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Math Help - Proof using Element Chasing the associative law of symetric difference

  1. #1
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    Exclamation Proof using Element Chasing the associative law of symetric difference

    Hi there,
    I have to prove using Element Chasing that (A delta B) delta C = A delta ( B delta C)
    I have the Venn drawn, and was told to start with:
    let x be an element of (A delta B) delta C
    if x is an element of (A delta B) union(and) C but not in x element (A delta B) intersection(or) C


    I am not just sure really where to go next in the proof.
    The help would be ever so appreciated.
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  2. #2
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    Re: Proof using Element Chasing the associative law of symetric difference

    suppose x is in (AΔB)ΔC.

    this means x is in AΔB, or x is in C, but not BOTH (it's in exactly ONE of these sets).

    case 1: x is in AΔB, x is not in C.

    case 1a: x is in A, x is not in B, x is not in C.

    since x is not in B, and not in C, x is not in BUC, and therefore not in BΔC, which is a subset of BUC.

    thus in this case, x is in A, but not in BΔC, so x is in AΔ(BΔC).

    case 1b: x is not in A, x is in B, x is not in C.

    in this case, x is in B-C, which is a subset of BΔC (BΔC = (B-C).

    so x is not in A, and is in BΔC, so x is in AΔ(BΔC).

    case 2: x is not in AΔB, x is in C.

    case 2a: x is is not in AUB, x is in C.

    in this case, x is not in A, and not in B, but in C, so x is in C-B, and thus in BΔC, so x is not in A, but in BΔC, so is in AΔ(BΔC).

    case 2b: x is in AUB, x is in C.

    here, x is clearly in A∩B (since A∩B = (AUB) - (AΔB)). thus x is in A,B and C, which means x is in B∩C, which means x is not in BΔC.

    thus, in this case, x is in A, but not in BΔC, so x is in AΔ(BΔC).

    ***************

    that's HALF the proof: (AΔB)ΔC is a subset of AΔ(BΔC). you do the other half.
    Thanks from Ragnarok
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