Strange isomorphism: $R/(A + B) \cong (R/B)/\bar{A}$. $R$ is a ring and $A, B$ are ideals of $R.$ 
I was playing around with some stuff and I was wondering if $R/(A + B) \cong (R/B)/\bar{A}.$ for $\bar{A}$ being the image of $A$ under $R \rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong? 

This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B \leq A+B \leq R$ implies that
$$
 R/(A+B) \cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $\pi : R \to R / B$
