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Boolean Algebra simplification question

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Lets say I have 4 inputs x1 x0 y1 y0

    if I have a sum-of-products say: x1x0'y1y0'+x0y1'y0

    can I simplify it to x0y1'y0 by pulling out x0y1'y0 giving x0y1'y0(y0'+1) knowing that 1+y0 is 1 and 1*signalz is signalz.

    Am I right in thinking this, I really dont want to do the truth table for this bad boy

    thanks
     
  2. jcsd
  3. Sep 20, 2012 #2
    Hi delta59,

    It's pretty straightforward to simplify short Boolean expressions having 2 or 3 variables using Boolean algebra (aka the switching algebra theorems). But when you get into 4 variables or higher, or when you're evaluating lengthy expressions, it's easier to use a Karnaugh Map.

    Here are some googled examples of K-maps for expressions of 4 variables.

    Try doing the K-map for your expression: [itex]x_1 \cdot x_0^' \cdot y_1 \cdot y_0^' + x_0 \cdot y_1^' \cdot y_0[/itex]. Does it simplify any further?
     
    Last edited: Sep 20, 2012
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