(related to Proposition: Multiplication of Natural Numbers Is Cancellative)
Because of the counter-example \[2\cdot 0=1\cdot 0\not\Rightarrow 2=1,\] the multiplication of natural numbers is obviously not cancellative for the natural number \(0\). However, we will prove it for all natural numbers \(z\in\mathbb N,~z\neq 0\). Because the multiplication of natural numbers "\( \cdot \)" is commutative, it is without loss of generality sufficient to show the right-cancellation property, i.e. \[x\cdot z=y\cdot z\Rightarrow x=y,~~~~~~(x,y,z\in\mathbb N,~z\neq 0).\] The proposition can be proven by induction of \(z\ge 1\).
For arbitrary \(x,y\in\mathbb N\), it follows from the definition of multiplication and the definition of addition that \[\begin{array}{rcl} x\cdot 1 &=&y\cdot 1\\ (x\cdot 0) + x&=&(y\cdot 0)+y\\ x&=&y \end{array}\]
Now, let assume that the inclusion \(x\cdot z_0=y\cdot z_0\Rightarrow x=y\) has been proven for all \(1\le z_0\le z\), where we use "\(\le\)" as the order relation of natural numbers. Then it follows again from the definition of multiplication of natural numbers that \[\begin{array}{rcl} x\cdot z^+ &=&(x\cdot z)+x,\\ y\cdot z^+ &=&(y\cdot z)+y. \end{array}\]
By assumption, if \(x\cdot z=y\cdot z\), then \(x=y\). The right sides of both equations equal each other and it follows \[x\cdot z^+=y\cdot z^+\Rightarrow x=y,\] which completes the proof that the multiplication of natural numbers is cancellative.
