Re: Implementable Set Theory and Consistency of ZFC

In article <6137b$471ef574$82a1e228$22995@xxxxxxxxxxxxxxxx>,
Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> wrote:

Jesse F. Hughes wrote:

Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> writes:

Jesse F. Hughes wrote:

Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> writes:

Jesse F. Hughes wrote:

Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> writes:

MoeBlee wrote:

Therefore it's
wise to snip the "Axiom of Infinity" section from the article, as I've
actually done now: it doesn't contribute anything that we didn't know
already. The funny thing is that none of you - who support set theory
soo much - has accomplished what I've accomplished here, quite on the
positive side. Moeblee, why don't you start to be thrilled about it ?

I don't see what you've accomplished that is of any note. We already
know that set theory without the axiom of infinity has a model.

No. You did't know that set theory without the axiom of infinity "has a
model", i.e. is implementable and is consistent.

What a silly claim.
Take any model of ZFC. It is a model of ZFC - Infinity. Hence ZFC
- Infinity is consistent.

Don't understand a word of what you say. Take, take ..

Weird. It's a fairly simple explanation. Something wrong on your
end, evidently.

No. You "take" everything, but you didn't show us anything. How can we
know that you "take" something then? (Oh well, I'm pretty sure that you
will come up again with something very, very trivial .. Surprise me)

You think there are no models of ZFC?

Educate me. I'm breathlessly awaiting.

That request presumes something not in evidence, that HdB is educable.
.

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