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Let Σ be an axiom system. Can there be a formula φ, s.t.

  • Con(Σ) does not imply Con(Σ + φ) AND
  • Con(Σ) does not imply Con(Σ + not φ)

If yes, can you give me an example for ZFC?

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To clarify: what's your metatheory? E.g. do you want these implications to be non-provable from, say, Peano Arithmetic? –  John Goodrick Oct 31 '09 at 16:31
    
It's a little confusing to change the question after you've accepted an answer. It makes it look like you already have an answer you're happy with, so people aren't as eager to help, and it makes the existing answers look like they're dodging the question. Consider just asking a new question with a link back to this one. –  Anton Geraschenko Nov 2 '09 at 21:19
    
You're right - I will change it. Thanks for pointing that out. –  Martin Lackner Nov 5 '09 at 15:39
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3 Answers

up vote 10 down vote accepted

No, it's impossible for any axiom system. If Σ is consistent, then by the Completeness theorem, it has some model M. In M, φ is either true or false. So M is a model of either (Σ+φ) or (Σ+not φ). 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.

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This is a great answer! –  Anton Geraschenko Oct 31 '09 at 19:12
    
Wait, there still seems to be a question here: namely, to give an example where the metatheory doesn't know which of Con(Sigma+phi) and Con(Sigma+~phi) is implied by Con(Sigma). –  Scott Aaronson Apr 24 '11 at 21:10
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Short answer: No.

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Which of the two questions are you answering? In either case, I'm not sure a straight yes/no answer is helpful here. –  John Goodrick Oct 31 '09 at 15:54
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Now that I know the answer, I've found my own simple proof. Probably it's interesting to someone else, so I post it:

I want to show that Con(Σ) is equivalent to ( Con(Σ + φ) OR Con(Σ + not φ) )

Proof: Con(Σ + φ) OR Con(Σ + not φ) iff

( Σ doesn't prove [φ -> FALSE] ) OR ( Σ doesn't prove [not φ -> FALSE] ) iff

Σ doesn't prove [(not φ -> FALSE) AND (φ -> FALSE)] iff

Σ doesn't prove [FALSE], which is Con(Σ).

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