Re: How many different unit fractions are lessorequal than all unit fractions?

Am 13.10.2024 um 00:46 schrieb joes:

Am Sat, 12 Oct 2024 18:51:10 +0200 schrieb WM:

On 10.10.2024 21:51, joes wrote:

Am Wed, 09 Oct 2024 15:24:21 +0200 schrieb WM:

On 09.10.2024 14:38, joes wrote:

Am Wed, 09 Oct 2024 10:11:20 +0200 schrieb WM:

Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.

Not a theorem, WM nonsense.
Using symbols:
     (AS e ENDSEG: E^oo n e IN: n e S) ->
     (E^oo n e IN: AS e ENDSEG: n e S)

Invalid quantifier shift.

Yes. @Mückenmann: See: https://en.wikipedia.org/wiki/Quantifier_shift

Valid quantifier shift.

lol. There is no such thing, Mückendepp!

You haven't proved your logical rule in general. In particular, you
can't use it here to argue it is valid, that would be circular.

Indeed.
Actually, this case is a counterexample concerning that "rule".
(AS e ENDSEG: E^oo n e IN: n e S) is true (easy, by definition)
but
(E^oo n e IN: AS e ENDSEG: n e S),
since
(En e IN: AS e ENDSEG: n e S)
already is false. [...]

I do not intend to prove quantifier shift in general. I prove: If every
endsegment is infinite, then infinitely many numbers are in all
endsegments [...]

It seems that Mückenheim does not know the difference between an unproven claim and a proof for this claim. :-)

Then you need to explain why the shifted proposition should be true.

Right.

[...] If every endsegment has an infinite subset, then
there exists one and the same infinite subset of every endsegment.

Oh, certainly not.
Actually, that's just a quantifier shift AGAIN. :-)

Proof: Inclusion monotony.

Oh, a single word/notion is a "proof" in Mückenheims world - fascinating! (Not!)

Not valid for [...]

Actually, just NONSENSE.