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Math Help - DeMorgan's theorem

  1. #1
    Junior Member
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    DeMorgan's theorem

    Hello,

    I need to use the first equation of E to derive the second equation for E. I've been messing around with this for over an hour, and can't seem to find a way to do it. Could someone provide some insight?



    Thanks so much,
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, centenial!


    E \;=\;(A\cdot B\cdot \overline{C}) + (A\cdot \overline{B}\cdot C) + (\overline{A}\cdot B\cdot C)

    \text{We can also write }E\text{ as a product of sums:}

    . . E \;=\;\overline{(\overline{A} + \overline{B} + C)\cdot (\overline{A} + B + \overline{C}) \cdot (A + \overline{B} + \overline{C})}

    \text{To derive this, you need to use DeMorgan's Theorems.}


    It's much easier to work "backwards" . . .


    We have: . E \;=\;\overline{(\overline{A} + \overline{B} + C)\cdot (\overline{A} + B + \overline{C}) \cdot (A + \overline{B} + \overline{C})}

    . . . . . . . . . . =\;\overline{(\overline A + \overline B + C)} + \overline{(\overline A + B + \overline C)} + \overline{(A + \overline B + \overline C)}

    . . . . . . . . . . =\; (A\cdot B\cdot\overline C) + (A\cdot\overline B\cdot C) + (\overline A\cdot B\cdot C)

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  3. #3
    Junior Member
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    Thanks! That makes sense now.
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