What does it mean that a set $S$ tautologically implies wff $ \tau$ ?
in Enderton introduction to mathematical logic , in page 23 ,
it define that a set $S$ tautologically implies wff $ \tau$ iff every truth assigment for the sentence symbols in $S$ and $ \tau$ that satisfies every member of $S$ also satisfies $ \tau$
i have some questions related to this definition ,
does this definition require that the elements of the wff $ \tau$ are in $S$ ?
or not ??
also , i think that giving an example about this definition is great !
if this definition doesn't require that the elements of $ \tau $ are in $S$ ,
does this mean that if we made a set $S$ whose intersection with the wff $ \tau$ if the empty set then S tautologically implies $ \tau$ in general ??
i find that the definition is not so clear for me !
thanx !
Best Answer
First, you need to understand the definition like this. Pool all the propositional letters (sentence symbols) which occur in wffs in the set $S$ together with the (same or different!) propositional letters which occur in $\tau$. So if $S$ contains the wffs $P$ and $(\neg P \lor Q)$, and $\tau$ is $(Q \lor R)$, then the relevant letters that occur somewhere or other are $P, Q, R$. Now consider valuations $V$ of these letters. $S \vdash \tau$ iff for every such $V$, if $V$ makes all the wffs in $S$ true, it makes $\tau$ true.
So, for example, we indeed have
$$\{P, (\neg P \lor Q)\} \vDash (Q \lor R),$$
even though the conclusion contains a propositional letter which isn't in the premisses. (And that's what we want! For the conclusion plainly does logically follow from the premisses!)
Second, suppose though that none of the propositional letters in $S$ are in $\tau$. To take a simple case, suppose $S$ contains the wff $P$, and $\tau$ is $Q$. Then we would not have
$$\{P\} \vDash Q$$
because the valuation $P \Rightarrow T$, $Q \Rightarrow F$ makes the sole premiss true and the supposed conclusion false, showing this isn't a tautological entailment.
In fact we'll have the following result. If $S$ and $\tau$ lack propositional letters in common,
$$S \vDash \tau$$
holds just when $\tau$ is already a tautology, or when the wffs $S$ are contradictory (cannot all be true together), but not otherwise.