User blog comment:Fejfo/Uncountable indexed veblen function/@comment-32213734-20180513045616

According rule 2 φΩ + 1(β) is β-th ordinal such as γ = φΩ(γ).

According rule 4 φΩ(β) = φβ(β)

Is φΩ + 1(β) Γ-function?

φΩ + 1 + α(β) = φ(1, α, β)

Let α = 0

φΩ + 1(β) = φ(1, 0, β) = Γβ

But if γ = φΩ(γ) = φγ(γ) are Γ numbers, then

Γβ = φ(Γβ, Γβ)

Particularly

Γ0 = φ(Γ0, Γ0)

But

Γ0 = φ(Γ0, 0) < φ(Γ0, Γ0)

Γβ = φ(Γβ, 0) < φ(Γβ, Γβ)

I think that φβ(β) does not have fixed points.