User blog comment:Mh314159/A hopefully powerful new system/@comment-35470197-20190628050713/@comment-35470197-20190628111544

The well-ordering corresponding to the recursion up to ns is \(\omega^{\omega\) as you wrote, but the rule set is actually different from that of FGH. If I am correct, then we have \begin{eqnarray*} [n][a] & \sim & f_{2n}(a) \\ [n][a,1] & \leq & f_{2 \omega}^a(\max\{n,a,1\}) \\ [n][a,2] & \leq & f_{2 \omega + 1}^a(\max\{n,a,2\}) \\ [n][a,b] & \leq & f_{2 \omega + b - 1}^a(\max\{n,a,b\}) \\ [n][a,b,1] & \leq & f_{2 \omega + \omega}^b(\max\{n,a,b,1\}) \\ [n][a,b,2] & \leq & f_{2 \omega + \omega + 1}^b(\max\{n,a,b,2\}) \\ [n][a,b,c] & \leq & f_{2 \omega + \omega + c - 1}^b(\max\{n,a,b,c\}) \\ & \vdots & \\ [n][\underbrace{n, \ldots, n}_n] & \leq & f_{2 \omega + \omega \times (n-1) + 1}(n), \end{eqnarray*} and hence the rule set ns corresponds to \(\omega^2\).