User blog comment:Bubby3/Walkthrough of BMS./@comment-35470197-20190216042718/@comment-30754445-20190220145152

The expression was supposed to be standard. Since it isn't, this means that I've botched the conversion somehow.

Let me try to do it one step at time:

(each row is a Hybird OCF → Buchholz OCF conversion)

ψ0(ψ1(ψ2(Ω3))) → ψ0(Ω3)

ψ0(ψ1(ψ2(Ω3))+1) → ψ0(Ω3+1)

ψ0(ψ1(ψ2(Ω3))+Ω) → ψ0(Ω3+Ω)

ψ0(ψ1(ψ2(Ω3))x2) → ψ0(Ω3+ψ1(Ω3))

ψ0(ψ1(ψ2(Ω3)+1)) → ψ0(Ω3+ψ1(Ω3+1))

ψ0(ψ1(ψ2(Ω3)+Ω)) → ψ0(Ω3+ψ1(Ω3+Ω))

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

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(ψ2(Ω3)+ψ1(Ω2))))

ψ0(ψ1(ψ2(Ω3)+Ω2)) → ψ0(Ω3+ψ1(Ω3+Ω2))

Hmmm... that's different from both my original result and yours. At least it is a standard expression now, right?

The only question that remains is which of our results is - indeed - equivalent to (0,0)(1,1)(2,2)(3,3)(2,2).

Well, let's try going the other way around:

According to my theory, Buchholz ψ0(Ω3+Ω2) is equivlaent to Hybrid ψ0(ψ1(ψ2(Ω3+Ω2))) which is equivalent to (0,0)(1,1)(2,2)(3,3)(3,2).

In your view, the same expression must correspond to a larger ordinal. So, what ordinal is equivalent, in your view, to (0,0)(1,1)(2,2)(3,3)(3,2)?