User blog:Vel!/Ordinal BEAF 2.0

Here is a simpler alteration to ordinal BEAF. It uses integers indexed by ordinals instead of ordinal polynomials.

Example: \(\{(0, 10), (1, 100), (\omega^\omega, 2)\}\)

Prime blocks
Define \(\Pi_p(\alpha)\) like so:


 * \(\Pi_p(0) = \emptyset\)
 * \(\Pi_p(\alpha + 1) = \{\alpha\} \cup \Pi_p(\alpha)\)
 * \(\Pi_p(\alpha) = \Pi_p(\alpha[p])\) if \(\alpha\) is a limit ordinal

Entries
Define \(E_\gamma(A)\) to be the entry in position \(\gamma\) in \(A\). It defaults to 1.

Also, define \(Q(A) = \{\gamma|E_\gamma(A) \neq 0\}\), which is the set of all positions with non-zero associated entries.

Pilots and copilots
Define \(P(A)\) (pilot) like so:

\[P(A) = \min\{\gamma > 1|E_\gamma(A) \neq 0\}\]

This is the first non-zero term after the prime in \(A\), reading terms from smallest to largest. \(P(\alpha)\) may not exist.

Define \(CP(A) = \{\gamma|\gamma + 1 = P(A)\}\). This is the set of copilots, which has one member iff the pilot is a successor ordinal, and no members otherwise.

The airplane is \(\Pi_p(P(A))\), and the passengers are \(\Pi_p(P(A)) \backslash \{P(A)\} \backslash CP(A)\).

The Three Rules
Let \(b = E_0(A)\) and \(p = E_1(A)\).

\[\alpha' := \bigcup_{\gamma \in Q(\alpha)}\left\{ \begin{array}{rl} \gamma = 1 : & p \\ \text{otherwise} : & E_\gamma(A) \\ \end{array} \right\}\] \[N(\alpha) = N\left(\bigcup_{\gamma \in Q(A)}\left\{ \begin{array}{rl} \gamma = P(A) : & E_{P(A)}(A) - 1 \\ \gamma \in CP(A) : & N(A') \\ \gamma \in \Pi_p(P(A)) : & b \\ \text{otherwise} : & E_\gamma(A) \\ \end{array} \right\}\right)\]
 * 1) The Base Rule: If \(A\) has no pilot, \(N(A) = (b + 1)^{p + 1}\).
 * 2) The Prime Rule: If \(p = 1\), \(N(A) = b\).
 * 3) The Catastrophic Rule: Otherwise:

Some notes about the last one:
 * In the definition of \(N(\alpha)\), the conditions are evaluated from top to bottom.
 * Since addition is not commutative over the ordinals, the \(\Sigma\)s should be evaluated with \(\gamma\) decreasing.
 * \(E_1(\alpha) - 1\) and \(E_{P(\alpha)}(\alpha) - 1\) are well-defined. In both situations, the left-hand term is greater than one. (If not, the first one would cause the prime rule to kick in, and the second one would cause the base rule to kick in).

That's it! Let me know about any errors or improvements that can be made to this formula.