Proving both distributive laws for Boolean algebra given by certain axioms.












1












$begingroup$


In the book "Introduction to mathematical logic", Elliott Mendelson gives the following axiomatization of Boolean algebra:



We call the triple $(B,cap,')$ a Boolean algebra whenever $B$ has at least two elements, $cap$ (meet) is two-argument operator and $'$ (complement) is one-argument operator and the following axioms are satisfied:




  1. $xcap y=ycap x$

  2. $xcap (ycap z)=(xcap y)cap z$

  3. $xcap y'=zcap z' iff xcap y=x$


I managed to obtain a few results like:




  1. $xcap x=x$


  2. $xcap x'=ycap y'$ (hence we can define $mathbb{0}$)

  3. $xcapmathbb{0}=mathbb{0}$

  4. $x''=x$


Next we define $xcup y:=(x'cap y')'$ and $mathbb{1}:=mathbb{0}'$ and I proved some more properties:




  1. $xcup y=ycup x$

  2. $(xcup y)cup z= xcup(ycup z)$


  3. $xcup(xcap y)=x$ (absorption)


  4. $xcap(xcup y)=x$ (absorption)

  5. $xcup x'=mathbb{1}$


What I can't prove are both distributive laws:




  1. $(xcap y)cup z=(xcup z)cap(ycup z)$

  2. $(xcup y)cap z=(xcap z)cup(ycap z)$


Can you help me?






boolean-algebra







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    In the book "Introduction to mathematical logic", Elliott Mendelson gives the following axiomatization of Boolean algebra:



    We call the triple $(B,cap,')$ a Boolean algebra whenever $B$ has at least two elements, $cap$ (meet) is two-argument operator and $'$ (complement) is one-argument operator and the following axioms are satisfied:




    1. $xcap y=ycap x$

    2. $xcap (ycap z)=(xcap y)cap z$

    3. $xcap y'=zcap z' iff xcap y=x$


    I managed to obtain a few results like:




    1. $xcap x=x$


    2. $xcap x'=ycap y'$ (hence we can define $mathbb{0}$)

    3. $xcapmathbb{0}=mathbb{0}$

    4. $x''=x$


    Next we define $xcup y:=(x'cap y')'$ and $mathbb{1}:=mathbb{0}'$ and I proved some more properties:




    1. $xcup y=ycup x$

    2. $(xcup y)cup z= xcup(ycup z)$


    3. $xcup(xcap y)=x$ (absorption)


    4. $xcap(xcup y)=x$ (absorption)

    5. $xcup x'=mathbb{1}$


    What I can't prove are both distributive laws:




    1. $(xcap y)cup z=(xcup z)cap(ycup z)$

    2. $(xcup y)cap z=(xcap z)cup(ycap z)$


    Can you help me?






    boolean-algebra







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      In the book "Introduction to mathematical logic", Elliott Mendelson gives the following axiomatization of Boolean algebra:



      We call the triple $(B,cap,')$ a Boolean algebra whenever $B$ has at least two elements, $cap$ (meet) is two-argument operator and $'$ (complement) is one-argument operator and the following axioms are satisfied:




      1. $xcap y=ycap x$

      2. $xcap (ycap z)=(xcap y)cap z$

      3. $xcap y'=zcap z' iff xcap y=x$


      I managed to obtain a few results like:




      1. $xcap x=x$


      2. $xcap x'=ycap y'$ (hence we can define $mathbb{0}$)

      3. $xcapmathbb{0}=mathbb{0}$

      4. $x''=x$


      Next we define $xcup y:=(x'cap y')'$ and $mathbb{1}:=mathbb{0}'$ and I proved some more properties:




      1. $xcup y=ycup x$

      2. $(xcup y)cup z= xcup(ycup z)$


      3. $xcup(xcap y)=x$ (absorption)


      4. $xcap(xcup y)=x$ (absorption)

      5. $xcup x'=mathbb{1}$


      What I can't prove are both distributive laws:




      1. $(xcap y)cup z=(xcup z)cap(ycup z)$

      2. $(xcup y)cap z=(xcap z)cup(ycap z)$


      Can you help me?






      boolean-algebra







      share|cite|improve this question









      $endgroup$




      In the book "Introduction to mathematical logic", Elliott Mendelson gives the following axiomatization of Boolean algebra:



      We call the triple $(B,cap,')$ a Boolean algebra whenever $B$ has at least two elements, $cap$ (meet) is two-argument operator and $'$ (complement) is one-argument operator and the following axioms are satisfied:




      1. $xcap y=ycap x$

      2. $xcap (ycap z)=(xcap y)cap z$

      3. $xcap y'=zcap z' iff xcap y=x$


      I managed to obtain a few results like:




      1. $xcap x=x$


      2. $xcap x'=ycap y'$ (hence we can define $mathbb{0}$)

      3. $xcapmathbb{0}=mathbb{0}$

      4. $x''=x$


      Next we define $xcup y:=(x'cap y')'$ and $mathbb{1}:=mathbb{0}'$ and I proved some more properties:




      1. $xcup y=ycup x$

      2. $(xcup y)cup z= xcup(ycup z)$


      3. $xcup(xcap y)=x$ (absorption)


      4. $xcap(xcup y)=x$ (absorption)

      5. $xcup x'=mathbb{1}$


      What I can't prove are both distributive laws:




      1. $(xcap y)cup z=(xcup z)cap(ycup z)$

      2. $(xcup y)cap z=(xcap z)cup(ycap z)$


      Can you help me?






      boolean-algebra




      boolean-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 20 at 1:17









      KulistyKulisty

      399116




      399116






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Having defined $0$, rewrite your third axiom as
          begin{equation}label{eq:0}
          x wedge y' = 0 Leftrightarrow x wedge y = x,tag{0}
          end{equation}
          and since you have already proven that the structure is a complemented lattice (and only need distributivity to conclude it's a Boolean algebra), then $x leq y$ iff $x wedge y = x$, which is notationally convenient.
          Thus
          begin{equation}label{eq:1}
          x wedge y' = 0 Leftrightarrow x leq ytag{1}
          end{equation}
          Now, with the equations that you claim you proved,
          $$x wedge (x wedge y)' wedge y = (x wedge y) wedge (x wedge y)' = 0,$$
          whence, by eqref{eq:0}
          $$x wedge (x wedge y)' wedge y' = x wedge (x wedge y)',$$
          and analogously,
          $$x wedge (x wedge z)' wedge z' = x wedge (x wedge z)',$$
          yielding
          $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z') = x wedge (x wedge y)' wedge (x wedge z)',$$
          or
          $$x wedge (x wedge y)' wedge (x wedge z)' leq y' wedge z',$$
          and, by eqref{eq:1},
          $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z')' = 0.$$
          From your definition of join and some identities you proved, this is
          $$x wedge ((x wedge y) vee (x wedge z))' wedge (yvee z) = 0,$$
          or, by eqref{eq:1},
          $$x wedge (y vee z) leq (x wedge y) vee (x wedge z).$$
          Since
          $$x wedge (y vee z) geq (x wedge y) vee (x wedge z)$$
          holds in every lattice, we conclude that
          $$x wedge (y vee z) = (x wedge y) vee (x wedge z),$$
          and therefore, $B$ is a Boolean algebra.

          (I suppose you know that in a lattice each distributive law follows from the other.)






          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%2f3080052%2fproving-both-distributive-laws-for-boolean-algebra-given-by-certain-axioms%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Having defined $0$, rewrite your third axiom as
            begin{equation}label{eq:0}
            x wedge y' = 0 Leftrightarrow x wedge y = x,tag{0}
            end{equation}
            and since you have already proven that the structure is a complemented lattice (and only need distributivity to conclude it's a Boolean algebra), then $x leq y$ iff $x wedge y = x$, which is notationally convenient.
            Thus
            begin{equation}label{eq:1}
            x wedge y' = 0 Leftrightarrow x leq ytag{1}
            end{equation}
            Now, with the equations that you claim you proved,
            $$x wedge (x wedge y)' wedge y = (x wedge y) wedge (x wedge y)' = 0,$$
            whence, by eqref{eq:0}
            $$x wedge (x wedge y)' wedge y' = x wedge (x wedge y)',$$
            and analogously,
            $$x wedge (x wedge z)' wedge z' = x wedge (x wedge z)',$$
            yielding
            $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z') = x wedge (x wedge y)' wedge (x wedge z)',$$
            or
            $$x wedge (x wedge y)' wedge (x wedge z)' leq y' wedge z',$$
            and, by eqref{eq:1},
            $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z')' = 0.$$
            From your definition of join and some identities you proved, this is
            $$x wedge ((x wedge y) vee (x wedge z))' wedge (yvee z) = 0,$$
            or, by eqref{eq:1},
            $$x wedge (y vee z) leq (x wedge y) vee (x wedge z).$$
            Since
            $$x wedge (y vee z) geq (x wedge y) vee (x wedge z)$$
            holds in every lattice, we conclude that
            $$x wedge (y vee z) = (x wedge y) vee (x wedge z),$$
            and therefore, $B$ is a Boolean algebra.

            (I suppose you know that in a lattice each distributive law follows from the other.)






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Having defined $0$, rewrite your third axiom as
              begin{equation}label{eq:0}
              x wedge y' = 0 Leftrightarrow x wedge y = x,tag{0}
              end{equation}
              and since you have already proven that the structure is a complemented lattice (and only need distributivity to conclude it's a Boolean algebra), then $x leq y$ iff $x wedge y = x$, which is notationally convenient.
              Thus
              begin{equation}label{eq:1}
              x wedge y' = 0 Leftrightarrow x leq ytag{1}
              end{equation}
              Now, with the equations that you claim you proved,
              $$x wedge (x wedge y)' wedge y = (x wedge y) wedge (x wedge y)' = 0,$$
              whence, by eqref{eq:0}
              $$x wedge (x wedge y)' wedge y' = x wedge (x wedge y)',$$
              and analogously,
              $$x wedge (x wedge z)' wedge z' = x wedge (x wedge z)',$$
              yielding
              $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z') = x wedge (x wedge y)' wedge (x wedge z)',$$
              or
              $$x wedge (x wedge y)' wedge (x wedge z)' leq y' wedge z',$$
              and, by eqref{eq:1},
              $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z')' = 0.$$
              From your definition of join and some identities you proved, this is
              $$x wedge ((x wedge y) vee (x wedge z))' wedge (yvee z) = 0,$$
              or, by eqref{eq:1},
              $$x wedge (y vee z) leq (x wedge y) vee (x wedge z).$$
              Since
              $$x wedge (y vee z) geq (x wedge y) vee (x wedge z)$$
              holds in every lattice, we conclude that
              $$x wedge (y vee z) = (x wedge y) vee (x wedge z),$$
              and therefore, $B$ is a Boolean algebra.

              (I suppose you know that in a lattice each distributive law follows from the other.)






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Having defined $0$, rewrite your third axiom as
                begin{equation}label{eq:0}
                x wedge y' = 0 Leftrightarrow x wedge y = x,tag{0}
                end{equation}
                and since you have already proven that the structure is a complemented lattice (and only need distributivity to conclude it's a Boolean algebra), then $x leq y$ iff $x wedge y = x$, which is notationally convenient.
                Thus
                begin{equation}label{eq:1}
                x wedge y' = 0 Leftrightarrow x leq ytag{1}
                end{equation}
                Now, with the equations that you claim you proved,
                $$x wedge (x wedge y)' wedge y = (x wedge y) wedge (x wedge y)' = 0,$$
                whence, by eqref{eq:0}
                $$x wedge (x wedge y)' wedge y' = x wedge (x wedge y)',$$
                and analogously,
                $$x wedge (x wedge z)' wedge z' = x wedge (x wedge z)',$$
                yielding
                $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z') = x wedge (x wedge y)' wedge (x wedge z)',$$
                or
                $$x wedge (x wedge y)' wedge (x wedge z)' leq y' wedge z',$$
                and, by eqref{eq:1},
                $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z')' = 0.$$
                From your definition of join and some identities you proved, this is
                $$x wedge ((x wedge y) vee (x wedge z))' wedge (yvee z) = 0,$$
                or, by eqref{eq:1},
                $$x wedge (y vee z) leq (x wedge y) vee (x wedge z).$$
                Since
                $$x wedge (y vee z) geq (x wedge y) vee (x wedge z)$$
                holds in every lattice, we conclude that
                $$x wedge (y vee z) = (x wedge y) vee (x wedge z),$$
                and therefore, $B$ is a Boolean algebra.

                (I suppose you know that in a lattice each distributive law follows from the other.)






                share|cite|improve this answer









                $endgroup$



                Having defined $0$, rewrite your third axiom as
                begin{equation}label{eq:0}
                x wedge y' = 0 Leftrightarrow x wedge y = x,tag{0}
                end{equation}
                and since you have already proven that the structure is a complemented lattice (and only need distributivity to conclude it's a Boolean algebra), then $x leq y$ iff $x wedge y = x$, which is notationally convenient.
                Thus
                begin{equation}label{eq:1}
                x wedge y' = 0 Leftrightarrow x leq ytag{1}
                end{equation}
                Now, with the equations that you claim you proved,
                $$x wedge (x wedge y)' wedge y = (x wedge y) wedge (x wedge y)' = 0,$$
                whence, by eqref{eq:0}
                $$x wedge (x wedge y)' wedge y' = x wedge (x wedge y)',$$
                and analogously,
                $$x wedge (x wedge z)' wedge z' = x wedge (x wedge z)',$$
                yielding
                $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z') = x wedge (x wedge y)' wedge (x wedge z)',$$
                or
                $$x wedge (x wedge y)' wedge (x wedge z)' leq y' wedge z',$$
                and, by eqref{eq:1},
                $$x wedge (x wedge y)' wedge (x wedge z)' wedge (y' wedge z')' = 0.$$
                From your definition of join and some identities you proved, this is
                $$x wedge ((x wedge y) vee (x wedge z))' wedge (yvee z) = 0,$$
                or, by eqref{eq:1},
                $$x wedge (y vee z) leq (x wedge y) vee (x wedge z).$$
                Since
                $$x wedge (y vee z) geq (x wedge y) vee (x wedge z)$$
                holds in every lattice, we conclude that
                $$x wedge (y vee z) = (x wedge y) vee (x wedge z),$$
                and therefore, $B$ is a Boolean algebra.

                (I suppose you know that in a lattice each distributive law follows from the other.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 20 at 11:39









                amrsaamrsa

                3,6702618




                3,6702618






























                    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%2f3080052%2fproving-both-distributive-laws-for-boolean-algebra-given-by-certain-axioms%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

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                    Read more

                    Antonio Litta Visconti Arese

                    Image
                    Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                    Read more