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Am Sun, 02 Mar 2025 18:51:05 +0100 schrieb WM:That is allowed if Zermelo csn do it.On 02.03.2025 13:22, Richard Damon wrote:Ah no, you're shifting the quantifier there.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:∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))On 28.02.2025 20:45, Richard Damon wrote:You have a source that uses the term "Induction" for the recursiveOn 2/28/25 11:44 AM, WM wrote:>Correct your qualifiers. Or look it up.No, Zermelo uses induction. I did not say that he uses the termNo, what you describe is NOT "induction".
induction.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite
inductive set F of FISONs.
iteration that builds the set?
Wikipedia>P gives a relationship, for instance "is element of F". The universal
And what SET did that build? P was a statement about a relationship.
quantifier ∀ proves that all FISONs belong to the set that can be
removed.
There are many sets ofOf course all FISONs are finite. All FISONs of Zermelo's set are finite. By induction Zermelo produces the set. Quantifier shift, obviously allowed by Zermelo. There is no other way to construct an infinite set.
FISONs that can be removed, one set per natural, containing all
FISONS of numbers less than that natural (also more sets of non-
contiguous FISONs). But those are all finite.
The set of sets ofIn exactly the same way as Z₀ is constructed by its elements, the set of removable FISONs is constructed by its elements.
FISONs that can be removed together does not contain the set of
all FISONs (although it does contain the infinite sets of the odd
or even FISONs).
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