Help with logical equivalence/proving not a contradiction












1












$begingroup$


The question is this:



Demonstrate using logical equivalences that $(p → q) ∧ (p → ¬q)$ is not
a contradiction. Identify all logical equivalences by name.



So far, I have



$(p → q) land (p → ¬q)$



a. $(¬plor q) land (¬plor¬q)$



b. $¬(¬plor q) lor(¬(¬plor¬q)$



c. $( pland¬q) lor (p lor q)$



I can't figure out where to go from here. Any help would be appreciated.






discrete-mathematics logic propositional-calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    The question is this:



    Demonstrate using logical equivalences that $(p → q) ∧ (p → ¬q)$ is not
    a contradiction. Identify all logical equivalences by name.



    So far, I have



    $(p → q) land (p → ¬q)$



    a. $(¬plor q) land (¬plor¬q)$



    b. $¬(¬plor q) lor(¬(¬plor¬q)$



    c. $( pland¬q) lor (p lor q)$



    I can't figure out where to go from here. Any help would be appreciated.






    discrete-mathematics logic propositional-calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      The question is this:



      Demonstrate using logical equivalences that $(p → q) ∧ (p → ¬q)$ is not
      a contradiction. Identify all logical equivalences by name.



      So far, I have



      $(p → q) land (p → ¬q)$



      a. $(¬plor q) land (¬plor¬q)$



      b. $¬(¬plor q) lor(¬(¬plor¬q)$



      c. $( pland¬q) lor (p lor q)$



      I can't figure out where to go from here. Any help would be appreciated.






      discrete-mathematics logic propositional-calculus







      share|cite|improve this question











      $endgroup$




      The question is this:



      Demonstrate using logical equivalences that $(p → q) ∧ (p → ¬q)$ is not
      a contradiction. Identify all logical equivalences by name.



      So far, I have



      $(p → q) land (p → ¬q)$



      a. $(¬plor q) land (¬plor¬q)$



      b. $¬(¬plor q) lor(¬(¬plor¬q)$



      c. $( pland¬q) lor (p lor q)$



      I can't figure out where to go from here. Any help would be appreciated.






      discrete-mathematics logic propositional-calculus




      discrete-mathematics logic propositional-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 14 at 9:58









      Mostafa Ayaz

      17.1k31039




      17.1k31039










      asked Jan 14 at 1:05









      Keith ReynoldsKeith Reynolds

      62




      62






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          It is not a contradiction becauss it is true when p is false.

          Truth tables show this quickly.

          a. perhaps is the best for showing that.

          a. is equivalent to: (not p) or (q and not q);

          which in turn is equivalent to; not p.

          So there's the answer - to show the orginal statement is

          equivalent to: not p.



          Little surprise, as the orginal statement is basically

          proof by contradiction.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            From :



            a) $(¬p lor q) land (¬p lor ¬q)$



            we have to apply the Distributive law to get :



            b) $lnot p lor (q land lnot q)$.



            Now we use the equivalence (often called : Negation law) : $(alpha land lnot alpha) equiv text F$ to get :



            c) $lnot p lor text F$.



            Finally, we use the equivalence (often called : Identity law) : $alpha lor text F equiv alpha$ to conclude with :




            d) $lnot p$




            that is not a contradiction.






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              An easy way to say that is: since $p$ concludes $q$ and $lnot q$, if it holds true then two controversy statements hold true which is a contradiction, but if $lnot p$ is true, the contradiction is removed.



              In a more formal way $$(pto q)land (pto lnot q)iff pto (qland lnot q)iff pto 0iff lnot pto 1$$which means that $lnot p$ is true.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Are you missing a $p$ here?
                $endgroup$
                – Max Herrmann
                Jan 14 at 11:19










              • $begingroup$
                Where do you mean I'm missing?
                $endgroup$
                – Mostafa Ayaz
                Jan 14 at 14:35










              • $begingroup$
                (...), but if $neg$ is true, (...)
                $endgroup$
                – Max Herrmann
                Jan 14 at 15:37










              • $begingroup$
                Thank you. Nice look through my answer :)
                $endgroup$
                – Mostafa Ayaz
                Jan 14 at 15:40



















              0












              $begingroup$

              Just an add-on to William Elliot's answer:



              $$
              begin{array}{c|c|c|c|c}
              p & q & p rightarrow q & p rightarrow neg q & (p rightarrow q) wedge (p rightarrow neg q) \ hline
              0 & 0 & 1 & 1 & 1 \
              0 & 1 & 1 & 1 & 1 \
              1 & 0 & 0 & 1 & 0 \
              1 & 1 & 1 & 0 & 0
              end{array}
              $$






              share|cite|improve this answer









              $endgroup$













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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                It is not a contradiction becauss it is true when p is false.

                Truth tables show this quickly.

                a. perhaps is the best for showing that.

                a. is equivalent to: (not p) or (q and not q);

                which in turn is equivalent to; not p.

                So there's the answer - to show the orginal statement is

                equivalent to: not p.



                Little surprise, as the orginal statement is basically

                proof by contradiction.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  It is not a contradiction becauss it is true when p is false.

                  Truth tables show this quickly.

                  a. perhaps is the best for showing that.

                  a. is equivalent to: (not p) or (q and not q);

                  which in turn is equivalent to; not p.

                  So there's the answer - to show the orginal statement is

                  equivalent to: not p.



                  Little surprise, as the orginal statement is basically

                  proof by contradiction.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    It is not a contradiction becauss it is true when p is false.

                    Truth tables show this quickly.

                    a. perhaps is the best for showing that.

                    a. is equivalent to: (not p) or (q and not q);

                    which in turn is equivalent to; not p.

                    So there's the answer - to show the orginal statement is

                    equivalent to: not p.



                    Little surprise, as the orginal statement is basically

                    proof by contradiction.






                    share|cite|improve this answer









                    $endgroup$



                    It is not a contradiction becauss it is true when p is false.

                    Truth tables show this quickly.

                    a. perhaps is the best for showing that.

                    a. is equivalent to: (not p) or (q and not q);

                    which in turn is equivalent to; not p.

                    So there's the answer - to show the orginal statement is

                    equivalent to: not p.



                    Little surprise, as the orginal statement is basically

                    proof by contradiction.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 14 at 3:14









                    William ElliotWilliam Elliot

                    8,7552820




                    8,7552820























                        1












                        $begingroup$

                        From :



                        a) $(¬p lor q) land (¬p lor ¬q)$



                        we have to apply the Distributive law to get :



                        b) $lnot p lor (q land lnot q)$.



                        Now we use the equivalence (often called : Negation law) : $(alpha land lnot alpha) equiv text F$ to get :



                        c) $lnot p lor text F$.



                        Finally, we use the equivalence (often called : Identity law) : $alpha lor text F equiv alpha$ to conclude with :




                        d) $lnot p$




                        that is not a contradiction.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          From :



                          a) $(¬p lor q) land (¬p lor ¬q)$



                          we have to apply the Distributive law to get :



                          b) $lnot p lor (q land lnot q)$.



                          Now we use the equivalence (often called : Negation law) : $(alpha land lnot alpha) equiv text F$ to get :



                          c) $lnot p lor text F$.



                          Finally, we use the equivalence (often called : Identity law) : $alpha lor text F equiv alpha$ to conclude with :




                          d) $lnot p$




                          that is not a contradiction.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            From :



                            a) $(¬p lor q) land (¬p lor ¬q)$



                            we have to apply the Distributive law to get :



                            b) $lnot p lor (q land lnot q)$.



                            Now we use the equivalence (often called : Negation law) : $(alpha land lnot alpha) equiv text F$ to get :



                            c) $lnot p lor text F$.



                            Finally, we use the equivalence (often called : Identity law) : $alpha lor text F equiv alpha$ to conclude with :




                            d) $lnot p$




                            that is not a contradiction.






                            share|cite|improve this answer











                            $endgroup$



                            From :



                            a) $(¬p lor q) land (¬p lor ¬q)$



                            we have to apply the Distributive law to get :



                            b) $lnot p lor (q land lnot q)$.



                            Now we use the equivalence (often called : Negation law) : $(alpha land lnot alpha) equiv text F$ to get :



                            c) $lnot p lor text F$.



                            Finally, we use the equivalence (often called : Identity law) : $alpha lor text F equiv alpha$ to conclude with :




                            d) $lnot p$




                            that is not a contradiction.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 14 at 8:51

























                            answered Jan 14 at 7:10









                            Mauro ALLEGRANZAMauro ALLEGRANZA

                            67.3k449115




                            67.3k449115























                                1












                                $begingroup$

                                An easy way to say that is: since $p$ concludes $q$ and $lnot q$, if it holds true then two controversy statements hold true which is a contradiction, but if $lnot p$ is true, the contradiction is removed.



                                In a more formal way $$(pto q)land (pto lnot q)iff pto (qland lnot q)iff pto 0iff lnot pto 1$$which means that $lnot p$ is true.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Are you missing a $p$ here?
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 11:19










                                • $begingroup$
                                  Where do you mean I'm missing?
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 14:35










                                • $begingroup$
                                  (...), but if $neg$ is true, (...)
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 15:37










                                • $begingroup$
                                  Thank you. Nice look through my answer :)
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 15:40
















                                1












                                $begingroup$

                                An easy way to say that is: since $p$ concludes $q$ and $lnot q$, if it holds true then two controversy statements hold true which is a contradiction, but if $lnot p$ is true, the contradiction is removed.



                                In a more formal way $$(pto q)land (pto lnot q)iff pto (qland lnot q)iff pto 0iff lnot pto 1$$which means that $lnot p$ is true.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Are you missing a $p$ here?
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 11:19










                                • $begingroup$
                                  Where do you mean I'm missing?
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 14:35










                                • $begingroup$
                                  (...), but if $neg$ is true, (...)
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 15:37










                                • $begingroup$
                                  Thank you. Nice look through my answer :)
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 15:40














                                1












                                1








                                1





                                $begingroup$

                                An easy way to say that is: since $p$ concludes $q$ and $lnot q$, if it holds true then two controversy statements hold true which is a contradiction, but if $lnot p$ is true, the contradiction is removed.



                                In a more formal way $$(pto q)land (pto lnot q)iff pto (qland lnot q)iff pto 0iff lnot pto 1$$which means that $lnot p$ is true.






                                share|cite|improve this answer











                                $endgroup$



                                An easy way to say that is: since $p$ concludes $q$ and $lnot q$, if it holds true then two controversy statements hold true which is a contradiction, but if $lnot p$ is true, the contradiction is removed.



                                In a more formal way $$(pto q)land (pto lnot q)iff pto (qland lnot q)iff pto 0iff lnot pto 1$$which means that $lnot p$ is true.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jan 14 at 15:39

























                                answered Jan 14 at 10:03









                                Mostafa AyazMostafa Ayaz

                                17.1k31039




                                17.1k31039












                                • $begingroup$
                                  Are you missing a $p$ here?
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 11:19










                                • $begingroup$
                                  Where do you mean I'm missing?
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 14:35










                                • $begingroup$
                                  (...), but if $neg$ is true, (...)
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 15:37










                                • $begingroup$
                                  Thank you. Nice look through my answer :)
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 15:40


















                                • $begingroup$
                                  Are you missing a $p$ here?
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 11:19










                                • $begingroup$
                                  Where do you mean I'm missing?
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 14:35










                                • $begingroup$
                                  (...), but if $neg$ is true, (...)
                                  $endgroup$
                                  – Max Herrmann
                                  Jan 14 at 15:37










                                • $begingroup$
                                  Thank you. Nice look through my answer :)
                                  $endgroup$
                                  – Mostafa Ayaz
                                  Jan 14 at 15:40
















                                $begingroup$
                                Are you missing a $p$ here?
                                $endgroup$
                                – Max Herrmann
                                Jan 14 at 11:19




                                $begingroup$
                                Are you missing a $p$ here?
                                $endgroup$
                                – Max Herrmann
                                Jan 14 at 11:19












                                $begingroup$
                                Where do you mean I'm missing?
                                $endgroup$
                                – Mostafa Ayaz
                                Jan 14 at 14:35




                                $begingroup$
                                Where do you mean I'm missing?
                                $endgroup$
                                – Mostafa Ayaz
                                Jan 14 at 14:35












                                $begingroup$
                                (...), but if $neg$ is true, (...)
                                $endgroup$
                                – Max Herrmann
                                Jan 14 at 15:37




                                $begingroup$
                                (...), but if $neg$ is true, (...)
                                $endgroup$
                                – Max Herrmann
                                Jan 14 at 15:37












                                $begingroup$
                                Thank you. Nice look through my answer :)
                                $endgroup$
                                – Mostafa Ayaz
                                Jan 14 at 15:40




                                $begingroup$
                                Thank you. Nice look through my answer :)
                                $endgroup$
                                – Mostafa Ayaz
                                Jan 14 at 15:40











                                0












                                $begingroup$

                                Just an add-on to William Elliot's answer:



                                $$
                                begin{array}{c|c|c|c|c}
                                p & q & p rightarrow q & p rightarrow neg q & (p rightarrow q) wedge (p rightarrow neg q) \ hline
                                0 & 0 & 1 & 1 & 1 \
                                0 & 1 & 1 & 1 & 1 \
                                1 & 0 & 0 & 1 & 0 \
                                1 & 1 & 1 & 0 & 0
                                end{array}
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Just an add-on to William Elliot's answer:



                                  $$
                                  begin{array}{c|c|c|c|c}
                                  p & q & p rightarrow q & p rightarrow neg q & (p rightarrow q) wedge (p rightarrow neg q) \ hline
                                  0 & 0 & 1 & 1 & 1 \
                                  0 & 1 & 1 & 1 & 1 \
                                  1 & 0 & 0 & 1 & 0 \
                                  1 & 1 & 1 & 0 & 0
                                  end{array}
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Just an add-on to William Elliot's answer:



                                    $$
                                    begin{array}{c|c|c|c|c}
                                    p & q & p rightarrow q & p rightarrow neg q & (p rightarrow q) wedge (p rightarrow neg q) \ hline
                                    0 & 0 & 1 & 1 & 1 \
                                    0 & 1 & 1 & 1 & 1 \
                                    1 & 0 & 0 & 1 & 0 \
                                    1 & 1 & 1 & 0 & 0
                                    end{array}
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Just an add-on to William Elliot's answer:



                                    $$
                                    begin{array}{c|c|c|c|c}
                                    p & q & p rightarrow q & p rightarrow neg q & (p rightarrow q) wedge (p rightarrow neg q) \ hline
                                    0 & 0 & 1 & 1 & 1 \
                                    0 & 1 & 1 & 1 & 1 \
                                    1 & 0 & 0 & 1 & 0 \
                                    1 & 1 & 1 & 0 & 0
                                    end{array}
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 14 at 7:30









                                    Max HerrmannMax Herrmann

                                    724419




                                    724419






























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                                        張江高科駅

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                                        張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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