Proof. The divisibility assumptions mean that $c = aa_1 = bb_1$ where $a_1$ and $b_1$ are some elements of $R$ . Because $R$ is a Bézout ring, there exist such elements $x$ and $y$ of $R$ that $\gcd(a,\,b) = 1 = xa+yb$ . This implies the equation $a_1 = xaa_1+yba_1 = xbb_1+yba_1$ which shows that $a_1$ is divisible by $b$ , i.e. $a_1 = bb_2$ , $b_2\in R$ . Consequently, $c = aa_1 = abb_2$ , or $ab \mid c$ Q.E.D.

Note 1. The theorem may by induction be generalized for several factors of $c$ .

Note 2. The theorem holds e.g. in all Bézout domains, especially in principal ideal domains, such as $\mathbb{Z}$ and polynomial rings over a field.