Re: The set of necessary FISONs

On 2/15/25 6:31 AM, WM wrote:

On 14.02.2025 18:32, joes wrote:
 

You claim that UF = X and also U({}) = X.

 No. You and others have claimed that there is a set of FISONs containing all natural numbers. I have accepted that a s premise.
 

If UF=N and N={}, then UF={}.

 No, learn what an implication is! In W. Mückenheim: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015) for instance it is explained for beginners.
 There is no FISON that is in the claimed set. Therefore
If UF = ℕ then ℕ = {}.
 This does *not* prove UF = {}. F contains sets and UF contains natural numbers, but not enough natural numbers, not ℕ.
 Regards, WM

And since you build your claimed "set" by Naive Set theory, your sets are worthless.
The set you show is empty, is the set of FISONs that are individually not necessary.
TO say that being empty says that you can't take the union of FISONs to bet to the set of Natural Numbers also says that we can not factor 36, as none of the factors we might use are actually requried.
Sorry, you are just proving that you don't know what real logic is.