Corollary 2.3.7 (Cancellation law). $Let \(a,b,c\) be natural numbers such that \(ac=bc\) and \(c\) is non-zero. Then \(a=b\).
\(Proof\). By the trichotomy of order (Proposition 2.2.1), we have three cases: \(a<b,a=b,a>b\). Suppose first that \(a<b\), then by Proposition 2.3.6 we have \(ac<bc\), a contradiction. We can obtain a similar contradiction when \(a>b\). Thus the only possibility is that \(a=b\), as desired. \(\square\)