(related to Proposition: Transitivity of the Order Relation of Natural Numbers)
Assume that for any natural numbers \(x,y,z\in\mathbb N\), the order relations * \(x < y\) and \(y < z\) are fulfilled. Then there exist natural numbers \(u\neq 0\) and \(v\neq 0\) with \(y=x+u\) and \(z=y+v\). From the associativity of adding natural numbers, it follows
\[z=y+v=(x+u)+v=x+(u+v).\] Since \(u+v\neq 0\), it follows \(x < z\).
It follows from \((i)\) and the transitivity of the equality relation "\(=\)".
Analogously to \((i)\), there exist natural numbers \(u\neq 0\) and \(v\neq 0\) with \(x=y+u\) and \(y=z+v\). From the associativity of adding natural numbers, it follows
\[x=y+u=(z+v)+u=z+(v+u).\] Since \(v+u\neq 0\), it follows \(x > z\).
It follows from \((iii)\) and the transitivity of the equality relation "\(=\)".