∼p∨(∼p∧q)≡∼p∧∼q , prove logical equivalence












0












$begingroup$


Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.



So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!



The theorems are:




  1. Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p


  2. Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)


  3. Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)


  4. Identity laws: p∧t ≡ p , p∨c ≡ p


  5. Negation laws: p∨∼p ≡ t , p∧∼p ≡ c


  6. Double negative law: ∼(∼p) ≡ p


  7. Idempotent laws: p∧p ≡ p , p∨p ≡ p


  8. Universal bound laws: p∨t≡t ,p∧c≡c


  9. De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q


  10. Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p


  11. Negations of t and c: ∼t ≡ c , ∼c ≡ t







discrete-mathematics logic







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$endgroup$












  • $begingroup$
    Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
    $endgroup$
    – Waqad Arshad
    Jan 12 at 8:55
















0












$begingroup$


Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.



So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!



The theorems are:




  1. Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p


  2. Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)


  3. Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)


  4. Identity laws: p∧t ≡ p , p∨c ≡ p


  5. Negation laws: p∨∼p ≡ t , p∧∼p ≡ c


  6. Double negative law: ∼(∼p) ≡ p


  7. Idempotent laws: p∧p ≡ p , p∨p ≡ p


  8. Universal bound laws: p∨t≡t ,p∧c≡c


  9. De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q


  10. Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p


  11. Negations of t and c: ∼t ≡ c , ∼c ≡ t







discrete-mathematics logic







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
    $endgroup$
    – Waqad Arshad
    Jan 12 at 8:55














0












0








0


1



$begingroup$


Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.



So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!



The theorems are:




  1. Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p


  2. Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)


  3. Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)


  4. Identity laws: p∧t ≡ p , p∨c ≡ p


  5. Negation laws: p∨∼p ≡ t , p∧∼p ≡ c


  6. Double negative law: ∼(∼p) ≡ p


  7. Idempotent laws: p∧p ≡ p , p∨p ≡ p


  8. Universal bound laws: p∨t≡t ,p∧c≡c


  9. De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q


  10. Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p


  11. Negations of t and c: ∼t ≡ c , ∼c ≡ t







discrete-mathematics logic







share|cite|improve this question











$endgroup$




Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.



So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!



The theorems are:




  1. Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p


  2. Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)


  3. Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)


  4. Identity laws: p∧t ≡ p , p∨c ≡ p


  5. Negation laws: p∨∼p ≡ t , p∧∼p ≡ c


  6. Double negative law: ∼(∼p) ≡ p


  7. Idempotent laws: p∧p ≡ p , p∨p ≡ p


  8. Universal bound laws: p∨t≡t ,p∧c≡c


  9. De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q


  10. Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p


  11. Negations of t and c: ∼t ≡ c , ∼c ≡ t







discrete-mathematics logic




discrete-mathematics logic






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edited Jan 12 at 8:24







Waqad Arshad

















asked Jan 12 at 8:19









Waqad ArshadWaqad Arshad

52




52












  • $begingroup$
    Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
    $endgroup$
    – Waqad Arshad
    Jan 12 at 8:55


















  • $begingroup$
    Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
    $endgroup$
    – Waqad Arshad
    Jan 12 at 8:55
















$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55




$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55










4 Answers
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$begingroup$

I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$






share|cite|improve this answer









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    2












    $begingroup$

    If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.



      So there no logical equivalence.






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$

        Hint



        Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.






        share|cite|improve this answer









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          4 Answers
          4






          active

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          4 Answers
          4






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          oldest

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          active

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          active

          oldest

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          3












          $begingroup$

          I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$






              share|cite|improve this answer









              $endgroup$



              I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 12 at 8:46









              PatricioPatricio

              3257




              3257























                  2












                  $begingroup$

                  If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.






                      share|cite|improve this answer









                      $endgroup$



                      If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 12 at 8:44









                      coffeemathcoffeemath

                      2,8451415




                      2,8451415























                          2












                          $begingroup$

                          If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.



                          So there no logical equivalence.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.



                            So there no logical equivalence.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.



                              So there no logical equivalence.






                              share|cite|improve this answer









                              $endgroup$



                              If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.



                              So there no logical equivalence.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 12 at 8:46









                              drhabdrhab

                              101k544130




                              101k544130























                                  2












                                  $begingroup$

                                  Hint



                                  Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    Hint



                                    Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      Hint



                                      Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Hint



                                      Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 12 at 8:48









                                      Mostafa AyazMostafa Ayaz

                                      15.6k3939




                                      15.6k3939






























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