# 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 $A$?

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If $A=\emptyset$, then every truth assignment $w$ satisfies $w(\psi)=1$ for every $\psi\in A$, because there are no such $\psi$. Thus $\emptyset\models\varphi$ if and only if $w(\varphi)=1$ for every truth assignment $w$. – walcher Sep 2 '13 at 23:10
@walcher, But if there is no such $\psi$ why does $\varphi$ have to be a tautology? – user92809 Sep 2 '13 at 23:17

## 2 Answers

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.

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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.

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