Show that $(AcapbarB) cup (barAcap barC) = (AcapbarB) cup (barAcap barC) cup (barB cap barC)$ by rewriting

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$A,B,C$ are subsets of the universal set $U$.

$barA$ is the absolute complement of $A$ = $Usetminus A$



Show that $(AcapbarB) cup (barAcap barC) = (AcapbarB) cup (barAcap barC) cup (barB cap barC)$ by rewriting either the left hand side or the right hand side.



I have been able to prove the similarity by showing that $barBcap barC subseteq (AcapbarB) cup (barAcap barC)$, but I want to show it by only rewriting either the RHS or the LHS by using rules like DeMorgan's laws, Absorption laws etc.



The problem I'm having is that in the RHS $barBcap barC$ is redundant; I already have the two sets that I want and if I try to "get rid of" $barBcap barC$ by taking the union of it and some of the other sets, like $(barBcap barC) cup (barAcap barC)$ then I don't have $(barAcap barC)$ on the RHS anymore. But I feel like it must be possible to always rewrite expressions of sets to all equivalent expressions by only using similarity laws like DeMorgan's laws. Maybe I'm wrong.




elementary-set-theory




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    up vote
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    down vote

    favorite












    $A,B,C$ are subsets of the universal set $U$.

    $barA$ is the absolute complement of $A$ = $Usetminus A$



    Show that $(AcapbarB) cup (barAcap barC) = (AcapbarB) cup (barAcap barC) cup (barB cap barC)$ by rewriting either the left hand side or the right hand side.



    I have been able to prove the similarity by showing that $barBcap barC subseteq (AcapbarB) cup (barAcap barC)$, but I want to show it by only rewriting either the RHS or the LHS by using rules like DeMorgan's laws, Absorption laws etc.



    The problem I'm having is that in the RHS $barBcap barC$ is redundant; I already have the two sets that I want and if I try to "get rid of" $barBcap barC$ by taking the union of it and some of the other sets, like $(barBcap barC) cup (barAcap barC)$ then I don't have $(barAcap barC)$ on the RHS anymore. But I feel like it must be possible to always rewrite expressions of sets to all equivalent expressions by only using similarity laws like DeMorgan's laws. Maybe I'm wrong.




    elementary-set-theory




    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $A,B,C$ are subsets of the universal set $U$.

      $barA$ is the absolute complement of $A$ = $Usetminus A$



      Show that $(AcapbarB) cup (barAcap barC) = (AcapbarB) cup (barAcap barC) cup (barB cap barC)$ by rewriting either the left hand side or the right hand side.



      I have been able to prove the similarity by showing that $barBcap barC subseteq (AcapbarB) cup (barAcap barC)$, but I want to show it by only rewriting either the RHS or the LHS by using rules like DeMorgan's laws, Absorption laws etc.



      The problem I'm having is that in the RHS $barBcap barC$ is redundant; I already have the two sets that I want and if I try to "get rid of" $barBcap barC$ by taking the union of it and some of the other sets, like $(barBcap barC) cup (barAcap barC)$ then I don't have $(barAcap barC)$ on the RHS anymore. But I feel like it must be possible to always rewrite expressions of sets to all equivalent expressions by only using similarity laws like DeMorgan's laws. Maybe I'm wrong.




      elementary-set-theory




      share|cite|improve this question












      $A,B,C$ are subsets of the universal set $U$.

      $barA$ is the absolute complement of $A$ = $Usetminus A$



      Show that $(AcapbarB) cup (barAcap barC) = (AcapbarB) cup (barAcap barC) cup (barB cap barC)$ by rewriting either the left hand side or the right hand side.



      I have been able to prove the similarity by showing that $barBcap barC subseteq (AcapbarB) cup (barAcap barC)$, but I want to show it by only rewriting either the RHS or the LHS by using rules like DeMorgan's laws, Absorption laws etc.



      The problem I'm having is that in the RHS $barBcap barC$ is redundant; I already have the two sets that I want and if I try to "get rid of" $barBcap barC$ by taking the union of it and some of the other sets, like $(barBcap barC) cup (barAcap barC)$ then I don't have $(barAcap barC)$ on the RHS anymore. But I feel like it must be possible to always rewrite expressions of sets to all equivalent expressions by only using similarity laws like DeMorgan's laws. Maybe I'm wrong.





      elementary-set-theory





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      share|cite|improve this question




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      asked Aug 26 at 6:44









      DancingIceCream

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