_{abnl;m}) = 0$
Index manipulating $g^{bn} (R^{m}_{~~~bmn;l} + R^{m}_{~~~blm;n} + R^{m}_{~~~bnl;m}) = 0$
Contracting $g^{bn} (R_{bn;l} - R_{bl;n} + R^{m}_{~~~bnl;m}) = 0$
Distributing and manipulating indices $R^n_{n;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0$
Summing $R_{;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0$
Sum W/ $g^{lp}$: $g^{lp} (R_{;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0$
Distributing $g^{lp} R_{;l} - g^{lp} R^n_{l;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0$
Summing $g^{lp} R_{;l} - R^{np}_{;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0$
Ricci Tensor (downstairs indices is symmetric, so I THINK it is valid to flip the 2 top indices here $g^{lp} R_{;l} - R^{pn}_{;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0$
$g^{lp} R_{;l} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = R^{pn}_{;n}$ becomes $R^{pn}_{;n} = g^{lp} R_{;l} + g^{lp} g^{bn} R^{m}_{~~~bnl;m}$
$R^{pn}_{;n} = g^{lp} R_{;l} - g^{lp} g^{bn} R^{m}_{~~~bln;m}$
$R^{pn}_{;n} = g^{lp} R_{;l} - R^{mnp}_{~~~~~~~~n;m}$
$R^{pn}_{;n} + R^{mnp}_{~~~~~~~~n;m} = g^{lp} R_{;l}$
The metric tensor and its inverse are symmetric $R^{pn}_{;n} + R^{mnp}_{~~~~~~~~n;m} = g^{pl} R_{;l}$
Summing over $n$, $R^{pn}_{;n} + R^{mp}_{;m} = g^{pl} R_{;l}$
I've seen others just replace indices like I am about to, so I am about 60% sure the following operation is valid. However, I am not sure. Any insight and corrections would be much appreciated here especially! $R^{mn}_{;n} + R^{mn}_{;m} = g^{mn} R_{;n}$
Combining like terms $2 R^{mn}_{;n} = g^{mn} R_{;n}$, which is just the result we're looking for, assuming I did it right.

Any help is much appreciated!
P.S., I'm not always great at articulating my thoughts, so my apologies if this question isn't clear. Also, I know this isn't high school material, but I am currently in high school, which is why I made my level "Basic/high school level."
Lastly, I know that this isn't the typical way one would go about proving this identity, but I was wondering if this was still a valid avenue to do so (I did it from scratch).

 

[PeterDonis]

Moderator's note: Thread moved to the Differential Geometry forum.