 Proof: Suppose that \(\mathit{Precedes}(n,a)\). Then, by definition, since \(n\) is a natural number, \(\mathit{Precedes}^{+}(0,n)\). So by Fact (3) about \(R^{+}\) (in the subsection on the Weak Ancestral in §4), it follows that \(\mathit{Precedes}^*(0,a)\), and so by the definition of \(\mathit{Precedes}^{+}\), it follows that \(\mathit{Precedes}^{+}(0,a)\); i.e., \(a\) is a natural number.