Hyper-Alpha Notation, Part IV Redone

(WORK IN PROGRESS)

See:

Part I

Part II: Arrows to Alphas

Part III: Full Circle

Original Part IV, a stupid attempt at an extension that will be ignored for this purpose

Let's get down to business. In the original Part IV, I defined <A,A> as an operator and allowed for operations on <A,A>. That sucked. I have a better idea.

A₁ = A

<a,b>★A_n = <a,a>★<A_n-1,A_n-1>^b (exponent denotes concatenation, NOT an operator)

A_n will still evaluate to <A_n-1,A_n-1>^b if placed after any number of up-arrows.

Ordinals below may change, especially after ψ(OFP).

Recursive progression

<a,b>A₂ ≈ ψ(Ω_ω)

<a,b>A₂A ≈ ψ(Ω_ω)+1

<a,b>A₂A₂ ≈ ψ(Ω_ω)2

<a,b>A₂^A ≈ ψ(Ω_ω)×ω

<a,b>A₂^A↑↑A ≈ ψ(Ω_ω)×ε₀

<a,b>A₂^A₂ ≈ ψ(Ω_ω)²

<a,b>A₂^A₂A ≈ ψ(Ω_ω)^ω

<a,b>A₂^A₂A↑↑A ≈ ψ(Ω_ω)^ε₀

<a,b>A₂↑↑3 ≈ ψ(Ω_ω)↑↑2

<a,b>A₂↑↑n ≈ ψ(Ω_ω)↑↑(n-1)

<a,b>A₂↑↑A ≈ ψ(Ω_ω+Ω)

<a,b>A₂↑↑A↑↑A ≈ ψ(Ω_ω+Ω2)

<a,b>A₂↑↑A↑↑↑A ≈ ψ(Ω_ω+Ω²)

<a,b>A₂↑↑A↑↑↑↑A ≈ ψ(Ω_ω+Ω³)

<a,b>A₂↑↑A ↑^A A ≈ ψ(Ω_ω+Ω^ω)

<a,b>A₂↑↑<A,A>A^AA ≈ ψ(Ω_ω+Ω^Ω)

<a,b>A₂↑↑<A,A>A↑↑A ≈ ψ(Ω_ω+Ω₂)

<a,b>A₂↑↑<A,A><A,A>A^AA ≈ ψ(Ω_ω+Ω₃)

<a,b>A₂↑↑A₂ ≈ ψ(Ω_ω×2)

<a,b>(A₂↑↑A₂)↑↑A ≈ ψ(Ω_ω×2+1)

<a,b>(A₂↑↑A₂)↑↑A↑↑A ≈ ψ(Ω_ω×2+Ω)

<a,b>(A₂↑↑A₂)↑↑A₂ = A₂↑↑A₂A ≈ ψ(Ω_ω×3)

<a,b>A₂↑↑A₂A^A ≈ ψ(Ω_ω×ω)

<a,b>A₂↑↑A₂(A↑↑A) ≈ ψ(Ω_ω×Ω)

<a,b>A₂↑↑A₂A₂ ≈ ψ(Ω_ω²)

<a,b>A₂↑↑A₂^A ≈ ψ(Ω_ω^ω)

<a,b>A₂↑↑A₂^A↑↑A ≈ ψ(Ω_ω^Ω)

<a,b>A₂↑↑A₂^A₂ ≈ ψ(Ω_ω^Ω_ω)

<a,b>A₂↑↑A₂↑↑A ≈ ψ(Ω_ω+1)

<a,b>A₂↑↑A₂↑↑A^A ≈ ψ(Ω_ω2)

<a,b>A₂↑↑A₂↑↑A↑↑A ≈ ψ(Ω_Ω)

<a,b>A₂↑↑A₂↑↑A↑↑↑A ≈ ψ(Ω_Ω²)

<a,b>A₂↑↑A₂↑↑<A,A>A^AA ≈ ψ(Ω_Ω2)

<a,b>A₂↑↑A₂↑↑A₂ ≈ ψ(Ω_Ωω)

<a,b>A₂↑↑A₂↑↑A₂↑↑A ≈ ψ(Ω_Ωω+1)

<a,b>A₂↑↑A₂↑↑A₂↑↑A₂ ≈ ψ(Ω_ΩΩω)

<a,b>A₂↑↑↑A ≈ ψ(OFP)

I don't know how to express ordinals beyond ψ(OFP) using an efficient notation. Until then I will use capital epsilon (E) to denote the αth omega fixed point as E_α, and continue using the Veblen hierarchy but with capital Greek letters. This is most likely ill-defined and should be left up to interpretation.

<a,b>A₂↑↑↑A ≈ ψ(E₀)

<a,b>(A₂↑↑↑A)↑↑A₂ ≈ ψ(Ω_E0+1)

<a,b>A₂↑↑↑AA ≈ ψ(E₁)

<a,b>A₂↑↑↑A^A ≈ ψ(ψ_I(E_ω)

<a,b>A₂↑↑↑A↑↑A ≈ ψ(E_ε0)

<a,b>A₂↑↑↑A↑↑↑A ≈ ψ(E_ζ0)

<a,b>A₂↑↑↑A ↑^A A ≈ ψ(E_φ(ω,0))

<a,b>A₂↑↑↑<A,A>A^AA ≈ ψ(E_ψ(Ω2))

<a,b>A₂↑↑↑A₂ ≈ ψ(E_ψ(Ωω))

<a,b>A₂↑↑↑A₂A ≈ ψ(E_ψ(Ωω)+1)

<a,b>A₂↑↑↑A₂↑↑↑A ≈ ψ(E_ψ(E0))

<a,b>A₂↑↑↑↑A ≈ ψ(E_Ω), the first fixed point of α ↦ ψ(E_α)

<a,b>(A₂↑↑↑↑A)↑↑↑A₂ ≈ ψ(E_Ω+1)

<a,b>A₂↑↑↑↑AA ≈ ψ(E_Ω+ω)

<a,b>A₂↑↑↑↑A↑↑A ≈ ψ(E_Ω+ε0)

<a,b>A₂↑↑↑↑A₂ ≈ ψ(E_Ω+ψ(E0))

<a,b>A₂↑↑↑↑A₂A^A ≈ ψ(E_Ω2)

<a,b>A₂↑↑↑↑A₂A₂ ≈ ψ(E_Ω×ψ(E0))

<a,b>A₂↑↑↑↑A₂^A ≈ ψ(E_Ω²)

<a,b>A₂↑↑↑↑A₂↑↑↑A ≈ ψ(E_E0)

<a,b>A₂↑↑↑↑A₂↑↑↑↑A ≈ ψ(E_EΩ)

<a,b>A₂↑↑↑↑↑A ≈ ψ(Z₀)

<a,b>A₂↑↑↑↑↑AA ≈ ψ(Z₁)

<a,b>A₂↑↑↑↑↑A₂ ≈ ψ(Z_ψ(Ωω))

<a,b>A₂↑↑↑↑↑A₂A^A ≈ ψ(Z_Ω)

<a,b>A₂↑↑↑↑↑A₂↑↑↑A ≈ ψ(Z_E0)

<a,b>A₂↑↑↑↑↑A₂↑↑↑↑↑A ≈ ψ(Z_Z0)

<a,b>A₂↑↑↑↑↑↑A ≈ ψ(H₀)

<a,b>A₂↑↑↑↑↑↑A₂ ≈ ψ(H_ψ(Ωω))

<a,b>A₂↑↑↑↑↑↑↑A ≈ ψ(Φ(4,0))

<a,b>A₂ ↑^A A ≈ ψ(Φ(ω,0))

<a,b>A₂ ↑^{A ↑A A} A ≈ ψ(Φ(φ(ω,0),0))

<a,b>A₂ ↑^A₂ A ≈ ψ(Φ(ψ(Ω_ω),0))

<a,b>A₂ ↑^{A₂ ↑A A} A ≈ ψ(Φ(ψ(Φ(ω,0)),0))

<a,b><A₂,A>A^AA ≈ ψ(Φ(Ω,0)), the first fixed point of α ↦ ψ(Φ(α,0))

<a,b><A₂,AA>A^AA ≈ ψ(Φ(Ω,1))

<a,b><A₂,A>A^AAA ≈ ψ(Φ(Ω+1,0))

<a,b><A₂,A>A₂ ≈ ψ(Φ(Ω+ψ(Ω_ω),0))

<a,b><A₂,A><A₂,A>A^AA ≈ ψ(Φ(Ω2,0))

Now for some recursive steps I skipped in Part 3.

Note that <a,b><A₂,A><A₂,A>A^AA solves like <a,b>(<A₂,A>(<A₂,A>A^AA)).

<a,b><A₂,A>(<A₂,A>A^AA)A ≈ ψ(Φ(Ω2+1,0))

<a,b><A₂,A>(<A₂,A>A^AA)(<A₂,A>A^AA) ≈ ψ(Φ(Ω3,0))

<a,b><A₂,A>(<A₂,A>A^AA)^A ≈ ψ(Φ(Ωω,0))

<a,b><A₂,A>(<A₂,A>A^AA) ↑^A A ≈ ψ(Φ(Ω×φ(ω,0),0))

<a,b><A₂,A><A₂,AA>A^AA ≈ ψ(Φ(Ω²,0))

<a,b><A₂,A><A₂,A^A>A^AA ≈ ψ(Φ(Ω^ω,0))

<a,b><A₂,A><A₂,A₂>A^AA ≈ ψ(Φ(Ω^ψ(Ω_ω),0))

<a,b><A₂,A><A₂,A>A^AAA ≈ ψ(Φ(Ω^Ω,0))

<a,b><A₂,A><A₂,A>A^AAAA ≈ ψ(Φ(Ω^Ω2,0))

<a,b><A₂,A><A₂,A>A^AA^A ≈ ψ(Φ(Ω^Ω2,0))

<a,b><A₂,A><A₂,A>A↑↑A ≈ ψ(Φ(E₀,0))

<a,b><A₂,A><A₂,A>A↑↑↑A ≈ ψ(Φ(Z₀,0))

<a,b><A₂,A><A₂,A>A ↑^A A ≈ ψ(Φ(Φ(ω,0),0))

<a,b><A₂,A><A₂,A>A₂ ≈ ψ(Φ(Φ(ψ(Ω_ω),0),0))

<a,b><A₂,A><A₂,A><A₂,A>A^AA ≈ ψ(Φ(Φ(Ω,0),0))

<a,b><A₂,A><A₂,A><A₂,A><A₂,A>A^AA ≈ ψ(Φ(Φ(Φ(Ω,0),0),0))

As you can see, A₂ is a very powerful extension by itself. But A₃ diagonalizes over it just the same.

After this point I tend to make some educated guesses based on previous related values, so take these values with quite a grain of salt.

<a,b>A₃ ≈ ψ(Φ(1,0,0)), in which Φ(1,0,0) is the first fixed point of α ↦ Φ(α,0).

<a,b>A₃A ≈ ψ(Φ(1,0,0))+1

<a,b>A₃A₂ ≈ ψ(Φ(1,0,0))+ψ(Ω_ω)

<a,b>A₃A₃ ≈ ψ(Φ(1,0,0))2

<a,b>A₃^A ≈ ψ(Φ(1,0,0))×ω

<a,b>A₃^A₂ ≈ ψ(Φ(1,0,0))×ψ(Ω_ω)

<a,b>A₃^A₃ ≈ ψ(Φ(1,0,0))²

<a,b>A₃^A₃A ≈ ψ(Φ(1,0,0))^ω

<a,b>A₃↑↑A ≈ ψ(Φ(1,0,0)+Ω)

<a,b>A₃↑↑A₂ ≈ ψ(Φ(1,0,0)+Ω_ω)

<a,b>A₃↑↑A₂↑↑A ≈ ψ(Φ(1,0,0)+Ω_ω+Ω)

<a,b>A₃↑↑A₂↑↑↑A ≈ ψ(Φ(1,0,0)+E₀)

<a,b>A₃↑↑A₂↑↑↑↑A ≈ ψ(Φ(1,0,0)+Z₀)

<a,b>A₃↑↑(<A₂,A>A^AA) ≈ ψ(Φ(1,0,0)+Φ(Ω,0))

<a,b>A₃↑↑A₃ ≈ ψ(Φ(1,0,0)×2)

<a,b>A₃↑↑A₃A ≈ ψ(Φ(1,0,0)×3)

<a,b>A₃↑↑A₃A^A ≈ ψ(Φ(1,0,0)×ω)

<a,b>A₃↑↑A₃(A↑↑A) ≈ ψ(Φ(1,0,0)×Ω)

<a,b>A₃↑↑A₃A₂ ≈ ψ(Φ(1,0,0)×Ω_ω)

<a,b>A₃↑↑A₃(A₂↑↑↑A) ≈ ψ(Φ(1,0,0)×E₀)

<a,b>A₃↑↑A₃A₃ ≈ ψ(Φ(1,0,0)²)

<a,b>A₃↑↑A₃^A ≈ ψ(Φ(1,0,0)^ω)

<a,b>A₃↑↑A₃^A₃ ≈ ψ(Φ(1,0,0)^{ψ(Φ(1,0,0))})

<a,b>A₃↑↑A₃↑↑A ≈ ψ(Φ(1,0,0)^Ω)

<a,b>A₃↑↑(A₃↑↑A)² ≈ ψ(Φ(1,0,0)^Φ(1,0,0)Ω)

<a,b>A₃↑↑(A₃↑↑A)^A ≈ ψ(Ω_Φ(1,0,0)+1)

<a,b>A₃↑↑A₃↑↑A₃ ≈ ψ(Ω_Φ(1,0,0)×2)

<a,b>A₃↑↑A₃↑↑A₃A ≈ ψ(Ω_{Φ(1,0,0)×ω})

<a,b>A₃↑↑A₃↑↑A₃↑↑A ≈ ψ(Ω_{ΩΦ(1,0,0)+1})

<a,b>A₃↑↑↑A ≈ ψ(E_Φ(1,0,0)+1)

<a,b>(A₃↑↑↑A)↑↑A₃ ≈ ψ(Ω_EΦ(1,0,0)+1)

<a,b>(A₃↑↑↑A)↑↑(A₃↑↑↑A) ≈ ψ(E_Φ(1,0,0)+2)

<a,b>A₃↑↑↑AA ≈ ψ(E_Φ(1,0,0)+ω)

<a,b>A₃↑↑↑A₂ ≈ ψ(E_{Φ(1,0,0)+Ωω})

<a,b>A₃↑↑↑A₃ ≈ ψ(E_Φ(1,0,0)×2)

<a,b>A₃↑↑↑A₃↑↑A ≈ ψ(E_{Φ(1,0,0)×Ω})

<a,b>A₃↑↑↑A₃↑↑A₃↑↑A ≈ ψ(E_{ΩΦ(1,0,0)+1})

<a,b>A₃↑↑↑A₃↑↑↑A ≈ ψ(E_EΦ(1,0,0)+1)

<a,b>A₃↑↑↑A₃↑↑↑A₃ ≈ ψ(E_{EΦ(1,0,0)×2})

<a,b>A₃↑↑↑↑A ≈ ψ(Z_Φ(1,0,0)+1)

<a,b>A₃↑↑↑↑↑A ≈ ψ(H_Φ(1,0,0)+1)

<a,b>A₃ ↑^A₂ A ≈ ψ(Φ(ψ(Ω_ω),Φ(1,0,0)+1))

<a,b>A₃ ↑^A₃ A ≈ ψ(Φ(Φ(1,0,0)+1,0))

Beyond this point, I honestly have very little idea as to how to handle my own simplistic, informal ordinal notation. Don't take anything below this seriously, I WILL revise this in due time.

<a,b><A₃,A>A^AA ≈ ψ(Φ(1,0,1))

<a,b><A₃,AA>A^AA ≈ ψ(Φ(1,0,2))

<a,b><A₃,A>A^AAA ≈ ψ(Φ(1,1,0))

<a,b><A₃,A>A^AA^A ≈ ψ(Φ(1,ω,0))

<a,b><A₃,A>A₂ ≈ ψ(Φ(1,Ω_ω,0))

<a,b><A₃,A>A₃ ≈ ψ(Φ(1,Φ(1,0,0),0))

<a,b><A₃,A><A₃,A>A^AA ≈ ψ(Φ(2,0,0))

<a,b><A₃,A><A₃,AA>A^AA ≈ ψ(Φ(2,1,0))

<a,b><A₃,A><A₃,A>A^AAA ≈ ψ(Φ(3,0,0))

<a,b><A₃,A><A₃,A>A₂ ≈ ψ(Φ(Ω_ω,0,0))

<a,b><A₃,A><A₃,A>A₃ ≈ ψ(Φ(Φ(1,0,0),0,0))

<a,b><A₃,A><A₃,A><A₃,A>A^AA ≈ ψ(Φ(1,0,0,0))

<a,b><A₃,A><A₃,A><A₃,A><A₃,A>A^AA ≈ ψ(Φ(1,0,0,0,0))

<a,b>A₄ ≈ ???

<a,b>A₅ ≈ ????

Limit: <a,b>A_A = <a,a>A_b. Defining (and especially approximating) A_AA, A_AA, "A↓↓A" (nested subscripts) and beyond will take much more work, and will absolutely enter Part V territory.