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The following is the start of basic corollary in my logic text:

For any set $\Gamma$ of sentences, $\Gamma\subseteq\text{Th}(\text{Mod}(\Gamma))$.

What happens when $\text{Mod}(\Gamma)$ is empty? Suppose $\Gamma={p_1,\neg p_1}$, so that there is no truth value $V$ which models $\Gamma$, or otherwise $V(p_1)=\text{T}=V(\neg{p_1})$, which doesn't respect logical connectives. Then I would have something like $\Gamma\subseteq\text{Th}(\emptyset)$. If I'm working under the definition that $\text{Th}(\emptyset)$ is the set of all tautologies, then this doesn't make any sense.

Should there be an extra condition that $\text{Mod}(\Gamma)$ be nonempty, or am I missing something?

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$\operatorname{Th}(\emptyset)$ consists of every sentence: $S \models \phi$ means that $M \models \phi$ for every model $M$ in $S$, but in the case of $S=\emptyset$ this is an empty condition.

Besides, the proof of $\Gamma \subseteq \operatorname{Th}(\operatorname{Mod}(\Gamma))$ is straight forward with the definitions and does not use $\operatorname{Mod}(\Gamma) \neq \emptyset$.

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  • $\begingroup$ Could you please show such a proof? I took some $\phi\in\Gamma$, and then said for all $V\in\text{Mod}(\Gamma)$, $V(\phi)=\text{T}$, and so $\phi\in\text{Th}\text{Mod}(\Gamma))$, but I felt I used the fact that $\text{Mod}\neq\emptyset$ in order to choose such a $V$. $\endgroup$
    – yunone
    Sep 6 '10 at 11:39
  • $\begingroup$ First you should solve the following exercise: $\emptyset \subseteq X$ for every set $X$. $\endgroup$ Sep 8 '10 at 10:13
  • $\begingroup$ @gebruiker: Bored? $\endgroup$ Jan 26 '16 at 18:31

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