Re: The set of necessary FISONs

On 2/9/25 9:48 AM, WM wrote:

On 09.02.2025 13:34, joes wrote:

Am Sat, 08 Feb 2025 22:54:46 +0100 schrieb WM:

On 08.02.2025 18:43, Jim Burns wrote:
>

Do you use ω-1 ?

>
Not in this proof.:
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.

Caveat: P is *not* satisfied by the set of all k, only by its elements.

 Peano is not satisfied by the set of all k either. But no exceptioncan be identified.
 Regards, WM
  

Peano being "satisfied" is a misuse of terminology.
Peano is a set of axioms (actually used to represent one of several diffent sets of them), which when we assume to be true we get the results of it.
If you don't accept the axioms, you don't get its results.
Note, if you mean the conditions of the induction axiom, you need to make sure you talk in the right terms.
And all that does is say that the set of all FISONs is a set where all the members are individually not needed to build a set whose union is the set of Natual Numbers.
There is nothing in the theory to let you jump from a property of the members to a property of the set as a whole, so your logic just breaks.