Showing $a setminus (a setminus b) = a cap b$ using only set notation.












0












$begingroup$


I need to prove that $a setminus (a setminus b) = a cap b$ only through set notations.



I have reached the fact that
$a setminus (a setminus b)$ = {x | x $in$ A $land$ x $notin$ (A $setminus$ B}



Which I then simplify using the De Morgan' formula to
{x | x $in$ A $land$ (x $notin$ A $lor$ x $in$ B)}.



It is evident that this is the same as A $cap$ B (by constructing diagrams and truth tables). But how do I get this expression to {x | x $in$ A $land$ x $in$ B}? Please help!






discrete-mathematics elementary-set-theory boolean-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use ${ X}$ for ${ X}$.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:50










  • $begingroup$
    Please edit the question accordingly.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:51
















0












$begingroup$


I need to prove that $a setminus (a setminus b) = a cap b$ only through set notations.



I have reached the fact that
$a setminus (a setminus b)$ = {x | x $in$ A $land$ x $notin$ (A $setminus$ B}



Which I then simplify using the De Morgan' formula to
{x | x $in$ A $land$ (x $notin$ A $lor$ x $in$ B)}.



It is evident that this is the same as A $cap$ B (by constructing diagrams and truth tables). But how do I get this expression to {x | x $in$ A $land$ x $in$ B}? Please help!






discrete-mathematics elementary-set-theory boolean-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use ${ X}$ for ${ X}$.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:50










  • $begingroup$
    Please edit the question accordingly.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:51














0












0








0





$begingroup$


I need to prove that $a setminus (a setminus b) = a cap b$ only through set notations.



I have reached the fact that
$a setminus (a setminus b)$ = {x | x $in$ A $land$ x $notin$ (A $setminus$ B}



Which I then simplify using the De Morgan' formula to
{x | x $in$ A $land$ (x $notin$ A $lor$ x $in$ B)}.



It is evident that this is the same as A $cap$ B (by constructing diagrams and truth tables). But how do I get this expression to {x | x $in$ A $land$ x $in$ B}? Please help!






discrete-mathematics elementary-set-theory boolean-algebra







share|cite|improve this question











$endgroup$




I need to prove that $a setminus (a setminus b) = a cap b$ only through set notations.



I have reached the fact that
$a setminus (a setminus b)$ = {x | x $in$ A $land$ x $notin$ (A $setminus$ B}



Which I then simplify using the De Morgan' formula to
{x | x $in$ A $land$ (x $notin$ A $lor$ x $in$ B)}.



It is evident that this is the same as A $cap$ B (by constructing diagrams and truth tables). But how do I get this expression to {x | x $in$ A $land$ x $in$ B}? Please help!






discrete-mathematics elementary-set-theory boolean-algebra




discrete-mathematics elementary-set-theory boolean-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 10:49









Shaun

9,442113684




9,442113684










asked Dec 8 '18 at 10:25









Leon VladimirovLeon Vladimirov

134




134












  • $begingroup$
    Use ${ X}$ for ${ X}$.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:50










  • $begingroup$
    Please edit the question accordingly.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:51


















  • $begingroup$
    Use ${ X}$ for ${ X}$.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:50










  • $begingroup$
    Please edit the question accordingly.
    $endgroup$
    – Shaun
    Dec 8 '18 at 10:51
















$begingroup$
Use ${ X}$ for ${ X}$.
$endgroup$
– Shaun
Dec 8 '18 at 10:50




$begingroup$
Use ${ X}$ for ${ X}$.
$endgroup$
– Shaun
Dec 8 '18 at 10:50












$begingroup$
Please edit the question accordingly.
$endgroup$
– Shaun
Dec 8 '18 at 10:51




$begingroup$
Please edit the question accordingly.
$endgroup$
– Shaun
Dec 8 '18 at 10:51










2 Answers
2






active

oldest

votes


















1












$begingroup$

Equivalent are:




  • $xin Awedge(xnotin Avee xin B)$

  • $(xin Awedge xnotin A)vee(xin Awedge xin B)$

  • $xin Awedge xin B$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer! Could you please explain how do I get from line 1 to line 2?
    $endgroup$
    – Leon Vladimirov
    Dec 8 '18 at 10:30










  • $begingroup$
    By applying distributivity of $wedge $ wrt $vee $. Not familiar with that yet? A Boolean algebra is distributive lattice.
    $endgroup$
    – drhab
    Dec 8 '18 at 10:33












  • $begingroup$
    Completely missed that! Understood. Thank you!
    $endgroup$
    – Leon Vladimirov
    Dec 8 '18 at 10:36










  • $begingroup$
    You are welcome
    $endgroup$
    – drhab
    Dec 8 '18 at 10:46



















0












$begingroup$

a - (a - b) = a - (a $cap$ b$^c$) =

a - (a$^c$ $cup$ b)$^c$ = a $cap$ (a$^c$ $cup$ b)

= (a $cap$ a$^c$) $cup$ (a $cap$ b) = a $cap$ b






share|cite|improve this answer









$endgroup$













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Equivalent are:




    • $xin Awedge(xnotin Avee xin B)$

    • $(xin Awedge xnotin A)vee(xin Awedge xin B)$

    • $xin Awedge xin B$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer! Could you please explain how do I get from line 1 to line 2?
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:30










    • $begingroup$
      By applying distributivity of $wedge $ wrt $vee $. Not familiar with that yet? A Boolean algebra is distributive lattice.
      $endgroup$
      – drhab
      Dec 8 '18 at 10:33












    • $begingroup$
      Completely missed that! Understood. Thank you!
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:36










    • $begingroup$
      You are welcome
      $endgroup$
      – drhab
      Dec 8 '18 at 10:46
















    1












    $begingroup$

    Equivalent are:




    • $xin Awedge(xnotin Avee xin B)$

    • $(xin Awedge xnotin A)vee(xin Awedge xin B)$

    • $xin Awedge xin B$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer! Could you please explain how do I get from line 1 to line 2?
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:30










    • $begingroup$
      By applying distributivity of $wedge $ wrt $vee $. Not familiar with that yet? A Boolean algebra is distributive lattice.
      $endgroup$
      – drhab
      Dec 8 '18 at 10:33












    • $begingroup$
      Completely missed that! Understood. Thank you!
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:36










    • $begingroup$
      You are welcome
      $endgroup$
      – drhab
      Dec 8 '18 at 10:46














    1












    1








    1





    $begingroup$

    Equivalent are:




    • $xin Awedge(xnotin Avee xin B)$

    • $(xin Awedge xnotin A)vee(xin Awedge xin B)$

    • $xin Awedge xin B$






    share|cite|improve this answer









    $endgroup$



    Equivalent are:




    • $xin Awedge(xnotin Avee xin B)$

    • $(xin Awedge xnotin A)vee(xin Awedge xin B)$

    • $xin Awedge xin B$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 8 '18 at 10:29









    drhabdrhab

    103k545136




    103k545136












    • $begingroup$
      Thank you for your answer! Could you please explain how do I get from line 1 to line 2?
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:30










    • $begingroup$
      By applying distributivity of $wedge $ wrt $vee $. Not familiar with that yet? A Boolean algebra is distributive lattice.
      $endgroup$
      – drhab
      Dec 8 '18 at 10:33












    • $begingroup$
      Completely missed that! Understood. Thank you!
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:36










    • $begingroup$
      You are welcome
      $endgroup$
      – drhab
      Dec 8 '18 at 10:46


















    • $begingroup$
      Thank you for your answer! Could you please explain how do I get from line 1 to line 2?
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:30










    • $begingroup$
      By applying distributivity of $wedge $ wrt $vee $. Not familiar with that yet? A Boolean algebra is distributive lattice.
      $endgroup$
      – drhab
      Dec 8 '18 at 10:33












    • $begingroup$
      Completely missed that! Understood. Thank you!
      $endgroup$
      – Leon Vladimirov
      Dec 8 '18 at 10:36










    • $begingroup$
      You are welcome
      $endgroup$
      – drhab
      Dec 8 '18 at 10:46
















    $begingroup$
    Thank you for your answer! Could you please explain how do I get from line 1 to line 2?
    $endgroup$
    – Leon Vladimirov
    Dec 8 '18 at 10:30




    $begingroup$
    Thank you for your answer! Could you please explain how do I get from line 1 to line 2?
    $endgroup$
    – Leon Vladimirov
    Dec 8 '18 at 10:30












    $begingroup$
    By applying distributivity of $wedge $ wrt $vee $. Not familiar with that yet? A Boolean algebra is distributive lattice.
    $endgroup$
    – drhab
    Dec 8 '18 at 10:33






    $begingroup$
    By applying distributivity of $wedge $ wrt $vee $. Not familiar with that yet? A Boolean algebra is distributive lattice.
    $endgroup$
    – drhab
    Dec 8 '18 at 10:33














    $begingroup$
    Completely missed that! Understood. Thank you!
    $endgroup$
    – Leon Vladimirov
    Dec 8 '18 at 10:36




    $begingroup$
    Completely missed that! Understood. Thank you!
    $endgroup$
    – Leon Vladimirov
    Dec 8 '18 at 10:36












    $begingroup$
    You are welcome
    $endgroup$
    – drhab
    Dec 8 '18 at 10:46




    $begingroup$
    You are welcome
    $endgroup$
    – drhab
    Dec 8 '18 at 10:46











    0












    $begingroup$

    a - (a - b) = a - (a $cap$ b$^c$) =

    a - (a$^c$ $cup$ b)$^c$ = a $cap$ (a$^c$ $cup$ b)

    = (a $cap$ a$^c$) $cup$ (a $cap$ b) = a $cap$ b






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      a - (a - b) = a - (a $cap$ b$^c$) =

      a - (a$^c$ $cup$ b)$^c$ = a $cap$ (a$^c$ $cup$ b)

      = (a $cap$ a$^c$) $cup$ (a $cap$ b) = a $cap$ b






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        a - (a - b) = a - (a $cap$ b$^c$) =

        a - (a$^c$ $cup$ b)$^c$ = a $cap$ (a$^c$ $cup$ b)

        = (a $cap$ a$^c$) $cup$ (a $cap$ b) = a $cap$ b






        share|cite|improve this answer









        $endgroup$



        a - (a - b) = a - (a $cap$ b$^c$) =

        a - (a$^c$ $cup$ b)$^c$ = a $cap$ (a$^c$ $cup$ b)

        = (a $cap$ a$^c$) $cup$ (a $cap$ b) = a $cap$ b







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 10:41









        William ElliotWilliam Elliot

        8,6922720




        8,6922720






























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