Re: The set of necessary FISONs

On 3/6/2025 12:05 PM, WM wrote:

On 06.03.2025 12:08, Jim Burns wrote:

On 3/6/2025 3:54 AM, WM wrote:

Am 06.03.2025 um 02:52 schrieb Jim Burns:

<<JB<WM>>>

>
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.

>
No.

<</JB<WM>>>
>
No BECAUSE
that doesn't DEFINE Z₀

>
It defines Z₀ precisely.

It does not exclude Bob.
In order to exclude Bob,
some clause must state or imply Bob isn't in Z₀
Induction says what's IN Z₀
More description is needed.

What does your inductive description permit in Z₀
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀

>
Induction excludes Bob.

I generalize 'end segment':
For all x, end.segment E(x) is
  the _emptiest_ set [more description!]
which holds x and which,
for each a in E(x), holds {a}
E(x) = {x,{x},{{x}},...}
⎛ I don't restrict x to being in
⎜  ℕ == its.own.only.inductive.subset
⎜ but if x isn't in
⎜  ℕ == its.own.only.inductive.subset
⎜ then nothing in E(x) is in
⎝  ℕ == its.own.only.inductive.subset
E(Bob) = {Bob,{Bob},{{Bob}},...}
∀a: a ∈ E(Bob) ⇒ {a} ∈ E(Bob)
E(Bob) is not inductive, but
only because {} ∉ E(Bob)
However,
E({})∪E(Bob) is inductive
because {} ∈ (E({})∪E(Bob))
A 'proof by induction' is not reliable
for E({})∪E(Bob)
even though E({})∪E(Bob) is inductive.
There are inductive predicates,
such as K(x) == 'is never King of England'
which have exceptions in E({})∪E(Bob)
https://www.youtube.com/watch?v=TjAg-8qqR3g
An unreliable argument is unacceptable.
E({})∪E(Bob) is unacceptable for proofs by induction,
because
E({})∪E(Bob) and E({}) are both
inductive subsets of E({})∪E(Bob)
which makes proof by induction unreliable.

Induction excludes Bob.

Induction alone does not exclude Bob.

It defines Z₀ precisely.

Induction alone describes both
Z₀ = E({}) and E({})∪E(Bob)

He is not in the empty set and
not in the set containing the empty set and
not in any set with more curly brackets.

Bob ∉ E({}) = Z₀ ('emptiest')
Bob ∈ E({})∪E(Bob)
{} ∈ E({})∪E(Bob)
∀a: a ∈ E({})∪E(Bob) ⇒ {a} ∈ E({})∪E(Bob)
Induction alone does not exclude Bob.