User blog comment:Mh314159/Alpha numbers (and beyond)/@comment-35470197-20191007022858/@comment-35470197-20191015090018

1) I guess that \(m = \alpha[\alpha[a-1,b,c],b-1\ldots,c]\) in Rule set 2 means \(m = \alpha[\alpha[a-1,b,\ldots,c],b-1\ldots,c]\). Also, Rule set 2 means right? It will be simpler (i.e. a little easier to analyse) and stronger if you apply the rule \(\alpha[a,b,\ldots,c] = \alpha^m[m]\) as long as it is applicable, but the ordinal in FGH does not change if I am correct. The ordinal roughly corresponding to \(\alpha[\underbrace{x,\ldots,x}_{x}]\) is \(\omega^{\omega}\), which is the boundary of level 8 and 9 with respect to my googological ruler.
 * 1) Apply the rule \(\alpha[a,b,\ldots,c] = \alpha^m[m]\) only when there is no zero.
 * 2) Otherwise, apply "Drop training zeros" or "Replace a non-zero trailing zero",

2) It is also a little complicated to precisely analyse \(f\) in ordinals in FGH, but the ordinal corresponding to \(f\{\underbrace{n,ldots,n}_{n}\}(x)\) is bounded by \(n^2+1\) if I correctly understand the recursion. In particular, it is bounded by \(\omega\).