User blog comment:Bubby3/Walkthrough of BMS./@comment-35470197-20190216042718/@comment-30754445-20190221023528

Looks like I need to brush up my on my Buchholz.

It also looks like my method does get the correct ordinals for the pair sequences. The problem seems to be solely in the conversion eto standard Buchholz.

And yes, looking at it again, I understand why ψ1(ψ2(Ω3)+ψ1(Ω2)) is not standard. You can't have ψ1(ψ2(Ω3)+...). That's where my error was.

It should have went:

ψ0(ψ1(ψ2(Ω3)+ψ1(Ω2))) → ψ0(Ω3+ψ1(Ω3+ψ1(Ω2)))

ψ0(ψ1(ψ2(Ω3)+ψ1(ψ2(Ω3)+ψ1(Ω2)))) → ψ0(Ω3+ψ1(Ω3+ψ1(Ω3+ψ1(Ω2))))

ψ0(ψ1(ψ2(Ω3)+ψ1(ψ2(Ω3)+ψ1(ψ2(Ω3)+ψ1(Ω2))))) → ψ0(Ω3+ψ1(Ω3+ψ1(Ω3+ψ1(Ω3+ψ1(Ω2)))))

ψ0(ψ1(ψ2(Ω3)+Ω2)) → ψ0(Ω3+Ω2) (yes!!!!)

And from here we continue to:

ψ0(ψ1(ψ2(Ω3)+Ω2)+1) → ψ0(Ω3+Ω2+1)

ψ0(ψ1(ψ2(Ω3)+Ω2)+Ω) → ψ0(Ω3+Ω2+Ω)

ψ0(ψ1(ψ2(Ω3)+Ω2)x2) → ψ0(Ω3+Ω2+ψ1(ψ2(Ω3)+Ω2))

ψ0(ψ1(ψ2(Ω3)+Ω2+1)) → ψ0(Ω3+Ω2+ψ1(ψ2(Ω3)+Ω2+1))

ψ0(ψ1(ψ2(Ω3)+Ω2+Ω)) → ψ0(Ω3+Ω2+ψ1(ψ2(Ω3)+Ω2+Ω))

ψ0(ψ1(ψ2(Ω3)+Ω2x2)) → ψ0(Ω3+Ω2x2)

ψ0(ψ1(ψ2(Ω3)+ψ2(Ω))) → ψ0(Ω3+ψ2(Ω))

ψ0(ψ1(ψ2(Ω3)+ψ2(Ω2))) → ψ0(Ω3+ψ2(Ω2))

ψ0(ψ1(ψ2(Ω3)x2))) → ψ0(Ω3+ψ2(Ω3))

ψ0(ψ1(ψ2(Ω3+1))) → ψ0(Ω3+ψ2(Ω3+1))

ψ0(ψ1(ψ2(Ω3+Ω2))) → ψ0(Ω3+ψ2(Ω3+Ω2))

PERFECT!

Thanks, p-bot! I think I'm finally getting the hang of Buchholz's system. This is something I've been trying to achieve for a long time now.