Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but,...












2












$begingroup$


Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?






logic propositional-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $sim$ do you mean $lnot$?
    $endgroup$
    – Yanko
    Jan 5 at 12:43






  • 1




    $begingroup$
    Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 12:44












  • $begingroup$
    But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
    $endgroup$
    – Alvin Carter
    Jan 5 at 12:53






  • 2




    $begingroup$
    Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
    $endgroup$
    – Yanko
    Jan 5 at 13:02












  • $begingroup$
    Got it. It helps a lot.
    $endgroup$
    – Alvin Carter
    Jan 5 at 19:19
















2












$begingroup$


Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?






logic propositional-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $sim$ do you mean $lnot$?
    $endgroup$
    – Yanko
    Jan 5 at 12:43






  • 1




    $begingroup$
    Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 12:44












  • $begingroup$
    But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
    $endgroup$
    – Alvin Carter
    Jan 5 at 12:53






  • 2




    $begingroup$
    Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
    $endgroup$
    – Yanko
    Jan 5 at 13:02












  • $begingroup$
    Got it. It helps a lot.
    $endgroup$
    – Alvin Carter
    Jan 5 at 19:19














2












2








2





$begingroup$


Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?






logic propositional-calculus







share|cite|improve this question











$endgroup$




Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?






logic propositional-calculus




logic propositional-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 20:43









Ruggiero Rilievi

179112




179112










asked Jan 5 at 12:41









Alvin CarterAlvin Carter

114




114












  • $begingroup$
    By $sim$ do you mean $lnot$?
    $endgroup$
    – Yanko
    Jan 5 at 12:43






  • 1




    $begingroup$
    Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 12:44












  • $begingroup$
    But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
    $endgroup$
    – Alvin Carter
    Jan 5 at 12:53






  • 2




    $begingroup$
    Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
    $endgroup$
    – Yanko
    Jan 5 at 13:02












  • $begingroup$
    Got it. It helps a lot.
    $endgroup$
    – Alvin Carter
    Jan 5 at 19:19


















  • $begingroup$
    By $sim$ do you mean $lnot$?
    $endgroup$
    – Yanko
    Jan 5 at 12:43






  • 1




    $begingroup$
    Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 12:44












  • $begingroup$
    But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
    $endgroup$
    – Alvin Carter
    Jan 5 at 12:53






  • 2




    $begingroup$
    Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
    $endgroup$
    – Yanko
    Jan 5 at 13:02












  • $begingroup$
    Got it. It helps a lot.
    $endgroup$
    – Alvin Carter
    Jan 5 at 19:19
















$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43




$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43




1




1




$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44






$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44














$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53




$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53




2




2




$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02






$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02














$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19




$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19










4 Answers
4






active

oldest

votes


















3












$begingroup$

If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.



I have no idea what you mean by $sim$ but it doesn't even matter.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The statement $pveeneg p$ is a true statement in Boolean algebra.



    Then:



    $$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$



    If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      $(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$



      $p∨(sim p∧q)=pvee q$ by the Absorption law.



      Try $p=1,q=0$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.



        The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.



          I have no idea what you mean by $sim$ but it doesn't even matter.






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.



            I have no idea what you mean by $sim$ but it doesn't even matter.






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.



              I have no idea what you mean by $sim$ but it doesn't even matter.






              share|cite|improve this answer









              $endgroup$



              If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.



              I have no idea what you mean by $sim$ but it doesn't even matter.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 5 at 12:44









              YankoYanko

              7,4801729




              7,4801729























                  2












                  $begingroup$

                  The statement $pveeneg p$ is a true statement in Boolean algebra.



                  Then:



                  $$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$



                  If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    The statement $pveeneg p$ is a true statement in Boolean algebra.



                    Then:



                    $$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$



                    If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      The statement $pveeneg p$ is a true statement in Boolean algebra.



                      Then:



                      $$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$



                      If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.






                      share|cite|improve this answer









                      $endgroup$



                      The statement $pveeneg p$ is a true statement in Boolean algebra.



                      Then:



                      $$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$



                      If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 5 at 12:49









                      drhabdrhab

                      102k545136




                      102k545136























                          1












                          $begingroup$

                          $(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$



                          $p∨(sim p∧q)=pvee q$ by the Absorption law.



                          Try $p=1,q=0$.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            $(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$



                            $p∨(sim p∧q)=pvee q$ by the Absorption law.



                            Try $p=1,q=0$.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              $(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$



                              $p∨(sim p∧q)=pvee q$ by the Absorption law.



                              Try $p=1,q=0$.






                              share|cite|improve this answer









                              $endgroup$



                              $(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$



                              $p∨(sim p∧q)=pvee q$ by the Absorption law.



                              Try $p=1,q=0$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 5 at 12:45









                              Shubham JohriShubham Johri

                              5,204718




                              5,204718























                                  0












                                  $begingroup$

                                  The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.



                                  The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.



                                    The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.



                                      The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.



                                      The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 5 at 12:45









                                      WuestenfuxWuestenfux

                                      4,7921513




                                      4,7921513






























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