(related to Proposition: Every Natural Number Is Greater or Equal Zero)
Let \(x\in\mathbb N\) be a given natural number. Since \(0\) is a natural number, we can have \(x=0\). Otherwise, we have \(x\neq 0\), and according to the trichotomy of natural numbers, it means that either \(x < 0\) or \(x > 0\). We will show that \(x > 0\).
Assume \(x < 0\). By the definition of the order relation, there is a natural number \(u\neq 0\) such that we have \(x+u=0\). But this is a contradiction, since, according to the addition of natural numbers, the sum of two natural numbers can only equal zero (\(x+ u=0\)), if both, \(x=0\) and \(u=0\). Therefore, the assumption \(x < 0\) is false and we must have \(x > 0\).
