Re: The set of necessary FISONs

On 2/8/25 5:27 AM, WM wrote:

On 08.02.2025 01:14, Richard Damon wrote:

On 2/7/25 11:39 AM, WM wrote:

On 07.02.2025 17:10, Jim Burns wrote:

On 2/6/2025 2:32 PM, WM wrote:

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I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).

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That's intended to be part of the definition of ℕ₁

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As well it is
the definition of the collection of all FISONs.

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Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.

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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.

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So all elements are individually not needed.

 All not needed elements belong to an inductive set. Inductive sets have no last element. Therefore the set of FISONs which could have the union ℕ has no first element. That means it is empty.
 Therefore U(A(n)) = ℕ ==> U{ } = { } = ℕ. This is false. By contraposition we get ~{ } = ℕ ==> ~ U(A(n)) = ℕ.
 Regards, WM

And thus you claim that 36 can not be factored, as all of its factors are not needed.
Sorry, you logic is based on misunderstanding.
You just are incapable of understanding what you are talking about.