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## Basic

A(α) = α+1

A(0,0) = sup(0, A(0), A(A(0)),...)

A(0,A(α)) = sup(A(0,α), A(A(0,α)), A(A(A(0,α))),...)

A(A(β),0) = sup(0, A(β,0), A(β,A(β,0)),...)

A(A(β),A(α)) = sup(A(A(β),α), A(β,A(A(β),α)), A(β,A(β,A(A(β),α))),...)

For α is a limit ordinal, A(β,α) = sup(A(β,α[1]), A(β,α[2]), A(β,α[3]),...)

For β is a limit ordinal, A(β,0) = sup(A(β[1],0), A(β[2],0), A(β[3],0),...)

For β is a limit ordinal, A(β,A(α)) = sup(A(β[1],A(β,0)), A(β[2],A(β,0)), A(β[3],A(β,0)),...)

## Introducing the Ω

A(Ω,0) = sup(0,A(0,0),A(A(0,0),0),...)

A(Ω,0) = A(A(Ω,0),0)

A(Ω,A(α)) = sup(A(Ω,α), A(A(Ω,α),1), A(A(A(Ω,α),1),1),...)

## Ωα

Ω1 = Ω

Ω0 = 0

To resolve an array with an array composed of Ωα+2's and Ωα+1's (except for arrays composed of Ω1's and Ω0's) one step, first make a copy of the array composed of Ωα+2's and Ωα+1's, and call it &. Then, change the Ωα+2's and Ωα+1's of & into Ωα+1's and Ωα's. Then, resolve this array one step. Change all of the Ωα+1's and Ωα's back into Ωα+2's and Ωα+1's, and then, back in the original array, replace the array composed of Ωα+2's and Ωα+1's (not &) into &. Then, delete &, but not in the original array.

For arrays composed of Ω1's and Ω0's, use the same rules as the previous 2 sections, but only one time, so it is only one step.

## Comparison with other ordinals

A ordinal notation Normal ordinals
$0$ $0$
$A(0)$ $1$
$A(A(0))$ $2$
$A(A(A(0)))$ $3$
$A(A(A(A(0))))$ $4$
$A(0,0)$ $\omega$
$A(A(0,0))$ $\omega+1$
$A(A(A(0,0)))$ $\omega+2$
$A(A(A(A(0,0))))$ $\omega+3$
$A(0,A(0))$ $\omega2$

$A(A(0,A(0)))$

$\omega2+1$
$A(A(A(0,A(0))))$ $\omega2+2$
$A(0,A(A(0)))$ $\omega3$
$A(A(0,A(A(0))))$ $\omega3+1$
$A(0,A(A(A(0))))$ $\omega4$
$A(0,A(A(A(A(0)))))$ $\omega5$
$A(0,A(0,0))$ $\omega^2$

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