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?
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.
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).