Why this is false?!

Question

$\Gamma \models (\alpha \vee \gamma)$ iff $(\Gamma \models \alpha$ or $\Gamma \models \gamma)$

[$\Longrightarrow$]

If $\Gamma \models (\alpha \vee \gamma)$ then ($\Gamma \models \alpha $ or $\Gamma \models \gamma$)

Suppose $\Gamma \models (\alpha \vee \gamma)$

By def. of logic consequence: $\forall\mathfrak{I}, \mathfrak{I}\models\Gamma \rightarrow \mathfrak{I}\models(\alpha \vee \gamma)$

By def. of relation of satisfaction of "$\vee$": $\forall\mathfrak{I}, \mathfrak{I}\models \Gamma \rightarrow \mathfrak{I} \models \alpha$ or $\mathfrak{I} \models \gamma$

By def. of logic consequence: $\Gamma \models \alpha$ ou $\Gamma\models\gamma$

[$\Longleftarrow$]

Not true! Why?

Answer 1

Consider $\Gamma=\varnothing$, $\alpha=\beta$ and $\gamma=\lnot\beta$. (where $\Gamma\nvDash \beta$ and $\Gamma\nvDash\lnot\beta$.)

More precisely, Take $\beta=\forall x P(x)$ where $P$ is unary predicate symbol. You can check that $\beta$ is true in the structure $\mathfrak{A}$ defined as $\mathfrak{A}=\{0\}$, $P^\mathfrak{A}=\{(0)\}$. But it is false in the structure $\mathfrak{B}$ defined as $\mathfrak{B}=\{0\}$, $P^\mathfrak{B}=\varnothing$. So $\nvDash \beta$ and $\nvDash \lnot\beta$ but $\vDash \beta\lor\lnot\beta$ because $\beta\lor\lnot\beta$ is tautology.

Answer 2

$ \models (\alpha \vee \lnot \alpha)$.