Re: The set of necessary FISONs

On 2/14/2025 7:52 AM, WM wrote:

On 13.02.2025 17:38, Jim Burns wrote:

On 2/12/2025 5:19 AM, WM wrote:

On 12.02.2025 10:46, joes wrote:

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Induction proves all elements and
even the set (Zermelo, Z).

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No.
Not the set.
Each member of the set.

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All members of the set.

The erroneous step is from „every finite number”
to „an infinite number”.

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Induction proves an infinite number.

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Induction is valid without any infinite sets.

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Induction creates infinite sets.

Zermelo set theory [I...VII] describes
a domain of discourse with
Z an inductiveᶻ set [VII]
𝒫(Z) set of subsets [IV]
𝒫ⁱⁿᵈ(Z) set of inductiveᶻ subsets [III]
⋂𝒫ⁱⁿᵈ(Z) intersection of inductiveᶻ subsets [III]
https://en.wikipedia.org/wiki/Zermelo_set_theory
⋂𝒫ⁱⁿᵈ(Z) is the minimal.inductiveᶻ set.
The only inductiveᶻ subset of ⋂𝒫ⁱⁿᵈ(Z) is ⋂𝒫ⁱⁿᵈ(Z)
Proving A is any inductive subset of ⋂𝒫ⁱⁿᵈ(Z)  is
proving A is the only inductive subset of ⋂𝒫ⁱⁿᵈ(Z), is
proving {n∈⋂𝒫ⁱⁿᵈ(Z):A(n)} = A is ⋂𝒫ⁱⁿᵈ(Z), is
proving ∀ᐢᴾⁱⁿᵈ⁽ᶻ⁾n: A(n)
...which is why
a proof by induction,
⎛ proving A is any inductive subset of
⎝ the minimal.inductive set
is
a proof.
... is a proof of
A(n) for all infinitely.many n in ⋂𝒫ⁱⁿᵈ(Z)
and not a proof of
A(⋂𝒫ⁱⁿᵈ(Z)) for infinite ⋂𝒫ⁱⁿᵈ(Z)

If i contruct x(1) and from x(n) follows x(n+1)
then the set of all x(n) is infinite.

then the set of all x(n) is
an inductive subset of the minimal.inductive set.

In ST+F, there are no infinite sets,
but there is induction.

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I apply mathematics.

In ST+F, each FISON exists.
If any counter.example FISON.number exists, ¬A(n)
then,
in the FISON ended by n
the first counter.example FISON.number exists
¬A(0) or ¬A(k+1) and ∀j≤k:A(j)
Induction follows not.first.falsely
for each FISON.number.
There is no set of all FISON.numbers in ST+F.