Re: The set of necessary FISONs

On 2/4/2025 5:11 AM, WM wrote:

On 03.02.2025 20:48, Jim Burns wrote:

On 2/3/2025 1:36 PM, WM wrote:

Therefore Peano, Zermelo, or v. Neumann
create ℕ as well as the set of all FISONs
for use in set theory.

>
Axioms describe.
Magic spells create.

>
To describe something
it must be existing.

To describe something existing,
it must be existing.
Describing what doesn't exist doesn't create it.
Most descriptions describe what doesn't exist.
We (mostly) avoid descriptions of what doesn't exist
by Herculean effort (Hercules exists?).
Sometimes we fail at
avoiding descriptions of what doesn't exist.
When that happens, we've described what doesn't exist.

If ℕ is existing,
we do not need axioms.

If ℕ had a contradictory description,
for example,
"shrinkable.by.one minimal.inductive set"
then ℕ would not exist.
We avoid contradictory descriptions for that reason,
but the descriptions exist.
See?
An existing description of a not.existing
shrinkable.by.one minimal.inductive set.
Describing it did not create it.

If ℕ is existing,
we do not need axioms.

If ℕ is described,
its description can have appended to it
not.first.false claims,
such as Q(k) is in ⟨ P(k) P(k)⇒Q(k) Q(k) ⟩
Because we see Q(k) preceded by P(k) and P(k)⇒Q(k)
we know that Q(k) is not.first.false.
Even if Q(k) is preceded by a false description,
which we know is false,
we know (we see) that Q(k) is not.first.false.
'Not.first.false' is NOT the same as 'true'.
True is what each claim in
  a finite sequence of
  only not.first.false claims
is.
For a finite sequence holding false descriptions,
that sequence is NOT only not.first.false.
For other claims in it,
perhaps they're true, perhaps they're false,
even the not.first.false claims.

If ℕ is existing,
we do not need axioms.

If the axioms are contradictory,
ℕ is not existing.
If you and I are thinking of different ℕ
axioms will make that difference evident.
If the axioms aren't contradictory
and you and I aren't thinking of different ℕ
axioms provide the claims P(k) and P(k)⇒Q(k)
to which we can append Q(k),
see the sequence is all not.first.false,
and know that Q(k) is also true.