User blog:Rgetar/Equal values of Veblen function

Some ordinals can be expressed through Veblen function in different ways.

For example, ε0. There are two such ways:

ε0 = φ(1, 0)

ε0 = φ(ε0)

And for ζ0 there are three such ways:

ζ0 = φ(2, 0)

ζ0 = φ(1, ζ0)

ζ0 = φ(ζ0)

А question may arise: how to get all such forms of a given ordinal (say, Γ0, or any other).

Here is algorithm how to get equal forms of an ordinal.

Algorithm
So, let we have an ordinal expressed through Veblen function, for example,

α = φ(β10, β9, β8, β7, β6, β5, β4, β3, β2, β1, β0)

Step 1. Select any two variables. (For example, β7 and β3).

Step 2. Left variable set to any value less than its current value. (β7 → γ < β7).

Step 3. Right variable set to α. (β3 → α).

Step 4. All variables located between them set to any values less than α. (β6, β5, β4 → δ6, δ5, δ4 < α).

Step 5. All variables located to the right of the right variable set to 0. (β2, β1, β0 → 0, 0, 0).

That's all!

So, from

α = φ(β10, β9, β8, β7, β6, β5, β4, β3, β2, β1, β0)

we get an equal form

α = φ(β10, β9, β8, γ, δ6, δ5, δ4, α, 0, 0, 0),

γ < β7,

δ6, δ5, δ4 < α.

(Here the selected variables are bolded).

Standard and non-standard forms
So, we have an algorithm how to get equal forms of an ordinal expressed in some given form.

But a question may arise: are they all such forms of an ordinal? Or, maybe, there are other equal forms, which cannot be obtained using this algorithm?

Answer: it depends on the given form.

It is an irreversible algorithm. If it performed twice, it is impossible to obtain the initial form.

Note: the result of this algorithm always contains the ordinal α itself as one (and only one) of the variables.

But there is a special equal form, which do not contain the ordinal α itself (all its variables are less than α). It may be named "standard form", and all other forms, which contain α as a variable, may be named "non-standard forms".

For example, ε0. Its standard form is φ(1, 0). (Indeed, 1, 0 < ε0). And its non-standard form is φ(ε0). (It contains one ε0 among variables).

And for ζ0 standard form is φ(2, 0). (Indeed, 2, 0 < ζ0). And its non-standard forms are φ(1, ζ0) and φ(ζ0). (They contain one ζ0 among variables).

So, if initial form of the ordinal is the standard form, all other (i. e. non-standard) equal forms of this ordinal may be obtained using this algorithm.

And, if initial form is a non-standard form (not all other equal forms may be obtained using this algorithm, but only some of them).

Examples
I will write ordinals initially in the standard for, then calculate all its non-standard forms.

φ(β)
ωβ = φ(β), β ≠ ωβ

There are no non-standard forms.

ε0
ε0 = φ(1, 0)

Non-standard form:

φ(ε0)

ε1
ε1 = φ(1, 1)

Non-standard form:

φ(ε1)

εβ
εβ = φ(1, β), β ≠ εβ

Non-standard form:

φ(εβ)

ζ0
ζ0 = φ(2, 0)

Non-standard forms:

φ(1, ζ0)

φ(ζ0)

ζ1
ζ1 = φ(2, 1)

Non-standard forms:

φ(1, ζ1)

φ(ζ1)

ζβ
ζβ = φ(2, β), β ≠ ζβ

Non-standard forms:

φ(1, ζβ)

φ(ζβ)

η0
η0 = φ(3, 0)

Non-standard forms:

φ(2, η0)

φ(1, η0)

φ(η0)

η1
η1 = φ(3, 1)

Non-standard forms:

φ(2, η1)

φ(1, η1)

φ(η1)

ηβ
ηβ = φ(3, β), β ≠ ηβ

Non-standard forms:

φ(2, ηβ)

φ(1, ηβ)

φ(ηβ)

φ(7, 0)
α = φ(7, 0)

Non-standard forms:

φ(6, α)

φ(5, α)

φ(4, α)

φ(3, α)

φ(2, α)

φ(1, α)

φ(α)

φ(7, 3)
α = φ(7, 3)

Non-standard forms:

φ(6, α)

φ(5, α)

φ(4, α)

φ(3, α)

φ(2, α)

φ(1, α)

φ(α)

φ(ω, 0)
α = φ(ω, 0)

Non-standard forms:

φ(n, α), n - natural number

φ(ω, 1)
α = φ(ω, 1)

Non-standard forms:

φ(n, α), n - natural number

φ(ω, β)
α = φ(ω, β), β ≠ α

Non-standard forms:

φ(n, α), n - natural number

φ(ω + 5, β)
α = φ(ω + 5, β), β ≠ α

Non-standard forms:

φ(ω + 4, α)

φ(ω + 3, α)

φ(ω + 2, α)

φ(ω + 1, α)

φ(ω, α)

φ(n, α), n - natural number

φ(ω2, β)
α = φ(ω2, β), β ≠ α

Non-standard forms:

φ(ω + n, α), n - natural number

φ(n, α), n - natural number

φ(β1, β0)
This is general case for two-variable Veblen function.

α = φ(β1, β0),

β1, β0 < α

Non-standard forms:

φ(γ, α),

γ < β1

Γ0
Γ0 = φ(1, 0, 0)

Non-standard forms:

φ(δ, Γ0), δ < Γ0

φ(Γ0, 0)

Γ1
Γ1 = φ(1, 0, 1)

Non-standard forms:

φ(δ, Γ1), δ < Γ1

φ(Γ1, 0)

Γβ
Γβ = φ(1, 0, β), β ≠ Γβ

Non-standard forms:

φ(δ, Γβ), δ < Γβ

φ(Γβ, 0)

φ(1, 1, 0)
α = φ(1, 1, 0)

Non-standard forms:

φ(δ, α), δ < α

φ(α, 0)

φ(1, 0, α)

φ(1, 1, β)
α = φ(1, 1, β), β ≠ α

Non-standard forms:

φ(δ, α), δ < α

φ(α, 0)

φ(1, 0, α)

φ(1, 5, 0)
α = φ(1, 5, 0)

Non-standard forms:

φ(δ, α), δ < α

φ(α, 0)

φ(1, 4, α)

φ(1, 3, α)

φ(1, 2, α)

φ(1, 1, α)

φ(1, 0, α)

φ(1, 5, β)
α = φ(1, 5, β), β ≠ α

Non-standard forms:

φ(δ, α), δ < α

φ(α, 0)

φ(1, 4, α)

φ(1, 3, α)

φ(1, 2, α)

φ(1, 1, α)

φ(1, 0, α)

φ(1, β1, β0)
α = φ(1, β1, β0),

β1, β0 < α

Non-standard forms:

φ(δ, α), δ < α

φ(α, 0)

φ(1, γ, α), γ < β1

φ(2, 0, 0)
α = φ(2, 0, 0)

Non-standard forms:

φ(1, δ, α), δ < α

φ(δ, α), δ < α

φ(1, α, 0)

φ(α, 0)

φ(2, 0, β)
α = φ(2, 0, β), β ≠ α

Non-standard forms:

φ(1, δ, α), δ < α

φ(δ, α), δ < α

φ(1, α, 0)

φ(α, 0)

φ(2, 5, 0)
α = φ(2, 5, 0)

Non-standard forms:

φ(1, δ, α), δ < α

φ(δ, α), δ < α

φ(1, α, 0)

φ(α, 0)

φ(2, 4, α)

φ(2, 3, α)

φ(2, 2, α)

φ(2, 1, α)

φ(2, 0, α)

φ(2, 5, β)
α = φ(2, 5, β), β ≠ α

Non-standard forms:

φ(1, δ, α), δ < α

φ(δ, α), δ < α

φ(1, α, 0)

φ(α, 0)

φ(2, 4, α)

φ(2, 3, α)

φ(2, 2, α)

φ(2, 1, α)

φ(2, 0, α)

φ(5, 0, 0)
α = φ(5, 0, 0)

Non-standard forms:

φ(4, δ, α), δ < α

φ(3, δ, α), δ < α

φ(2, δ, α), δ < α

φ(1, δ, α), δ < α

φ(δ, α), δ < α

φ(4, α, 0)

φ(3, α, 0)

φ(2, α, 0)

φ(1, α, 0)

φ(α, 0)

φ(5, 3, 0)
α = φ(5, 3, 0)

Non-standard forms:

φ(4, δ, α), δ < α

φ(3, δ, α), δ < α

φ(2, δ, α), δ < α

φ(1, δ, α), δ < α

φ(δ, α), δ < α

φ(4, α, 0)

φ(3, α, 0)

φ(2, α, 0)

φ(1, α, 0)

φ(α, 0)

φ(5, 2, α)

φ(5, 1, α)

φ(5, 0, α)

φ(β2, β1, β0)
This is general case for three-variable Veblen function.

α = φ(β2, β1, β0),

β2, β1, β0 < α

Non-standard forms:

φ(γ, δ, α), γ < β2, δ < α

φ(γ, α, 0), γ < β2

φ(β2, γ, α), γ < β1

φ(1, 0, 0, 0)
α = φ(1, 0, 0, 0)

Non-standard forms:

φ(δ2, δ1, α), δ2, δ1 < α

φ(δ2, α, 0), δ2 < α

φ(α, 0, 0)

φ(β3, β2, β1, β0)
This is general case for four-variable Veblen function.

α = φ(β3, β2, β1, β0),

β3, β2, β1, β0 < α

Non-standard forms:

φ(γ, δ2, δ1, α), γ < β3, δ2, δ1 < α

φ(γ, δ2, α, 0), γ < β3, δ2 < α

φ(γ, α, 0, 0), γ < β3

φ(β2, γ, δ1, α), γ < β2, δ1 < α

φ(β2, γ, α, 0), γ < β2

φ(β3, β2, γ, α), γ < β1

φ(1, 0, 0, 0, 0)
α = φ(1, 0, 0, 0, 0)

Non-standard forms:

φ(δ3, δ2, δ1, α), δ3, δ2, δ1 < α

φ(δ3, δ2, α, 0), δ3, δ2 < α

φ(δ3, α, 0, 0), δ3 < α

φ(α, 0, 0, 0)

φ(β4, β3, β2, β1, β0)
This is general case for five-variable Veblen function.

α = φ(β4, β3, β2, β1, β0),

β4, β3, β2, β1, β0 < α

Non-standard forms:

φ(γ, δ3, δ2, δ1, α), γ < β4, δ3, δ2, δ1 < α

φ(γ, δ3, δ2, α, 0), γ < β4, δ3, δ2 < α

φ(γ, δ3, α, 0, 0), γ < β4, δ3 < α

φ(γ, α, 0, 0, 0), γ < β4

φ(β4, γ, δ2, δ1, α), γ < β3, δ2, δ1 < α

φ(β4, γ, δ2, α, 0), γ < β3, δ2 < α

φ(β4, γ, α, 0, 0), γ < β3

φ(β4, β3, γ, δ1, α), γ < β2, δ1 < α

φ(β4, β3, γ, α, 0), γ < β2

φ(β4, β3, β2, γ, α), γ < β1

SVO
Non-standard forms of SVO:

φ(natural number of ordinals < SVO, SVO, natural number of zeros)

(Similar is also true for LVO, BHO, TFB, Ω, Ω2, ..., but for such ordinals there are also other equal forms in multi-dimensional generalization of Veblen function)