User blog:Rgetar/Idea for FGH for larger transfinite ordinals

In order to generalize FGH for transfinite ordinals in my previous blog I proposed that
 * fI(x) is next Ω number after x for x < I

and Wythagoras in March 9, 2014 blog proposed that
 * fI(x) = Ωx

where I is least weakly inaccessible cardinal. That is in both cases fI increases cardinality in infinite x.

Increasing cardinality by large cardinal L
Yesterday I made up a better way (it is applicable not only to transfinite FGH, but also to my "booster-base" notation and, maybe, to other notations): fα(x) should not increase cardinality of infinite x, if α and x are less than some very large regular cardinal L L1. Least ordinal, increasing cardinality of infinite x < L, should be L itself:
 * fL(x) is next Ω number after x for x < L

Examples:
 * fL(ω) = Ω
 * fL(Ω) = Ω2
 * fL(Ω2) = Ω3

FGH with Ω numbers
How fα(x) for cof(α) = Ωβ + 1 should work?

If Ωβ = card(x) then
 * fα(x) = fα[x](x)

And if Ωβ > card(x) then
 * fα(x) = ff α(Ωβ) (x)

Other L cardinals
For x ≥ L fα(x) should not increase cardinality of x, if α and x are less than some very large regular cardinal L2 > L. Least ordinal, increasing cardinality of x such as L ≤ x < L2, should be L2 itself:
 * fL 2 (x) is next Ω number after x for L ≤ x < L2

Examples:
 * fL 2 (L) = ΩL + 1
 * fL 2 (ΩL + 1) = ΩL + 2
 * fL 2 (ΩL + 2) = ΩL + 3

Generally, for x ≥ Ln fα(x) should not increase cardinality of x, if α and x are less than some very large regular cardinal Ln + 1 > Ln. Least ordinal, increasing cardinality of x such as Ln ≤ x < Ln + 1, should be Ln + 1 itself:
 * fL n + 1 (x) is next Ω number after x for Ln ≤ x < Ln + 1

And for x < Ln
 * fL n + 1 (x) = ff L n + 1 (Ln) (x)

General idea
General idea: for Ln ≤ x < Ln + 1 (it can be considered that L0 = ω) α in fα(x) such as Ln + 1 ≤ cof(α) < Ln + 2 correspond to some function of x: α = Ln + 1 corresponds to increasing Ω (that is output is next Ω number after x); α = Ln + 12 corresponds to increasing I (that is output is next weakly inaccessible cardinal after x); etc.

Examples
I do not understand large cardinals very well, so I used Scorcher007' s ordinal list to explore them.
 * fL(ω) = Ω


 * fL2(ω) = I


 * fL3(ω) = I(2, 0)


 * fL4(ω) = I(3, 0)


 * fL5(ω) = I(4, 0)


 * fL2(ω) = I(1, 0, 0)


 * fL2 + L(ω) = I(1, 1, 0)


 * fL2 + L2(ω) = I(1, 2, 0)


 * fL2 + L3(ω) = I(1, 3, 0)


 * fL2 + L4(ω) = I(1, 4, 0)


 * fL2 + L5(ω) = I(1, 5, 0)


 * fL22(ω) = I(2, 0, 0)


 * fL22 + L(ω) = I(2, 1, 0)


 * fL22 + L2(ω) = I(2, 2, 0)


 * fL22 + L3(ω) = I(2, 3, 0)


 * fL22 + L4(ω) = I(2, 4, 0)


 * fL22 + L5(ω) = I(2, 5, 0)


 * fL23(ω) = I(3, 0, 0)


 * fL24(ω) = I(4, 0, 0)


 * fL25(ω) = I(5, 0, 0)


 * fL3(ω) = I(1, 0, 0, 0)


 * fL4(ω) = I(1, 0, 0, 0, 0)


 * fL5(ω) = I(1, 0, 0, 0, 0, 0)


 * fL 2 (ω) = ff L 2 (L) (ω) = fΩ L + 1 (ω) = M


 * fΩ L + 1 + L (ω) = M-I(1, 0)


 * fΩ L + 1 + L2 (ω) = M-I(2, 0)


 * fΩ L + 1 + L3 (ω) = M-I(3, 0)


 * fΩ L + 1 + L4 (ω) = M-I(4, 0)


 * fΩ L + 1 + L5 (ω) = M-I(5, 0)


 * fΩ L + 1 + L2 (ω) = M-I(1, 0, 0)


 * fΩ L + 1 + L3 (ω) = M-I(1, 0, 0, 0)


 * fΩ L + 1 + L4 (ω) = M-I(1, 0, 0, 0, 0)


 * fΩ L + 1 + L5 (ω) = M-I(1, 0, 0, 0, 0, 0)


 * fΩ L + 12 (ω) = M(1, 0)


 * fΩ L + 12 + L (ω) = M(1, α)-I(1, 0)


 * fΩ L + 12 + L2 (ω) = M(1, α)-I(2, 0)


 * fΩ L + 12 + L3 (ω) = M(1, α)-I(3, 0)


 * fΩ L + 12 + L4 (ω) = M(1, α)-I(4, 0)


 * fΩ L + 12 + L5 (ω) = M(1, α)-I(5, 0)


 * fΩ L + 12 + L2 (ω) = M(1, α)-I(1, 0, 0)


 * fΩ L + 12 + L3 (ω) = M(1, α)-I(1, 0, 0, 0)


 * fΩ L + 12 + L4 (ω) = M(1, α)-I(1, 0, 0, 0, 0)


 * fΩ L + 12 + L5 (ω) = M(1, α)-I(1, 0, 0, 0, 0, 0)


 * fΩ L + 13 (ω) = M(2, 0)


 * fΩ L + 14 (ω) = M(3, 0)


 * fΩ L + 15 (ω) = M(4, 0)


 * fΩ L + 12 (ω) = M(1, 0, 0)


 * fΩ L + 12 + ΩL + 1 (ω) = M(1, 1, 0)


 * fΩ L + 12 + ΩL + 2 (ω) = M(1, 2, 0)


 * fΩ L + 12 + ΩL + 3 (ω) = M(1, 3, 0)


 * fΩ L + 12 + ΩL + 4 (ω) = M(1, 4, 0)


 * fΩ L + 12 + ΩL + 5 (ω) = M(1, 5, 0)


 * fΩ L + 13 (ω) = M(1, 0, 0, 0)


 * fΩ L + 14 (ω) = M(1, 0, 0, 0, 0)


 * fΩ L + 15 (ω) = M(1, 0, 0, 0, 0, 0)


 * fΩ L + 2 (ω) = M(2; 0)


 * fΩ L + 2 + L (ω) = M(2; α)-I(1, 0)


 * fΩ L + 2 + L2 (ω) = M(2; α)-I(2, 0)


 * fΩ L + 2 + L3 (ω) = M(2; α)-I(3, 0)


 * fΩ L + 2 + L4 (ω) = M(2; α)-I(4, 0)


 * fΩ L + 2 + L5 (ω) = M(2; α)-I(5, 0)


 * fΩ L + 2 + L2 (ω) = M(2; α)-I(1, 0, 0)


 * fΩ L + 2 + L3 (ω) = M(2; α)-I(1, 0, 0, 0)


 * fΩ L + 2 + L4 (ω) = M(2; α)-I(1, 0, 0, 0, 0)


 * fΩ L + 2 + L5 (ω) = M(2; α)-I(1, 0, 0, 0, 0, 0)


 * fΩ L + 2 + ΩL + 1 (ω) = M(2; α)-M(1, 0)


 * fΩ L + 2 + ΩL + 12 (ω) = M(2; α)-M(2, 0)


 * fΩ L + 2 + ΩL + 13 (ω) = M(2; α)-M(3, 0)


 * fΩ L + 2 + ΩL + 14 (ω) = M(2; α)-M(4, 0)


 * fΩ L + 2 + ΩL + 15 (ω) = M(2; α)-M(5, 0)


 * fΩ L + 2 + ΩL + 12 (ω) = M(2; α)-M(1, 0, 0)


 * fΩ L + 2 + ΩL + 13 (ω) = M(2; α)-M(1, 0, 0, 0)


 * fΩ L + 2 + ΩL + 14 (ω) = M(2; α)-M(1, 0, 0, 0, 0)


 * fΩ L + 2 + ΩL + 15 (ω) = M(2; α)-M(1, 0, 0, 0, 0, 0)


 * fΩ L + 3 (ω) = M(3; 0)


 * fΩ L + 4 (ω) = M(4; 0)


 * fΩ L + 5 (ω) = M(5; 0)


 * fΩ L2 (ω) = M(1, 0; 0)


 * fΩ L2 + 1 (ω) = M(1, 1; 0)


 * fΩ L2 + 2 (ω) = M(1, 2; 0)


 * fΩ L2 + 3 (ω) = M(1, 3; 0)


 * fΩ L2 + 4 (ω) = M(1, 4; 0)


 * fΩ L2 + 5 (ω) = M(1, 5; 0)


 * fΩ L3 (ω) = M(2, 0; 0)


 * fΩ L4 (ω) = M(3, 0; 0)


 * fΩ L5 (ω) = M(4, 0; 0)


 * fΩ L2 (ω) = M(1, 0, 0; 0)


 * fΩ L3 (ω) = M(1, 0, 0, 0; 0)


 * fΩ L4 (ω) = M(1, 0, 0, 0, 0; 0)


 * fΩ L5 (ω) = M(1, 0, 0, 0, 0, 0; 0)

From this point I'm not sure. Yesterday I thought that fI L + 1 (ω) = K, but now I wonder: what is then fΩ Ω L + 1 (ω), fΩ Ω Ω L + 1   (ω), fΦ(1, L + 1)(ω)..? Anyway, let for now assume that fI L + 1 (ω) = K.


 * fL 22 (ω) = ff L 22 (L) (ω) = fI L + 1 (ω) = K


 * fI L + 1 + L (ω) = K-I(1, 0)


 * fI L + 1 + L2 (ω) = K-I(2, 0)


 * fI L + 1 + L3 (ω) = K-I(3, 0)


 * fI L + 1 + L4 (ω) = K-I(4, 0)


 * fI L + 1 + L5 (ω) = K-I(5, 0)


 * fI L + 1 + ΩL + 1 (ω) = K-M(1, 0)


 * fI L + 1 + ΩL + 12 (ω) = K-M(2, 0)


 * fI L + 1 + ΩL + 13 (ω) = K-M(3, 0)


 * fI L + 1 + ΩL + 14 (ω) = K-M(4, 0)


 * fI L + 1 + ΩL + 15 (ω) = K-M(5, 0)


 * fI L + 1 + ΩL + 2 (ω) = K-M(2; 1, 0)


 * fI L + 1 + ΩL + 3 (ω) = K-M(3; 1, 0)


 * fI L + 1 + ΩL + 4 (ω) = K-M(4; 1, 0)


 * fI L + 1 + ΩL + 5 (ω) = K-M(5; 1, 0)


 * fI L + 12 (ω) = K(1, 0)


 * fI L + 13 (ω) = K(2, 0)


 * fI L + 14 (ω) = K(3, 0)


 * fI L + 15 (ω) = K(4, 0)


 * fI L + 12 (ω) = K(1, 0, 0)


 * fI L + 13 (ω) = K(1, 0, 0, 0)


 * fI L + 14 (ω) = K(1, 0, 0, 0, 0)


 * fI L + 15 (ω) = K(1, 0, 0, 0, 0, 0)


 * fΩ I L + 1 + 1 (ω) = K~M(1; 0)


 * fΩ I L + 1 + 2 (ω) = K~M(2; 0)


 * fΩ I L + 1 + 3 (ω) = K~M(3; 0)


 * fΩ I L + 1 + 4 (ω) = K~M(4; 0)


 * fΩ I L + 1 + 5 (ω) = K~M(5; 0)


 * fI L + 2 (ω) = K(1; 0)


 * fI L + 3 (ω) = K(2; 0)


 * fI L + 4 (ω) = K(3; 0)


 * fI L + 5 (ω) = K(4; 0)


 * fI L2 (ω) = K(1, 0; 0)


 * fI L3 (ω) = K(2, 0; 0)


 * fI L4 (ω) = K(3, 0; 0)


 * fI L5 (ω) = K(4, 0; 0)


 * fI L2 (ω) = K(1, 0, 0; 0)


 * fI L3 (ω) = K(1, 0, 0, 0; 0)


 * fI L4 (ω) = K(1, 0, 0, 0, 0; 0)


 * fI L5 (ω) = K(1, 0, 0, 0, 0, 0; 0)


 * fL 23 (ω) = ff L 23 (L) (ω) = fI(2, L + 1)(ω) = K(2|1 0; 0)


 * fI(2, L + 2)(ω) = K(2|1 1; 0)


 * fI(2, L + 3)(ω) = K(2|1 2; 0)


 * fI(2, L + 4)(ω) = K(2|1 3; 0)


 * fI(2, L + 5)(ω) = K(2|1 4; 0)


 * fL 24 (ω) = ff L 24 (L) (ω) = fI(3, L + 1)(ω) = K(3|1 0; 0)


 * fL 25 (ω) = ff L 25 (L) (ω) = fI(4, L + 1)(ω) = K(4|1 0; 0)


 * fL 26 (ω) = ff L 26 (L) (ω) = fI(5, L + 1)(ω) = K(5|1 0; 0)


 * fL 22 (ω) = ff L 22 (L) (ω) = fI(1, 0, L + 1)(ω) = K(1, 0|1 0; 0)


 * fL 23 (ω) = ff L 23 (L) (ω) = fI(1, 0, 0, L + 1)(ω) = K(1, 0, 0|1 0; 0)


 * fL 24 (ω) = ff L 24 (L) (ω) = fI(1, 0, 0, 0, L + 1)(ω) = K(1, 0, 0, 0|1 0; 0)


 * fL 25 (ω) = ff L 25 (L) (ω) = fI(1, 0, 0, 0, 0, L + 1)(ω) = K(1, 0, 0, 0, 0|1 0; 0)


 * fL 3 (ω) = ff L 3 (L2) (ω) = fΩ L 2 + 1 (ω) = ff Ω L 2 + 1  (L) (ω) = fM L + 1(L) (ω) = K[2]


 * fL 4 (ω) = fΩ L 3 + 1 (ω) = fM L 2 + 1 (L) (ω) = fK[2] L + 1(L) (ω) = K[3] (maybe K[4], not K[3]: what is then fK L + 1(L) (ω)..?)


 * fL 5 (ω) = fΩ L 4 + 1 (ω) = fM L 3 + 1 (L) (ω) = fK[2] L 2 + 1 (L) (ω) = fK[3] L + 1(L) (ω) = K[4]


 * fL ω (ω) = K[ω] = Ξ


 * fL Ξ (ω) = K[Ξ]


 * fL K[Ξ] (ω) = K[K[Ξ]]


 * fL K[K[Ξ]]|undefined (ω) = K[K[K[Ξ]]]


 * fL K[K[K[Ξ]]]|undefined (ω) = K[K[K[K[Ξ]]]]


 * fL K[K[K[K[Ξ]]]]|undefined (ω) = K[K[K[K[K[Ξ]]]]]


 * fL K[K[K[K[K[Ξ]]]]]|undefined (ω) = K[K[K[K[K[K[Ξ]]]]]]


 * fL L (ω) = K[K[K[K[K[...]]]]] = ϒ = fL ϒ (ω)

So, this notation may go up to Large Stegert ordinal (#24 in "few more jumps" table of Scorcher007's list of ordinals) and even beyond (if this analysis is correct).

FGH with I numbers
So, how fα(x) for cof(α) = Iβ + 1 should work?

Let
 * x < L

If Iβ + 1 is next I number after x then
 * Iβ + 1 = fL2(x)

Let
 * α = cof(α) = fL2(x)

then
 * fα(x) = ff L2(x) (x)

Let's define that
 * ff L2(x) (x) = ff L + x(x) (x)

And now general case (that is maybe α ≠ cof(α)):
 * fα(x) = fα[f L + x(x)] (x)

For Ln ≤ x < Ln + 1 "L" should be replaced with "Ln + 1":
 * fα(x) = fα[f L n + 1 + x (x)] (x)

And if Iβ + 1 is larger than next I number after x then
 * fα(x) = fα[f α(Iβ)] (x)

cof(α) = fβγ + 1(x), cof(β) = Ln + 1
Now more general case for cof(α) = fβ(x), cof(β) = L, Ln ≤ x < Ln + 1:
 * fα(x) = fα[f β[x](x)] (x)

For γ > 0:
 * fα(x) = fα[f α(fβγ(x))] (x)

FGH with M numbers

 * α = fL 2 (x) = ff L 2 (L) (x) = fΩ L + 1 (x)


 * fα(x) = ff Ω L + 1 (x) (x) = ff f L 2 (L) (x) (x) = ff f x(L) (x) (x)