In Discrete Mathematics, is there a difference between $(neg P wedge neg Q)$ and $neg (P wedge Q)$?












0












$begingroup$


I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.



My query comes from a practice problem in a book:




Either John and Bill are both telling the truth, or neither of them is.




Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.



Thank you.






discrete-mathematics logic propositional-calculus logic-translation







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:46












  • $begingroup$
    @Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:48












  • $begingroup$
    Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:48












  • $begingroup$
    @Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:49












  • $begingroup$
    OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:51


















0












$begingroup$


I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.



My query comes from a practice problem in a book:




Either John and Bill are both telling the truth, or neither of them is.




Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.



Thank you.






discrete-mathematics logic propositional-calculus logic-translation







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:46












  • $begingroup$
    @Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:48












  • $begingroup$
    Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:48












  • $begingroup$
    @Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:49












  • $begingroup$
    OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:51
















0












0








0





$begingroup$


I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.



My query comes from a practice problem in a book:




Either John and Bill are both telling the truth, or neither of them is.




Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.



Thank you.






discrete-mathematics logic propositional-calculus logic-translation







share|cite|improve this question











$endgroup$




I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.



My query comes from a practice problem in a book:




Either John and Bill are both telling the truth, or neither of them is.




Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.



Thank you.






discrete-mathematics logic propositional-calculus logic-translation




discrete-mathematics logic propositional-calculus logic-translation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 21:22









Bram28

64.5k44793




64.5k44793










asked Dec 27 '18 at 18:42









Jamie CorkhillJamie Corkhill

1178




1178








  • 3




    $begingroup$
    Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:46












  • $begingroup$
    @Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:48












  • $begingroup$
    Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:48












  • $begingroup$
    @Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:49












  • $begingroup$
    OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:51
















  • 3




    $begingroup$
    Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:46












  • $begingroup$
    @Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:48












  • $begingroup$
    Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:48












  • $begingroup$
    @Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
    $endgroup$
    – Jamie Corkhill
    Dec 27 '18 at 18:49












  • $begingroup$
    OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
    $endgroup$
    – Bram28
    Dec 27 '18 at 18:51










3




3




$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46






$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46














$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48






$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48














$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48






$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48














$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49






$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49














$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51






$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51












3 Answers
3






active

oldest

votes


















3












$begingroup$

Neither John nor Bill telling the truth is:



$$neg P land neg Q$$



or, equivalently:



$$neg (P lor Q)$$



But that is not the same as:



$$neg (P land Q)$$



because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.



In short: 'both not' is not the same as 'not both'






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Try this for yourself:



    Suppose $P$ is true and $Q$ is false. Then:




    • Is $(neg P wedge neg Q)$ true or false?

    • Is $neg(Pwedge Q)$ true or false?






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      These two statements does not have the same truth value as
      By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.



      Here,
      John and Bill are both telling the truth corresponds to $p wedge q$.



      The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).



      It is not the same as neither of them is telling the truth($neg p vee neg q$).






      share|cite|improve this answer











      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Neither John nor Bill telling the truth is:



        $$neg P land neg Q$$



        or, equivalently:



        $$neg (P lor Q)$$



        But that is not the same as:



        $$neg (P land Q)$$



        because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.



        In short: 'both not' is not the same as 'not both'






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Neither John nor Bill telling the truth is:



          $$neg P land neg Q$$



          or, equivalently:



          $$neg (P lor Q)$$



          But that is not the same as:



          $$neg (P land Q)$$



          because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.



          In short: 'both not' is not the same as 'not both'






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Neither John nor Bill telling the truth is:



            $$neg P land neg Q$$



            or, equivalently:



            $$neg (P lor Q)$$



            But that is not the same as:



            $$neg (P land Q)$$



            because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.



            In short: 'both not' is not the same as 'not both'






            share|cite|improve this answer









            $endgroup$



            Neither John nor Bill telling the truth is:



            $$neg P land neg Q$$



            or, equivalently:



            $$neg (P lor Q)$$



            But that is not the same as:



            $$neg (P land Q)$$



            because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.



            In short: 'both not' is not the same as 'not both'







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 27 '18 at 18:55









            Bram28Bram28

            64.5k44793




            64.5k44793























                1












                $begingroup$

                Try this for yourself:



                Suppose $P$ is true and $Q$ is false. Then:




                • Is $(neg P wedge neg Q)$ true or false?

                • Is $neg(Pwedge Q)$ true or false?






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Try this for yourself:



                  Suppose $P$ is true and $Q$ is false. Then:




                  • Is $(neg P wedge neg Q)$ true or false?

                  • Is $neg(Pwedge Q)$ true or false?






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Try this for yourself:



                    Suppose $P$ is true and $Q$ is false. Then:




                    • Is $(neg P wedge neg Q)$ true or false?

                    • Is $neg(Pwedge Q)$ true or false?






                    share|cite|improve this answer









                    $endgroup$



                    Try this for yourself:



                    Suppose $P$ is true and $Q$ is false. Then:




                    • Is $(neg P wedge neg Q)$ true or false?

                    • Is $neg(Pwedge Q)$ true or false?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 27 '18 at 18:48









                    TonyKTonyK

                    44.1k358137




                    44.1k358137























                        0












                        $begingroup$

                        These two statements does not have the same truth value as
                        By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.



                        Here,
                        John and Bill are both telling the truth corresponds to $p wedge q$.



                        The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).



                        It is not the same as neither of them is telling the truth($neg p vee neg q$).






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          These two statements does not have the same truth value as
                          By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.



                          Here,
                          John and Bill are both telling the truth corresponds to $p wedge q$.



                          The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).



                          It is not the same as neither of them is telling the truth($neg p vee neg q$).






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            These two statements does not have the same truth value as
                            By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.



                            Here,
                            John and Bill are both telling the truth corresponds to $p wedge q$.



                            The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).



                            It is not the same as neither of them is telling the truth($neg p vee neg q$).






                            share|cite|improve this answer











                            $endgroup$



                            These two statements does not have the same truth value as
                            By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.



                            Here,
                            John and Bill are both telling the truth corresponds to $p wedge q$.



                            The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).



                            It is not the same as neither of them is telling the truth($neg p vee neg q$).







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 27 '18 at 18:54

























                            answered Dec 27 '18 at 18:48









                            mathpadawanmathpadawan

                            2,021422




                            2,021422






























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