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Math Help - Two proofs I've done...

  1. #1
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    Two proofs I've done...

    Hi guys I did two proofs for a logic class that I'm in. Could someone please check these to make sure I'm on the right track?

    For all sets A and B, (A\B)U(AnB)=A.

    Let x E A
    if x E B, then x E (AnB)
    if x not E B, then x E (A\B)
    therefore x E (A\B)U(AnB)
    therefore (A\B)U(AnB) = A

    Let S be a subset of a universal set U. The characteristic function fs is the function from U to the set {0,1} defined by

    fs(x)={1, if x E S
    {0, if x not E S
    Let A and B be sets. Show that for all x, fAnB(x) = fA(x) * fB(x)

    Let x E U
    Case 1: If x E A ^ x E B, then fsAnB(x) = 1 = 1*1 = fsA(x) * fsB(x)
    Case 2: If x E A but x not E B, then fsAnB(x) = 0 = 1*0 = fsA(x) * fsB(x)
    Case 3: If x not E A but x E B, then fsAnB(x) = 0 = 0*1 = fsA(x) * fsB(x)
    Case 4: If x not E A ^ x not E B, then fsAnB(x) = 0 = 0*0 = fsA(x) * fsB(x)
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  2. #2
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    Quote Originally Posted by Benderbrau View Post
    Hi guys I did two proofs for a logic class that I'm in. Could someone please check these to make sure I'm on the right track?

    For all sets A and B, (A\B)U(AnB)=A.

    Let x E A
    if x E B, then x E (AnB)
    if x not E B, then x E (A\B)
    therefore x E (A\B)U(AnB)
    therefore (A\B)U(AnB) = A
    No, you only proved that that A is contained in the union. You have to show the reverse direction.
    Quote Originally Posted by Benderbrau View Post

    Let S be a subset of a universal set U. The characteristic function fs is the function from U to the set {0,1} defined by

    fs(x)={1, if x E S
    {0, if x not E S
    Let A and B be sets. Show that for all x, fAnB(x) = fA(x) * fB(x)

    Let x E U
    Case 1: If x E A ^ x E B, then fsAnB(x) = 1 = 1*1 = fsA(x) * fsB(x)
    Case 2: If x E A but x not E B, then fsAnB(x) = 0 = 1*0 = fsA(x) * fsB(x)
    Case 3: If x not E A but x E B, then fsAnB(x) = 0 = 0*1 = fsA(x) * fsB(x)
    Case 4: If x not E A ^ x not E B, then fsAnB(x) = 0 = 0*0 = fsA(x) * fsB(x)
    Yes this is valid since you enumerated all cases.
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  3. #3
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    So in order to complete the first proof, could I do something like this?

    Let x E (A\B)U(AnB)
    if x E (A\B), then x not E B
    if x E (AnB), then x E B
    therefore x E A
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  4. #4
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    Quote Originally Posted by Benderbrau View Post
    So in order to complete the first proof, could I do something like this?

    Let x E (A\B)U(AnB)
    if x E (A\B), then x not E B
    if x E (AnB), then x E B
    therefore x E A
    The steps are correct, but the two then's are wrong. You should conclude x\in A in both cases.
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