User blog comment:Ubersketch/CNFlex/@comment-34876209-20190910033450

It was a little difficult to understand your notation, the way it was written out, but, I think I got it now. Okay, so if I understand right...


 * "There are four rules*...
 * Base Case: \(A=0\)
 * Additive Case: \(\alpha+\beta=\alpha \beta C\)
 * Exponentive Case: \(\alpha B=\omega^{\alpha}\)
 * Epsilonic Case: \(\varepsilon_{\alpha}=D\alpha\).

*In CNFlex + epsilon that is, which I'll be covering because it covers more.


 * ...and six standard ways to apply those rules...
 * Trivial Form: \(A\)
 * Exponentive Form: \(\alpha B\) where \(\alpha\) is standard.
 * Combinative Form: \(\alpha \blacksquare \beta \blacksquare B\) where \(\blacksquare\) denotes either the \(B\) or \(C\) functions, and \(\alpha\) and \(\beta\) are standard. In addition, \(\alpha\blacksquare\geq\beta\blacksquare\).
 * Epsilonic Form: \(D\alpha\) where \(\alpha\) is standard.
 * Epsilonomegic Form: \(D\alpha\beta B\) where \(\alpha\) and \(\beta\) are standard.
 * Epsepsilonic Form: \(D\alpha D\beta\) where \(\alpha\geq\beta\) and \(\alpha\) and \(\beta\) are standard.


 * This not only generates all ordinals beneath \(\zeta_0\) but provides a way to alphabetically sift through them all."

Do I understand this right?

Because, although \(A\) [0], \(AB\) [1], \(ABB\) [\(\omega\)], \(ABBB\) [\(\omega^{\omega}\)], and so forth would be valid expressions, there seems to be no way to justify an expression that represents 2. The Combinative Form is concerning because if the expression must end in 'B' to be valid then no matter what numbers you can add up, they can only be in the exponents place of an \(\omega\). For example we can make ABABB [\(\omega^2\)] but not 2 according to this!

I think this whole problem could be solved by simplifying that form rule a little:


 * "Combinative Form: \(\alpha \blacksquare \beta \blacksquare B\) where \(\blacksquare\) denotes either the \(B\) or \(C\) functions, and \(\alpha\) and \(\beta\) are standard. In addition, \(\alpha\blacksquare\geq\beta\blacksquare\)."

Can now turn into:


 * Combinative Form: \(\alpha \beta C\) if \(\alpha\geq\beta\) and both \(\alpha\) and \(\beta\) are in standard form.

Now \(ABABC\) [2] is a valid expression, and so is \(ABABCABC\) [3], and \(ABABCABCABC\) [4], and so on for all the natural numbers.*

I feel like this would cover things, but I'm only confident up until \(\varepsilon_0\), and it's been like a year since I've tried to analyze anything like this so pardon any mistakes.

This looks cool though! Reminds me of the SKI calculus a bit.