Line in proof: $\gcd(a,b) \le \gcd(b,r)$ - why? Looking at the following GCD proof, I'm unsure I understand why line 4 and 8 are necessarily true. How is it that $\text{gcd}(a,b) \leq \text{gcd}(b,r)$?
$$\begin{align}
&\text{gcd}(a,b) \setminus a \wedge \text{gcd}(a,b) \setminus b \tag{1}\\
&\text{gcd}(a,b) \setminus (a - qb) \tag{2}\\
&\text{gcd}(a,b) \setminus r \tag{3}\\
&\text{gcd}(a,b) \leq \text{gcd}(b,r) \tag{4}\\
\\
&\text{gcd}(b,r) \setminus b \wedge \text{gcd}(b,r) \setminus r \tag{5}\\
&\text{gcd}(b,r) \setminus (qb+r) \tag{6}\\
&\text{gcd}(b,r) \setminus a \tag{7}\\
&\text{gcd}(b,r) \leq \text{gcd}(a,b) \tag{8}\\
\\
&\text{gcd}(a,b) = \text{gcd}(b,r) \tag{9}
\end{align}$$
Edit: Please do not edit the question to change $\setminus$ to |. Both are accepted notation, even if $\setminus$ is less common.
 A: Here I am assuming that $a = qb + r$, where each are integers. For line (4), since $\text{gcd}(a, b)$ divides $b$ and $r$, then it divides $\text{gcd}(b,r)$. We can see this because any common divisor of $b$ and $r$ must also divide its greatest common divisor. So it follows that $\text{gcd}(a,b) \leq \text{gcd}(b,r)$. We can then apply the same argument to line (8).
Edit: As Bill in the comments noted, this proof in most contexts is circular and so should be avoided. Look at Bill's answer for better clarity.
A: The key idea of the proof is greatly clarified if we combine both directions. The first three lines of both directions combine to yield the following divisibility equivalence
$$ \color{#c00}d\mid a,b\iff \color{#c00}d\mid b,r,\ \ \rm therefore $$
$a,b\,$ & $\,b,r\,$ have same set $S$ of common divisors $\,\color{#c00}d,\,$ so same greatest common divisor $=\max S$
Remark $ $ We can simplify the equivalence by rewriting it as follows
$$\begin{align} {\rm if}\, \ d\mid b\ \ \ {\rm then}\,\ \ d\mid a\ &\iff d\mid a-qb\\[.2em]
{\rm said}\,\bmod d\!:\ {\rm if}\ b\equiv 0\,\ {\rm then}\,\ a\equiv 0&\iff a-qb\equiv 0\end{align}\qquad\qquad$$
