User blog comment:LittlePeng9/FORMAL comparison of Hydra battles and Hardy hierarchy/@comment-10429372-20141017061807/@comment-1605058-20141018201256

Okay, Deedlit asked me about this, so here is my explanation:

We need to show that if we start with number \(\alpha[n+1]\) and we keep taking fundamental sequences of ordinals (using CNF) we will get to number \(\alpha[n]+1\). Suppose \(\alpha=\beta+\omega^{\gamma+1}\). Then \(\alpha[n+1]=\beta+\omega^\gamma n+\omega^\gamma=\alpha[n]+\omega^\gamma\). As long as the latter term in the last sum is not 0, we will be reducing it independently of \(\alpha[n]\). Because we will always reach 0, and the only possible last step before 0 is 1, we see that it's necessary that \(\alpha[n]+1\) is reached eventually.

Now suppose that \(\alpha=\beta+\omega^\gamma\) with \(\gamma\) limit. Then \(\alpha[n+1]=\beta+\omega^{\gamma[n+1]}\). Now, when taking further fundamental sequences, we will at all times have numbers of the form \(\beta+\omega^\delta+\epsilon\). The last term in sum will be reduced independently, and after that we will be taking FS of \(\delta\) in the exponent. If we employ transfinite induction, we can see that eventually we will reach the ordinal \(\beta+\omega^{\gamma[n]+1}\). Taking fundamental sequence, we will get \(\beta+\omega^{\gamma[n]}+\omega^{\gamma[n]}k=\alpha[n]+\omega^{\gamma[n]}k\) for some k. Again, the last term will reach 0, and before that it will reach 1, so we will have \(\alpha[n]+1\) as we wanted.