Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reduction of Boolean Expression to its Lowest From

  1. May 21, 2013 #1
    Hey Guys

    I am Currently doing my degree in information systems. At the moment the subject is PLC(Processing and Logic Concepts)

    I understand all of it, I am not sertain on how to start reducing a boolean expression. Once the first step is done I normally get it and can complete the process of reduction using the Laws.

    Does anyone know if i hint or a trick on where to start reducing a Boolean Expression

    EX. F = AB'C + ABC + A'B'C' + A'BC'

    I am not looking for the answer to the expression only a tip on how to start reducing these tipes of Expressions.

    Thank You in Advance
  2. jcsd
  3. May 21, 2013 #2


    User Avatar
    Science Advisor

    I'm no expert on this but the first thing I would do is start looking for "similarities" in the terms. For example, I see both A and C in the first two, both A' and C' in the second two:

    F= AB'C+ ABC+ A'B'C'+ A'BC'= A(B+ B')C+ A'(B+ B')C'
    and, of course, those two now have the same "B+ B' " so is the same as
    F= (B+ B')(AC+ A'C')_

    You haven't said what you consider "reduced to its lowest form".
  4. May 21, 2013 #3

    Thank You for your reply. It helped me alot. What I mean by Reduced to its lowest form is - The Boolean Expression is reduced to a point where no more Laws or Rules can be applied to the remainder of the original expression.

    Would I be correct if I reduced the expression from where you stopped to :

    F = (B+B')(AC+A'C')
    = (1)(AC+A'C')
    = AC + A'C'
    = AC

    Thank You for your help
  5. May 21, 2013 #4


    User Avatar
    Science Advisor

    Stop at F = AC + A'C', you can't justify the last line.
  6. May 22, 2013 #5
    Thanks I see what you mean. I went to class last night and the same was said to me on how to start these expressions. Look for similarities, Thank You again for your assistance I am almost sure I got this now
  7. May 22, 2013 #6
    The last line can actually be done using an XNOR gate.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook