$\omega$-completeness and $\omega$-consistency

Question

I am studying some model theory at the moment. Now I came into a confusion about the terms $\omega$-consistent and $\omega$-complete.

Let $\mathcal{L} = \{ +, \cdot, S, 0 \}$ be the language of arithmetic. The book "model theory" by Chang and Keisler states:

1. A theory $T$ in $\mathcal{L}$ is said to be $\omega$-consistent iff there is no formula $\varphi(x)$ of $\mathcal{L}$ such that $T \models \varphi(0), T \models \varphi(1), T \models \varphi(2),...$ and $T \models (\exists x) \lnot \varphi(x)$.
2. $T$ is said to be $\omega$-complete iff for every formula $\varphi(x)$ of $\mathcal{L}$ we have $T \models \varphi(0), T \models \varphi(1), T \models \varphi(2),...$ implies $T \models (\forall x) \varphi(x)$.

It seems to me that they are equivalent, for (in 1.) that there is no formula $\varphi(x)$ with the property means that for all formulas $\varphi(x)$ the negation holds, which is exactly the property in 2.

What is my mistake? Thank you in advance for any of your help!

Answer 1

I think the confusion might be caused by you conflating $T \nvDash (\exists x) \neg \phi (x)$ and $T \vDash \neg ((\exists x) \neg \phi (x))$

1) says that if $T \vDash \phi(0)$ and so on then $T \nvDash (\exists x) \neg \phi (x)$, and 2) says if $T \vDash \phi(0)$ and so on then $T \vDash \neg ((\exists x) \neg \phi (x))$