Question about proof of Bézout's lemma

Question

Can someone please explain me why

$r=|a|-qd=|a|-q(|a|\bar{s}+bt)=|a|(1-q\bar{s})-bqt\in S\cup \{0\}$

?

Answer 1

$r $ is of the form $xa+yb $ with $x=\pm (1-q\bar{s})$ and $y=-qt $. Hence $r\in S\cup \{0\}$.