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







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    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






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      asked Jan 20 at 1:17









      KulistyKulisty

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          $begingroup$

          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.)






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            $begingroup$

            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






























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