HW Help: Boolean Simplification

Discussion in 'Homework Help' started by rtran2015, Oct 3, 2015.

  1. rtran2015

    Thread Starter New Member

    Oct 3, 2015
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    I have a couple questions (#) for my hw, below is what I worked on with things to help better understand the questions

    (-[_]) = "not" [_]
    AB = A*B = A & B
    A+B = A or B

    Given: Boolean Theorems T1-T12

    Simplify each Boolean Theorem

    (a) Y = (-A)BC + (-A)B(-C)
    (b) Y = (-A)(-B)(-C) + A(-B)
    (c) Y = ABC(-D) + A(-B)(-C)(-D) + (-(A + B + C + D))

    Work:
    (a): = (-A)B; T5 Complements
    (b): = How should I begin this process?
    Using the truth tables I was able to derive

    =(-A)(-B)(-C) + A(-B)(-C) + A(-B)C
    = (-A)(-B)(-C) + A(-B)C = (-B); using T5
    = A(-B)(-C) + A(-B)C = A(-B); using T5
    =(-B) + A(-B);
    Though I'm not even sure this is right
    (1) How should I simplify if I didn't use the truth table?

    (c): Used the same logic as (b)
    = (-B)(-C)(-D) + (-D)

    Some needed clarification:

    (2) B(-B) = 0; T5, but is it possible with multiple variables? AB+A(-B) = A?
    (3) Can we use single variable theorem (T1-T5) with multiple variables?
    (4) AB + (-A)(-B)D; Can this be simplified more?

    Note:

    Let me know if I need to explain more on a certain area (i.e. if you need all the boolean theorems)
    Give me simple equations to solve to better convey an idea
    Let me also know if you prefer the work to be at the top of the page or bottom if you care
    Book: Digital Design and computer Architecture by Harris and Harris, Exercise 2.14
     
  2. rtran2015

    Thread Starter New Member

    Oct 3, 2015
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    Additional:
    (-(ABCD)) = (-A)+(-B)+(-C)+(-D); via deMorgan's

    (5) (-(A(-B))) = -AB; is this possible?

    (6) is there a way to make it so a character can have a line above it i.e. A_var, or a cleaner way to portray "NOT"
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    For b I got:
    y=a'b'c'+ab' distributive law
    =b'(a'c'+a) t10: a+a'b=a+b
    =b'(a+c') distributive law
    =ab'+b'c'
     
  4. WBahn

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    Mar 31, 2012
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    It's hard to know what you are and aren't allowed to do when your assignment says to use theorems T1 through T12 and you haven't told us what theorems T1 through T12 are.
     
  5. WBahn

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    Mar 31, 2012
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    What happens if you "factor out" (-B) from both terms?

    Now consider the following:

    XY + (-X) = XY + (-X)(Y + (-Y))
    = XY + [(-X)Y] + (-X)(-Y)
    = XY + [(-X)Y + (-X)Y] + (-X)(-Y)
    = [XY + (-X)Y] + [(-X)Y + (-X)(-Y)]
    = [X + (-X)]Y + (-X)[Y + (-Y)]
    = Y + (-X)

    This is merely

    X = X(1) combined with 1 = (Y + (-Y))

    Look at this carefully. What you are claiming is that

    (-A)(-C) + AC = 1

    Does that ring true?

    This is just the reverse of the step before the last one.

    XY + X(-Y) = X(Y + (-Y)) = X(1) = X

    Assuming that this was correct to this point, does the value of A matter? What happens if you "factor out" (-B) from both terms.

    Though I'm not even sure this is right
    (1) How should I simplify if I didn't use the truth table?
    [/QUOTE]

    Using Boolean Algebra, as I've been doing.
     
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  6. rtran2015

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    Oct 3, 2015
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    for:
    =b'(a'c'+a) t10: a+a'b=a+b

    I understand if (a'c' +ac') = c'

    How does (a'c' +a) = (a+c') through combining?
    How are you removing a' but still keep a?
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I told you, it is t10, go look up how t10 is derived.
     
  8. rtran2015

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    Oct 3, 2015
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    T1: Identity B*1=B; T1': B+0=B
    T2: Null Element B*0=0; T2': B+1=1
    T3: Idempotency BB=1; T3': B+B=1
    T4: Involution B''=B
    T5: Complements BB'=0; T5': B+B'=1
    T6: Commute BC=CB; T6': B+C=C+B
    T7: Associate (BC)D=B(CD); T7': (B+C)+D=B+(C+D)
    T8: Distribute (BC)+(BD)=B(C+D); T8': (B+C)(B+D)=B+(CD)
    T9: Covering B(B+C)=B; T9':B+(BC)=B
    T10: Combining (BC)+(BC')=B T10': (B+C)(B+C')=B

    T11: Consensus (BC)+(B'D)+(CD) T11': (B+C)(B'+D)(C+D)
    =BC+B'D =(B+C)(B'+D)

    T12: DeMorgan's [B[0]*B[1]*...]' T12': [B[0]+B[1]+...]'
    =(B'[0]+B'[1]+...) =(B'[0]*B'[1]...)

    Modrators note: added a space between : and ( to avoid conversion to smilies
     
    Last edited by a moderator: Oct 3, 2015
  9. rtran2015

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    Oct 3, 2015
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    the frowny face is ': ('
     
  10. shteii01

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    Feb 19, 2010
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    T10a here: http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/
    The first example has the proof of T10.

    [​IMG]
     
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  11. rtran2015

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    Oct 3, 2015
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  12. rtran2015

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    Oct 3, 2015
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    I understand now

    thanks
     
  13. rtran2015

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    Oct 3, 2015
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    Given:
    ABCD' + A'B'C'D';

    D'(ABC +A'B'C'); T8
    D'(1); T5

    Is this allowed?
     
  14. WBahn

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    Just think about it for a moment.

    You are claiming that

    ABC + A'B'C' = 1

    Does this make any more sense then when I asked if

    (-A)(-C) + AC = 1

    rang true back in Post #5.

    Can you think of at least one combination of inputs for which

    ABC + A'B'C' = 0

    If you can, then you KNOW that the claim you are making is not correct.

    Well, what if A = 0. That guarantees that the first term, ABC, is 0 regardless of what we choose for B and C. So now is there a choice we can make for B (or C) that will make the second term, A'B'C", 0 despite the fact that we chose A to be 0? If so, your claim is toast.

    Get in the habit of ALWAYS asking if your results make sense. If you can't tell whether or not they make sense, then think about them that much harder because you will learn a lot in the process.
     
  15. rtran2015

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    Oct 3, 2015
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    Oh okay so

    ABC = 111
    A'B'C' = 000

    Due to the Axiom of A1 where [ B=0 if B != 1] and [B'=1 if B=0]

    Hence
    if ABC = 111, then A'B'C' = 000
    = 1 + 0 = 1 through Axiom 5' "AND/OR"
     
  16. WBahn

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    Mar 31, 2012
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    It is NOT sufficient to show that you get what you want for one particular choice of inputs. You have to PROVE that you get what you want for ALL possible combinations of inputs.

    I, on the other hand, only need to demonstrate that you DON'T get what you want for at least ONE combination of inputs.

    What is the result of ABC + A'B'C' if A=0 and B = 1 ?

    What does that tell you about your claim that ABC + A'B'C' = 1?
     
  17. rtran2015

    Thread Starter New Member

    Oct 3, 2015
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    oh alright,
    Then it isn't sufficient to say ABC + A'B'C' is equal to 1
    I will try a different approach to the question then

    The given equation:

    = ABCD' + A[BCD]' + ([A+B+C+D]')
    = ABCD' + A(B'+C'+D') + ([A+B+C+D]'); T12 DeMorgan's
    = ABCD' + A(B'+C'+D') + (A'B'C'D'); T12 DeMorgan's
    = ABCD' + AB' + AC' + AD' + A'B'C'D'; T8 Distribute <---
    -----------------------------------------------------------------------
    =A(BCD'+B') + ...; T8
    =A(B'+CD')+...; T6, T10
    =AB'+ACD'+AC'+...;T8
    =AB'+AC'+AD'+AD'+...; T8,T6,T10
    -----------------------------------------------------------------------
    =AB'+AC'+AD'+A'B'C'D'; T3
    =A[BCD]'+[A+B+C+D]'; T8,T12, T12

    This is as far as I have gotten
     
  18. WBahn

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    Mar 31, 2012
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    This is NOT the given equation!

    The given equation is:

    Which is

    Y = ABCD' + AB'C'D' + (A + B + C + D)'

    You are now trying to claim that

    B'C'D' = (BCD)'

    Again, ask yourself if that makes sense!

    The left hand side is 0 if ANY of the three variables is 1. But on the right hand side, in order to be 0 ALL of the three variables must be 1.

    Stop being so sloppy!
     
  19. rtran2015

    Thread Starter New Member

    Oct 3, 2015
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    Bear with me on this

    The given equation IS:
    ABCD' + A[BCD]' + ([A+B+C+D]')


    This equation is wrong:
    Y = ABCD' + AB'C'D' + (A + B + C + D)'

    I understand:
    B'C'D' != (BCD)'
    B'C'D' == (B+C+D)';T12 DeMorgan's

    would it change if there was another variable:

    A*(B'+C'+D') = A*(BCD)'; T12?

    I apologize for any confusion
    I'm still a beginner

    *Added more instructions for clarity*
    = ABCD' + A[BCD]' + ([A+B+C+D]')
    = ABCD' + A(B'+C'+D') + (A+B+C+D)'; T12 <---- Stop me if I am wrong here
    = ABCD' + A(B'+C'+D') + A'B'C'D'; T12
    = ABCD' + AB' + AC' + AD' + A'B'C'D'; T8
    -----------------------------------------------------------------------
    =A(BCD'+B')+AC'+AD'+A'B'C'D'; T8
    =A(B'+BCD')+AC'+AD'+A'B'C'D';T6
    =A(B'+CD')+AC'+AD'+A'B'C'D';T10
    =AB' +ACD' + AC' +AD' +A'B'C'D';T8
    =AB'+A(CD'+C')+AD'+A'B'C'D';T8
    =AB'+A(C'+CD')+AD'+A'B'C'D';T6
    =AB'+A(C'+D')+AD'+A'B'C'D';T10
    =AB'+AC'+AD'+AD'+A'B'C'D';T8
    =AB'+AC'+AD'+A'B'C'D'; T3
    =A(B'+C'+D')+A'B'C'D';T8
    ---------------------------------------------------
    =A[BCD]'+A'B'C'D';T12
    =A[BCD]'+[A+B+C+D]'; T12
     
  20. WBahn

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    Mar 31, 2012
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    Okay, so if I understand you correctly you are saying that the original equation for part (c) in your original post was incorrect and that the actual given starting equation is:

    ABCD' + A[BCD]' + ([A+B+C+D]')

    Good!

    Nope. You have A*(something) = A*(something) where you have merely expressed (something) using two different, but equivalent, forms.

    Not a problem -- you learn this stuff by fighting with this stuff.

    Fine to this point. Before proceeding as you do from here, see if you can't simplify this expression a bit more directly.

    For instance, what does ABCD' + AD' simply to?

    Without saying whether this is valid or not, there is the question of what it means to "simplify" an expression. There really needs to be some metric by which you can determine that one expression is "simpler" than another. In problems like this, it is usually implied that "simplest" means a SOP form with as few terms as possible.
     
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