User blog comment:Nedherman1/Arrow Destruction (AD)/@comment-31966679-20180903190222/@comment-35470197-20180904011942

I think that \([a;b]\) is well-defined. (Sorry if I am incorrect.)

Put \(f(n) = [n]\). Then the recursive definition implies \begin{eqnarray*} [a;1] & = & f^{f(a)}(a) \\ [a;b] & = & f^{f(b)}([f^{f^{f(b)}(b)}(a);b-1]), \end{eqnarray*} if I correctly understand the meaning of "\(\underbrace{] \ldots]}_{[b]}\)" and something like that. (Or "\(\underbrace{] \ldots]}_{[b]}\)" might be a typo of "\(\underbrace{;b-1] \ldots ;b-1]}_{[b]}\)".)