The multiplication of natural numbers is cancellative, if the number we want to "cancel out" does not equal \(0\)^{1}, i.e. for all \(x,y,z\in\mathbb N\) with \(z\neq 0\) the following laws (both) are fulfilled:
Left cancellation property: If the equation \(z \cdot x=z \cdot y\) holds, then it implies \(x=y\).
Right cancellation property: If the equation \(x \cdot z=y \cdot z\) holds, then it implies \(x=y\).
Conversely, the equation \(x=y\) implies
for all \(x,y,z\in\mathbb N\) with \(z\neq 0\).
Proofs: 1
Please note that if we allow \(z=0\), then the proposition is wrong: e.g. \(0\cdot 5=0\cdot 3\), but \(5\neq 3\). ↩