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Proof of Completeness Theorem in Enderton's Logic, satisfiability of $Gamma cup Theta cup Lambda$

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 $negbeta$.

Is my reasoning correct?

logic propositional-calculus

share|cite|improve this question

asked Mar 28 at 0:57

MuchToLearnMuchToLearn

1189

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

add a comment | 

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 $negbeta$.

Is my reasoning correct?

logic propositional-calculus

share|cite|improve this question

asked Mar 28 at 0:57

MuchToLearnMuchToLearn

1189

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

add a comment | 

$begingroup$

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 $negbeta$.

Is my reasoning correct?

logic propositional-calculus

share|cite|improve this question

asked Mar 28 at 0:57

MuchToLearnMuchToLearn

1189

$endgroup$

share|cite|improve this question

asked Mar 28 at 0:57

MuchToLearnMuchToLearn

1189

share|cite|improve this question

asked Mar 28 at 0:57

MuchToLearnMuchToLearn

1189

asked Mar 28 at 0:57

MuchToLearnMuchToLearn

1189

asked Mar 28 at 0:57

MuchToLearnMuchToLearn

1189