How show $ S \models \forall x ( \alpha \Leftrightarrow \beta)$? I read some notes on Logic Course. I read that we can conclude:

$$ S \models \forall x ( \alpha   \Leftrightarrow  \beta)$$ if and only if $ S \models \forall\, x\, \alpha$ has conclusion $ S \models\, \forall\, x\, \beta $

S be a model. $ \alpha$ and $ \beta$ is a formula. I misunderstand why this is  true? 
 A: The answer to this question is similar to the one I gave here.
Take the model S whose domain is $\{a,b \}$, and let $\alpha:=F(x)$ and $\beta:=G(x)$, and finally let $F=\{a \}$ and $G=\{b \}$.
Recall that for finite domains $(\forall x)\varphi$ is equivalent to the conjunction $\varphi a_1 \land \varphi a_2 \,...\land \,\varphi a_n \; $ for an $n$ sized domain.
I'll show that S is a countermodel to the following statement: 
$$(1)\;\;\;\;\; For\;all\;models\; M \;\;\;\;\; M\vDash \forall x(\alpha \to  \beta) \;\;iff\;\; M\vDash \forall x(\alpha) \Rightarrow \;M\vDash \forall x(\alpha)$$
Which is sufficient to show that the posted statement is false, since the material biconditional is equivalent to the conjunction of conditionals, so if one of the conditionals is false, then so is the biconditional. Hence, if $A\to B$ is false, then $A\iff B$ is false (here $\iff$ denotes the material, object linguistic biconditional; I'm using the notational convention in the question, although $\equiv$ would have been more appropriate, lest one risks conflation with the meta-linguistic implication).
$(1)$ fails from right to left. 
For suppose  $S\vDash (Fa\land Fb)\Rightarrow S\vDash(Ga\land Gb)\, $, which is true vacuously, since the antecedent is false. That is, $b\notin F$, so $(Fa\land Fb)$ is false, which makes $S\vDash (Fa\land Fb)$ false. 
But $S\vDash (Fa\to Ga)\land (Fa\to Ga)\,$ is false, since in the first conjunct the antecedent is true, but the consequent is false. That is, $a\in F$ but $a\notin G$, so the conditional $Fa\to Ga\,$ is false, which makes the conjunction $(Fa\to Ga)\land (Fa\to Ga)\,$ false. 
