prove the equivalence of a+b=b and a*b=a in the boolean algebra theorem [closed]












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This is a question in the boolean algebra and their equivalence should be proved i the theorem 5 of the boolean algebra.Am capable of proving the theorem 5 but i still dont ahve a clue on how to start answering this question.Ihave tried to look for similar questions but i have not found any...the one that i have found odoesn't have the the solutions that i can refer to.It is a new topic in my life so please help me solve this.






boolean-algebra







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closed as unclear what you're asking by José Carlos Santos, drhab, verret, Shailesh, Lee David Chung Lin Jan 31 at 4:17


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 4




    $begingroup$
    Welcome to MSE. Are we all supposed to know what theorem 5 is?
    $endgroup$
    – José Carlos Santos
    Jan 30 at 8:36
















1












$begingroup$


This is a question in the boolean algebra and their equivalence should be proved i the theorem 5 of the boolean algebra.Am capable of proving the theorem 5 but i still dont ahve a clue on how to start answering this question.Ihave tried to look for similar questions but i have not found any...the one that i have found odoesn't have the the solutions that i can refer to.It is a new topic in my life so please help me solve this.






boolean-algebra







share|cite|improve this question









$endgroup$



closed as unclear what you're asking by José Carlos Santos, drhab, verret, Shailesh, Lee David Chung Lin Jan 31 at 4:17


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 4




    $begingroup$
    Welcome to MSE. Are we all supposed to know what theorem 5 is?
    $endgroup$
    – José Carlos Santos
    Jan 30 at 8:36














1












1








1


0



$begingroup$


This is a question in the boolean algebra and their equivalence should be proved i the theorem 5 of the boolean algebra.Am capable of proving the theorem 5 but i still dont ahve a clue on how to start answering this question.Ihave tried to look for similar questions but i have not found any...the one that i have found odoesn't have the the solutions that i can refer to.It is a new topic in my life so please help me solve this.






boolean-algebra







share|cite|improve this question









$endgroup$




This is a question in the boolean algebra and their equivalence should be proved i the theorem 5 of the boolean algebra.Am capable of proving the theorem 5 but i still dont ahve a clue on how to start answering this question.Ihave tried to look for similar questions but i have not found any...the one that i have found odoesn't have the the solutions that i can refer to.It is a new topic in my life so please help me solve this.






boolean-algebra




boolean-algebra






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asked Jan 30 at 8:32









RhinoRhino

112




112




closed as unclear what you're asking by José Carlos Santos, drhab, verret, Shailesh, Lee David Chung Lin Jan 31 at 4:17


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by José Carlos Santos, drhab, verret, Shailesh, Lee David Chung Lin Jan 31 at 4:17


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 4




    $begingroup$
    Welcome to MSE. Are we all supposed to know what theorem 5 is?
    $endgroup$
    – José Carlos Santos
    Jan 30 at 8:36














  • 4




    $begingroup$
    Welcome to MSE. Are we all supposed to know what theorem 5 is?
    $endgroup$
    – José Carlos Santos
    Jan 30 at 8:36








4




4




$begingroup$
Welcome to MSE. Are we all supposed to know what theorem 5 is?
$endgroup$
– José Carlos Santos
Jan 30 at 8:36




$begingroup$
Welcome to MSE. Are we all supposed to know what theorem 5 is?
$endgroup$
– José Carlos Santos
Jan 30 at 8:36










2 Answers
2






active

oldest

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1












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If $a+b=b$ then $asubset b$, which gives $ab=a$.



The second part is the same.






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$endgroup$





















    1












    $begingroup$

    $a + b = b Longrightarrow a + ab = ab Longrightarrow a = a1 = a(1 + b) = a + ab = ab; tag 1$



    $ab = a Longrightarrow a = ab + a = a(b + 1) = a1 = a. tag 2$






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      If $a+b=b$ then $asubset b$, which gives $ab=a$.



      The second part is the same.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        If $a+b=b$ then $asubset b$, which gives $ab=a$.



        The second part is the same.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          If $a+b=b$ then $asubset b$, which gives $ab=a$.



          The second part is the same.






          share|cite|improve this answer









          $endgroup$



          If $a+b=b$ then $asubset b$, which gives $ab=a$.



          The second part is the same.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 8:37









          Michael RozenbergMichael Rozenberg

          109k1896201




          109k1896201























              1












              $begingroup$

              $a + b = b Longrightarrow a + ab = ab Longrightarrow a = a1 = a(1 + b) = a + ab = ab; tag 1$



              $ab = a Longrightarrow a = ab + a = a(b + 1) = a1 = a. tag 2$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $a + b = b Longrightarrow a + ab = ab Longrightarrow a = a1 = a(1 + b) = a + ab = ab; tag 1$



                $ab = a Longrightarrow a = ab + a = a(b + 1) = a1 = a. tag 2$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $a + b = b Longrightarrow a + ab = ab Longrightarrow a = a1 = a(1 + b) = a + ab = ab; tag 1$



                  $ab = a Longrightarrow a = ab + a = a(b + 1) = a1 = a. tag 2$






                  share|cite|improve this answer









                  $endgroup$



                  $a + b = b Longrightarrow a + ab = ab Longrightarrow a = a1 = a(1 + b) = a + ab = ab; tag 1$



                  $ab = a Longrightarrow a = ab + a = a(b + 1) = a1 = a. tag 2$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 30 at 11:25









                  Robert LewisRobert Lewis

                  48.6k23167




                  48.6k23167















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