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?
 A: In your attempt, the problem is with the line  

Since $\alpha \vDash \beta$ and $\beta \vDash \alpha$, we have for every truth assignment $v$ that $\overline{v}(\alpha) = \overline{v}(\beta) = T$.  

We do not have $v(\alpha) = v(\beta) = T$ for every $v$. We only know If $v(\alpha) = T$ then $v(\beta) = T$ and vice versa. As you notice, this excludes the case where an assignment does not satisfy $\alpha$ or $\beta$.  
Instead, we need to consider two cases:  
Consider an arbitrary assignment $v$.
Case 1: $v(\alpha) = T$. Then by $\alpha \vDash \beta$, we also have $v(\beta) = T$, since for all $v$, if $v(\alpha) = T$ then $v(\beta) = T$.
Case 2: $v(\alpha) = F$. Then by $\beta \vDash \alpha$, $v(\beta) = F$, since otherwise there would be an assignment $v$ s.t. $v(\beta) = T$ but $v(\alpha) = F$ which contradicts the premise that if for all $v$, if $v(\beta) = T$ then $v(\alpha) = T$.
So in any of the cases, $v(\alpha) = v(\beta)$, thus by definition of $\leftrightarrow$, $v(\alpha \leftrightarrow \beta) = T$. Since $v$ arbitrary, the above holds for all assignments $v$, therefore $\vDash \alpha \leftrightarrow \beta$.  
(The proof could be shortened a bit; i.m.o. the "since"-parts in the two cases are rather self-evident.)  

Another possiblity would be to use the import-export theorem (sometimes referred to as the deduction theorem) which states that  
$$A \vDash B\ \Longleftrightarrow\ \vDash A \to B$$
With that, we have
$$\alpha \vDash \beta \text{ and } \beta \vDash \alpha$$
$$\Longrightarrow\ \vDash \alpha \to \beta \text{ and } \vDash \beta \to \alpha$$
i.e.
$$\text{"For all valuations $v$, $v(\alpha \to \beta) = T$ and $v(\beta \to \alpha) = T$"}$$
from which
$$\vDash \alpha \leftrightarrow \beta$$
follows more or less immediately by definition of $\to$ and $\leftrightarrow$.  
A: 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.
