User blog:Bubby3/Proof of termination of pair sequence system

In this blog post, I am going to show that BMS pair sequence systems terminate in all versions of BM except for BM1, and BM2.2

Primitive sequence system
I am going to show the system equilvlent to Kirby-Paris hydra. I am going to start by converting a BMS expression to a hydra. A zero in the expression corresponds to a node directly attached to the root. A non-zero number n corresponds to a node at level n, and with it's parent the last  entry which is less than n. Lets look at a simplifed version of the rules of primative sequence system To prove that the rules are the same, look at the rules for kirby-paris hydra and how the compare to BMS. This even works with BM1, because all the entries are less than the last entry.
 * If the last entry is 0, cut it off and replace n with f(n)
 * If the last entry is not 0, the bad part is defined as the array between the last entry that is strictly less than the last entry and the second-to-last entry, and the good part is defined as the part before the bad part. We then replace the array, GBn, where n is the last entry with GB...BBB with f(n)+1 copies of itself.
 * If the array ends in 0, we cut off the entry, and do something to the value of n. This is the same as Kirby-paris hydra when the rightmost leaf has no grandparent, or is directly attached to the node. Good so far
 * The bad part is the parent of the last entry and everything that is a child of that entry, with the last entry removed. And we copy the bad part, which corresponds to copying the parent of the entry after we remove it. So we did it.

Pair sequence system
Pair sequence system can be rewritten as labeled nodes, with the label as their second entry, so an entry (n,m) has label m. Entries in the form (n,0), or with label 0, still behave like before, searching for a entyr with the last entry less than it. That is because the bad root finder algrothim called the upper branch ignoring, when the last entry is 0, it acts like it did in primative sequence system. becasue the bad root searching algrhim searches for the parentr, of the last nonzero subelement of the last entry, and when there is only 1 nonzero entry, it acts like primative sequence.

The notation gets intersting when the last entry is in the form (a,n), for n>0. To find the bad root, it searches for the ancestor in the tree, or the nodes in the path connecting it to the root node, which is the closest to the leaf, or has the highest  level (distance from the root), that has a lower label than n. That is the same as the least nested node that is an ancestor of the last entry with a lower label. It then recursively subsitutes the leaf for a copy of the the subtree starting with the bad root for the leaf entry, with n nests. The reason is that that to find the ascending matrix, we find the difference between the first entry of the leaf and the bad root, which is how much higher the bad root is than the root, which puts the bad part in the place of the former last entry when we copy the bad part increased by matrix. If you keep copying the bad part, it nests the bad part where the last entry used to be. It is repeated with f(n) nests, or n nests.

This is exactly the same as Bucholoz hydra, without the adding another node with label k-1 if the label is less than k-1. This difference doesn't matter in the final strength becasue in the sequences that come from (0,0,0)(1,1,1), we aren't going to deal with a node with a bad root with label less than 1 less than 1 less than that of the last entry, because all the labels from 0 to n-1 are ansectors of the last entry in a sequence that comes from (0,0,0)(1,1,1). This is (mostly) equlivent to R function up to {0,{0}}. So, the burden of proof has switched from pair sequence system to Bucholoz hydra, where you make n nests, or to R function. People don't double the consistence of those functions, right?

Please comment with any problems with the proof.