User blog comment:Syst3ms/A sketch for an — actually — formal definition of UNOCF/@comment-35470197-20180803231131/@comment-25601061-20180810161843

@PsiCubed2

The idea of well-defining UNOCF as an actual OCF has been pretty much been dropped. It's now being well-defined as an ordinal notation instead (see here). So now it can have any strength we want it to have (as long as we define things correctly), and therefore be able to go beyond Deedlit's weakly compact OCF.

As for how UNOCF compares with Deedlit's weakly compact OCF (which I will refer to as KOCF), here's a (very) simplified version of a proof that ψT(TTsup>T ) in UNOCF corresponds to K in KOCF: As I said, this is a very simplified version of the proof, so there are many specific details missing, such as various cases for limit ordinals, but it still works as a sketch for the full proof.
 * I assume you understand why UNOCF and KOCF are roughly the same up to M(1,0).
 * The first point in which UNOCF and KOCF differ is M(1,0). This is because, while in KOCF M(1,0) in χM(1,0)(_,0) produces regulars, in UNOCF M(1,0) in ψM(1,0)(_) only produces fixed points. This is the same with all M(a,b,c,...)s - and, since M(1,0) in χM(1,0)(_) in KOCF produces regulars diagonalizing over M_x and M(a,b,c,...)s (and therefore ψN(x)s) in UNOCF are those regulars themselves, M(1,0) in KOCF = N in UNOCF.
 * Furthermore, if C(1+x;0) = M(x,0), then C(1+x;y) = M(x,y), and since M(x+1,0) in χM(x+1,0)(_,0) produces regulars that diagonalize over M(x,y) while ψC(1+x+1;0)(_) are those ordinals themselves, M(x+1,0) = C(1+x+1;0).
 * However, C(1;0;0) is not M(1,0,0) as some might've expected - even though M(1,0,0) in χM(1,0,0)(_,0) produces ordinals that diagonalize over M(x,0) and C(a,b,c,...,0;0)s diagonalize over C(x;0) themselves, C(1,0;a,b,c,...)s (and therefore ψC(1,1;0)(x)s) also diagonalize over C(x;0), so M(1,0,0) is actually C(1,1;0).
 * Let A is an array. Then, for any A, M(A,b+1,0) can be calculated using the procedure for M(x+1,0) (but starting with M(A,b,0) instead of M(x,0)), and M(A,0,0) can be calculated using the procedure for M(1,0,0) (but using χM(A,0,0) instead of χM(1,0,0)). By doing this, we get that K = Ξ(2,0).
 * As (n+1)-Mahloness is to n-Mahloness as Mahloness is to inaccessibility and a comma in C(...;_;0) generates different levels of Mahloness while a comma in C(...;_) generates different levels of inaccessibility, a comma in C(...;_;0;0) generates different levels of 1-Mahloness, a comma in C(...;_;0;0;0) generates different levels of 2-Mahloness, etc. This means that ψT(TT x ) is the first (2+x)-Mahlo, i.e Ξ(2+x,0).
 * Finally, if Ξ(x,0) corresponds to ψψ T(TT T ) (y), Ξ(x+1,0) corresponds to ψψ T(TT T ) (y+1) by the bullet point above, and a K in x in Ξ(x,0) makes Ξ(x,0) in Ψ^x_Ξ(x,0)(_) diagonalize over Ψ^y_Ξ(x,0)(_) for y < x, so K corresponds to ψT(TT T ) in ψψ T(TT T ) (_).