Re: Implementable Set Theory and Consistency of ZFC

Jesse F. Hughes wrote:

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

lwalke3@xxxxxxxxx wrote:

Notice that if T is a theory and phi is any
axiom, then if M is a model of T+phi, then
M is a model of T. In other words, any model
of a theory is a model of any subtheory.
ZF-Infinity is a subtheory of ZF, since the
former is a subset of the latter. And so any
model of ZF is a model of ZF-Infinity.

Yes. And a "model" of (ZF-Infinity) is not per se a model of ZFC, which
is the very difference between an "if" and an "iff".

Everyone else seems to know the difference between "if" and "iff".
But you wrote the following:

It's impossible to have infinity without Infinity. If not, show us
such a model, please.

What on earth did you mean, if not: Show us a model of ZF - Infinity
with infinite sets in the model?

Yeah, just do it! Surprise me!

Han de Bruijn

.

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