Can an implication be a tautology? I have a problem understanding this proof:
$ F →G \equiv \top \implies F \models G $. 
My textbook proceeds as follows: 
$ \implies $ : we assume $ F \to G \equiv \top $ . This means that the implication $ F \to G $ holds for every possible truth assignment. Looking at the definition of the implication, we see that this is the case if either both propositions are true, both are false or the first one is false  and the second one is true , but never when $ F = 1 $ and $ G = 0 $, because that case would render the implication false. Looking at those three cases, we see that whenever $ G = 1 $ , $F$ is also $1 $ . This is the definition of the logical consequence, so we have $ F \models G $ .  
Now my question is: Isn't the definition of the tautology that it is always true? Which means no matter what truth assignments we give the propositions involved in a formula, the formula is always true. But now we said that G can never be false while F is true, so isn't this a contradiction? How can an implication be a tautology? 
 A: It looks like you're missing the fact that $F$ and $G$ must be placeholders for entire formulas. This means that the same propositional variable might occur in both $F$ and $G$, which can restrict what the possible truth values of $F$ and $G$ in combination are.
An example where the theorem makes a claim is if we let $F$ be the formula $A\land B$ and $G$ be the formula $A\lor B$. Then this instance of the theorem says,

If $(A\land B)\to(A\lor B)\equiv \top$, then $A\land B \vDash A\lor B$.

It is actually the case that $(A\land B)\to(A\lor B)$ is a tautology -- you can see by a truth table that it is impossible for $A\land B$ to evaluate to $1$ and at the same time $A\lor B$ evaluates to $0$.
So in this case the premise of the theorem holds, so it does actually claim that $A\land B\vDash A\lor B$. And this conclusion is indeed true too.
On the other hand, if we swap the formulas around and let $F$ be $A\lor B$ and $G$ be $A\land B$, then we get the claim

If $(A\lor B)\to(A\land B)\equiv \top$ then $A\lor B\vDash A\land B$.

In this case it is possible for $F$ to be true and at the same time $G$ be false, such as in the truth assignment $A=0$, $B=1$. But this means that $F\to G\equiv \top$ is false for this choice of $F$ and $G$. So the theorem doesn't actually promise us anything in this case, because the premise doesn't hold.
