3
$\begingroup$

I'm reading the proof of the Completeness Theorem from Enderton's "A Mathematical Introduction to Logic". I'm having issues seeing how the following highlighted sentence actually holds (excerpt from page 137).

Let $\Lambda$ be the set of logical axioms for the expanded language. Since $\Gamma \cup \Theta$ is consistent, there is no formula $\beta$ such that $\Gamma \cup \Theta \cup \Lambda$ tautologically implies both $\beta$ and $\neg \beta$. (This is by Theorem 24B; the compactness theorem of sentential logic is used here.) Hence there is a truth assignment $v$ for the set of all prime formulas that satisfies $\Gamma \cup \Theta \cup \Lambda$.

I have tried to reason "by contrapositive". That is, suppose a set of (sentential) formulas $\Sigma$ is unsatisfiable. Then, vacuously, every truth assignment that satisfies $\Sigma$ will also satisfy any formula at all. Hence $\Sigma$ tautologically implies any formula. In particular, for any given formula $\beta$, $\Sigma$ tautologically implies both $\beta$ and $\neg\beta$.

Is my reasoning correct?

$\endgroup$
  • 1
    $\begingroup$ Looks right to me. Remember it well, cause "consistent implies satisfiable" is used just as often, if not more than "valid implies provable". $\endgroup$ – spaceisdarkgreen Mar 28 at 1:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.