Wolfgang Schwarz :: Suppose ZFC proves its own inconsistency

Suppose ZFC proves its own inconsistency

Posted on Friday, 15 Apr 2005.

Suppose we find a proof, in ZFC, that ZFC is inconsistent. Does it follow that ZFC is inconsistent?

On the one hand, if we could infer from ZFC $$m[1]$ ~Con(ZFC) that ZFC is inconsistent, we could contrapositively infer the consistency of ZFC & Con(ZFC) from Con(ZFC); and since ZFC & Con(ZFC) obviously entails Con(ZFC), ZFC & Con(ZFC) would thereby entail its own consistency. Which it only can if it is inconsistent (GÃ¶del's second incompleteness theorem). So it seems that we can only infer that ZFC is inconsistent from the observation that ZFC entails its own inconsistency if we presupposes that ZFC & Con(ZFC) is inconsistent.

(~Con(ZFC) says that some number encodes a ZFC-proof of 1=0. So assume ZFC is consistent but entails ~Con(ZFC). By the consistency assumption, there is no proof of 1=0 from ZFC, and no number that encodes such a proof. The number 17, for example, does not encode such a proof. Moreover, ZFC entails that the number 17 does not encode a proof of 1=0, for "encodes a (ZFC-)proof of" is representable (definable) in ZFC. Similarly for all other numbers. So for all numbers n, ZFC entails that n is not a proof of 1=0, and yet by assumption ZFC entails that some number is a proof of 1=0. So if ZFC & Con(ZFC) is inconsistent, ZFC itself is omega-inconsistent. Hence if we find a ZFC-proof for ~Con(ZFC), we can at least conclude that ZFC is omega-inconsistent (and false, of course).)

However, if we know that ZFC $$m[1]$ ~Con(ZFC), we also know that ZFC & Con(ZFC) is inconsistent. And surely we can presuppose what we know! So doesn't the above argument break down and we can after all conclude that ZFC is inconsistent?

Comments

# Richard Zach on 15 April 2005, 21:06

Um, how do you get from "ZFC + Con(ZFC) is inconsistent" to "ZFC is inconsistent"?

# wo on 16 April 2005, 13:27

Sorry, it shows that this entry was written in a hurry.

I don't have any positive reason to believe that
one can infer "ZFC is inconsistent" from "ZFC + Con(ZFC) is inconsistent". In fact, I'm sure one cannot infer this, because I believe that ZFC + Con(ZFC) is consistent, and then it follows from G?del's second incompleteness theorem that the inference is invalid.

On the other hand, the inference is clearly valid if ZFC is inconsistent. So whether or not the inference is valid depends on certain assumptions about ZFC. My question is whether the inference can be shown to be invalid without assuming that ZFC cannot prove its own inconsistency. I don't see how.

# J.Foukzon on 23 December 2005, 21:37

ZFC IS INCONSISTENT!!!
Proof see:
http://planetmath.org/?op=getobj&from=papers&id=329

# J. Foukzon on 15 July 2006, 19:24

Set theory ZFC is inconsistent
Theorem.Set theory ZFC is inconsistent.
The proof of it the unexpected fact, leans on that standard assumption
(SA), that: set of all formulas of the canonical set theory ZFC is an
infinite countable ZFC-set.
http://www.geocities.com/jaykovf1/Jakov.pdf
http://geocities.com/jaykovf1/blog.html

# J. Foukzon on 03 May 2015, 07:35

Inconsistent countable set in second order ZFC and not existence of the strongly inaccessible cardinals.

https://www.researchgate.net/publication/274960585_Inconsistent_countable_set_in_second_order_ZFC_and_not_existence_of_the_strongly_inaccessible_cardinals

DESCRIPTION: In this article we derived an importent example of the inconsistent countable set in second order ZFC with the full second-order semantic.

# david c cooke on 04 August 2022, 12:59

Fateology develops a consistent theory of ZFC in which a mathematical Universe
entails a global field, hence, the existence of evolution as a dual Universe,
is presented in a radically different approach to time.

Wolfgang Schwarz

Suppose ZFC proves its own inconsistency

Comments

Add a comment