User blog comment:B1mb0w/Fundamental Sequences/@comment-5529393-20151113121456/@comment-5529393-20151114145947

You rules are not complete. For example, they do not address anything of the form $$\varphi(\ldots,\beta)$$ where $$\beta$$ is a limit ordinal, nor anything of the form $$\varphi(\ldots, \alpha, 0)$$ where $$\alpha$$ is a limit ordinal, nor anything of the form $$\varphi(\ldots, \alpha, 0_{[m+1]})$$ where $$\alpha$$ is not 1. And of course, as I have mentioned previously your third rule does not work.

LittlePeng9 gives a good proof of why $$\varphi(\alpha,\beta)\uparrow\uparrow \omega = \varepsilon_{\varphi(\alpha,\beta)+1}$$. If your prefer phis, this is  $$\varphi(\alpha,\beta)\uparrow\uparrow \omega =  \varphi(1, \varphi(\alpha,\beta)+1)$$.

I don't know what you mean by "using the diagonal at zero". My point was that your rule and the previous rule are identical for $$\varphi(\alpha+1,0)$$, only the presentation is different. The old rule starts from 0, then for each increment of the parameter you apply $$\varphi(\alpha,_)$$. Your rule says that $$\varphi(a+1,0) [n] = \varphi^n(a,0_*)$$ which one can easily is saying the same thing. Saying that $$\varphi(\alpha+1,0)[0] = 0$$ exactly matches the 0 in your expression.