(related to Proposition: Order Relation for Natural Numbers, Revised)
If \(y > x\), then it follows from the definition of order relation that there exist a natural number \(u\neq 0\) with \(y=x+u\). According to the fact that every natural number is greater or equal zero, we must have \(u > 0\). Because \(1=0^+\) is the unique successor of \(0\), it is also the smallest number \(u\), for which \(u > 0\). Therefore, if \(u=1\), then \(y = x +1\). Otherwise \(y > x+1\). Altogether, it follows \(y \ge x+1\).
Assume \( y \ge x +1\) and \( y > x +1\). Then we have \(x + 1 > x\) and by the transitivity of the order relation, \(y > x\).
Assume \( y \ge x +1\) and \( y = x +1\). Then we have \(x + 1 > x\) and we have trivially \(y > x\).
The proof is analogous to the proof of \((1)\), for symmetry reasons.
