User blog comment:Emlightened/Early Birthday Present For Deedlit/@comment-5529393-20170730212008/@comment-5529393-20170731021930

Just to make sure we are on the same page (so forgive me for saying stuff that you already know), there is the cardinal that we are collapsing below, and also the Mahlo level of the ordinal we are collapsing to. So in the K notation the ordinal collapsing function is written $$\Psi^\xi_\pi (\alpha)$$, where $$\pi$$ is the cardinal that is being collapsed, $$\xi$$ is the type of ordinal that the collapsed ordinal is allowed to be (0 means any ordinal, so the normal ordinal collapsing function; 1 means weakly inaccessible, 2 means weakly Mahlo, 3 means 2-weakly Mahlo, and so on), and $$\alpha$$ represents the inductive level. Now, it seems the main purpose of collapsing our weakly Mahlo cardinals is to get weakly inaccessible cardinals, so perhaps it is enough to always have $$\xi = 1$$ when $$\pi$$ is weakly Mahlo but not 2-weakly Mahlo. Then we could always have $$\xi = 2 $$ when $$\pi$$ is 2-weakly Mahlo but not 3-weakly Mahlo, and so on. But then, if $$\pi$$ is $$\omega$$-weakly Mahlo, we can't just take the predecessor of $$\omega$$, so we need \xi to at least range over the finite ordinals. So it looks like we can't get rid of $$\xi$$ in our K notation.

I would think we would also need the $$\xi$$ parameter in your notation as well. Does your notation work out so that weakly Mahlo cardinals always collapse to weakly inaccessible cardinals, 2-weakly Mahlo cardinals always collapse to weakly Mahlo cardinals, and so on? What happens to $$\omega$$-weakly Mahlo cardinals when they are collapsed?