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Noope.
Read again.
It's wrong.
ψ(ψI(0)) is FAR greater than ω.
And fω+1(G) beats your ''fψ(ψI(0))(G)''
The fundamental sequences for limit ordinals can produce larger values than fundamental sequences for smaller limit ordinals.
Really.
My point is that they don't necessarily
have to, because the definition of f depends on the choice of all your fundamental sequences (which need not be related at all), and since the
only requirement for a fundamental sequence for a limit ordinal is that it strictly increases and converges to that limit ordinal, you can very well choose such sequences for ω and ψ(ψ
I(0)) such that the first few elements are the same, e.g. ψ(ψ
I(0))[G] = G and ω[G] = G.
Now remember the definition of fast-growing hierarchy, which says that if α is a limit ordinal, f
α(n) is defined as f
α[n](n). So in this case, f
ω+1(G) = f
ωG(G) > f
ω(G) = f
ω[G](G) = f
G(G) = f
ψ(ψI(0))[G](G) = f
ψ(ψI(0))(G).
Now as I said before, different choices of fundamental sequences lead to different fast-growing hierarchies, and it may very well be possible (for appropriate choice of the fundamental sequences) for your number to be larger than Fallen's. But until you provide a choice of fundamental sequence for all the limit ordinal numbers up to ψ(ψ
I(0)), your number is not even well-defined, as opposed to Fallen's perfectly well-defined A
AG(G)(G).
My number is Ω_A(NiB), nested NiB times.
Well, I hereby define a function Z that takes an integer and adds the imaginary unit. Now your definition of A(x) is broken forever, because chained arrows don't operate on complex numbers. So there.