Taranovsky’s notation is a simple yet strong ordinal notation, introduced by Dmytro Taranovsky.

# Definition

First, Taranovsky’s notation is actually made up of many systems, called 1st system, 2nd system, 3rd system, and so on.

In the n-th system, we uses a binary function: C(α,β), and two constants: 0 and Ω_{n}. Here the 0 is not an ordinal or a number – it’s just a notation for “something”. And the same, Ω_{n} is not an ordinal – it’s just “something”.

Here’s the definition of *standard form*. First, 0 and Ω_{n} are in standard form. Then C(α,β) is in standard form iff it fits all those shown below:

- α and β are in standard form
- β = 0, or β = Ω
_{n}, or β = C(γ,δ) with α ≤ γ - α is n-built from below from <C(α,β)

But what’s “≤” and what’s “n-built from below from”? To answer this question, we need to define some more things.

First we need to define “n-built from below from” as follows.

- α is 0-built from below from <β iff α < β.
- α is (k+1)-built from below from β iff for all subterm γ of α, γ ≤ α or there is such a subterm δ of α that γ is subterm of δ and δ is k-built from below from β.
- The “subterm” in α can be defined as follows.
- In any part of expression of α, η is subterm of η itself.
- In any part of expression of α, if η = C(γ,δ), and x is a subterm of γ or δ, then x is a subterm of η.

To define binary relation “<” and “≤”, we need another form of the notation – *postfix form*. We can make a postfix form of every normal expression in such a way. First delete all the (′s, )′s and commas, which means we’ll get a string only containing 0, Ω_{n} and C’s. Then reverse the string. Now we get the postfix form. For **expressions in standard form**, we make the postfix form of them, then the comparisons (“≤”, “≥”, “<” and “>”) are done in the lexicographical order where C < 0 < Ω_{n}.

Property: In n-th system, for b < Ω_{n}, Ω_{n} is 1-built from below from b but not 0-built from below from b.

So why is it an ordinal notation? Because the “<” relation of expressions in standard form fits the ordinal axioms.

Now, we know that 0 is truly the ordinal 0. But unlike in θ function, Ω_{n} is not the n-th uncountable cardinal – it’s just some very large ordinal, but they’re also such common fixed point that , etc.

## One variable function

Any ordinal α can be expressed as or in standard form (k can also be 0). The former one is less than Ω_{n} (we call it *low-ordinal* here) while the latter one is greaterequal to Ω_{n} (we call it *high-ordinal* here).

Define the one variable function C_{1} as follows, it’s actually another representation of ordinals: for low-ordinal α, .

For high-ordinal α, , which is the same as ordinal addition and exponentiation.

It turns out that, if , then C_{1}(α) = α, and C_{1}(β+α) = C_{1}(β)+α.

All the 1st system, 2nd system, 3rd system, n-th systems can be combined into one notation as follows: the constants are 0 and Ω_{n} (for every positive integer n), and binary function C. Ω_{i} = C(Ω_{i+1}, 0) and the standard form always uses Ω_{i} instead of C(Ω_{i+1}, 0). And, to check for standard form and compare ordinals, use Ω_{i} = C(Ω_{i+1}, 0) to convert each Ω to Ω_{n} for a single positive integer n (it does not matter which n) and then use the n-th ordinal notation system.

Here‘s a calculator in Python program, which convert ordinals into Cantor’s normal form, binary φ function, and C_{1}-representation. Just note that it uses “C” instead of “C_{1}” for the one variable function, and “C2” instead of “C” for the binary function.

# Analysis

Below , doesn’t appear. For expressions without any Ω_{n}′s, α is 1-built from below from β, so “α is n-built from below from β” is always true.

So C(0,0) = 1, C(0,C(0,0)) = 2, C(0,C(0,C(0,0))) = 3, and C(0,α) = α+1.

Then and .

Then and .

and .

and .

and .

and .

and .

It turns out that, for , if C(α,β) is standard.

Here’re more results in 1st system.

. Note that ordinals above Ω_{1} are high-ordinals and C(0,Ω_{1}) = Ω_{1}+1 in 1st system.

It seems that the Ω_{1} in C function works very similar to the Ω in θ function. And they are also very similar to my EAN: Ω_{1} or Ω expands into “ω layers of” the C or θ where they are, and the results are the supremum of any finite layers of the C or θ where the Ω_{1} or Ω is.

The 1st system only handles ordinals up to the Bachmann-Howard ordinal. So we need 2nd system now. In 2nd system, Ω_{1} = C(Ω_{2},0), and the Bachmann-Howard ordinal is the supremum of all the C(C(Ω_{2},0),0), C(C(C(Ω_{2},0),C(Ω_{2},0)),0), C(C(C(C(Ω_{2},0),C(Ω_{2},0)),C(Ω_{2},0)),0), C(C(C(C(C(Ω_{2},0),C(Ω_{2},0)),C(Ω_{2},0)),C(Ω_{2},0)),0), etc. Thus the BHO equals .

. This ordinal (C(C(C(C(Ω_{2},Ω_{1}),Ω_{1}),C(Ω_{2},Ω_{1})),0)) is the supremum of all the C(C(C(Ω_{1},Ω_{1}),C(Ω_{2},Ω_{1})),0), C(C(C(C(Ω_{1},Ω_{1}),Ω_{1}),C(Ω_{2},Ω_{1})),0), C(C(C(C(C(Ω_{1},Ω_{1}),Ω_{1}),Ω_{1}),C(Ω_{2},Ω_{1})),0), etc. It’s not C(C(Ω_{2},Ω_{1})2,0)=C(C(C(Ω_{2},C(Ω_{2},0)),C(Ω_{2},C(Ω_{2},0))),0)! And is standard in 2nd system.

. Note again that C(0,C(Ω_{2},Ω_{1})) = C(Ω_{2},Ω_{1})+1.

. This ordinal (C(C(C(Ω_{2},Ω_{1}),C(Ω_{2},Ω_{1})),0)) is the supremum of all the C(C(Ω_{2},Ω_{1}),0), C(C(C(C(Ω_{2},Ω_{1}),Ω_{1}),C(Ω_{2},Ω_{1})),0), C(C(C(C(C(C(Ω_{2},Ω_{1}),Ω_{1}),C(Ω_{2},Ω_{1})),Ω_{1}),C(Ω_{2},Ω_{1})),0), etc.

.

. This ordinal (C(C(C(C(Ω_{2},Ω_{1}),C(Ω_{2},Ω_{1})),C(Ω_{2},Ω_{1})),0)) is the supremum of all the C(C(C(Ω_{2},Ω_{1}),C(Ω_{2},Ω_{1})),0), C(C(C(C(C(C(Ω_{2},Ω_{1}),C(Ω_{2},Ω_{1})),Ω_{1}),C(Ω_{2},Ω_{1})),C(Ω_{2},Ω_{1})),0), C(C(C(C(C(C(C(C(C(Ω_{2},Ω_{1}),C(Ω_{2},Ω_{1})),Ω_{1}),C(Ω_{2},Ω_{1})),C(Ω_{2},Ω_{1})),Ω_{1}),C(Ω_{2},Ω_{1})),C(Ω_{2},Ω_{1})),0), etc.

The supremum of all the C(C(C(0,Ω_{2}),0),0), C(C(C(C(C(0,Ω_{2}),0),Ω_{2}),0),0), C(C(C(C(C(C(C(0,Ω_{2}),0),Ω_{2}),0),Ω_{2}),0),0), etc. is C(C(C(Ω_{2},Ω_{2}),0),0) = C(C(Ω_{2}2,0),0). And that’s the limit of θ function.

The fundamental sequences in Taranovsky’s notation can be simply defined. First, we define a function: L(α), it’s the amount of C’s in the standard form of ordinal α. Then we define:

That’s it! So we get a very fast-growing function: with n C’s. It marks the strength of Taranovsky’s notation.

The full strength of Taranovsky’s notation is unknown yet. It might be as weak as second-order arithmetic (Z_{2}), and might be as strong as second-order arithmetic with projective determinacy (Z_{2}+PD), according to Taranovsky’s page.

“The full strength of Taranovsky’s notation is unknown yet. It might be as weak as second-order arithmetic (Z2), and might be as strong as second-order arithmetic with projective determinacy (Z2+PD), according to Taranovsky’s page.”

Actually Taranovsky has claimed it to be stronger than even ZFC.

I personally don’t believe that,but I’m just telling you the fact that he has said it.

http://web.mit.edu/dmytro/www/other/OrdinalNotation.htm#A6.2

Also “as weak as second-order arithmetic” is a funny sounding statement,because SOA is ridiculously strong.

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How much is C(C(W_2*2+1,0),0)?

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See the pDAN vs TON page. It is approx to psi(W_(psi_I(0)w)).

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θ(α↦Ω_α) = θ(Ω_Ω_Ω_Ω_Ω_…) ω-times – is limit of θ() function.

But I can continue:

θ(Ω_Ω_Ω_Ω_Ω_…) ω-times = α_0

θ(Ω_Ω_Ω_Ω_Ω_…) α_0-times = α_1

θ(Ω_Ω_Ω_Ω_Ω_…) α_1-times = α_2

e.t.c.

α_ω

α_θ(Ω)

α_θ(Ω^Ω)

α_θ(Ω_Ω)

α_θ(Ω_Ω_Ω_Ω_Ω_…) = α_α_0

e.t.c.

α_α_0

α_α_α_0

α_α_α_α_… ω-times = β↦α_β = β_0

How to write in Taranovsky’s notation or pDAN values:

α_1

α_2

α_ω

α_α_0

β_0

e.t.c.

?

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Using the 2nd system:

Your α_0 = C(C(Ω2,0),0)

α_1 = C(C(Ω2+C(Ω+C(Ω2,0)+α_0,0),0),0)

α_2 = C(C(Ω2+C(Ω+C(Ω2,0)+α_1,0),0),0)

α_ω = C(C(Ω2+C(Ω+C(Ω2,0)+C(Ω,0),0),0),0)

α_α_0 = C(C(Ω2+C(Ω+C(Ω2,0)+C(Ω,0)×α_0,0),0),0)

β_0 = C(C(Ω2+C(Ω+C(Ω2,0)+C(Ω,0)^2,0),0),0)

The results seem erratic, because things like C(C(Ω+C(Ω2,0),0),C(C(Ω2,0),0)) aren’t standard.

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I’m confused as compared…

C(C(Ω2+C(Ω+C(Ω2,0)+C(Ω,0),0),0),0) > C(C(Ω2,C(Ω2*2,0)),0) ~ {1,,2{1,,1,,2}2}

or

C(C(Ω2+C(Ω+C(Ω2,0)+C(Ω,0),0),0),0) < C(C(Ω2,C(Ω2*2,0)),0) ~ {1,,2{1,,1,,2}2}

?

I just want to compare α_1, α_ω, α_α_0, β_0, etc with pDAN

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OK. Compared with pDAN, we have:

{1,,1{1,,1,,2}2} has recursion level α_0

{1,,1{1,,1,,2}1{1{1,,1{1,,1,,2}2}2}2} has recursion level α_1

{1,,1{1,,1,,2}1{1{1,,1{1,,1,,2}1{1{1,,1{1,,1,,2}2}2}2}2}2} has recursion level α_2

{1,,1{1,,1,,2}1`2} has recursion level α_ω

{1,,1{1,,1,,2}1`1{1{1,,1{1,,1,,2}2}2}2} has recursion level α_α_0

{1,,1{1,,1,,2}1`1`2} has recursion level β_0

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Do I understand correctly that if I compare

{1,,1{1,,1,,2}1`2} – ε_0 level of 1`2

{1,,1{1,,1,,2}1`1`2} – ζ_0 level of 1`1`2

{1,,1{1,,1,,2}1,,1{1,,1,,2}2} – θ(α↦Ω_α) level of 1,,1{1,,1,,2}2

сan it be said that the transition from {1,,1{1,,1,,2}2} to {1,,1{2,,1,,2}2} will be:

( use some function Ф() )

α_0 = Ф(ω)

α_1 = Ф(ω^ω)

α_2 = Ф(ω^ω^ω)

α_ω = Ф(φ(1, 0))

β_0 = Ф(φ(2, 0))

β_ω = Ф(φ(3, 0))

e t.c.

{1,,1{1,,1,,2}1,,1{1,,1,,2}2} – Ф(θ(α↦Ω_α)) = Ф(α) ?

after:

Ф(α)

Ф(Ф(α))

Ф(Ф(Ф(α)))

Ф(Ф(Ф(Ф(…)))) – {1,,1{2,,1,,2}2} ?

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You are quite wrong.

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So lexicographical order of C represent ordinals? How do you evaluate the value of C’s?

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Just see C, 0 and Ω as simply symbols (don’t have “value”), and let C < 0 < Ω.

The values of "C-represented ordinals" come from comparisons.

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And I recalled Reverse Polish Notation.

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I don’t get how to compare in postfix form.

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It’s in lexicographical order. For example, “” < "C" < "CC" < "CCC" < "CCCCCCCCCC" < "C0" < "C0C" < "C0CCCCC" < "C0CC0" < "C0C0" < "C0CΩ" < "C00" < "C0Ω" < "CΩ" < "CΩC" < "CΩ0" < "CΩΩ" < "CΩΩΩ" < "0" < "0C" < "0CC" < "00" < "00C" < "0Ω" < "Ω".

Generally, the empty string is the least one, "C" is larger, then "0" is larger, then "Ω" is larger. And, if string a < string b, then ca < cb for any string c.

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Could someone explain to me Aarex Tiaokhiao’s number Otopersuperoogol?

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My Old Googology Is Closed!!!!

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C(1,1) is stronger, which is the 2nd fixed point of C(0,C(0,…))

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1 is C(0,0), so your C(1,1) is C(C(0,0),C(0,0)). But it’s not in standard form, because C(0,0) > 0, and it’s against the 2nd rule of standard form.

In Taranovsky’s notation, we only see expressions in standard form as ordinals.

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Oh. I got that. By the way. What C() have level {1,,1{1,,1,,2}3}?

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