Re: The set of necessary FISONs

On 2/12/2025 4:19 AM, WM wrote:

On 12.02.2025 01:38, Jim Burns wrote:

On 2/11/2025 2:23 PM, WM wrote:

What is wrong in my application in your opinion?

>
You (WM) currently are ignoring that,
for each two FISON.numbers j′ and i′
there exists a FISON.number maximum k′ of
i′ and the successor j′+1 of j′

∀ᴺj:∀ᴺi:
   ∃ᴺk = max{i,j+1}
∀⁽ꟳ⁾Fⱼ:∀⁽ꟳ⁾Fᵢ:
   ∃⁽ꟳ⁾Fₖ = ⋃{Fᵢ,Fⱼ∪{Fⱼ}}

which means
you ignore that
for each FISON F'
the union of FISONs.after F' are equal.
which contradicts U{F} = U{}

>
Proofs by induction have no reason to observe your claim.

Modal fallacies do not obey
∀ᴺj:∀ᴺi: ∃ᴺk = max{i,j+1}
but then,
they're _fallacies_

If Zermelo's induction is valid for
his set Z including the set ℕ,
then my proof is  valid for the set F too.

𝒫ⁱⁿᵈ(X)  =  the set of inductiveᶻ subsets of set X
𝕀 is inductiveᶻ  ⇔  0∈𝕀 ∧ ∀a:a∈𝕀⇒{a}∈𝕀
𝕄 is minimal.inductiveᶻ  ⇔  𝒫ⁱⁿᵈ(𝕄) = {𝕄}
A proof by induction:
⎛
⎜⎛ Insert lemma that P(0) and that,
⎜⎝ for each k ∈ 𝕄, P(k) ⇒ P({k})
⎜
⎜ Consider {n∈𝕄:P(n)}
⎜
⎜⎛ Assume k ∈ {n∈𝕄:P(n)}
⎜⎜⎛ k ∈ 𝕄
⎜⎜⎝ {k} ∈ 𝕄
⎜⎜⎛ P(k)
⎜⎜⎝ P({k})
⎜⎝ {k} ∈ {n∈𝕄:P(n)}
⎜
⎜ ∀k: k ∈ {n∈𝕄:P(n)} ⇒ {a} ∈ {n∈𝕄:P(n)}
⎜
⎜ P(0)
⎜ 0 ∈ {n∈𝕄:P(n)}
⎜
⎜ 𝕄 is minimal.inductiveᶻ
⎜ 𝒫ⁱⁿᵈ(𝕄) = {𝕄}
⎜ {n∈𝕄:P(n)} is inductiveᶻ
⎜ {n∈𝕄:P(n)} ⊆ 𝕄
⎜ {n∈𝕄:P(n)} ∈ 𝒫ⁱⁿᵈ(𝕄)
⎜ {n∈𝕄:P(n)} ∈ {𝕄}
⎜ {n∈𝕄:P(n)} = 𝕄
⎝ ∀n∈𝕄:P(n)