User blog comment:Ynought/Extended array hierarchy/@comment-35470197-20190117222337/@comment-35470197-20190119081827

> why both use \(k-1\)

Ah, now I understand. I thought that the following rule is applied to the case where \(k = 0\) because it looks like a portion of rule 2.

> and \((\bullet)^k_{(\bullet_2)} = (\bullet)^{k-1}_{(\bullet_2)^{k-1}_{(\bullet)^{k-1}_{(\bullet_2)^{k-1} cdots}}}\)

It might be better to write the condition when \(k > 0\) after  and in order to help us to understand that this rule is not contained in rule 2 but in a new line.

By the way, applying the rules, I have the following computation. Is that right?

Input : \((2,1)^0_{(1,1)}\)

If \(k=0\) then -> Ok, \(k=0\) now.

1. remove it -> Ok, the current array is \((2,1)_{(1,1)}\) now.

'''2. replace \(\bullet_2\) with the array gained after rule 1. \(a\)-times''' -> Ok, the first replacement yields \((2,1)_{(2,1)_{(1,1)}}\), and the second replacement yields \((2,1)_{(2,1)_{(2,1)_{(1,1)}}}\)

Then the result differs from what you wrote.