User blog:Rgetar/Veblen functions with rules φ(α) = ωα; φ(α) = ω + α; φ(α) = 1 + α

There is a rule in Veblen function:

φ(α) = ωα

where α is ordinal.

I was wandering, what if replace this rule with some another rule, for example,

φ(α) = ωα

or

φ(α) = ω + α

or

φ(α) = 1 + α

(Note: lately I usually write arrays of ordinals as single "larger" ordinals, but here I will write arrays as arrays).

Two variables
Two-variable Veblen function for rule φ(α) = ωα is

φ(a, b) = ωω a b

Note:

φ(a, 0) = 0

(unlike Veblen function for rule φ(α) = ωα, where φ(a, 0) > 0)

Three variables
α = φ(1, 0, a) must satisfy the equation

α = φ(α, 0) = 0

α = 0

So, there is only one fixed point of φ(α, 0). It is 0.

So, φ(1, 0, 0) = 0, and φ(1, 0, a) for a > 0 does not exist.

α = φ(1, 1, a) must satisfy the equation

α = φ(1, 0, α)

α = φ(1, 0, 0)

α = 0

So, φ(1, 1, 0) = 0, and φ(1, 1, a) for a > 0 does not exist.

Apparently, for any array X φ(X, 0) = 0, and φ(X, a) for a > 0 does not exist.

Another definition of Veblen function
To overcome the issue with nonexistent values of the Veblen function with rule φ(α) = ωα I made up another definition of the many variable Veblen function:

So, let φ(..., α9, α8, α7, α6, α5, α4, α3, α2, α1, β) is many variable Veblen function.

Set S0 consists of 0 and all φ(..., α9, α8, α7, α6, α5, α4, α3, α2, α1, γ) + 1, where γ < β.

Definition of set Sn + 1, where n is natural number:

Choose any variable αi (say, α5). Replace it with any δ < αi. Replace all variables right to it with any elements of Sn. We get

φ(..., α9, α8, α7, α6, δ, λ4, λ3, λ2, λ1, λ0)

where δ < αi (δ can be instead of any αi, not only α5), all λj ∈ Sn.

Sn + 1 in union of Sn and set of all φ(..., α9, α8, α7, α6, δ, λ4, λ3, λ2, λ1, λ0).

Definition of set S:

S is union of all Sn.

φ(..., α9, α8, α7, α6, α5, α4, α3, α2, α1, β) = sup(S).

For Veblen function (with one variable rule φ(α) = ωα) this definition produce the same function, but for rule φ(α) = ωα it produces different function: now φ(X, a) for a > 0 exists, and

φ(1, 0, 1 + a) = εa

φ(1, 1, 1 + a) = ζa

Further in left part of equation is Veblen function for φ(α) = ωα, and in right part - for φ(α) = ωα.

φ(1, 1 + b, 1 + a) = φ(b, a)

φ(2, 0, 1 + a) = Γa

φ(1 + c, b, 1 + a) = φ(c, b, a), where c > 0

For any array with non-zero variable left to the third from right variable

φ(X, 0) = 0

φ(X, 1 + a) = φ(X, a)

φ(α) = ω + α
Two-variable Veblen function for rule φ(α) = ω + α is

φ(1 + a, b) = ωa + b

Three variables:

φ(1, 0, a) = εa

φ(1, 1, a) = ζa

φ(1, b, a) = φ(b, a)

φ(2, 0, a) = Γa

φ(1 + c, b, a) = φ(c, b, a)

For any array with non-zero variable left to the third from right variable

φ(X) = φ(X)

(coincides with Veblen function for ωα)

φ(α) = 1 + α
Two-variable Veblen function for rule φ(α) = 1 + α is

φ(a, b) = ωa + b

For three and more variables is just as for rule φ(α) = ω + α.

Reverse order
What about rules φ(α) = αω; φ(α) = α + ω; φ(α) = α + 1?

For Veblen function with definition using fixed points we need fixed points. But φ(α) = αω has only one fixed point (0), and φ(α) = α + ω, φ(α) = α + 1 have no fixed points.

So, for any array X with at least one non-zero variable

for rule φ(α) = αω

φ(X, 0) = 0

φ(X, a), where a > 0, does not exist;

for rules φ(α) = α + ω, φ(α) = α + 1

φ(X, a) does not exist.

But for definition using S set, apparently, function with rule φ(α) = αω coincides with function with rule φ(α) = ωα, function with rule φ(α) = α + ω coincides with function with rule φ(α) = ω + α, and function with rule φ(α) = α + 1 coincides with function with rule φ(α) = 1 + α.