(related to Proposition: Addition of Natural Numbers Is Cancellative With Respect To Inequalities)
Let \(x,y,z\) be arbitrary natural numbers.
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 associativity and commutativity laws for adding natural numbers Therefore we have \[y+z=(x+u)+z=x+(u+z)=x+(z+u)=(x+z)+u.\]
It follows, \(x+z < y + z\).
Assume, \(x + z < y + z\) , 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 adding natural numbers that \(x + z = y + z\), 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 \((y+v) + z < y + z\), or equivalently (applying the associativity and commutativity of adding natural numbers) \((y+z) +v < y+z\). This is again a contradiction, since \((y + z) + v > y + z\). Thus, we must have \(x < y\).
Follows from \((1)\) and the commutativity of adding 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 adding natural numbers.
The proof is analogous to the proof of \((3)\), for symmetry reasons.
