Re: Implementable Set Theory and Consistency of ZFC

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:

Even assuming that there are four models rather than one, so what?
The point is that you have *not* given a proof of (5)-(8) using only
axioms (1)-(4). Thus, you have not done what you said.
Even if you show that there are seven models for (1)-(4) in which
(5)-(8) are also true, you haven't done what you said. Even if you
show there are *infinitely* many models satisfying this condition, you
haven't done what you said. So, perhaps you should either do what
you said or change your claim.

No. Because _nothing_ sensible ever counts as a proof in your conception
of mathematics.

Quite wrong. I've told you what counts as a proof of the claim that
(5)-(8) are theorems of (1)-(4). Namely, a proof of each of (5)-(8)
using only (1)-(4) as axioms.
What is so controversial about that?

Nothing. I've done just _that_ in my article.

Weird. How come no one else can recognize that you've done that?

You forget that 'sci.math' is not "no one else". It's a relatvely small
Internet Cafe, with some really weird inhabitants.

Also, if you *had* done that, then axioms (1)-(4) + (9) would be
sufficient for ZFC. Don't you find it a touch odd that no one else
has noticed this fascinating fact?

The fact that Infinity X is an axiom of standard ZFC makes it necessary,
it seems, to include the axioms (5-9), in order to make infinite sets
make more "look alike" finite sets. Which wouldn't have been necessary
with a more realistic approach (I mean, e.g. Choice is provable within
the realm of finite sets, as is well known).

You *do* know that if (1)-(4) prove (5)-(8), then so do (1)-(4) + (9),
right?

I can't do any sensible reasoning with Infinity (9 = X right?) included.

Han de Bruijn

.

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