Re: how

On 6/10/2024 9:33 AM, WM wrote:

Le 08/06/2024 à 20:16, Jim Burns a écrit :

On 6/8/2024 8:42 AM, WM wrote:

If j ∈ Defble then j^j^j ∈ Defble.
Nevertheless j^j^j^j^j is finite, but
there are ℵo undefinable natural numbers.

>
No.

>
Then define natural numbers with
less or no undefinable successors.

Proposal 1.
Definitions are only
statements of _what the definer means_
Without evidence to the contrary,
the definer is presumed to be
honest and aware of what they mean,  and
definitions are presumed to be
true statements of what they mean.
On questions beyond what the definer means,
their definitions do not receive
a presumption of truth,
but they remain free to argue their POV.

There are 0 first undefinable natural numbers.
There are 0 undefinable natural numbers.

There are 0 natural numbers with
fewer numbers.after than its predecessor.
There are 0 natural numbers which are
first with fewer numbers.after than 0
There are 0 natural numbers which are
with fewer numbers.after than 0
----
A natural number with fewer numbers.after than 0
can be _defined_
Under proposal 1,
it's presumed that the definer honestly means that,
but no more than that is presumed.
Everything else is what arguments are for.
For example,
from EAX (Empty Adjunct eXtensionality)
there's an argument that
the natural number defined with fewer numbers.after than 0
not.exists.

This could only be disproved by defining them.

>
No.
It has been disproved by
giving a description of an individual number j in ℕ⁺

>
j is only collectively.
Proof:
You can replace it by every individual like 4711 or 10​^10​^10000.

Whatever we learn
by finite only not.first.false claim.sequence about j
we learn about 4711 and 10​^10​^10000.