User blog:Koteitan/Explanation of BMS mechanism by the row-wise ordinal notation and its greedy extension

The row-wise ordinal notation of Bashicu matrix system
[http://googology.wikia.com/wiki/User_blog:Bubby3/BM2_pair_sequence_system_proof_of_termination. Bubby3 showed] that (*,b) is the diagonalizer of (*,b-1). It can also be said as below;

$$ A[n]= \begin{pmatrix} G_0(B_0(1))\\ G_1(B_1(1))\\ \vdots\\ G_{k-1}(B_{k-1}(1))\\ 0\\ \vdots\\ 0 \end{pmatrix}[n]=\begin{pmatrix} G_0(\overbrace{B_0(B_0(\cdots B_0(0) \cdots))}^{n}\\ G_1(B_1(B_1(\cdots B_1(0) \cdots))\\ \vdots\\ G_{k-2}(B_{k-2}(B_{k-2}(\cdots B_{k-2}(0) \cdots))\\ G_{k-1}(\underbrace{B_{k-1}(0)+B_{k-1}(0)+\cdots+B_{k-1}(0)}_{n})\\ 0\\ \vdots\\ 0 \end{pmatrix} $$

$$ =\begin{pmatrix} G_0({B_0}^n(0))\\ G_1({B_1}^n(0))\\ \vdots\\ G_{k-2}({B_{k-2}}^n(0))\\ G_{k-1}({B_{k-1}}(0)\cdot n)\\ 0\\ \vdots\\ 0 \end{pmatrix} $$

$$G_k(\alpha)$$ is the ordinal function of the good part of the $$k$$ th row of the matrix $$A$$. $$B_0(\alpha)$$ is the ordinal function of the bad part of the $$k$$ row of the matrix $$A$$. The 1 in $$B_k(1)$$ indicates the cut head in this turn. $$G_k(\alpha)$$ and $$B_k(\alpha)$$ are represented with $$1$$ and $$\omega$$ and their additions and exponentiations as like unsorted Cuntor normal form, moreover, a $$1$$ is replaced into $$\alpha$$. $$G_k(\alpha)$$ and $$B_k(\alpha)$$ are decided in every turn. The rules for decision of $$G_k(\alpha)$$ and $$B_k(\alpha)$$ were proposed as BM1, BM2, [http://googology.wikia.com/wiki/User_blog:Bubby3/BM1_is_weaker_than_expected. Bubby3's] and BM2.2(Concestor approach).

For example, when $$A=(0,0)(1,1)(1,1)$$ ,

$$G_0(\alpha)=G_1(\alpha)=\alpha$$

$$B_0(\alpha)=B_1(\alpha)=\omega^{1+\alpha}$$

$$\begin{pmatrix} 0,&1,&1\\ 0,&1,&1 \end{pmatrix}[3]=\begin{pmatrix} \omega^{1+\omega^{1+\omega^{1+\omega^0}}}\\ \omega^{1+0}+\omega^{1+0}+\omega^{1+0}+\omega^{1+0} \end{pmatrix}=\begin{pmatrix} 0,&1,&1,&2,&2,&3,&3,&4\\ 0,&1,&0,&1,&0,&1,&0,&1\\ \end{pmatrix} $$

With this, the diagonalizer in the 0 to k-2 th rows works as the repetition of the function composition. In the lowermost row the well founded reduction was done for the guarantee of its termination.

Extension to Tuple ordinal system
If the termination guarantee is given by the last row, why we have to decide bad part equally in all rows? The upper row can be diagonalized with the largest bad part every turn with decision in which the bad root is the root node and the bad part is the whole row. In the last row, we can select bad part in which the reduction takes the longest time. It means the taking the fundamental sequence.

Below is the definition of the Tuple ordinal system using the greedy modification as avobe, and the setting $$(\omega, \cdots, \omega)^{\rm T}[n]$$ for its start gives large ordinal in the first row at the turn in which the second row becomes zero.



\begin{pmatrix} \omega\uparrow\uparrow x\\ \omega\uparrow\uparrow x\\ \vdots\\ \omega\uparrow\uparrow x\\ \alpha\\0\\\vdots\\0 \end{pmatrix}[n]=\begin{pmatrix} \omega\uparrow\uparrow(x+n)\\ \omega\uparrow\uparrow(x+n)\\ \vdots\\ \omega\uparrow\uparrow(x+n)\\ \alpha[n]\\0\\\vdots\\0 \end{pmatrix} $$

In this equation, $$\alpha[n]$$ means the element of the fundamental sequence of the ordinal $$\alpha$$.

However, I don't think that is strong than original BMS, because the $$k$$ th row form the bottom seem to grow in the grow rate $$(H_{\epsilon_0})^k(n)=H_{\epsilon_0 \cdot k}(n)$$, so its grow rate is limited in $$H_{\epsilon_0 \cdot \omega}(n)$$.

Can you find what is wrong?