02/19/2008, 09:41 PM
In my view, it is not a matter of proof, but a matter of choice. If we choose to hold \( a [N-1] b = a [N] (\text{hy}N\text{log}_a(b) + 1) \) to be true for all N, then the natural consequence is that \( a [0] b = b + 1 \) for all a (because \( a [1] (\text{sub}_a(b) + 1) = a + ((b - a) + 1) = b + 1 \)). If we choose to hold \( a [N-1] a = a [N] 2 \) true for all N, then we get \( a [0] b = b + 2 \) for a == b. Both cannot be true, so we must make a choice.
Andrew Robbins
Andrew Robbins