User:12AbBa/TON analysis

In this article, I will be analyzing TON.

This page is a LaTeX-free page, to discourage slow loading. = General =

Non-confusing definition
First, TON, short for Taranovsky's ordinal notation, is divided into an infinite number of systems, namely the 1st system, the 2nd system, the 3rd system, etc. Expressions in TON look like this: C(α,β). The nth system uses only a few symbols, namely "C", "(", ")", ",", "0", "Ωn". Note that unlike in other notations, Ωn is not the nth uncountable. It isn't even an ordinal, though it could be treated as an ordinal fixed point. It's just a symbol.

The easiest thing about TON is comparing the size of two ordinals. To do this, we use the postfix form. To compare ordinals this way, first delete all of the "("s, the ")"s, and the ","s. Then reverse the string. Now, compare them in alphabetical order, where the "alphabet" here is C0Ωn. For example, to confirm that C(C(0,0),C(C(0,C(0,0)),0))=ω2+ω is smaller than C(C(0,C(0,C(0,C(0,0)))),0)=ω4, we write the postfix forms:

000C0CC00CC

000C0C0C0CC

Now we see that the second is bigger.

Next, we define the standard form of an ordinal. Note that "standard" is an understatement, since we only treat ordinals in standard form as real ordinals.

Now, of course, 0 and Ωn are ordinals. Also, obviously, for C(α,β) to be an ordinal, α and β have to be ordinals.

The second rule is that if the β equals C(γ,δ), γ≥α. Thus, if a ordinal is written as C(α1,C(α2,...C(αk-1,C(αk,0 or Ω_n))...)), α1≤α2≤...≤αk-1≤αk. This rule makes things like C(C(0,0),C(0,0)) non-standard.

The third rule is the hardest to understand. First, we have to understand what is the "subterm". Basically, the subterm of some expression is a part of it. Formally this could be expressed as: Now we define "n-built from below from": Finally, we can express the third rule! It is: α is n-built from below from <C(α,β).
 * η is a subterm of η;
 * if η=C(α,β), then the subterms of α and β are the subterms of η.
 * α is 0-built from below from <β means that α<β;
 * α is (k+1)-built from below from <β means that: for every subterm γ of α,
 * γ≤α, or
 * there exists a δ "between" γ and α (i.e. γ is a subterm of δ, which is a subterm of α) such that δ is k-built from below from β.

Now, we've defined the standard form! But there's one more thing before we can fully define TON. This one is easy:

C(α1,C(α2,...C(αk-1,C(αk,Ωn))...))=Ωn+ωαk+ωαk-1+...+ωα2+ωα1

That's it!

Explanation
Now, you might say: "You haven't actually defined the notation. You only gave the definition of the standard form!" Well, this is the full definition. Remember that all ordinals can be expressed in TON. Therefore, there is a one-to-one correspondence: TON ordinals <--> Normal (countable) ordinals (up to the limit of TON) Since we can compare the size of TON ordinals, we can "pack" them to correspond them to the normal ordinals.

Think of it this way. TON is like an infinite library. Each book in the TON library has a title with the alphabet "C0Ωn", which is the postfix form of a TON ordinal. The library only accepts standard books. The books are ordered in their alphabetical order. Now, if the library does all of this, then the books will each correspond to its order in the library, its ordinal. No reordering of books is possible, nor will any new book fit in. = Analysis =

Before First System
Since there is no "0th system", pretend that this is just the 1st system.

Before I start the table, notice that for expressions without Ωs, any ordinal α will be larger than all of its subterms, so the third rule of the standard form doesn't matter.

Turns out, if α<ε0, then C(α,β)=β+ωα. Actually, this addition principle works for larger α too, but you have to be careful about standardness.

First System
C(ψ(0)+ω,ψ(0)) C(C(C(0,0),C(Ω1,0)),C(Ω1,0)) 0Ω1C0Ω1C00CCC ψ(0)ω^ω C(ψ(0)+ω^ω,ψ(0) C(C(C(C(0,0),0),C(Ω1,0)),C(Ω1,0)) 0Ω1C0Ω1C000CCCC ψ(0)ω^ω^ω C(ψ(0)2,ψ(0)) C(C(C(Ω1,0),C(Ω1,0)),C(Ω1,0)) 0Ω1C0Ω1C0Ω1CCC ψ(0)^2 C(ψ(0)3,ψ(0)) C(C(C(Ω,0),C(C(Ω,0),C(Ω,0))),C(Ω,0)) 0ΩC0ΩC0ΩCC0ΩCCC ψ(0)^3 C(ψ(0)ω,ψ(0)) C(C(C(0,C(Ω,0)),C(Ω,0)),C(Ω,0)) 0ΩC0ΩC0ΩC0CCC ψ(0)^ω C(ψ(0)ω^ω,ψ(0)) C(C(C(C(0,0),C(Ω,0)),C(Ω,0)),C(Ω,0)) 0ΩC0ΩC0ΩC00CCCC ψ(0)^ω^ω C(ψ(0)^2,ψ(0)) C(C(C(C(Ω,0),C(Ω,0)),C(Ω,0)),C(Ω,0)) 0ΩC0ΩC0ΩC0ΩCCCC ψ(0)^ψ(0) C(ψ(0)^3,ψ(0)) C(C(C(C(Ω,0),C(C(Ω,0),C(Ω,0))),C(Ω,0)),C(Ω,0)) 0ΩC0ΩC0ΩC0ΩCC0ΩCCCC ψ(0)^ψ(0)^2 C(ψ(0)^ω,ψ(0)) C(C(C(C(0,C(Ω,0)),C(Ω,0)),C(Ω,0)),C(Ω,0)) 0ΩC0ΩC0ΩC0ΩC0CCCC ψ(0)^ψ(0)^ω C(ψ(0)^ψ(0),ψ(0)) C(C(C(C(C(Ω,0),C(Ω,0)),C(Ω,0)),C(Ω,0)),C(Ω,0)) 0ΩC0ΩC0ΩC0ΩC0ΩCCCCC ψ(0)^ψ(0)^ψ(0) C(ψ(0)^ψ(0)^ψ(0),ψ(0)) C(C(C(C(C(C(Ω,0),C(Ω,0)),C(Ω,0)),C(Ω,0)),C(Ω,0)),C(Ω,0)) 0ΩC0ΩC0ΩC0ΩC0ΩC0ΩCCCCCC ψ(0)^ψ(0)^ψ(0)^ψ(0) C(Ω,ψ(0)) C(Ω,C(Ω,0)) 0ΩCΩC ψ(1) C(ψ(1),ψ(1)) C(C(Ω,C(Ω,0)),C(Ω,C(Ω,0))) 0ΩCΩC0ΩCΩCC ψ(1)2 C(ψ(1)+1,ψ(1)) C(C(0,C(Ω,C(Ω,0))),C(Ω,C(Ω,0))) 0ΩCΩC0ΩCΩC0CC ψ(1)ω

New pattern: C(Ω1,α)=α↑↑ω. Again, watch out for standardness!

testing 123