  It does not seem possible to prove that \(x+y\) is a number of order \(n\) if \(x,y\) are numbers of order \(n\). On the other hand, it is possible to show that if \(x\) is a number of order \(n+1\), and \(y\) a number of order \(n+1\) then \(x+y\) is a number of order \(n\). Indeed, the property \(P(z)\) that “\(x+z\) is a number of order \(n\)” is an inductive property of order \(n+1: P\)(0) holds since \(x+0 = x\) is a number of order \(n+1\), and hence of order \(n\), and if \(P(z)\) holds, that is if \(x+z\) is a number of order \(n\), then so is \(x+(z+1) = (x+z)+1\), and so \(P(z+1)\) holds. Since \(y\) is a number of order \(n+1\), and \(P(z)\) is an inductive property of order \(n+1, P(y)\) holds and so \(x+y\) is a number of order \(n\). This example illustrates well the complexities introduced by the ramified hierarchy. 