User blog comment:B1mb0w/Fundamental Sequences/@comment-5529393-20151113121456/@comment-5529393-20151115095627

(1) Yes, you are correct; phi(1,a) ^^ w = phi(1, a+1), I messed that up. However, this still shows that your rule 3 is incorrect. My previous statement that zeta_0 ^^ w = epsilon_{zeta_0 + 1} is still correct, as zeta_0 = epsilon_zeta_0. epsilon_{zeta_0 + 1} is not equal to zeta_1, so it is not true that zeta_0 ^^ w = zeta_1 as your rule 3 claims.

(2) epsilon_zeta_0 is well-defined. Recall their definitions: epsilon_a is the a'th fixed point of the function f(x) = w^x, and zeta_a is the a'th fixed point of the function f(x) = epsilon_x. So by definition, epsilon_zeta_0 = zeta_0. The "contradiction" that you reach does not show that epsilon_zeta_0 is not well-defined, rather it shows that your rule for fundamental sequences does not give the same values for the expressions zeta_0 and epsilon_zeta_0. The solution is to always represent an ordinal in a certain normal form, and only apply the fundamental sequence rule for ordinals in that form.

A nice standard form for ordinals below the SVO is the following:

Define an additively principal ordinal as an ordinal of the form w^a.

1. Always represent an ordinal a as a_1 + a_2 + ... + a_n, where the a_i are additively principal ordinals and a_1 >= a_2 >= ... >= a_n.

2. If a is an additively principal ordinal, then represent it in the form a = phi (b_1, b_2, ..., b_n) where b_i < a for all i.  (It takes some doing, but one can show that for each additively principal ordinal below the SVO, there is a unique phi(b_1, b_2, ..., b_n) such that b_i < a for all i.)

These rules give a unique normal representation for each ordinal, so there is no inconsistenty stemming from applying the fundamental sequence rules to different representations of the same ordinal.

(3) It's a minor thing, but you need to add u_[m] at the beginning of all of your rules, otherwise there is no rule for something like phi(a,b,c,d,0,0) for example.

(4) Also, it seems a the first rule seems a bit arbitrary and inconsistent with the second rule by starting with 1. It would be more natural and succinct to simply use the second rule, so that phi(1,0_[m+1]) = phi(0_*, 0_[m]).

(5) You really are making that "artificial or non-unique" decision of defining [0], since you define [n] for all values of n.

For example, take your rule phi(1,0) [n] = phi^n (0_*). This is simply shorthand for

phi(1,0) [0] = 0

phi(1,0) [1] = phi(0)

phi(1,0) [2] = phi^2(0)

phi(1,0) [3] = phi^3(0)

...

This can be condensed into your formula, or it can be expressed as

phi(1,0) [0] = 0

phi(1,0) [n+1] = phi( phi(1,0) [n])

So both rules are shorthand for the same set of individual assignments, and both rules "arbitrarily" define phi(1,0)[0] = 0.