If F satisfiable then ¬F is unsatisfiable.












0












$begingroup$


If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.



[A]α = ¬(¬[A]))
= ¬([A])
= ¬ (0)
= 1









share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.



    [A]α = ¬(¬[A]))
    = ¬([A])
    = ¬ (0)
    = 1









    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.



      [A]α = ¬(¬[A]))
      = ¬([A])
      = ¬ (0)
      = 1









      share|cite|improve this question









      $endgroup$




      If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.



      [A]α = ¬(¬[A]))
      = ¬([A])
      = ¬ (0)
      = 1






      logic propositional-calculus first-order-logic satisfiability






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      share|cite|improve this question










      asked Jan 16 at 15:55









      OctavianOctavian

      32




      32






















          1 Answer
          1






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          1












          $begingroup$

          In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).



          This does not implies that its negation is unsatisfiable.



          Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.



          Clearly $F$ is sat, and also $lnot F$ is sat.



          But there are many more : $p lor q, p to q, p land q, ldots$





          The interesting relation is :




          if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what do you mean by that example p is a sentential letter?
            $endgroup$
            – Octavian
            Jan 16 at 17:02










          • $begingroup$
            @Octavian - it is the formula $p$ (made of one only symbol).
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 16 at 19:31











          Your Answer





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          $begingroup$

          In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).



          This does not implies that its negation is unsatisfiable.



          Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.



          Clearly $F$ is sat, and also $lnot F$ is sat.



          But there are many more : $p lor q, p to q, p land q, ldots$





          The interesting relation is :




          if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what do you mean by that example p is a sentential letter?
            $endgroup$
            – Octavian
            Jan 16 at 17:02










          • $begingroup$
            @Octavian - it is the formula $p$ (made of one only symbol).
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 16 at 19:31
















          1












          $begingroup$

          In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).



          This does not implies that its negation is unsatisfiable.



          Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.



          Clearly $F$ is sat, and also $lnot F$ is sat.



          But there are many more : $p lor q, p to q, p land q, ldots$





          The interesting relation is :




          if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what do you mean by that example p is a sentential letter?
            $endgroup$
            – Octavian
            Jan 16 at 17:02










          • $begingroup$
            @Octavian - it is the formula $p$ (made of one only symbol).
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 16 at 19:31














          1












          1








          1





          $begingroup$

          In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).



          This does not implies that its negation is unsatisfiable.



          Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.



          Clearly $F$ is sat, and also $lnot F$ is sat.



          But there are many more : $p lor q, p to q, p land q, ldots$





          The interesting relation is :




          if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).







          share|cite|improve this answer











          $endgroup$



          In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).



          This does not implies that its negation is unsatisfiable.



          Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.



          Clearly $F$ is sat, and also $lnot F$ is sat.



          But there are many more : $p lor q, p to q, p land q, ldots$





          The interesting relation is :




          if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 19:41

























          answered Jan 16 at 16:14









          Mauro ALLEGRANZAMauro ALLEGRANZA

          65.8k449114




          65.8k449114












          • $begingroup$
            what do you mean by that example p is a sentential letter?
            $endgroup$
            – Octavian
            Jan 16 at 17:02










          • $begingroup$
            @Octavian - it is the formula $p$ (made of one only symbol).
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 16 at 19:31


















          • $begingroup$
            what do you mean by that example p is a sentential letter?
            $endgroup$
            – Octavian
            Jan 16 at 17:02










          • $begingroup$
            @Octavian - it is the formula $p$ (made of one only symbol).
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 16 at 19:31
















          $begingroup$
          what do you mean by that example p is a sentential letter?
          $endgroup$
          – Octavian
          Jan 16 at 17:02




          $begingroup$
          what do you mean by that example p is a sentential letter?
          $endgroup$
          – Octavian
          Jan 16 at 17:02












          $begingroup$
          @Octavian - it is the formula $p$ (made of one only symbol).
          $endgroup$
          – Mauro ALLEGRANZA
          Jan 16 at 19:31




          $begingroup$
          @Octavian - it is the formula $p$ (made of one only symbol).
          $endgroup$
          – Mauro ALLEGRANZA
          Jan 16 at 19:31


















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