User:Alemagno12/AAAgoogology's TON analysis

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Introduction
In this page I will show my analysis results of Taranovsky’s C notation. There could be mistakes in this page. Also, there are many parts in which the explanation shown is not very complete, so please comment if you don’t understand something shown in this page.

It is hard to predict the rules to find the fundamental sequence (FS), so I am changing them many times by comparing it to FS computed by my program. The current rule set is verified to be correct for ordinals containing \(\Omega_2\) and up to 9 \(C\)‘s, ordinals containing \(\Omega_3\) and up to 8 \(C\)’s, ordinals containing \(\Omega_4\) and up to 8 \(C\)’s, and ordinals containing \(\Omega_5\) and up to 7 \(C\)’s using the program, but the program can be wrong, and there might be mistakes in ordinals more complicated than the ones that I checked. If you find a mistake, please tell me in the comment.

The three things I think are significant about this page are:
 * 1) Made the nested bracket (which are hard to see) into tree structures which are easier to see.
 * 2) Combined the definition of \(n\)-built from below for all \(n\)-th ordinal notation system.
 * 3) Made a conjecture about the method of finding FS that is not brute forcing.

For 1., there are people that tried doing this previously (expressing C structure using trees), but it is easier to understand because I fixed the direction of the edges. Also, we can write tree structures using the method of writing matrices in LaTeX. For 2., now we can understand the notation without having to think about which ordinal notation system it is. The “numbering of nodes” in this article corresponds to \(n\)-built from below. For 3., a brute force method of finding the FS is already given by Hyp cos, but to better understand the structure of ordinals using Taranovsky’s C, I think it is important to make a method to find the FS without brute forcing. 解析結果は十分な検証がされていないため、私のユーザーページの下に用意したページで公開しますが、 内容は他のページに転載していただいて構いません. 更新情報についてはユーザーブログの方に記載します. が訳されていません.

Taranovsky’s C
In this page I will introduce my results of Main Ordinal Notation System published here. I will talk about the “Combined System” which combines \(n\)-th ordinal notation system for all n.

Correspondence with tree structure
In this notation there are constants \(\0), \(\Omega_n\) (\(n\ge1\)) and a binary function \(C(a,b)\). In order to make the complicated structure easier to understand, I will write the structure using trees. \[ C(a,b)= \begin{array}{ccc} a& & \\ C&-&b \end{array} \] \(C\) corresponds to a node with two children. The first argument goes up, and the second argument goes to the right. The root node of the tree is at bottom left. The following is true: \[ \Omega_n=C(\Omega_{n+1},0)= \begin{array}{ccc} \Omega_{n+1}& & \\ C&-&0 \end{array} \]

Standard form
Multiple combinations/structures of \(0\), \(\Omega_n\) and \(C\) sometimes equal to the same ordinal. In Taranovsky’s C every ordinal has only 1 standard form. Expressions that are not standard can be turned into stanard form expressions using three kinds of transformations: replacement of \(\Omega\), minimizing the second argument and maximizing the first argument.

Replacement of \(\Omega\)
\[ \begin{array}{ccc} \Omega_{n+1}& & \\ C&-&0 \end{array} \rightarrow\Omega_n \] Do the replacement above.

Minimizing the second argument
If \(C(b,c)\) is standard and \(a>b\), do the replacement below: \[ \begin{array}{ccccc} a& &b& & \\ C&-&C&-&c \end{array} \rightarrow \begin{array}{ccccc} a& & \\ C&-&c \end{array} \]

大小比較
Consider comparing \[ a= \begin{array}{ccccccc} a_m& &a_2& &a_1& & \\ C &\cdots&C &-&C &-&0 \end{array} \] and \[ b= \begin{array}{ccccccc} b_n& &b_2& &b_1& & \\ C &\cdots&C &-&C &-&0 \end{array \] If \(a_1=b_1,\ldots,a_{i-1}=b_{i-1},a_i<b_i\) or \(m<n,a_1=b_1,\ldots,a_m=b_m\), \(a<b\) is true. \(a=0\) corresponds to \(m=0\). If \(m<n\) then \(\Omega_m<\Omega_n\). If two expressions can’t be compared using the rule above, do replacement of \(\Omega\) and compare.

Successor ordinal
The success ordinal of \(a), which is \(a+1\), can be expressed as: \[ a+1=C(0,a)= \begin{array}{ccc} 0& & \\ C&-&a \end{array} \]

Limit ordinals and fundamental sequences 1
Limit ordinals (that have FS) can have multiple FS’s, but to define fast-growing hierarchhy we need to specify one FS for every limit ordinal (that have FS). \[ \begin{array}{ccc} 0& & \\ C&-&a\\ C&-&b \end{array} \] The above ordinal is a limit ordinal and it has a FS. The nth term of its FS is \[ \begin{array}{ccc} 0& & \\ C&-&a\\ C&-&b \end{array} \]

If the nth term of the FS of \(a\) is \(a_n\), The \(n\)th term of the FS of the ordinal below \[ C(a,b)= \begin{array}{ccc} a& & \\ C&-&b \end{array} \] is \[ C(a_n,b)= \begin{array}{ccc} a_n& & \\ C &-&b \end{array} \]

\(\Omega_n\) are limit ordinals but they do not have FS. I will write about the FS of \(C(a,b)\) when \(a\) is a limit ordinal that does not have FS later in this page.

Numbering of the nodes
For ordinal \(C(a,b)\), label numbers to the nodes of the tree of \(C(a,b)\) following there rules.
 * 1) The value of every label must be an integer.
 * 2) When you go right or up an edge, the number must not decrease.
 * 3) Nodes smaller than \(C(a,b)\) and their children don’t get labels.
 * 4) If a node does not have a label in either of its two children, it does not get a label.
 * 5) All other nodes must have a label.
 * 6) If the ordinal corresponding to node \(c\) is the largest in the nodes from the root to \(c\) (including the root and \(c\)), the label of node \(c\) must be bigger than the label of the node below \(c\).
 * 7) If \(\Omega_n\) is the largest ordinal in the nodes from the root to \(\Omega_n\), \(\Omega_n\) has label \(n\).
 * 8) If \(\Omega_n\) is not the largest ordinal in the nodes from the root to \(\Omega_n\), replace \(\Omega_n\) with \(C(\Omega_{n+1},0)\).
 * 9) All nodes must have the largest label it can have without violating any other rules.

Maximizing the first argument
If \(a\) and \(b\) in \\(C(a,b)\) are standard, label the nodes of \(C(a,b)\). If the label of the root is not negative the first argument is maximized. If not, repeat the following procedure to get the first argument maximized. ??????
 * 1) Find the \(\Omega_n\) at the most right and bottom which is determining the label of the root and replace it with \(\Omega_{n+1}\).