Re: The set of necessary FISONs

On 2/14/2025 11:40 AM, WM wrote:

On 14.02.2025 16:42, Jim Burns wrote:

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

Induction creates infinite sets.

>
Zermelo set theory [I...VII] describes
a domain of discourse with
Z an inductiveᶻ set

>
Where all elements are covered by induction.

No,
we don't know that, in inductiveᶻ Z,
each element is covered by inductionᶻ.
However,
we know that
Z has a subset ⋂𝒫ⁱⁿᵈ(Z) of which
we know that, in minimal.inductiveᶻ ⋂𝒫ⁱⁿᵈ(Z)
each element is covered by inductionᶻ
Being covered by induction is
being in the only.inductive.subset.
We don't know Z has an only.inductiveᶻ.subset.
We know ⋂𝒫ⁱⁿᵈ(Z) has an only.inductiveᶻ.subset.
The set ⋂𝒫ⁱⁿᵈ(Z) of all elements in ⋂𝒫ⁱⁿᵈ(Z)
is not.in only.inductiveᶻ.subset ⋂𝒫ⁱⁿᵈ(Z)
⋂𝒫ⁱⁿᵈ(Z) is not covered by inductionᶻ.

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.

>
And no element is outside.

And then
no element of the minimal.inductive set
is outside of the set of all n such that x(n)
And then, we know that,
for infinitely.many n in the minimal.inductive set,
x(n)
This reasoning only applies to
elements of the minimal.inductive set.
The minimal.inductive set is not
an element of the minimal.inductive set.
This reasoning does not apply to it.

There is no set of all FISON.numbers in ST+F.

>
But it is the result of the proof.

Your alleged proof takes an unjustified leap
from "each FISON is omissible"
to "each set of omissible FISONs is ommissible".
Each FISON is.
Some sets are and some sets aren't.
{F} is a set of omissible FISONs.
{F} isn't an omissible set of FISONs.
Without the leap, there is no conflict.