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On 2/27/2025 5:01 PM, WM wrote:
Zermelo's approach does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)It does. Zermelo says it, and it is easy to prove it:
How is Z accomplished?By Zermelo's approach:Zermello's Infinity guarantees a superset Z of Z₀>
How is that accomplished?
Z ⇒ 𝒫(Z)
It's not.even.wrong.Where does he get is Z from?
And it's not Zermelo's approach,
Z
{ } and a ==> {a}.His Z is ensured by induction.Nope.
No.We don't even need the intersectionConsider Robinson arithmetic.
if we reduce Zermelo's approach to
Lorenzen's approach:
I is a natural number, and
if x is a natural numbers then x+1 is a natural number.
Proofs by induction are unreliable in Robinson arithmetic.Irrelevant.
Wherever you got Lorenzen's approach from,Then you admit to stand outside of mathematics. Lorenzen uses the same induction as Cantor, Dedekind, Peano, Schmidt, Zermelo, or v. Neumann.
send it back.
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