Theory is $\omega$-inconsistent if, for some property P of natural numbers, T proves P(0), P(1), P(2), and so on (that is, for every standard natural number n, T proves that P(n) holds), but T also proves that there is some (necessarily nonstandard) natural number n such that P(n) fails. http://en.wikipedia.org/wiki/Omega-consistent
How I understand this is that those two sentences:
(1): $\forall_n P(n)$
(2): $\exists_n (not P(n)$
may be true at the same time and that does not lead to condtradiction. How is it possible?
Isn't proving second sentence not enought to prove that first one is false? In other words what is wrong with this proof:
$\exists_n P(n) \iff not~\forall_n(not~P(n))$ -- This is De Morgan's law
so $\exists_n (not P(n)) \iff not~\forall_nP(n)$ so last term is contradiction of (1).