User blog comment:Alemagno12/An extremely fast-growing OCF/@comment-5029411-20170726203426/@comment-25601061-20170727011017

I am not done defining this. Also, L(ω) is well-defined.

L(x) for x<ω = Ωx, and since ω[x] = x, ΨLL(ω)(x) for x<ω = L(x) = Ωx.

To get from ω[n] to ω[n+1], we just add 1 to ω[n]. So ΨLL(ω)(x+1) = Ωx+1, and for limit ordinal x, ΨLL(ω)(x) = sup(ΨLL(ω)(x[1]),ΨLL(ω)(x[2]),ΨLL(ω)(x[3]),...) = sup(Ωx[1],Ωx[2],Ωx[3],...)

So ΨLL(ω)(x) = Ωx, and L(ω) is the omega-limit of ΨLL(ω)(x) (which is the same Ωx) as for all x, which is the same as I.