Show that if $\alpha \vDash \beta$ and $\beta \vDash \alpha$, then $\vDash (\alpha \leftrightarrow \beta)$ I would like to show that if $\alpha \vDash \beta$ and $\beta \vDash \alpha$, then $\vDash (\alpha \leftrightarrow \beta)$, and I'm thinking that the argument below fails.
Since $\alpha \vDash \beta$ and $\beta \vDash \alpha$, we have for every truth assignment $v$ that $\overline{v}(\alpha) = \overline{v}(\beta) = T$. Therefore, $\overline{v}(\alpha \leftrightarrow \beta) = T$. 
However, I think it would be incorrect to conclude that $\vDash(\alpha \leftrightarrow \beta)$, because I have only shown that $v$ satisfies $\alpha \leftrightarrow \beta$ if $v$ also satisfies $\alpha$ and $\beta$. 
Thoughts?

If you didn't have any information about the formulas $\alpha$ and $\beta$, there would be four options for their truth values: (1) both true, (2) both false, (3) $\alpha$ true but $\beta$ false, and (4) $\beta$ true but $\alpha$ false. (These are essentially the four lines of a truth table.) 
But you do have some information about $\alpha$ and $\beta$. You have that $\alpha\models\beta$, which prevents option (3). And you have that $\beta\models\alpha$, which prevents option (4). 
So, you must have option (1) or option (2). In both of these situations, $\alpha\leftrightarrow\beta$ is true.
