Ya, I always used to get in trouble for that. I would jump too fast. I do most of the math in my head, I only write down things if I have to. I'll give you some examples of what I mean by this notation:
\[
\begin{align}
\alpha \uparrow^0 z &= \alpha \cdot z\\
\alpha \uparrow^{0,\circ m} z &= \alpha^m \cdot z\\
\alpha \uparrow^1 z &= \alpha^z\\
\alpha \uparrow^{1,\circ n} z &= \exp_{\alpha}^{\circ n} z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^1 z &= \alpha^m \cdot \alpha^z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^{1, \circ n} z &= \alpha^m \cdot \exp_{\alpha}^{\circ n} z\\
\end{align}
\]
Where my point was, when we use regular iteration (the schroder iteration/the fractional calculus iteration), we get your rule:
\[
\alpha \uparrow^{n,\circ y} \alpha \uparrow^{n+1} z = \alpha \uparrow^{n+1} z+y\\
\]
But for god's sake! Do we need all these uparrows? Let's get rid of one:
\[
\alpha \uparrow^{(n,\circ y),n+1} z=\alpha \uparrow^{n+1} z+y\\
\]
So there's a correspondence between this notation and a translation algebra.
So I write:
\[
\alpha \uparrow^{(k_1,\circ m_1),...,(k_n,m_n)} z = \alpha \uparrow^{(k_1,\circ m_1)}\cdots \alpha \uparrow^{(k_n,\circ m_n)} z\\
\]
Which just saves time because it gets rid of \(n-1\) up arrows and I'm not constantly retyping that out, lol. It's not the prettiest, but it gets the job done.
Additionally, if you were confused; \(s\Big{|}_{\mathbb{N}}\) means the variable \(\Re(s) >0\) is now restricted to \(s \in \mathbb{N}\)--which is standard notation, though not very popular. I like it though.
EDIT:
I really think the fractional calculus is the approach for you though. It works off of \(\mathbb{N}\), but constructs it for \(T = \mathbb{R}^+\). Which is to say, your PEANO idea, is totally satisfied by the bounded analytic hyper-operators. Albeit, I don't get everything you're saying. But the points you seem to be iterating, seem to point exactly to the bounded analytic operators. The sole difference, I'd say, is that we start with the seed functions:
\[
\begin{align}
&\alpha+x\\
&\alpha \cdot x
\end{align}
\]
Which are analytic for \(1 < \alpha < \eta\) and \(x > 0\). From here we can construct hyper operators; and they are bounded and analytic. Additionally, I believe they are an example of the monoid goodstein sequence you keep saying. Correct me if I'm wrong. I apologize, as confused you are by my work; I can empathize with the feeling I have for some of your work
But I think the bounded analytic hyper-operators are at least an example of "something that looks like this monoid/\(\mathbb{N}\)/goodstein thing you have". But maybe I'm misinterpreting....
I think this is super important because; everything in this scenario relates to Mellin transforms and Generating functions. And you know what's next door to the mellin transform... the fourier transform. What's next door to that is Hilbert spaces. Next thing you know we're doing categorical fourier transforms, lmao (Which are a thing!).
\[
\begin{align}
\alpha \uparrow^0 z &= \alpha \cdot z\\
\alpha \uparrow^{0,\circ m} z &= \alpha^m \cdot z\\
\alpha \uparrow^1 z &= \alpha^z\\
\alpha \uparrow^{1,\circ n} z &= \exp_{\alpha}^{\circ n} z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^1 z &= \alpha^m \cdot \alpha^z\\
\alpha \uparrow^{0, \circ m} \alpha \uparrow^{1, \circ n} z &= \alpha^m \cdot \exp_{\alpha}^{\circ n} z\\
\end{align}
\]
Where my point was, when we use regular iteration (the schroder iteration/the fractional calculus iteration), we get your rule:
\[
\alpha \uparrow^{n,\circ y} \alpha \uparrow^{n+1} z = \alpha \uparrow^{n+1} z+y\\
\]
But for god's sake! Do we need all these uparrows? Let's get rid of one:
\[
\alpha \uparrow^{(n,\circ y),n+1} z=\alpha \uparrow^{n+1} z+y\\
\]
So there's a correspondence between this notation and a translation algebra.
So I write:
\[
\alpha \uparrow^{(k_1,\circ m_1),...,(k_n,m_n)} z = \alpha \uparrow^{(k_1,\circ m_1)}\cdots \alpha \uparrow^{(k_n,\circ m_n)} z\\
\]
Which just saves time because it gets rid of \(n-1\) up arrows and I'm not constantly retyping that out, lol. It's not the prettiest, but it gets the job done.
Additionally, if you were confused; \(s\Big{|}_{\mathbb{N}}\) means the variable \(\Re(s) >0\) is now restricted to \(s \in \mathbb{N}\)--which is standard notation, though not very popular. I like it though.
EDIT:
I really think the fractional calculus is the approach for you though. It works off of \(\mathbb{N}\), but constructs it for \(T = \mathbb{R}^+\). Which is to say, your PEANO idea, is totally satisfied by the bounded analytic hyper-operators. Albeit, I don't get everything you're saying. But the points you seem to be iterating, seem to point exactly to the bounded analytic operators. The sole difference, I'd say, is that we start with the seed functions:
\[
\begin{align}
&\alpha+x\\
&\alpha \cdot x
\end{align}
\]
Which are analytic for \(1 < \alpha < \eta\) and \(x > 0\). From here we can construct hyper operators; and they are bounded and analytic. Additionally, I believe they are an example of the monoid goodstein sequence you keep saying. Correct me if I'm wrong. I apologize, as confused you are by my work; I can empathize with the feeling I have for some of your work
But I think the bounded analytic hyper-operators are at least an example of "something that looks like this monoid/\(\mathbb{N}\)/goodstein thing you have". But maybe I'm misinterpreting....
I think this is super important because; everything in this scenario relates to Mellin transforms and Generating functions. And you know what's next door to the mellin transform... the fourier transform. What's next door to that is Hilbert spaces. Next thing you know we're doing categorical fourier transforms, lmao (Which are a thing!).