Logical equivalence of ¬p→q










1












$begingroup$


Just wondering what other ways $neg p to q$ can be expressed.



I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    Just wondering what other ways $neg p to q$ can be expressed.



    I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      Just wondering what other ways $neg p to q$ can be expressed.



      I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.










      share|cite|improve this question











      $endgroup$




      Just wondering what other ways $neg p to q$ can be expressed.



      I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.







      propositional-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 11 '18 at 18:41









      Mutantoe

      604513




      604513










      asked Nov 11 '18 at 8:23









      Jordan SolomonsJordan Solomons

      197




      197




















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.



          But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.



          Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanks for confirming!
            $endgroup$
            – Jordan Solomons
            Nov 11 '18 at 10:55


















          2












          $begingroup$

          You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.



          In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
            Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.






            share|cite|improve this answer









            $endgroup$












              Your Answer





              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("mathjaxEditing", function ()
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              );
              );
              , "mathjax-editing");

              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader:
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              ,
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













              draft saved

              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2993594%2flogical-equivalence-of-%25c2%25acp%25e2%2586%2592q%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.



              But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.



              Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                Thanks for confirming!
                $endgroup$
                – Jordan Solomons
                Nov 11 '18 at 10:55















              5












              $begingroup$

              As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.



              But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.



              Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                Thanks for confirming!
                $endgroup$
                – Jordan Solomons
                Nov 11 '18 at 10:55













              5












              5








              5





              $begingroup$

              As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.



              But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.



              Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.






              share|cite|improve this answer









              $endgroup$



              As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.



              But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.



              Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 11 '18 at 8:49









              Peter SmithPeter Smith

              40.6k340120




              40.6k340120











              • $begingroup$
                Thanks for confirming!
                $endgroup$
                – Jordan Solomons
                Nov 11 '18 at 10:55
















              • $begingroup$
                Thanks for confirming!
                $endgroup$
                – Jordan Solomons
                Nov 11 '18 at 10:55















              $begingroup$
              Thanks for confirming!
              $endgroup$
              – Jordan Solomons
              Nov 11 '18 at 10:55




              $begingroup$
              Thanks for confirming!
              $endgroup$
              – Jordan Solomons
              Nov 11 '18 at 10:55











              2












              $begingroup$

              You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.



              In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.



                In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.



                  In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.






                  share|cite|improve this answer









                  $endgroup$



                  You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.



                  In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 11 '18 at 8:34









                  MacRanceMacRance

                  1226




                  1226





















                      1












                      $begingroup$

                      As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
                      Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
                        Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
                          Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.






                          share|cite|improve this answer









                          $endgroup$



                          As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
                          Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 11 '18 at 8:32









                          WuestenfuxWuestenfux

                          4,2691413




                          4,2691413



























                              draft saved

                              draft discarded
















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid


                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.

                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2993594%2flogical-equivalence-of-%25c2%25acp%25e2%2586%2592q%23new-answer', 'question_page');

                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

                              Edmonton

                              Crossroads (UK TV series)