Question about Logical implication Suppose $A$ is a set of propositional formulas, and suppose $\varphi$ is a propositional formula. In my textbook they write $A \models \varphi$, if for every truth assignment $w$ such that $w(\psi) = 1$ for every $\psi \in A$, we also have $w(\varphi)=1$.
They now ask me to show that $\emptyset\models \varphi$ if and only if $\varphi$ is a tautology. But how can I show this? The definition tells us nothing about empty $A$?
 A: Suppose $A = \varnothing$. Then, for any truth value assignment $w$, $w(\psi) = 1$ (vacuously) for each and every $\psi \in A = \varnothing$  
Why? 
Because there are no $\psi \in A = \varnothing$. So the hypothesis is satisfied vacuously, because it is not possible for there to exist any $\psi \in \varnothing$ such that $w(\psi) \neq 1$.
Hence $\varnothing \models \varphi$ if and only if for each truth value assignment $w$, $w(\varphi) =  1$, too. This holds if and only if $\varphi$ is a tautology.
A: This is a nice opportunity to illustrate for the novice the role of proof strategies in reasoning mathematically about logical claims.
Notice, first of all, that the definition of entailment $\models$, if put in negative terms, says that to prove that $A\models\varphi$ fails to hold one has to find some truth-assignment $v$ that simultaneously satisfies $A$ and does not satisfy $\varphi$.
Furthermore, recall that $v$ is said to satisfy a given set of formulas $A$ if $v(\psi)=1$ for every $\psi\in A$; thus, contrapositively, we may say that $A$ is not satisfied by $v$ if there is some $\psi\in A$ that is not satisfied by $v$, that is, such that $v(\psi)=0$.
[Lemma] Any truth-assignment $v$ satisfies the empty set of formulas.
Proof. To assert that the empty set $\emptyset$ is not satisfied by a given truth-assignment $v$ would require us first of all to find some formula $\psi$ in $\emptyset$ with a certain property (namely, the property of not being satisfied by $v$).  But this is an impossible task, as there are no formulas in the empty set! 
(Incidentally, this is precisely what it means to say that a certain property is vacuously true.)
[Main Claim: one direction] ($\emptyset\models \varphi$) implies that ($\varphi$ is a tautology).
Proof. Suppose, by contraposition, that $\varphi$ is not a tautology, that is, suppose that there is some truth-assignment $w$ such that $w(\varphi)=0$.  By the previous Lemma, we alread know that $w$ satisfies the empty set $\emptyset$.  Thus, by the definition of entailment, $\emptyset\models \varphi$ fails to hold.
[Main Claim: other direction] ($\varphi$ is a tautology) implies that ($\emptyset\models \varphi$).
Proof. Assume that $\varphi$ is a tautology, that is, assume that $v(\varphi)=1$ for every truth-assignment $v$.  Suppose now by reductio that $\emptyset\models \varphi$ fails to hold.  It follows, in particular that there is some truth-assignment $w$ such that $w(\varphi)=0$ (the failure of $\emptyset\models \varphi$ also entails, in particular, that such $w$ satisfies $\emptyset$, but that is innocuous and known anyway to be true, from the above Lemma), that is, we are given a truth-assignment that does not satisfy $\varphi$.  This immediately contradicts the assumption about the tautological character of $\varphi$.
PS: The exhaustive amount of detail in the above proofs usually reduce to something like one- or two-line arguments as students get more and more experienced in producing them.
