User blog comment:Rgetar/Idea for FGH for larger transfinite ordinals/@comment-35470197-20190702232339/@comment-32213734-20190707084102

I wrote a program for FGH-based ordinal notation for cofinalities ≤ Ω (or rather I modified my previous program Ordinal Explorer v3.1 and created Ordinal Explorer v4.0). The program works, but expansion are often too large (I got the idea of how to reduce them). I also suspected that my system of fundamental sequences is "cut" (in my terminology), I checked it, and no, it is not "cut". But it is ok. (If it were "cut", then comparison with lesser elements of the list during expansions would not be needed). Also I redefined superscript of f: now
 * fα(x) = fα0(x)

instead of
 * fα(x) = fα1(x)

Also I got an idea concerning well-foundness (but I failed). Let S0 is well-founded set. Let FGH expression is an element of S0 or fαβ(x), where α, β, x are FGH expressions.

General form of an FGH expression is
 * fα k βk(...(fα 2 β2(fα 1 β1(x)))...)

where x, αi, βi are FGH expressions, 0 ≤ k < ω. Comparison:
 * fαβ(x) > x for any α, β, x

For other cases:
 * expression with larger x is larger
 * if x are equal, then expression with larger α1 is larger
 * if α1 are equal, then expression with larger β1 is larger
 * if β1 are equal, then expression with larger α2 is larger
 * if α2 are equal, then expression with larger β2 is larger
 * if β2 are equal, then expression with larger α3 is larger

Set of all FGH expressions is not well-founded. Example: let
 * 0, 1 ∈ S0

Infinite descending chain:
 * f10(0)
 * f10(f00(0))
 * f10(f00(f00(0)))
 * f10(f00(f00(f00(0))))
 * f10(f00(f00(f00(f00(0)))))

Let Si + 1 is set of all FGH expressions such as
 * x, αi, βi ∈ Si

and
 * α1 > α2 > ... > αk

Let S is union of Si for all natural i. I proved that Si are well-founded, and I thought that S is well-founded too, but then I proved that S is not well-founded.

(Union of well-founded sets may be not well-founded. Example: let S is union of Si for all natural i, where Si is set of all integers ≥ -i. Si are well-founded, but S is not).

This is because a descending chain of element of S may be not a subset of any Si.

So, other conditions are needed. For example, if all elements of Si + 1, which are not elements of Si, are larger than Si, then any descending chain of an element of S is subset of some Si, so, S will be well-founded. But I quess that this condition is not very useful.