(related to Proposition: Addition of Natural Numbers Is Cancellative)
Because the addition of natural numbers "\( + \)" is commutative, it is without loss of generality sufficient to show the right cancellation property, i.e. \[x+z=y+z\Rightarrow x=y,~~~~~~(x,y,z\in\mathbb N).\] The proposition can be proven by induction.
For arbitrary \(x,y\in\mathbb N\), it follows from the definition of addition that \[x+0=y+0\Rightarrow x=y.\]
Note that \(x+0=x\) and \(y+0=y\) are both equalities being equivalence relations. Thus, we can replace the implication sign "\(\Rightarrow\)" by the equivalence sign "\(\Leftrightarrow\)", and the reasoning is still logically correct:
\[x+0=y+0\Leftrightarrow x=y.\]
Now, let assume that the inclusion \(x+a_0=y+a_0\Leftrightarrow x=y\) has been proven for all \(z_0\le z\), where we use "\(\le\)" as the order relation of natural numbers. Then it follows again from the definition of addition that \[x+z^+=y+z^+\Leftrightarrow (x+z)^+=(y+z)^+.\] Because both successors are unique and because \(x+z=y+z\) is equivalent to \(x=y\) by assumption, it follows that \[(x+z)^+=(y+z)^+\Leftrightarrow x=y.\] Altogether, it follows from \(x + z=y + z\) that \(x=y\) for all natural numbers \(x,y,z\). Thus, the addition of natural numbers is cancellative. \[ x+z=y+z \Rightarrow x=y,\]
and we have proven the conversion of the cancellative law:
\[ x=y\Rightarrow x+z=y+z\]
for all natural numbers \(x,y,z\).
