User blog comment:Bubby3/Walkthrough of BMS./@comment-35470197-20190216042718/@comment-35470197-20190221014932

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

I noticed that the standard expression of your result ψ_0(Ω_3+ψ_1(Ω_3+ψ_2(Ω_3)+Ω_2)) is ψ_0(Ω_3+Ω_2). Therefore


 * 1) Our results coincide with each other as long as we allow non-standard expression.
 * 2) Your assumption that the expression is standard was incorrect.
 * 3) My explanation that \(ψ_0(Ω_3+Ω_2))\) is greater than your result is incorrect. (Sorry, this was a silly mistake in computation.)

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

The last expression ψ_0(Ω_3+ψ_1(Ω_3+Ω_2)) is not standard. Its standard expression is ψ_0(Ω_3+Ω_2).

At least, the expression ψ_0(Ω_3+ψ_1(Ω_3+ψ_1(Ω_2))) and others above are standard. On the other hand, the next expression ψ_0(Ω_3+ψ_1(Ω_3+ψ_1(ψ_2(Ω_3)+ψ_1(Ω_2)))) is not standard, because its subterm ψ_1(ψ_2(Ω_3)+ψ_1(Ω_2)) is not either.

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

Then both results are indeed equivalent to it.

> According to my theory, Buchholz ψ0(Ω3+Ω2) is equivalent 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)?

In my view, (0,0)(1,1)(2,2)(3,3)(3,2) corresponds to ψ_0(Ω_3+ψ_2(Ω_3+Ω_2)).