User blog comment:Nayuta Ito/faketest/e0/@comment-30754445-20180804174000/@comment-32697988-20180805101605

@PsiCubed2

Ok. I will continue this.

Again ψ is normal OCF and U is UNOCF.

ψ(ψ1(0))=ψ(εΩ+1)=U(Ω2)

ψ(ψ1(1))=ψ(εΩ+2)=U(Ω2×2)

As you said, an additional Ω2 in UNOCF increases the α in ψ(ψ1(α)) by 1.

ψ(ψ1(2))=ψ(εΩ+3)=U(Ω2×3)

ψ(ψ1(3))=ψ(εΩ+4)=U(Ω2×4)

ψ(ψ1(ω))=ψ(εΩ+ω)=U(Ω2×ω)

In general, for countable α<ψ(εΩ×2),

ψ(ψ1(α))=ψ(εΩ+1+α)=U(Ω2×(1+α)).

So,

ψ(ψ1(Ω))=ψ(εΩ×2)=U(Ω2×Ω).

Then,

ψ(ψ1(Ω+1))=ψ(εΩ×2+1)=U(Ω2×(Ω+1))

ψ(ψ1(Ω+ω))=ψ(εΩ×2+ω)=U(Ω2×(Ω+ω))

ψ(ψ1(Ω×2))=ψ(εΩ×3)=U(Ω2×Ω×2)

ψ(ψ1(Ω×ω))=ψ(εΩ×ω)=U(Ω2×Ω×ω)

ψ(ψ1(Ω2))=ψ(εΩ2)=U(Ω2×Ω2)

A pattern is found here: ψ(ψ1(α)) corresponds to ψ(εα) and U(Ω2×α). (But this is not true for some α)

Using this pattern,

ψ(ψ1(Ωω))=ψ(εΩω)=U(Ω2×Ωω)

ψ(ψ1(ΩΩ))=ψ(εΩΩ)=U(Ω2×ΩΩ)

ψ(ψ1(ψ1(0)))=ψ(εε Ω+1 )=U(Ω2×U1(Ω2))

ψ(ψ1(ψ1(1)))=ψ(εε Ω+2 )=U(Ω2×U1(Ω2×2))

ψ(ψ1(ψ1(ω)))=ψ(εε Ω+ω )=U(Ω2×U1(Ω2×ω))

ψ(ψ1(ψ1(Ω)))=ψ(εε Ω×2 )=U(Ω2×U1(Ω2×Ω))

ψ(ψ1(ψ1(ψ1(0))))=ψ(εε ε Ω+1 )=U(Ω2×U1(Ω2×U1(Ω2)))

And repeating ψ1 makes Ω2 in normal OCFs, so:

ψ(Ω2)=ψ(ζΩ+1)=U(Ω22).