Re: Implementable Set Theory and Consistency of ZFC

Jesse F. Hughes wrote:

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

Jesse F. Hughes wrote:

I did not ask about your needs. I asked whether the formula ~Infinity
is also a theorem of (1)-(4), where ~Infinity is the negation of the
axiom of Infinity.

(~Infinity) is _not_ a theorem of (1)-(4), in this article:

http://hdebruijn.soo.dto.tudelft.nl/jaar2007/set_theory.pdf

Why not?

You show that foundation is true in your model. You conclude (1)-(4)
entail foundation.

~Infinity is also true in your model. Why do you not conclude that
(1)-(4) entail ~Infinity?

It is remarkably difficult to get an answer from you sometimes.

Really?

Really.

I simply cannot "prove" anything (constructively) about something I can
not understand (constructively).

Han de Bruijn

.

Relevant Pages

  • Re: Ultimate debunking of Cantors Theory
    ... absent that axiom must also be consistent, as its theorems are a subset, ... *negation* of the axiom of infinity, ...
    (sci.math)
  • Re: Cantor Confusion
    ... for the tenth time: Without the axiom of infinity it is ... For the eleventh time: Infinity is NOWHERE. ... the negation of the axiom of choice is a theorem of ZF. ... "Foundations of Set Theory", North Holland, Amsterdam. ...
    (sci.math)
  • Re: Ultimate debunking of Cantors Theory
    ... absent that axiom must also be consistent, as its theorems are a subset, ... *negation* of the axiom of infinity, ...
    (sci.math)
  • Re: Ultimate debunking of Cantors Theory
    ... If any set theory with a given axiom is consistent, ... *negation* of the axiom of infinity, ...
    (sci.math)
  • Re: Implementable Set Theory and Consistency of ZFC
    ... is also a theorem of -, where ~Infinity is the negation of the ... axiom of Infinity. ... entail foundation. ... Then you cannot prove that adding an axiom to a system of axioms can ...
    (sci.math)
Loading