User blog comment:DrCeasium/Hyperfactorial array notation: Analysis part 2/@comment-5529393-20130527105642/@comment-5529393-20130530163334

Okay, right off the bat, [1...1,2]w/[2] is not at level \(\phi(\omega+1,0\), or even at level \(\phi(\omega,0) + 1\).

[1...1,2] w/ [2] = [1,1,2] w/ [1] -> [1] w/ [1] = [1...1,2] w/ n = [1...1,2] with n 1's.

So you evaluate [1,1,2], then you evaluate [1...1,2] with n 1's. This is less than level \(\phi(\omega,0) + 1\), which would be composing [1...1,2] with n 1's n times in succession.

Similarly, [1...1,2] w/ [k] is less than \(\phi(\omega,0) + 1\). So is [1...1,2] w/ [1,2]:

[1...1,2] w/ [1,2] = [1...1,2] w/ [n] = [1...1,2] (with n 1's) w/ [1] -> [1] w/ [1] = [1...1,2] w/ n = [1...1,2] with n 1's.

So you evaluate [1...1,2] twice, which is still less than evaluating [1...1,2] n times.

To reach \(\phi(\omega,0) + 1\), we need [1...1,2] w/ [1,1,2]:

[1...1,2] w/ [1,1,2] = [1...1,2] w/ [1,n] = [1...1,2] w/ [n, n-1] = [1...1,2] (with n 1's) w/ [1, n-1] -> [1] w/ [1, n-1] -> [1...1,2] w/ [1, n-1] = [1...1,2] w/ [n, n-2] = etc.

so we see that we evaluate [1...1,2] n times, which is precisely level \(\phi(\omega,0) + 1\). This is a rather far cry from \(\phi(\omega^2, 0)\).

More later.