User blog comment:Alemagno12/Quick poll/@comment-30754445-20180823131111

What's the difference between "coming up" with the growth rate and actually justifying it?

There's no way to even guess the growth-rate of a notation without doing some kind of analysis first. A superficial visual inspection of how a notation works is completely unreliable: Even a tiny change in one of the expansion rules can alter the notation's strength considerably (usually for the worse).

At any rate, my personal answer to your question is this:

I would start with directly evaluating the low level expressions in that notation, and expressing them a "standard system" I'm already familiar with (like the FGH, or my own letter notation).

At some point, after doing this for some time, I'll start noticing patterns which will allow me to make an educated guess of how to generalize things.

For example, let's say I'm completely unfamiliar with BEAF and I won't to investigate how strong it is.

I could start with the known equality of {a,b,c} = a [c arrows] b ~ fω(c), and proceed from there (explicitly using the given rules) to evaluate {a,b,1,2}:

(I'm doing everything here with the FGH for sake of clarity. In my personal notes, though, I would use my own notation which I've already verified):

{a,1,1,2} = a

{a,2,1,2} = {a,a,{a,1,1,2}} = {a,a,a} ~ fω(a)

{a,3,1,2} = {a,a,{a,2,1,2}} ~ {a,a, fω(a)} ~ fω( fω(a))

{a,4,1,2} = = {a,a,{a,3,1,2}} ~ {a,a, fω( fω(c))} ~ fω( fω(fω(a)))

And so on. By now the pattern is clear, and I can simply write {a,b,1,2} ~ fωb-1 (a) ~  fω+1(b)

(note that I'm ignoring things like the "-1". We're only interested in the general ball park, and none of this is supposed to be a formal proof anyway)

Then I go to {a,b,2,2}. Following a similar progression, I can quickiy deduce that {a,b,2,2} ~ fω+2(b). now I can clearly see that {a,b,c+1,2} is always a simple repetition of {a,b,c,2} (which is iterated b times), and this leads to the guess that {a,b,c,2} ~ fω+c(b) ~ fω2(c).

At this point, it would still be too early to even guess how {a,b,c,d} works in general. So I'll start with analyzing {a,b,c,3}. After noticing that the process is virtually identical to what happened with {a,b,c,2}, I'll be able to conjecture that {a,b,c,3} ~ fω3(c) and that {a,b,c,d} ~ fω2(c).

Then I'll go to 5-entry arrays. Again, I'll start carefully, but it won't take too long for me to realize that {a,b,c,d,e} corresponds to fω2(e-1)+ω(d-1)+c(b) ~ fω3(e-1) ~ fω3(e). By now it would also be clear to me that n-entry arrays seen to correspond to fωn-2(n) ~ fωω(n).

Then I'll have to slow down again and tread carefully again, is I venture into two-row arrays and beyond. I'm not going to bore you with the details of my reasoning at every single level, especially when you can probably fill them out yourself.

To summarize:

I don't see how you can even guess the strength of a notation without a step-by-step thorough analysis. It doesn't have to be a formal proof, but it does have to be quite detailed.