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Let $T$ be a theory which is consistent and $\omega$-inconsistent, how to prove that $T$ is $\omega$-incomplete?

Recall: A theory $T$ is $\omega$-incomplete if there is a formula $A$ with only one free variable such that the theoy $T$ proves $A(\bar{n})$ for all natural number $n$ but it doesn't prove $\forall x A(x)$. ($\bar{n}$ is the numeral of $n$)

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This is a trivial consequence of the definition of an ω-inconsistent theory. There is a formula $A$ such that $T$ proves $A(\overline n)$ for every natural $n$, as well as $\exists x\,\neg A(x)$. Then $T$, being consistent, can't prove $\forall x\,A(x)$. – Emil Jeřábek Mar 9 2011 at 15:18

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