most recent 30 from http://mathoverflow.net 2013-05-25T14:25:03Z http://mathoverflow.net/feeds/question/3528 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3528/is-there-a-formula-phi-s-t-phi-and-not-phi-have-a-stronger-consistency Martin Lackner 2009-10-31T10:46:02Z 2009-11-05T15:39:38Z

<p>Let &Sigma; be an axiom system. Can there be a formula &phi;, s.t. </p> <ul> <li>Con(&Sigma;) does not imply Con(&Sigma; + &phi;) AND</li> <li>Con(&Sigma;) does not imply Con(&Sigma; + not &phi;)</li> </ul> <p>If yes, can you give me an example for ZFC?</p>

http://mathoverflow.net/questions/3528/is-there-a-formula-phi-s-t-phi-and-not-phi-have-a-stronger-consistency/3537#3537 Carsten Schultz 2009-10-31T12:31:32Z 2009-10-31T12:31:32Z

<p>Short answer: No.</p>

http://mathoverflow.net/questions/3528/is-there-a-formula-phi-s-t-phi-and-not-phi-have-a-stronger-consistency/3576#3576 Richard Dore 2009-10-31T18:18:20Z 2009-10-31T20:15:46Z

<p>No, it's impossible for any axiom system. If &Sigma; is consistent, then by the Completeness theorem, it has some model M. In M, &phi; is either true or false. So M is a model of either (&Sigma;+&phi;) or (&Sigma;+not &phi;). So at least one of them is consistent. It might be that your metatheory doesn't know which one is consistent, but it knows that at least one of them is.</p>

http://mathoverflow.net/questions/3528/is-there-a-formula-phi-s-t-phi-and-not-phi-have-a-stronger-consistency/3850#3850 Martin Lackner 2009-11-02T21:12:52Z 2009-11-02T21:12:52Z

<p>Now that I know the answer, I've found my own simple proof. Probably it's interesting to someone else, so I post it:</p> <p>I want to show that Con(&Sigma;) is equivalent to ( Con(Σ + φ) OR Con(Σ + not φ) )</p> <p>Proof: Con(Σ + φ) OR Con(Σ + not φ) iff</p> <p>( Σ doesn't prove [φ -> FALSE] ) OR ( Σ doesn't prove [not φ -> FALSE] ) iff</p> <p>Σ doesn't prove [(not φ -> FALSE) AND (φ -> FALSE)] iff</p> <p>Σ doesn't prove [FALSE], which is Con(Σ).</p>