Boolean simplification trouble

Discussion in 'Homework Help' started by joshf, Nov 14, 2006.

  1. joshf

    Thread Starter New Member

    Nov 14, 2006
    Ok where to start. First you guys are a godsend. Who knew circuits would be harder than programming?

    I read your tutorial and am still having trouble with Boolean simplification.

    What I am going to do is write out my attempts at simplification of several problems and ask for any input on A. whether I got the correct answer and if not where I went wrong. and B. if there is a simpler logic. Two of these are homework problems the rest are just problems I am using to increase my knowledge of proper boolean process

    Problem number one:

    A'B + AB + A'B'

    READ AS WITH MY Logic A not and B OR A and B OR A not and B not

    ((A + B') (A' + B') ( A + B))' Here I applied involution

    (A + B')' After (A' + B') cancels (A + B)

    A' + B my final answer

    Problem number two:

    AB + A'B + AB'

    ((A+B)' (A'+B)' (A+B')')' involution

    ((A'+B') (A+B') (A'+B))'

    (A'+B')' after cancelation

    A + B My final answer

    In the above two problems I tried to adopt my logic from an internet example that used a k map.... but now my second inverse seems wrong according to my professor's lecture notes:eek:

    Ok the tough ones:

    Problem number three:

    (A + B) (A' + C) (B' + C')

    ( (AB)' +(A'C)' +(B'C')' )'

    ( A'B' + AC' + BC )'

    This is the point that I get stuck! every attempt to simplify further leads to everything cancelling out everything else!

    my final answer is...

    AB + A'C + B'C'

    If the consensus theorem applies - (A + B)(A' + C)(B + C) = (A + B)(A' + C) - or - A B + A'C + BC = AB +A'C - how would I apply it? it seems to me in my example all the variables cancel each other out ie. A to A', C to C' , B to B'

    Problem number four:

    A'B + AB'

    ( (A + B') (A' + B) )'

    ( AA' + AB + B'A' + B'B)'

    ( 0 + AB + B'A' + 0 )'

    ( AB + A'B' )'

    (A' + B') ( A + B) my final answer

    Problem number five:

    (a+b)’ + c

    c + (a + b)'

    which could be ....

    c + a' + b'


    ( (c)' (a'b')')'


    c+a'+b' My final answer.

    Ok, I realize I probably got most of these problems wrong so any help would be much appreciated. I did read the tutorial and will look at it again. My problem is the examples seem to make sense but my application is off.
  2. cat

    New Member

    Sep 25, 2006
    Hi JOSHF,

    I worked out all of the problems and from what i can see, you have used very complicated ways to arrive at the answer. You've gotten some questions right but i got different answers for others. Here are the ways i did all the questions.

    =A'B + AB + A'B + A'B' (I duplicated the A'B term)
    =B(A'+A) + A'(B+B') (pull out the common terms)
    =A' + B (Your answer is right!)

    AB + A'B + AB'
    = AB + A'B + AB + AB' (duplicate the AB term, This won't do anything to the boolean equation)
    = B(A+A') + A(B+B')
    = A+B (you have the correct answer)

    (A+B)(A'+C)(B'+C') (well, my first step was to multiply the whole thing out)
    = (AA' + AC + BA' +BC)(B'+C') (its like (a+b)(x+y) you do this thing in the same way as you would do in a math class)

    notice that we have AA', which is 0. so you actually have
    (AC + BA' +BC)(B'+C') then do the multiplication again

    B'AC +B'BA' + B'BC + C'AC + C'BA' + C'BC
    next step would be to try to identify the terms in which you have a variable and its compliment multiplied (AND) togather so we can eliminate that term. the rule is xx'=0

    from there you can get B'AC + C'BA'
    all the other terms equal to 0 because you have a particular variable and its compliment AND ed together which will produce 0

    I think that is the simplified form of the original equation.


    A'B + AB'
    I don't think you can simplify this equation any further because this is the definition of an XOR gate. What you have done is complicated the given original equation which is....... not really good.


    (a+b)' +c
    a'b' + c
    i guess that is the simplified version. You have to use DeMorgan theorem more carefully.

    Hope this helps.
  3. joshf

    Thread Starter New Member

    Nov 14, 2006
    thanks 4 your help

    it gave me confidence and helped me grasp the concept better.