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Proof of A-(BnC)=(A-B)U(A-C)

  1. Jun 16, 2013 #1

    If A-B={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]B}
    A-C={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]C}

    then (A-B)[itex]\bigcup[/itex](A-C)={xlx[itex]\in[/itex]A, x[itex]\notin[/itex](B and C)

    Let X=A and Y=(B[itex]\bigcap[/itex]C)

    X-Y={xlx[itex]\in[/itex]X and x[itex]\notin[/itex]Y}
    x[itex]\notin[/itex](B and C)

    Therefore, A-(B[itex]\bigcap[/itex]C)=(A-B)[itex]\bigcup[/itex](A-C).

    Is this anywhere close to being right? I want to self study some rigorous math this summer but the hard part is that I don't have anything to check against to see if my answers are correct or not.
  2. jcsd
  3. Jun 17, 2013 #2


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    Science Advisor

    (A-B)[itex]\bigcup[/itex](A-C)={xlx[itex]\in[/itex]A, x[itex]\notin[/itex](B and C)
    has to be justified.

    also "and" and "[itex]\bigcap[/itex]" are the same.
  4. Jun 17, 2013 #3


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    Staff: Mentor

    I always find it fascinating that people use latex just for a single symbol, instead of formatting whole expression.

    [tex]A-(B\cap C)=(A-B)\cup(A-C)[/tex]
  5. Jun 18, 2013 #4
    Ok so I worked through all the exercises up to this one. Here's what I came up with.


    Let S=B[itex]\cap[/itex]C={xlx[itex]\in[/itex]B, and x[itex]\in[/itex]C} (By the definition of intersection)

    Thus A-S={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]S}
    ={xlx[itex]\in[/itex]A,x[itex]\notin[/itex]B, and x[itex]\notin[/itex]C} (see justification 1)

    Next, let S[itex]_{1}[/itex]=A-B={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]B}
    let S[itex]_{2}[/itex]=A-C={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]C}

    Thus, S[itex]_{1}[/itex][itex]\cup[/itex]S[itex]_{2}[/itex]={xlx[itex]\in[/itex]A. x[itex]\notin[/itex]B, and x[itex]\notin[/itex]C} (by the definition of union - see 2)

    Therefore: A-S=S[itex]_{1}[/itex][itex]\cup[/itex]S[itex]_{2}[/itex]


    1)A-B={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]B}
    2) Union is defined as the set of those elements which are in A, in B, or in both. Therefore if A={xlx satisfies P and x[itex]\notin[/itex]C} and B={xlx satisfies Q and x[itex]\notin[/itex]D} (where P and Q are arbitrary conditions and C and D are other sets) then A[itex]\cup[/itex]B= {xlx satisfies P,Q, x[itex]\notin[/itex]C, and x[itex]\notin[/itex]D}.

    What do you guys think?
  6. Jun 19, 2013 #5


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    Wrong: should be {xlx∈ A and (x∉ B or x∉ C)}
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