Re: The set of necessary FISONs

On 2/18/2025 3:54 AM, WM wrote:

On 17.02.2025 20:59, Jim Burns wrote:

On 2/17/2025 2:27 PM, WM wrote:

Proofs by induction cover all FISONs.

>
Proofs by induction prove that
some property A(k) describes each element of
some inductive

>
set.

some inductive subset of
an inductive set with an only.inductive.subset.
In your (WM's) posts, you try to show that
our (matheologists') reasoning is incorrect.
You would find your posts greatly improved
by trying to prove _our_ reasoning, not
what you mistakenly think is ours,
is incorrect.
For inductive sets with multiple.inductive.subsets,
it's inadequate to prove that a subset is inductive
in order to conclude that subset is the whole set.
Therefore, we matheologists do not conclude that,
for inductive sets with multiple.inductive.subsets.
For inductive sets with an only.inductive.subset,
proving that a subset is inductive is sufficient.
Therefore, we conclude that.

Here this property is that
FISON F(n) can be removed without changing the premise
U({F(1),F(2),F(3),...}\{F(1),F(2),...,F(n)}) = ℕ.
 (*)

The subset of {F} with that property
is inductive.
Showing that a subset is inductive
is called proof by induction.
{F} is the only.inductive.subset of {F}.
In the case of {F}, a proof by induction shows
that any subset of {F} with that property is {F},
because that subset can't be anything else..
That's shown in the rest of the proof,
a part that's rarely seen, because,
if we're shown once that {F} is
an inductive set with an only.inductive subset,
we will continue to know that about {F}.

That reasoning is silent about
whether the _set_ (not its elements) has A(k).

That reasoning is silent about
whether the _set_ (not its elements) has A(k).

Therefore I gave you an example
that you should be able to understand:
If every human has ended,
then the human race has ended.

Imagine
the baton Bob is passed from human to human,
through all time and all space. To every human.
Imagine Philip José Farmer's Riverworld.
Whatever holds Bob, ever, is a human.
No human is the human race.
There is no Bob.holder human race.
Each FISON holds Bob.
The set of Bob.holders does not hold Bob.

Analogously:
If every FISON has been removed
without changing the union, then the set
{F(1), F(2), F(39, ...}
has been removed without changing the union.

Aᴬᴬᴬ(F) == "The FISON are Almost All After F"
The set {F:Aᴬᴬᴬ(F)} of almost.all.after FISONs
is
the set {F:omissible} of FISONs which,
removed from {F} with its priors,
leaves the union unchanged.
Almost all FISONs are NOT after {F}
None are.
For that and many other reasons,
{F} is not.in {F:Aᴬᴬᴬ(F)}
{F} is not.in {F:omissible}
{F} is not.in {F}
⎛ ∀F′ ∈ {F}: ⋃{F:F′<F} = ⋃{F}
⎝ ⋃{F:{F}<F} ≠ ⋃{F}
is not a contradiction.