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Boolean Algebra Simplification

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]R=(A \cdot \overline{B})+(\overline{A} \cdot B)[/itex]​

    Using Boolean algebra, including the complement relation [itex]D\cdot \overline{D}=0[/itex], convert R to a form that uses one NAND, one AND and one OR gate (a total of three gates).

    2. Relevant equations

    de Morgan's theorem...

    [itex]\overline{A+B}=\overline{A}\cdot\overline{B}[/itex]
    [itex]\overline{A \cdot B}=\overline{A} + \overline{B}[/itex]

    3. The attempt at a solution

    By some trial and error I arrived at this;

    [itex](\overline{A}+\overline{B})\cdot(A+B)[/itex]

    And then using the second of de Morgan's equations on the first OR I got to this;

    [itex]\overline{A\cdot B}\cdot(A+B)[/itex]

    Which works as a solution, it has one NAND, one AND and one OR, however, I can't just skip to that bit, it's the working out in between that I'm missing, I'm not sure how to fill in the gaps.

    Thanks for any help you can offer!
     
  2. jcsd
  3. Jun 3, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You can show that the last statement is equivalent to the original one if you work backwards - it is easier in that direction, but the other direction works as well.
     
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