(note that all terms and things used in this are in the set of Positive Whole Numbers)
First, we define the @: a@b = \(a\uparrow^{a\uparrow^{a^b}b}b\)
Then,\(a@^{c}b = a@^{c-1}(a@^{c-1}b)\) until c is equal to one, in that case, \(a@^{c}b = a@b\)
Then, we introduce a shorthand for the @ notation: \(a[b]c = a@^{b}c\)
I use the modified FIH, hereby I will declare it's base rule as "\(f_0(n) =\) "insert function name"
Suppose a large countable ordinal such to that every finite number a satisfies a < u. A fast iteration hierarchy of functions, \(f_a:\mathbb N \rightarrow \mathbb N\)
\(f_0(n)=\)"insert function"
\(f_{n+1}(n) = f_{n}\underbrace{\rightarrow n \cdots \rightarrow n}_{\text{n copies of 'n', where array only has n.}}(n)\) until n = 0, in which case, refer to the top rule.
Where the \(\uparrow\) is defined as follows:
\begin{eqnarray*} f_x \uparrow^n b(c) = \left\{ \begin{array}{ll} {f_x}^b(c) & (n = 1) \\ f_x(c) & (n > 1, b = 1) \\ f_x \uparrow^{n-1} (f_x \uparrow^{n-1} (f_x \uparrow^{n-1} (f_x \uparrow^n (b-1)(c))(c) & (n > 1, b > 1) \end{array} \right. \end{eqnarray*}
\(a \rightarrow b = a^{b}\)
Where the chained arrows is as follows:
\(f_x \rightarrow b \rightarrow c = a\uparrow^cb = f_x\underbrace{\uparrow\ldots\uparrow}_cb\) (in which up-arrow notation is used) = a[c+2]b (in which the bracket notation of hyper operator is used)
\(f_x \rightarrow\ldots\rightarrow b \rightarrow 1 = f_x \rightarrow\ldots\rightarrow b\) — When last entry is 1, it will be ignored.
\(f_x \rightarrow\ldots\rightarrow b \rightarrow 1 \rightarrow c = f_x \rightarrow\ldots\rightarrow b\)
\(f_x \rightarrow\ldots\rightarrow b \rightarrow (c + 1) \rightarrow (d + 1) = f_x \rightarrow\ldots\rightarrow b \rightarrow (f_x \rightarrow\ldots\rightarrow b \rightarrow c \rightarrow (d + 1) ) \rightarrow d\)
\(a[1]^{c}b = f_{c-1}(a@b),f_0(n) = a[b]c\)
\(a[b]^{d}c = f_{d-1}(a@^{b}c)), f_0(n) = a[b]^{d-1}c\), until d = 1, in which case refer below
\(a[b]^{1}c = a[b]c\)
Then, we define two rules: \(a[b]c = [a,b,c]\) and \(a[b]^{d}c = [a,b,c,d]\)
After that, I continue to extend for arrays of length n.
\([\underbrace{x_1,x_2,\cdots,x_{n-1},x_n}_{n},1] = [\underbrace{x_1,x_2,\cdots,x_{n-1},x_n}_{n}]\)
\([1,\underbrace{x_1,x_2,\cdots,x_{n-1},x_n}_{n}] = [\underbrace{x_1,x_2,\cdots,x_{n-1},x_n}_{n}]\)
\([\underbrace{x_1,x_2,\cdots,x_{n-1},x_n}_{n}] = [\underbrace{x_1-1,x_2*(n-1),\cdots,x_{n-1}*(n-1),x_n*(n-1)}_{n}]\)(apply only if \(x_1\) is larger than 1.
\([x_1,x_2,\underbrace{1,\cdots, 1}_{p\,1s},x_{n-1},x_n] = [x_1*(n-1),x_2*(n-1),\underbrace{1,\cdots,1}_{(p-1)1s},x_{n-1}*(n-1),x_n*(n-1)]\)
The * operator is defined below:
\(a*b = [\underbrace{a,a,\cdots,a,a}_{\text {b copies of a, and no other term other than a.}}]\)