Dual of each of these compound propositions.












1












$begingroup$


I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:



a) $p∧¬q∧¬r$



b) $(p∧q∧r)∨s$



c) $(p∨F)∧(q∨T)$



My answers:



a) $¬p∨q∨r$



b) $(¬p∨¬q∨¬r)∧¬s$



c) $(¬p∧T)∨(¬q∧F)$



Did I got it right?
Thank you!






discrete-mathematics logic proof-verification







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  • $begingroup$
    What is your question?
    $endgroup$
    – Yeah..
    Feb 17 '16 at 2:22










  • $begingroup$
    @Yeah. The question appears to be : "Did I get it right?"
    $endgroup$
    – Graham Kemp
    Feb 17 '16 at 2:23










  • $begingroup$
    @GrahamKemp YES! :P
    $endgroup$
    – Krin Yongvongphaiboon
    Feb 17 '16 at 2:24
















1












$begingroup$


I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:



a) $p∧¬q∧¬r$



b) $(p∧q∧r)∨s$



c) $(p∨F)∧(q∨T)$



My answers:



a) $¬p∨q∨r$



b) $(¬p∨¬q∨¬r)∧¬s$



c) $(¬p∧T)∨(¬q∧F)$



Did I got it right?
Thank you!






discrete-mathematics logic proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question?
    $endgroup$
    – Yeah..
    Feb 17 '16 at 2:22










  • $begingroup$
    @Yeah. The question appears to be : "Did I get it right?"
    $endgroup$
    – Graham Kemp
    Feb 17 '16 at 2:23










  • $begingroup$
    @GrahamKemp YES! :P
    $endgroup$
    – Krin Yongvongphaiboon
    Feb 17 '16 at 2:24














1












1








1





$begingroup$


I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:



a) $p∧¬q∧¬r$



b) $(p∧q∧r)∨s$



c) $(p∨F)∧(q∨T)$



My answers:



a) $¬p∨q∨r$



b) $(¬p∨¬q∨¬r)∧¬s$



c) $(¬p∧T)∨(¬q∧F)$



Did I got it right?
Thank you!






discrete-mathematics logic proof-verification







share|cite|improve this question











$endgroup$




I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:



a) $p∧¬q∧¬r$



b) $(p∧q∧r)∨s$



c) $(p∨F)∧(q∨T)$



My answers:



a) $¬p∨q∨r$



b) $(¬p∨¬q∨¬r)∧¬s$



c) $(¬p∧T)∨(¬q∧F)$



Did I got it right?
Thank you!






discrete-mathematics logic proof-verification




discrete-mathematics logic proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 17 '16 at 2:25







Krin Yongvongphaiboon

















asked Feb 17 '16 at 2:17









Krin YongvongphaiboonKrin Yongvongphaiboon

64




64












  • $begingroup$
    What is your question?
    $endgroup$
    – Yeah..
    Feb 17 '16 at 2:22










  • $begingroup$
    @Yeah. The question appears to be : "Did I get it right?"
    $endgroup$
    – Graham Kemp
    Feb 17 '16 at 2:23










  • $begingroup$
    @GrahamKemp YES! :P
    $endgroup$
    – Krin Yongvongphaiboon
    Feb 17 '16 at 2:24


















  • $begingroup$
    What is your question?
    $endgroup$
    – Yeah..
    Feb 17 '16 at 2:22










  • $begingroup$
    @Yeah. The question appears to be : "Did I get it right?"
    $endgroup$
    – Graham Kemp
    Feb 17 '16 at 2:23










  • $begingroup$
    @GrahamKemp YES! :P
    $endgroup$
    – Krin Yongvongphaiboon
    Feb 17 '16 at 2:24
















$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22




$begingroup$
What is your question?
$endgroup$
– Yeah..
Feb 17 '16 at 2:22












$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23




$begingroup$
@Yeah. The question appears to be : "Did I get it right?"
$endgroup$
– Graham Kemp
Feb 17 '16 at 2:23












$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24




$begingroup$
@GrahamKemp YES! :P
$endgroup$
– Krin Yongvongphaiboon
Feb 17 '16 at 2:24










2 Answers
2






active

oldest

votes


















2












$begingroup$

What you have done is wrong. Why you have negated all the proposition and the then find the dual?



Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).



Here is the definition of dual of a compound proposition-
"The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s∗."
(Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)



Example:

S =(p ∧ q)∨ (¬p ∨ q)∨ F

S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)



So the correct answer to your question is:



a) p∧¬q∧¬r



b) (p∧q∧r)∨s



c) (p∨F)∧(q∨T)



answers:



a) p∨¬q∨¬r



b) (p∨q∨r)∧s



c) (p∧T)∨(q∧F)






share|cite|improve this answer











$endgroup$





















    -1












    $begingroup$

    Yes.   All three check: okay.



    You've negated all the propositions and dualled all the operators.



    That is what you needed to do.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for confirming :)
      $endgroup$
      – Krin Yongvongphaiboon
      Feb 17 '16 at 2:25











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    What you have done is wrong. Why you have negated all the proposition and the then find the dual?



    Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).



    Here is the definition of dual of a compound proposition-
    "The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
    by F, and each F by T. The dual of s is denoted by s∗."
    (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)



    Example:

    S =(p ∧ q)∨ (¬p ∨ q)∨ F

    S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)



    So the correct answer to your question is:



    a) p∧¬q∧¬r



    b) (p∧q∧r)∨s



    c) (p∨F)∧(q∨T)



    answers:



    a) p∨¬q∨¬r



    b) (p∨q∨r)∧s



    c) (p∧T)∨(q∧F)






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      What you have done is wrong. Why you have negated all the proposition and the then find the dual?



      Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).



      Here is the definition of dual of a compound proposition-
      "The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
      by F, and each F by T. The dual of s is denoted by s∗."
      (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)



      Example:

      S =(p ∧ q)∨ (¬p ∨ q)∨ F

      S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)



      So the correct answer to your question is:



      a) p∧¬q∧¬r



      b) (p∧q∧r)∨s



      c) (p∨F)∧(q∨T)



      answers:



      a) p∨¬q∨¬r



      b) (p∨q∨r)∧s



      c) (p∧T)∨(q∧F)






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        What you have done is wrong. Why you have negated all the proposition and the then find the dual?



        Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).



        Here is the definition of dual of a compound proposition-
        "The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
        by F, and each F by T. The dual of s is denoted by s∗."
        (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)



        Example:

        S =(p ∧ q)∨ (¬p ∨ q)∨ F

        S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)



        So the correct answer to your question is:



        a) p∧¬q∧¬r



        b) (p∧q∧r)∨s



        c) (p∨F)∧(q∨T)



        answers:



        a) p∨¬q∨¬r



        b) (p∨q∨r)∧s



        c) (p∧T)∨(q∧F)






        share|cite|improve this answer











        $endgroup$



        What you have done is wrong. Why you have negated all the proposition and the then find the dual?



        Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).



        Here is the definition of dual of a compound proposition-
        "The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
        by F, and each F by T. The dual of s is denoted by s∗."
        (Reference: Discrete Mathematics (7th Edition) Kenneth H. Rosen)



        Example:

        S =(p ∧ q)∨ (¬p ∨ q)∨ F

        S* =(p ∨ q)∧ (¬p ∧ q)∧T (dual of S is denoted by S*)



        So the correct answer to your question is:



        a) p∧¬q∧¬r



        b) (p∧q∧r)∨s



        c) (p∨F)∧(q∨T)



        answers:



        a) p∨¬q∨¬r



        b) (p∨q∨r)∧s



        c) (p∧T)∨(q∧F)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 18 at 16:57

























        answered Jan 18 at 16:39









        Sumit RanjanSumit Ranjan

        315




        315























            -1












            $begingroup$

            Yes.   All three check: okay.



            You've negated all the propositions and dualled all the operators.



            That is what you needed to do.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for confirming :)
              $endgroup$
              – Krin Yongvongphaiboon
              Feb 17 '16 at 2:25
















            -1












            $begingroup$

            Yes.   All three check: okay.



            You've negated all the propositions and dualled all the operators.



            That is what you needed to do.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for confirming :)
              $endgroup$
              – Krin Yongvongphaiboon
              Feb 17 '16 at 2:25














            -1












            -1








            -1





            $begingroup$

            Yes.   All three check: okay.



            You've negated all the propositions and dualled all the operators.



            That is what you needed to do.






            share|cite|improve this answer











            $endgroup$



            Yes.   All three check: okay.



            You've negated all the propositions and dualled all the operators.



            That is what you needed to do.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            answered Feb 17 '16 at 2:21


























            community wiki





            Graham Kemp













            • $begingroup$
              Thank you for confirming :)
              $endgroup$
              – Krin Yongvongphaiboon
              Feb 17 '16 at 2:25


















            • $begingroup$
              Thank you for confirming :)
              $endgroup$
              – Krin Yongvongphaiboon
              Feb 17 '16 at 2:25
















            $begingroup$
            Thank you for confirming :)
            $endgroup$
            – Krin Yongvongphaiboon
            Feb 17 '16 at 2:25




            $begingroup$
            Thank you for confirming :)
            $endgroup$
            – Krin Yongvongphaiboon
            Feb 17 '16 at 2:25


















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