Now, assuming PA is really consistent, it follows that PA + ¬Con(PA) is also consistent, for if it were not, then PA would prove Con(PA) ("reductio"), contradicting Gödel's second incompleteness theorem. However, PA + ¬Con(PA) is "not" ω-consistent. This is because, for any particular natural number "n", PA + ¬Con(PA) proves that "n" is not the Gödel number of a proof that 0=1 (PA itself proves that fact; the extra assumption ¬Con(PA) is not needed). However, PA + ¬Con(PA) proves that, for "some" natural number "n", "n" "is" the Gödel number of such a proof (this is just a direct restatement of the claim ¬Con(PA) ).