User blog comment:Scorcher007/Analysis DAN up to Z2/@comment-11227630-20180919121102/@comment-11227630-20180919145140

I see. But S[Ωa+1] is still stronger than you listed.

From your list, the notation using S[Ωa+1] seems to be as strong as the notation using S[εa+1].

However, we can do more below S[Ωa+1]. Think of a hypothesis binary OCF θ such that θ(a,b) is always below the next admissible after b (just as the normal θ works), and extend it to somewhere below S[Ωa+1]. Now think of this sequence: {S[a+1], S[θ(S[b+1],a+1)], S[θ(S[θ(S[c+1],b+1)],a+1)], ...} notice that all these things, their limits, the limits of limits, and so on, are all below S[Ωa+1], so S[Ωa+1] should work as the diagonalizer and the collapsing function using it will enumerate limit points of them.

In your list, the a in S[function(a)] approximately corresponds with `,, or {1{1,,,3}2,,,2} in sDAN. So The S[Ωa+1] will correspond with {1{1,,,1,2}2,,,2} (note that the {1,,,k} doesn't search out of the outside { ____ ,,,2} until it reduces to k=3).
 * ψ(S[εa+1]) ~ s(n,n{1{1{1,,,3}3,,,2}2{1,,,3}2,,,2}2) = s(n,n{1{1,,,3}3,,,2}2)
 * ψ(S[θ(εΩ+1,a+1)]) ~ s(n,n{1{1,,,3}4,,,2}2)
 * ψ(S[θ(Ωω,a+1)]) ~ s(n,n{1{1,,,3}1,2,,,2}2)
 * ψ(S[θ(I,a+1)]) ~ s(n,n{1{1,,,3}1{1,,,3}2,,,2}2)
 * ψ(S[θ(M,a+1)]) ~ s(n,n{1{1{1,,,3}2,,,3}2,,,2}2) = s(n,n{1{1,,,3}2,,,3}2)
 * ψ(S[θ(S[b+1],a+1)]) ~ s(n,n{1,,,4}2)
 * ψ(S[θ(S[θ(S[c+1],b+1)],a+1)]) ~ s(n,n{1,,,5}2)
 * ψ(S[Ωa+1]) ~ s(n,n{1,,,1,2}2)