I have a problem understanding the proof for the following theorem in my textbook
Theorem Let $R$ be an Euclidian ring. Let $a,b\in R$ with $b\neq 0$ and see the result of the Euclidian algorithm
$a=q_0b+r_0$
$b=q_1r_0+r_1$
$r_0=q_2r_1+r_2$
...
$r_{n-2}=q_nr_{n-1}+r_n$
$r_{n-1}=q_{n+1}r_n$
Then $(a,b)$ equals the principal ideal $(r_n)$
Where the proof goes
Proof Set $r_{-2}=a$, $r_{-1}=b$ and $r_{n+1}=0$. Then we have that $r_{i-2}=q_ir_{i-1}+r_i$ for all $i=0,...,n+1$. Thus $(r_{i-2},r_{i-1})=(r_{i-1},r_i)$ for all $i=0,...,n+1$ and therefore $(a,b)=(r_n)$
Now I get that $(r_{i-2},r_{i-1})=(r_{i-1},r_i)$ would imply $(a,b)=(r_n)$ and therefore $r_n=gcd(a,b)$, but I don't the part with $(r_{i-2},r_{i-1})=(r_{i-1},r_i)$ for all $i=0,...,n+1$. Why does that hold from the fact that $r_{i-2}=q_ir_{i-1}+r_i$ for all $i=0,...,n+1$?
There are two parts:
(1) $d=r_n$ is a common divisor of $a$ and $b$. To see this, notice that $d$ divides $r_n$ and so $d$ divides $r_{n-1}$ from which it follows that $d$ divides $r_{n-2}$. Continuing this way, the result follows.
(2) $d$ is the largest common divisor (largest in terms of divibility relation). Suppose $c$ is a common divisor of $a$ and $b$. Then $c$ is a common divisor of $b$ and $r_0$. Continuing this way shows that $c$ is divisor of $d=r_n$. Done.