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De Morgan's Laws Question

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove ##A\setminus(B\cap C)=(A\setminus B)\cup(A\setminus C)##

    2. Relevant equations
    3. The attempt at a solution

    We will show that every element in ##A\setminus(B\cap C)## is contained in ##(A\setminus B)## or ##(A\setminus C)##. If ##x\in A\setminus (B\cup C)## then ##x\in A##, ##x\notin B## and ##x\notin C##, thus ##x\in(A\setminus B)## and ##x\in(A\setminus C)## now I'm assuming the next step is to say that this shows that ##x\in(A\setminus B)\cup(A\setminus C)## but I'm confused as to why we this is a union instead of an intersection. If ##x## is not in B and C then why do we use and/or instead of just and? I am obviously interpreting this incorrectly so hopefully someone can clarify.
     
  2. jcsd
  3. Sep 21, 2015 #2
    Okay so since it's possible that x is in either B or C we have ##x\in A##, ##x\in (B\cup C)##?
     
  4. Sep 21, 2015 #3

    verty

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    It might be helpful to "de morgan" each side. For example, the left side becomes ##\overline{A} \cup (B \cap C)##.
     
  5. Sep 21, 2015 #4
    So I should also try to show that ##(A\setminus B)\cup(A\setminus C)=A\setminus(B\cap C)##? Also I haven't yet learnt what ##\overline{A}## represents.
     
  6. Sep 21, 2015 #5

    verty

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    Oh right, in that case do it the way you were doing it but be careful not to get the symbols wrong. I thought you had to apply De Morgan's laws somehow, so the obvious thing was to take the complement of each side.
     
  7. Sep 21, 2015 #6

    Ray Vickson

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    If ##E^c## denotes the complement ("not-##E##) of a set ##E##, then ##D \backslash E = D \cap E^c##, essentially by definition. Do you know how one of the DeMorgan laws relates ##(B \cap C)^c## to ##B^c## and ##C^c##?
     
    Last edited: Sep 22, 2015
  8. Sep 21, 2015 #7
    So for the other side: If ##x\in (A\setminus B)\cup (A\setminus C)## then ##x\in A## and ##x\in (B\cup C)##
    In my statistics class we had ##P(B\cap C)^c=P(B^c)+P(C^c)-P(B\cup C)##
     
  9. Sep 22, 2015 #8

    Ray Vickson

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    This is irrelevant: the question has nothing to do with statistics or probability. It is just a question about set manipulation.
     
  10. Sep 22, 2015 #9
    Hmm I think I have solved this question, if ##x\in A\setminus(B\cap C)## then ##x\in A## and ##x\notin (B\cap C)##, it follows that ##x\notin B## or ##x\notin C## which leads to ##x\in (A\setminus B)\cup (A\setminus C)##
     
  11. Sep 22, 2015 #10

    Dick

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    I think you have solved it as well.
     
  12. Sep 22, 2015 #11
    I would really appreciate it if someone could look over this one as well, I'm supposed to prove that ##A\cup (B\cap C)=(A\cup B)\cap (A\cup C)##, In order to prove this I must show that that x can go from an element of the LHS to the RHS (Not really sure how to word this suggestions would be appreciated) ##x\in (A\cup B)\cap (A\cup C)##, Starting from the LHS I have that if ##x\in A\cup (B\cap C)## then ##x\in A## and ##x\in (B\cap C)##, this leads to two cases, the first case is when ##x\in A##, this implies that ##x\in (A\cup B)\cap (A\cup C)## since if ##x\in A## this statement holds true. The second case is when ##x\in (B\cap C)##, this also implies the RHS since if ##x\in (B\cap C)## then the statement ##(A\cup B)\cap (A\cup C)## also holds, therefore ##A\cup (B\cap C)=(A\cup B)\cap (A\cup C)##. I'm curious as to whether or not this is sufficient or if I have to manipulate the expressions more.
     
  13. Sep 23, 2015 #12

    Ray Vickson

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    You have gone one way (proving that ##A\setminus(B \cap C) \subset (A\setminus B) \cup (A\setminus C)##. You also need to establish the reverse.
     
  14. Sep 23, 2015 #13
    Is it possible that the proof changes when going the other way around or is it always the same steps just proving them in reverse?
     
  15. Sep 23, 2015 #14

    Ray Vickson

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    You tell me.
     
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