Show that p ∧ q → r and p ∨ q → r are not logically equivalent.












1












$begingroup$


I tried to use truth tables but couldn't get it, I used logical equivalences and got it, can someone show me how to do it with a truth table?






discrete-mathematics logic







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  • 2




    $begingroup$
    How did you use equivalences and 'got it' when they are not equivalent?
    $endgroup$
    – Bram28
    Jan 24 at 2:17
















1












$begingroup$


I tried to use truth tables but couldn't get it, I used logical equivalences and got it, can someone show me how to do it with a truth table?






discrete-mathematics logic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How did you use equivalences and 'got it' when they are not equivalent?
    $endgroup$
    – Bram28
    Jan 24 at 2:17














1












1








1





$begingroup$


I tried to use truth tables but couldn't get it, I used logical equivalences and got it, can someone show me how to do it with a truth table?






discrete-mathematics logic







share|cite|improve this question











$endgroup$




I tried to use truth tables but couldn't get it, I used logical equivalences and got it, can someone show me how to do it with a truth table?






discrete-mathematics logic




discrete-mathematics logic






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edited Jan 24 at 7:12









Mauro ALLEGRANZA

66.9k449115




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asked Jan 24 at 2:09









Usama GhawjiUsama Ghawji

465




465








  • 2




    $begingroup$
    How did you use equivalences and 'got it' when they are not equivalent?
    $endgroup$
    – Bram28
    Jan 24 at 2:17














  • 2




    $begingroup$
    How did you use equivalences and 'got it' when they are not equivalent?
    $endgroup$
    – Bram28
    Jan 24 at 2:17








2




2




$begingroup$
How did you use equivalences and 'got it' when they are not equivalent?
$endgroup$
– Bram28
Jan 24 at 2:17




$begingroup$
How did you use equivalences and 'got it' when they are not equivalent?
$endgroup$
– Bram28
Jan 24 at 2:17










4 Answers
4






active

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0












$begingroup$

You probably just miscalculated the truth table.
Here is the complete truth table for the two statements:



$$begin{array}{ccc|c|c}
p & q & r & pwedge q to r & pvee qto r \ hline
T & T & T & T & T \
T & T & F & F & F \
T & F & T & T & T \
T & F & F & T & F \
F & T & T & T & T \
F & T & F & T & F \
F & F & T & T & T \
F & F & F & T & T
end{array} $$



It is easy to see that the two statements are not equivalent.



An easy way to work out the truth table for a statement of the form $Ato B$ is to note that this will only be false when $A$ is true but $B$ is false, and is true everywhere else. This fact is what I used to calculate the tables above.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $p=1,q=0,r=0, pwedge q = 0, pvee q=1, pwedge qrightarrow r = 1, pvee qrightarrow r = 0$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      begin{array}{ccc|ccc|ccc}
      p&q&r&(p land q)& rightarrow &r&(plor q)& rightarrow &r\
      hline
      T&T&T&T&T&T&T&T&T\
      T&T&F&T&F&F&T&F&F\
      T&F&T&F&T&T&T&T&T\
      T&F&F&F&color{red}T&F&T&color{red}F&F\
      F&T&T&F&T&T&T&T&T\
      F&T&F&F&color{red}T&F&T&color{red}F&F\
      F&F&T&F&T&T&F&T&T\
      F&F&F&F&T&F&F&T&F\
      end{array}






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$


        can someone show me how to do it with a truth table?




        If and only if they were logically equivalent, then you could construct a table such that $pland qto r$ and $plor qto r$ are given identical truth values by each assignment of $p,q,$ and $r$.



        So you would have to construct a table for the eight assignments of the literals and verify that that happens.



        However, that can be a lot of effort.   We only need one counterexample to show that they are not equivalent.   That is, one assignment of $p,q,r$ which gives differing values for the conditionals.



        So if we could see a way for this to happen, then it would save all that work.   Now how might that happen?



        Why now, a conditional is only falsified when the antecedent is true while the consequent is false.   So let us assign $r$ to be false and think about which evaluations of $p,q$ are needed to make one of $pland q$ or $plor q$ false while the other is true (falsifying one conditional but not the other).






        share|cite|improve this answer









        $endgroup$













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          4 Answers
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          0












          $begingroup$

          You probably just miscalculated the truth table.
          Here is the complete truth table for the two statements:



          $$begin{array}{ccc|c|c}
          p & q & r & pwedge q to r & pvee qto r \ hline
          T & T & T & T & T \
          T & T & F & F & F \
          T & F & T & T & T \
          T & F & F & T & F \
          F & T & T & T & T \
          F & T & F & T & F \
          F & F & T & T & T \
          F & F & F & T & T
          end{array} $$



          It is easy to see that the two statements are not equivalent.



          An easy way to work out the truth table for a statement of the form $Ato B$ is to note that this will only be false when $A$ is true but $B$ is false, and is true everywhere else. This fact is what I used to calculate the tables above.






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            You probably just miscalculated the truth table.
            Here is the complete truth table for the two statements:



            $$begin{array}{ccc|c|c}
            p & q & r & pwedge q to r & pvee qto r \ hline
            T & T & T & T & T \
            T & T & F & F & F \
            T & F & T & T & T \
            T & F & F & T & F \
            F & T & T & T & T \
            F & T & F & T & F \
            F & F & T & T & T \
            F & F & F & T & T
            end{array} $$



            It is easy to see that the two statements are not equivalent.



            An easy way to work out the truth table for a statement of the form $Ato B$ is to note that this will only be false when $A$ is true but $B$ is false, and is true everywhere else. This fact is what I used to calculate the tables above.






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              You probably just miscalculated the truth table.
              Here is the complete truth table for the two statements:



              $$begin{array}{ccc|c|c}
              p & q & r & pwedge q to r & pvee qto r \ hline
              T & T & T & T & T \
              T & T & F & F & F \
              T & F & T & T & T \
              T & F & F & T & F \
              F & T & T & T & T \
              F & T & F & T & F \
              F & F & T & T & T \
              F & F & F & T & T
              end{array} $$



              It is easy to see that the two statements are not equivalent.



              An easy way to work out the truth table for a statement of the form $Ato B$ is to note that this will only be false when $A$ is true but $B$ is false, and is true everywhere else. This fact is what I used to calculate the tables above.






              share|cite|improve this answer









              $endgroup$



              You probably just miscalculated the truth table.
              Here is the complete truth table for the two statements:



              $$begin{array}{ccc|c|c}
              p & q & r & pwedge q to r & pvee qto r \ hline
              T & T & T & T & T \
              T & T & F & F & F \
              T & F & T & T & T \
              T & F & F & T & F \
              F & T & T & T & T \
              F & T & F & T & F \
              F & F & T & T & T \
              F & F & F & T & T
              end{array} $$



              It is easy to see that the two statements are not equivalent.



              An easy way to work out the truth table for a statement of the form $Ato B$ is to note that this will only be false when $A$ is true but $B$ is false, and is true everywhere else. This fact is what I used to calculate the tables above.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 24 at 2:19









              AJFarmarAJFarmar

              15910




              15910























                  2












                  $begingroup$

                  $p=1,q=0,r=0, pwedge q = 0, pvee q=1, pwedge qrightarrow r = 1, pvee qrightarrow r = 0$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    $p=1,q=0,r=0, pwedge q = 0, pvee q=1, pwedge qrightarrow r = 1, pvee qrightarrow r = 0$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      $p=1,q=0,r=0, pwedge q = 0, pvee q=1, pwedge qrightarrow r = 1, pvee qrightarrow r = 0$






                      share|cite|improve this answer









                      $endgroup$



                      $p=1,q=0,r=0, pwedge q = 0, pvee q=1, pwedge qrightarrow r = 1, pvee qrightarrow r = 0$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 24 at 2:13









                      lightxbulblightxbulb

                      980311




                      980311























                          0












                          $begingroup$

                          begin{array}{ccc|ccc|ccc}
                          p&q&r&(p land q)& rightarrow &r&(plor q)& rightarrow &r\
                          hline
                          T&T&T&T&T&T&T&T&T\
                          T&T&F&T&F&F&T&F&F\
                          T&F&T&F&T&T&T&T&T\
                          T&F&F&F&color{red}T&F&T&color{red}F&F\
                          F&T&T&F&T&T&T&T&T\
                          F&T&F&F&color{red}T&F&T&color{red}F&F\
                          F&F&T&F&T&T&F&T&T\
                          F&F&F&F&T&F&F&T&F\
                          end{array}






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            begin{array}{ccc|ccc|ccc}
                            p&q&r&(p land q)& rightarrow &r&(plor q)& rightarrow &r\
                            hline
                            T&T&T&T&T&T&T&T&T\
                            T&T&F&T&F&F&T&F&F\
                            T&F&T&F&T&T&T&T&T\
                            T&F&F&F&color{red}T&F&T&color{red}F&F\
                            F&T&T&F&T&T&T&T&T\
                            F&T&F&F&color{red}T&F&T&color{red}F&F\
                            F&F&T&F&T&T&F&T&T\
                            F&F&F&F&T&F&F&T&F\
                            end{array}






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              begin{array}{ccc|ccc|ccc}
                              p&q&r&(p land q)& rightarrow &r&(plor q)& rightarrow &r\
                              hline
                              T&T&T&T&T&T&T&T&T\
                              T&T&F&T&F&F&T&F&F\
                              T&F&T&F&T&T&T&T&T\
                              T&F&F&F&color{red}T&F&T&color{red}F&F\
                              F&T&T&F&T&T&T&T&T\
                              F&T&F&F&color{red}T&F&T&color{red}F&F\
                              F&F&T&F&T&T&F&T&T\
                              F&F&F&F&T&F&F&T&F\
                              end{array}






                              share|cite|improve this answer









                              $endgroup$



                              begin{array}{ccc|ccc|ccc}
                              p&q&r&(p land q)& rightarrow &r&(plor q)& rightarrow &r\
                              hline
                              T&T&T&T&T&T&T&T&T\
                              T&T&F&T&F&F&T&F&F\
                              T&F&T&F&T&T&T&T&T\
                              T&F&F&F&color{red}T&F&T&color{red}F&F\
                              F&T&T&F&T&T&T&T&T\
                              F&T&F&F&color{red}T&F&T&color{red}F&F\
                              F&F&T&F&T&T&F&T&T\
                              F&F&F&F&T&F&F&T&F\
                              end{array}







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 24 at 2:30









                              Bram28Bram28

                              63.2k44793




                              63.2k44793























                                  0












                                  $begingroup$


                                  can someone show me how to do it with a truth table?




                                  If and only if they were logically equivalent, then you could construct a table such that $pland qto r$ and $plor qto r$ are given identical truth values by each assignment of $p,q,$ and $r$.



                                  So you would have to construct a table for the eight assignments of the literals and verify that that happens.



                                  However, that can be a lot of effort.   We only need one counterexample to show that they are not equivalent.   That is, one assignment of $p,q,r$ which gives differing values for the conditionals.



                                  So if we could see a way for this to happen, then it would save all that work.   Now how might that happen?



                                  Why now, a conditional is only falsified when the antecedent is true while the consequent is false.   So let us assign $r$ to be false and think about which evaluations of $p,q$ are needed to make one of $pland q$ or $plor q$ false while the other is true (falsifying one conditional but not the other).






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$


                                    can someone show me how to do it with a truth table?




                                    If and only if they were logically equivalent, then you could construct a table such that $pland qto r$ and $plor qto r$ are given identical truth values by each assignment of $p,q,$ and $r$.



                                    So you would have to construct a table for the eight assignments of the literals and verify that that happens.



                                    However, that can be a lot of effort.   We only need one counterexample to show that they are not equivalent.   That is, one assignment of $p,q,r$ which gives differing values for the conditionals.



                                    So if we could see a way for this to happen, then it would save all that work.   Now how might that happen?



                                    Why now, a conditional is only falsified when the antecedent is true while the consequent is false.   So let us assign $r$ to be false and think about which evaluations of $p,q$ are needed to make one of $pland q$ or $plor q$ false while the other is true (falsifying one conditional but not the other).






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$


                                      can someone show me how to do it with a truth table?




                                      If and only if they were logically equivalent, then you could construct a table such that $pland qto r$ and $plor qto r$ are given identical truth values by each assignment of $p,q,$ and $r$.



                                      So you would have to construct a table for the eight assignments of the literals and verify that that happens.



                                      However, that can be a lot of effort.   We only need one counterexample to show that they are not equivalent.   That is, one assignment of $p,q,r$ which gives differing values for the conditionals.



                                      So if we could see a way for this to happen, then it would save all that work.   Now how might that happen?



                                      Why now, a conditional is only falsified when the antecedent is true while the consequent is false.   So let us assign $r$ to be false and think about which evaluations of $p,q$ are needed to make one of $pland q$ or $plor q$ false while the other is true (falsifying one conditional but not the other).






                                      share|cite|improve this answer









                                      $endgroup$




                                      can someone show me how to do it with a truth table?




                                      If and only if they were logically equivalent, then you could construct a table such that $pland qto r$ and $plor qto r$ are given identical truth values by each assignment of $p,q,$ and $r$.



                                      So you would have to construct a table for the eight assignments of the literals and verify that that happens.



                                      However, that can be a lot of effort.   We only need one counterexample to show that they are not equivalent.   That is, one assignment of $p,q,r$ which gives differing values for the conditionals.



                                      So if we could see a way for this to happen, then it would save all that work.   Now how might that happen?



                                      Why now, a conditional is only falsified when the antecedent is true while the consequent is false.   So let us assign $r$ to be false and think about which evaluations of $p,q$ are needed to make one of $pland q$ or $plor q$ false while the other is true (falsifying one conditional but not the other).







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 24 at 2:39









                                      Graham KempGraham Kemp

                                      86.4k43478




                                      86.4k43478






























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