User blog comment:Bubby3/Walkthrough of BMS./@comment-35470197-20190216042718/@comment-30754445-20190220125348

Well, it's a "walkthrough" rather than a proof or even a sold justification. Not really adequete, but it's still better than nothing.

As for your second question:

Pair sequences do seem to be exactly equivalent to expressions in a Buchhloz-Madore hybrid OCF (that is: an OCF which uses the rules of Buchhloz, but defines the ψ's inductively like Madore).

So in that OCF:

(0,0)(1,1)(2,2)(3,3)(2,2) = ψ0(ψ1(ψ2(ψ3(0))+ψ2(0))) = ψ0(ψ1(ψ2(ψ3(0))+ Well, it's a "walkthrough" rather than a proof or even a sold justification. Not really adequete, but it's still better than nothing.

As for your second question:

Pair sequences do seem to be exactly equivalent to expressions in a Buchholz-Madore hybrid OCF (that is: an OCF which uses the rules of Buchholz, but defines the ψ's inductively like Madore).

So in that OCF:

(0,0)(1,1)(2,2)(3,3)(2,2) = ψ0(ψ1(ψ2(ψ3(0))+ψ2(0))) = ψ0(ψ1(ψ2(Ω3)+Ω2))

And

(0,0)(1,1)(2,2)(3,3)(2,2)(2,1) = ψ0(ψ1(ψ2(ψ3(0))+ψ2(0)+ψ1(0))) = ψ0(ψ1(ψ2(Ω3)+Ω2+Ω1))

Converting to standard Buchholz, we get (I think. I'm far more well-versed in the Madore style):

(0,0)(1,1)(2,2)(3,3)(2,2) = ψ0( Ω3+ ψ1( Ω3+ ψ2(Ω3)+Ω2))

(0,0)(1,1)(2,2)(3,3)(2,2)(2,1) = ψ0( Ω3+ ψ1( Ω3+ ψ2(Ω3)+Ω2+Ω1))

To summarize, in pair-sequences (x,n) always corresponds to an additional ψn. If it's the last pair in the sequence, then it will be parsed as ψn(0) which is equal to Ωn. So Bubby was correct (if a bit imprecise).

Of-course, this direct correspondence cannot work beyond pair-sequences, because the inductive Madore-style definition is limited to ψ0(ψ1(ψ2(ψ3(ψ4(...))))) (what we'd normally call ψ(Ωω).