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Homework Help: Laws of Boolean Algebra

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data and attempted solutions
    Simply each of the following Boolean expressions:
    i)
    http://img69.imageshack.us/img69/4438/equa1.jpg [Broken]

    ii)
    http://img69.imageshack.us/img69/9907/equa2.jpg [Broken]

    iii)Expand the following SOP form to proper miniterms (don't forget to delete identical terms):
    http://img69.imageshack.us/img69/1740/equa3.jpg [Broken]

    2. Relevant equations
    Laws of Boolean Algebra:

    A + A = A A . A = A NOT NOT A = A
    A + 0 = A A . 0 = 0
    A + 1 = 1 A . 1 = A
    A + NOT A = 1 A . NOT A = 0

    (A + B) + C = A + (B + C) (A . B) . C = A . (B . C)
    (A + B) = (B + A) (A . B) = (B . A)

    A . (B + C) = (A . B) + (A . C) A + (B . C) = (A + B) . (A + C)

    NOT (A . B) = NOT A + NOT B
    NOT (A + B) = NOT A . NOT B

    X . Y + X . NOTY

    P.S
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 4, 2010 #2
    Re: Boolean

    in 1) you made the first step incorrectly
    NOT (A + B) = NOT A . NOT B

    same in ii)
     
  4. Feb 4, 2010 #3
    Re: Boolean

    so for question i) would the first line be NOT A . NOT A . B then i went A . NOT A(B) = B because A . NOT A = 0
    i don't think this is correct.
     
  5. Feb 5, 2010 #4
    Re: Boolean

    Try to be very punctual about what you do, do one step at a time, and it helps to note the relevant equality while you are not sure.
    You can check each step by drawing the Karnaugh table for it. I do the first step for you to see what I mean.

    [tex]\overline{\overline{A} + A*\overline{B}}[/tex]
    Its table:
    [tex]
    \halign{\hfil # & # & # & #\hfil \cr
    & & A & \cr
    & & 0 & 1 \cr
    B & 0& 0 & 0 \cr
    & 1 & 0 & 1 \cr }[/tex]

    Now using [tex] \overline{x+y} = \overline{x} * \overline{y}[/tex] we got:
    [tex]\overline{\overline{A}} * (\overline{A*\overline{B}})[/tex]
    Now if you fill the table for it, you will see whether the transformation is done right.

    Of course you can deliver a normal form, and a simplified expression directly from the table. However I recommend to master the algebraic solution also, because there are cases where the algebraic way is much easier.
     
  6. Feb 5, 2010 #5
    Re: Boolean

    Well by using the kmap or truth table i found out that

    0 * 0 equals [tex](A * \overline{A}B)[/tex]
    which equals to 0 so answer is [tex](A* B)[/tex]

    Now how would you find this without using a kmap or truth table?
     
  7. Feb 5, 2010 #6
    Re: Boolean

    I believe you violated DeMorgan's Theorem in step one, like magwas said.

    I see in i) you complemented the overall output, starting with the + as the outermost operation. Yet you complement A AND NOT B individually, instead of as a whole. Remember that (A . NOTB) is an input of itself. Once you change the operation inside of (A . NOT B) to (A + NOT B), then you can complement the inputs A and B individually.

    Additionally, remember that when you complement output and inputs, you go from OR to AND, yet I see you never changed the OR.

    Perhaps my explanation is confusing, so here is a picture.
    http://img519.imageshack.us/img519/5554/boolean.jpg [Broken]

    Hopefully someone will back me up, as I am learning this as well.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 6, 2010 #7
    Re: Boolean

    Using algebra. Bool algebra is very similar to usual algebra. Only we have one additional operator above + and *. Remember that NOT() is a function. For arbitrary f, you cannot do f(a + b) = f(a) + f(b), or any other meaningful thing like that.
    With f being NOT, we have the luxury of having the DeMorgan rules:
    [tex]\neg(a+b)=\neg(a)*\neg(b) [/tex]
    and
    [tex]\neg(a*b) = \neg(a) + \neg(b)[/tex].

    So all you have to do is do simple algebra, and adhere to the rules.
     
  9. Feb 6, 2010 #8
    Re: Boolean

    yeh i got the answer thanks for help.
     
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