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On 02.03.2025 13:22, Richard Damon wrote:Which proves that P is true for all Natural Numbers.On 3/1/25 1:09 PM, WM wrote:Induction is the nucleus: P(0) /\ ∀k(P(k) ==> P(k+1)On 01.03.2025 17:25, Richard Damon wrote:On 3/1/25 9:31 AM, WM wrote:>On 28.02.2025 20:45, Richard Damon wrote:>On 2/28/25 11:44 AM, WM wrote:>>No, Zermelo uses induction. I did not say that he uses the term induction.>
No, what you describe is NOT "induction".
Correct your qualifiers. Or look it up.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite inductive set F of FISONs.
You have a source that uses the term "Induction" for the recursive iteration that builds the set?
∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
Wikipedia
No, it means the set of REQUIRED FISONs is empty, not that we can't use a set of FISONs to make N.>P gives a relationship, for instance "is element of F". The universal quantifier ∀ proves that all FISONs belong to the set that can be removed.
And what SET did that build? P was a statement about a relationship.
But you didn't. You showed that there are no specific FISON requried to be in that set. You didn't show that you couldn't use FISONs to make the set.You can try to claim that the system it creates is inconsistent, but to do so you need to show the error using the system as defined.I did: UF = ℕ ==> Ø = ℕ.
Regards, WM
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