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SAT for a formula using Tableaux Propositional Logic (precedence of operators)


Need help with solving proposition logic formula, should be a tautologyPropositional logic and distributive lawHelp with natural deduction (Propositional logic)Motivation for signed tableaux rules for propositional intuitionistic logiclogic: derive a formula using lawsclause “elimination” using assumptions in a propositional formulaHow to prove this propositional logic equation?Initialization in the tableaux method for first order logicPropositional logic proof checkGiven is a set of clauses. Find a logic formula in CNF such that..













1












$begingroup$


My doubt is in check if the following formula $phi$ is SAT or not using the Tableaux Method. Let me write formula:



$phi = neg left ( p vee q supset left ( left ( neg p wedge q right ) vee p vee neg q right ) right )$



How start rules of tableaux method in this case?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    My doubt is in check if the following formula $phi$ is SAT or not using the Tableaux Method. Let me write formula:



    $phi = neg left ( p vee q supset left ( left ( neg p wedge q right ) vee p vee neg q right ) right )$



    How start rules of tableaux method in this case?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      My doubt is in check if the following formula $phi$ is SAT or not using the Tableaux Method. Let me write formula:



      $phi = neg left ( p vee q supset left ( left ( neg p wedge q right ) vee p vee neg q right ) right )$



      How start rules of tableaux method in this case?










      share|cite|improve this question











      $endgroup$




      My doubt is in check if the following formula $phi$ is SAT or not using the Tableaux Method. Let me write formula:



      $phi = neg left ( p vee q supset left ( left ( neg p wedge q right ) vee p vee neg q right ) right )$



      How start rules of tableaux method in this case?







      logic propositional-calculus formal-languages satisfiability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 12 at 21:02







      mayday24

















      asked Mar 12 at 15:35









      mayday24mayday24

      83




      83






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          You have to apply the rule corresponding to the principal connective of the negated formula.



          Assuming the usual convention for omitting parentheses, we have that conjunction and disjunction symbols apply to as little as possible.



          Thus, the formula will be read as :




          $¬[(p∨q) to ((¬p∧q)∨p∨¬q)]$.




          In this case, the formula is the negation of a conditional; thus, you have to start using the rule :




          $lnot (alpha to beta)$.







          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Thank you! I apply the rest of Tableaux Method if can be useful for someone.



            $p vee q$ $[mu_1]$from rule of AND. I apply tableaux method for this point later..



            $ neg ( (neg p wedge q) vee p vee neg q)$ from rule of AND, associative property



            $ neg (neg p wedge q) wedge neg(p vee neg q)$ de Morgan property



            $neg (neg p wedge q) $ rule of AND



            $neg(p vee neg q)$ rule of OR



            $p vee neg q $ $[mu_2]$



            $neg p wedge q $



            $neg p$



            $q$



            ..apply first OR $[mu_1]$



            $beta_1$ branch: $p$ clash! Branch closed.



            $beta_2$ branch: $q$



            ..apply second OR $[mu_2]$



            $beta_{2_1}$ branch: $p$ clash! Branch closed.



            $beta_{2_2}$ branch: $neg q$ clash! Branch closed.



            We can say that $phi$ is UNSAT.






            share|cite|improve this answer











            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

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              active

              oldest

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              0












              $begingroup$

              You have to apply the rule corresponding to the principal connective of the negated formula.



              Assuming the usual convention for omitting parentheses, we have that conjunction and disjunction symbols apply to as little as possible.



              Thus, the formula will be read as :




              $¬[(p∨q) to ((¬p∧q)∨p∨¬q)]$.




              In this case, the formula is the negation of a conditional; thus, you have to start using the rule :




              $lnot (alpha to beta)$.







              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                You have to apply the rule corresponding to the principal connective of the negated formula.



                Assuming the usual convention for omitting parentheses, we have that conjunction and disjunction symbols apply to as little as possible.



                Thus, the formula will be read as :




                $¬[(p∨q) to ((¬p∧q)∨p∨¬q)]$.




                In this case, the formula is the negation of a conditional; thus, you have to start using the rule :




                $lnot (alpha to beta)$.







                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You have to apply the rule corresponding to the principal connective of the negated formula.



                  Assuming the usual convention for omitting parentheses, we have that conjunction and disjunction symbols apply to as little as possible.



                  Thus, the formula will be read as :




                  $¬[(p∨q) to ((¬p∧q)∨p∨¬q)]$.




                  In this case, the formula is the negation of a conditional; thus, you have to start using the rule :




                  $lnot (alpha to beta)$.







                  share|cite|improve this answer











                  $endgroup$



                  You have to apply the rule corresponding to the principal connective of the negated formula.



                  Assuming the usual convention for omitting parentheses, we have that conjunction and disjunction symbols apply to as little as possible.



                  Thus, the formula will be read as :




                  $¬[(p∨q) to ((¬p∧q)∨p∨¬q)]$.




                  In this case, the formula is the negation of a conditional; thus, you have to start using the rule :




                  $lnot (alpha to beta)$.








                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 12 at 15:45

























                  answered Mar 12 at 15:39









                  Mauro ALLEGRANZAMauro ALLEGRANZA

                  67.2k449115




                  67.2k449115























                      0












                      $begingroup$

                      Thank you! I apply the rest of Tableaux Method if can be useful for someone.



                      $p vee q$ $[mu_1]$from rule of AND. I apply tableaux method for this point later..



                      $ neg ( (neg p wedge q) vee p vee neg q)$ from rule of AND, associative property



                      $ neg (neg p wedge q) wedge neg(p vee neg q)$ de Morgan property



                      $neg (neg p wedge q) $ rule of AND



                      $neg(p vee neg q)$ rule of OR



                      $p vee neg q $ $[mu_2]$



                      $neg p wedge q $



                      $neg p$



                      $q$



                      ..apply first OR $[mu_1]$



                      $beta_1$ branch: $p$ clash! Branch closed.



                      $beta_2$ branch: $q$



                      ..apply second OR $[mu_2]$



                      $beta_{2_1}$ branch: $p$ clash! Branch closed.



                      $beta_{2_2}$ branch: $neg q$ clash! Branch closed.



                      We can say that $phi$ is UNSAT.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Thank you! I apply the rest of Tableaux Method if can be useful for someone.



                        $p vee q$ $[mu_1]$from rule of AND. I apply tableaux method for this point later..



                        $ neg ( (neg p wedge q) vee p vee neg q)$ from rule of AND, associative property



                        $ neg (neg p wedge q) wedge neg(p vee neg q)$ de Morgan property



                        $neg (neg p wedge q) $ rule of AND



                        $neg(p vee neg q)$ rule of OR



                        $p vee neg q $ $[mu_2]$



                        $neg p wedge q $



                        $neg p$



                        $q$



                        ..apply first OR $[mu_1]$



                        $beta_1$ branch: $p$ clash! Branch closed.



                        $beta_2$ branch: $q$



                        ..apply second OR $[mu_2]$



                        $beta_{2_1}$ branch: $p$ clash! Branch closed.



                        $beta_{2_2}$ branch: $neg q$ clash! Branch closed.



                        We can say that $phi$ is UNSAT.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Thank you! I apply the rest of Tableaux Method if can be useful for someone.



                          $p vee q$ $[mu_1]$from rule of AND. I apply tableaux method for this point later..



                          $ neg ( (neg p wedge q) vee p vee neg q)$ from rule of AND, associative property



                          $ neg (neg p wedge q) wedge neg(p vee neg q)$ de Morgan property



                          $neg (neg p wedge q) $ rule of AND



                          $neg(p vee neg q)$ rule of OR



                          $p vee neg q $ $[mu_2]$



                          $neg p wedge q $



                          $neg p$



                          $q$



                          ..apply first OR $[mu_1]$



                          $beta_1$ branch: $p$ clash! Branch closed.



                          $beta_2$ branch: $q$



                          ..apply second OR $[mu_2]$



                          $beta_{2_1}$ branch: $p$ clash! Branch closed.



                          $beta_{2_2}$ branch: $neg q$ clash! Branch closed.



                          We can say that $phi$ is UNSAT.






                          share|cite|improve this answer











                          $endgroup$



                          Thank you! I apply the rest of Tableaux Method if can be useful for someone.



                          $p vee q$ $[mu_1]$from rule of AND. I apply tableaux method for this point later..



                          $ neg ( (neg p wedge q) vee p vee neg q)$ from rule of AND, associative property



                          $ neg (neg p wedge q) wedge neg(p vee neg q)$ de Morgan property



                          $neg (neg p wedge q) $ rule of AND



                          $neg(p vee neg q)$ rule of OR



                          $p vee neg q $ $[mu_2]$



                          $neg p wedge q $



                          $neg p$



                          $q$



                          ..apply first OR $[mu_1]$



                          $beta_1$ branch: $p$ clash! Branch closed.



                          $beta_2$ branch: $q$



                          ..apply second OR $[mu_2]$



                          $beta_{2_1}$ branch: $p$ clash! Branch closed.



                          $beta_{2_2}$ branch: $neg q$ clash! Branch closed.



                          We can say that $phi$ is UNSAT.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Mar 12 at 16:20

























                          answered Mar 12 at 16:15









                          mayday24mayday24

                          83




                          83






























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