User blog comment:MilkyWay90/Help with understanding Veblen array notation/@comment-30754445-20180811202716/@comment-30754445-20180819105559

@MilkyWay90

Look at rule 4 more carefully:

φ(a,b, ... ,c+1,0,0, ... ,0) = sup (xN) where:

x1 = 0

xN+1 = φ(a,b, ... ,c,xN,0, ..., 0)

Notice the difference in the trailing zeros between line 1 and line 3:

Line 1: "c+1,0,0, ..., 0"

Line 3: "c,xN,0, ... ,0"

See the difference? Line 1 has an additional 0. So when using rule 4, the string of trailing zeros in the xN's will be one zero shorter. IOW, for φ(1,0,0) you'll get xN+1=φ(0,xN,0) rather than φ(0,xN,0,0).

Still confused? Then let us expand φ(1,0,0) step by step:

1. φ(1,0,0) = φ(0,0,1,0,0) (by rule 1)

2. To calculate φ(0,0,1,0,0) we use rule 4 with a=b=c=0.

Note that in order to write φ(0,0,1,0,0) as φ(a,b,c+1,0,0, ..., 0), we must interpert the final "0, ... , 0" as a single 0! So using rule 4, we get:

φ(0,0,1,0,0) = sup(xN) where:

x1 = 0

xN+1 = φ(a,b,c,xN,0, ... ,0) = φ(a,b,c,xN,0) = φ(0,0,0,xN,0) = φ(xN,0)

Which is, of-course, precisely the definition of gamma-0.

As for φ(0,...,0), that's easy. Just use rule 1 to drop the leading zeros, and you get (by rule 2):

φ(0,...,0) = φ(0) = ω0 = 1.

@Fefjo

"The main thing I actually strugled with was how to practicaly compute these ordinals (like in a program). And it may just be a single tricky concept that when explained well is understandable in a few hours but finding such a good explantion isn't easy. "

The trick is to forget about all the set-theoretic definitions and work in reverse:

Instead of looking at OCF expressions and trying to figure out which ordinals they correspond to, we just retrace the ordinals we already know how to "count up" to, and treat Ω as a wild-card that we use whenever we get stuck.

To simplify things further, we can remember that the ψ's count the ε-numbers. So we start simply with ψ(n)=εn, and keep in mind that we are only allowed to use ordinals that we've already constructed before.

Using this method, it is quite easy to see that we get stuck at ζ0, and therefore we have to use our wild-card (the Ω). So we write ζ0 as ψ(Ω).

(note to those who got used to UNOCF, before they complain that the above is wrong: UNOCF works differently than most OCFs. The above is the usual way to do things. UNOCF up to the BHO, though, can be seen as following the exactly same idea with one exception: It counts powers of ω, instead of ε-numbers)

Now, if you try to jump from here to understanding how the system works all the way up to the BHO or even the SVO, you're guarenteed to get hopelessly lost. The trick is to proceed one small step at a time: If ψ(Ω) is ζ0, then what is ψ(Ω+1)? What's ψ(Ω+2)? What's the general rule for ψ(Ω+x) where x is a countable ordinal we've already constructed?

The limit of all these would be ψ(Ω+Ω)=ψ(Ωx2). And if you followed the steps carefully, you should have no problem deducing that ψ(Ωx2)=ζ1.

Continue like this, step-by-step (perhaps asking a question or two when something doesn't seem to make sense), and you should have no problem finding the Madore-Psi expression for any ordinal you are already familiar with. And by the time you've exhausted these, you already gathered enough experience to continue without the "training wheels" of another notation.

I maintain that none of this is particularly difficult. It is time-consuming, and it may seem like a daunting journey at first. But anybody who has enough background to understand Veblen functions up to the SVO, should have no problem following the above technique and reach the BHO. The main obstacle is that of motivation: You gotta somehow accept, pretty much on faith, that all these little steps will add up to a coherent (and quite elegant) "stairway" in the end.

Note that for this trick to work, a strong familiarity with ordinals up to the SVO is recommended. And a strong familiarity with everything up to Γ0 is absolutely essential. This is because Γ0 = ψ(ΩΩ) and SVO = ψ(ΩΩ ω ). Without a good understanding of these milestones, you're going to have a very difficult time trying to understand how power-towers of Ω actually work (which is - of-course - the exact point where the ψ function takes off).