Show the logical equivalence of $p to q to perp$ and $p wedge q to perp$. [on hold]












-1














I'm asked to show the logical equivalence of $p to q to perp$ and $p wedge q to perp$. When I make the truth tables, the two statements seem to come out to two very different outcomes, perhaps I'm interpreting the question wrong and the answer is that they aren't logically equivalent?






propositional-calculus







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put on hold as off-topic by Mauro ALLEGRANZA, amWhy, José Carlos Santos, KReiser, Shailesh yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
    – 5xum
    yesterday






  • 3




    Where are parentheses ? In order to anwer correctly, you have to put them...
    – Mauro ALLEGRANZA
    yesterday












  • We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
    – Mauro ALLEGRANZA
    yesterday










  • And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
    – Mauro ALLEGRANZA
    yesterday










  • You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
    – Mark Bennet
    yesterday
















-1














I'm asked to show the logical equivalence of $p to q to perp$ and $p wedge q to perp$. When I make the truth tables, the two statements seem to come out to two very different outcomes, perhaps I'm interpreting the question wrong and the answer is that they aren't logically equivalent?






propositional-calculus







share|cite|improve this question









New contributor




Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by Mauro ALLEGRANZA, amWhy, José Carlos Santos, KReiser, Shailesh yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
    – 5xum
    yesterday






  • 3




    Where are parentheses ? In order to anwer correctly, you have to put them...
    – Mauro ALLEGRANZA
    yesterday












  • We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
    – Mauro ALLEGRANZA
    yesterday










  • And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
    – Mauro ALLEGRANZA
    yesterday










  • You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
    – Mark Bennet
    yesterday














-1












-1








-1







I'm asked to show the logical equivalence of $p to q to perp$ and $p wedge q to perp$. When I make the truth tables, the two statements seem to come out to two very different outcomes, perhaps I'm interpreting the question wrong and the answer is that they aren't logically equivalent?






propositional-calculus







share|cite|improve this question









New contributor




Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I'm asked to show the logical equivalence of $p to q to perp$ and $p wedge q to perp$. When I make the truth tables, the two statements seem to come out to two very different outcomes, perhaps I'm interpreting the question wrong and the answer is that they aren't logically equivalent?






propositional-calculus




propositional-calculus






share|cite|improve this question









New contributor




Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday









metamorphy

3,5721521




3,5721521






New contributor




Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









Richard Cameron

11




11




New contributor




Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Richard Cameron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by Mauro ALLEGRANZA, amWhy, José Carlos Santos, KReiser, Shailesh yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Mauro ALLEGRANZA, amWhy, José Carlos Santos, KReiser, Shailesh yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
    – 5xum
    yesterday






  • 3




    Where are parentheses ? In order to anwer correctly, you have to put them...
    – Mauro ALLEGRANZA
    yesterday












  • We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
    – Mauro ALLEGRANZA
    yesterday










  • And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
    – Mauro ALLEGRANZA
    yesterday










  • You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
    – Mark Bennet
    yesterday


















  • Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
    – 5xum
    yesterday






  • 3




    Where are parentheses ? In order to anwer correctly, you have to put them...
    – Mauro ALLEGRANZA
    yesterday












  • We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
    – Mauro ALLEGRANZA
    yesterday










  • And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
    – Mauro ALLEGRANZA
    yesterday










  • You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
    – Mark Bennet
    yesterday
















Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
– 5xum
yesterday




Which values of $p$ and $q$ did you take to get the "two very different outcomes"?
– 5xum
yesterday




3




3




Where are parentheses ? In order to anwer correctly, you have to put them...
– Mauro ALLEGRANZA
yesterday






Where are parentheses ? In order to anwer correctly, you have to put them...
– Mauro ALLEGRANZA
yesterday














We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
– Mauro ALLEGRANZA
yesterday




We have that $lnot (p to q)$ is equivalent to $(p land lnot q)$
– Mauro ALLEGRANZA
yesterday












And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
– Mauro ALLEGRANZA
yesterday




And also that $(p to lnot q)$ is equiv to $lnot (p land q)$.
– Mauro ALLEGRANZA
yesterday












You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
– Mark Bennet
yesterday




You need brackets to identify what goes with what. When you are combining binary operators which need not be associative, this is crucial.
– Mark Bennet
yesterday










1 Answer
1






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0














First, note that $x to perp = lnot x lor perp = lnot x$. So



$p wedge q to perp = lnot(p land q)$



$(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$



$p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$



So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    First, note that $x to perp = lnot x lor perp = lnot x$. So



    $p wedge q to perp = lnot(p land q)$



    $(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$



    $p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$



    So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.






    share|cite|improve this answer


























      0














      First, note that $x to perp = lnot x lor perp = lnot x$. So



      $p wedge q to perp = lnot(p land q)$



      $(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$



      $p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$



      So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.






      share|cite|improve this answer
























        0












        0








        0






        First, note that $x to perp = lnot x lor perp = lnot x$. So



        $p wedge q to perp = lnot(p land q)$



        $(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$



        $p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$



        So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.






        share|cite|improve this answer












        First, note that $x to perp = lnot x lor perp = lnot x$. So



        $p wedge q to perp = lnot(p land q)$



        $(p to q) to perp = lnot(p to q) = lnot(lnot p lor q) = p land lnot q$



        $p to (q to perp) = p to lnot q = lnot p lor lnot q = lnot (p land q)$



        So, as pointed out in the comments above, it all depends where you put the parentheses when you interpret $p to q to perp$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        gandalf61

        7,703623




        7,703623















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