User blog comment:Billicusp/So I don't know if anybody noticed.../@comment-27173506-20160318160910/@comment-27173506-20160318163323

I'd like to mainly point out the fact that he claims to reach X(MM) based on a pattern that took him to \(\omega^{\omega}\) or even \(\psi(\Omega^{\Omega})\). Big ordinals just aren't that simple. X(MM) is greater than any recursive extension of \(\alpha \mapsto I(1 @ \alpha)\). That's incredibly huge. I find it hard to believe that it can be so relatively easy to reach.

I will supply a few quotes from the article that support my point: "...and so will be represented using the χ function, which works just like the ψ function except it uses M, the first weakly-Mahlo cardinal as a diagoniser, rather than Ω." The \(χ\) function is not just the \(\psi\) function using \(M\), and most definitely does not work just like it.

"You may have noticed some pretty large jumps in the final sections of analysis against the χ function. This is because the [1(1)[1](1)[1]] inside the EFAN array works just like the M in the χ function, and this kind of analysis has already been done, so we don't need to go through it again." Just because it works in one place doesn't mean it works in another.

For example, using the fundamental sequences \(\Gamma_0[0] = 0 \text{and} \Gamma_0[n+1] = \varphi(\Gamma_0[n],0)\) and \(\varepsilon_0[0] = 0 \text{and} \varepsilon_0[n+1] = \omega^{\varepsilon_0[n]}\) we have \(f_{\text{SVO}}(2) = f_{\varphi(1 @ \omega)}(2) = f_{\Gamma_0}(2) = f_{\varepsilon_0}(2) = f_\omega(2) = f_2(2)\). Does this mean that the SVO is equal to 2? Since it can be shown as well that \(f_{\text{SVO}}(1) = f_1(1)\), can we take the pattern as obvious and self evident that \(f_{\text{SVO}}(n) = f_n(n)\)? No, it means that under certain circumstances we can the SVO with some other ordinal. This does not mean that we can always do that, nor that this is a strict rule. Of course, if proved it can be true (though in this case it quite clearly isn't) it's true, but I will remain slightly skeptical until then.