User blog:GamesFan2000/EEF (Part 2: Transfinite Ordinals 1 - Alpha Part 1)

Rule 6: EEα(n) is the first expression in the function that uses a transfinite ordinal subscript. For alpha, the subscript becomes the answer of EEn(n), and the base also becomes the answer of EEn(n). This is a huge leap in power from the finite subscripts. This is probably beyond epsilon-naught in the FGH, possibly higher, but maybe I’m wrong. Exact growth rates are appreciated from those who are more familiar with the FGH.

Example: EEα(2) = EEEE_{2}(2)(EE2(2))

Rule 7: Transfinite ordinal subscripts can be extended with hyper-operations. The first of these is EEα+1(n). Alpha plus one means to nest the answer of EEα(n) EEα(n) times within EEα functions. For that matter, EEα+b(n) means to nest the answer of EEα+(b-1)(n) EEα+(b-1)(n) times within EEα+(b-1) functions.

Rule 8: EEα+α(n) is the next major growth leap in the function. The subscript and base both become EEα+n(n). For any length subscript of adding alphas, the subscript and base become the answer of EEα+α+…n(n). For any non-transfinite being added, EEα+α+…b(n), the answer of EEα+α+…(b-1)(n) is nested EEα+α+…(b-1)(n) times within EEα+α+…(b-1) functions. If the non-transfinite number becomes 0, it will be cropped. I have a very poor understanding of the FGH beyond epsilon-naught, so I can’t provide growth rates beyond this point. Once again, exact growth rates are appreciated.