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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