I don't know the name of what I'm seeking, but maybe someone could help point me in the right direction. I do know that what I'm dealing with, at surface, presents as a programming problem but fundamentally is a math--possibly a combinatorics--problem. I'll as best I can lay it out programmatically.
Let's say I have an array containing 500 arrays, each containing unique combinations of precisely 6 values between 1 and 12.
var myArray = [ [1,2,3,4,5,6], [7,8,9,10,11,12], [2,3,4,5,6,7], ... ];
Now, let's say that I want to reduce this set of 500 arrays to a set of 20. But here's the catch: I …

Associativity:
\begin{align}
\alpha[m<3](\beta[m<3]\gamma)=(\alpha[m<3]\beta)[m<3]\gamma
\end{align}

For $\alpha,n,\gamma < \omega$
\begin{align}
\alpha[0]\beta & =S(\alpha)=\alpha[1]1\\
\alpha[1]0 & = \alpha\\
\alpha[2]0 & = 0\\
\alpha[m>2]0 & =1\\
\alpha[m](\beta+1) & = \alpha[m-1]^{\beta}\alpha\\
\end{align}

For $\alpha,\beta,\gamma \geq \omega$
\begin{align}
\alpha[0]\beta & =S(\alpha)=\alpha[1]1\\
\alpha[1]0 & = \alpha\\
\alpha[2]0 & = 0\\
\alpha[m>2]0 & =1\\
\alpha[m](\beta+1) & = \alpha[m-1]^{\beta+1}\alpha = \alpha[m-1]^{\beta}(\alpha[m-1]\alpha)\\

$\alpha[m](\omega+1) \overset{\textrm{def}}{\not\equiv} \sup (\alpha[m-1]\beta|\beta < \omega+1)$

Because why not, I only introduce a pointwise discontinuity in this system.
what I did is basically force any supremum sequence that has exponential towers all of the same element except the topmost one, to become undefined

Now hopefully the layering functions $a_{\gamma}(\alpha[m-1]\alpha)$ actually form a well ordering and then comparison with the epsilon numbers should be possible by comparing the $a$s and the $\omega$

$\textbf{Tier 4: Tetration (More carefully)}$

Recall that there are two roadblocks to defining ordinal tetration:

\begin{align}
\omega^{\epsilon_{\alpha}} & =\epsilon_{\alpha}\tag{1}\\
\alpha^{\beta\gamma} & =(\alpha^{\beta})^{\gamma}\tag{2}
\end{align}

This means, if we define tetration in the usual (bottom up) way:

\begin{align}
{}^{0}\alpha & =1\\
{}^{\beta+1}\alpha & = \alpha^{{}^{\beta}\alpha}\\
\end{align}

we will quickly hit Roadblock $(1)$ for example:

$\sup(\{{}^n\omega \leq \omega^{{}^n\omega} = {}^{n+1}\omega|n\in\Bbb{N}\})=\{{}^{\omega}\omega \leq \omega^{{}^{\omega}\omega…