(related to Proposition: Multiplication of Natural Numbers Is Cancellative With Respect to the Order Relation)
Let \(x,y,z\) be natural numbers with \(z\neq 0\).
According to the definition of order relation of natural numbers, we there exists a natural number \(u\neq 0\) with \(y=x+u\). By virtue of the distributivity law for natural numbers we get \[yz=(x+u)z=xz+uz.\]
Since \(uz\neq 0\) by hypothesis, it follows, \(xz < yz\).
Assume, \(xz < yz\), but not \(x < y\). According to the trichotomy of the order relation for natural numbers, we must have otherwise \(x = y\) or \(x > y\). If \(x = y\), it would follow from the cancellation law for multiplying natural numbers that \(xz = yz\), which is a contradiction to the assumption \(x z < y z\). If \(x > y\), then it would exist a natural number \(v\neq 0\) with \(x=y+v\). Then we would get \((y+v)\cdot z < y z\), or equivalently (applying the distributivity law for natural numbers) \(yz+vz < yz\). This is again a contradiction, since \(yz+vz > yz\).Thus, we must have \(x < y\).
Follows from \((1)\) and the commutativity of multiplying natural numbers.
The proof is analogous to the proof of \((1)\) and \((1a)\), for symmetry reasons.
In the "\( < \)" case, the proof is identical to the proof \((1)\) or \((1a)\), for symmetry reasons. For the "\( = \)" case, the proof is identical to the proof of the cancellation law for multiplying natural numbers.
The proof is analogous to the proof of \((3)\), for symmetry reasons.
