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Math Help - Boolean Algebra and proofs

  1. #1
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    Arrow Boolean Algebra and proofs

    Prove using Prove algebraically : 1) x'′⊕ y = x⊕y' = (x⊕y)'
    2) x⊕1 = x'
    3) x⊕x' = 1
    4) (A+B)(A'C'+C)(B'+AC') = A'B (A+B)(A.C+C)(B+AC)=AB
    I know x⊕y = xy'+x'y But how do i deal with X'⊕Y? What does that become?
    And for 4) I did:
    (A+B)[(X+Y)'+C] [B.(AC)']
    (A+B)[(A+B)'+C] [(B.(A'+B')]
    then what do I do from here?
    http://i.imgur.com/CPlRB91.png[/FONT][/COLOR]
    Been having trouble solving these questions of my homework. Thank You!
    Last edited by Plato; February 2nd 2013 at 04:17 AM.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Boolean Algebra and proofs

    I tried to do the monster one, i have no experience with boolean functions or circuit design but here is what i came up with


    if a_1a_0 and  b_1 b_0 represent 2, 2-digit binary numbers, then
    addition can be done as follows with the result being  m_2m_1m_0 a 3 bit binary number

     m_0 = a_0 XOR b_0
     m_1 = (a_0 AND b_0) XOR (a_1 XOR b_1)
     m_2 = [(b_0 AND a_0) XOR (a_1 XOR b_1)] OR (a_1 AND b_1) (OR  m_2 = m_1 OR (a_1 AND b_1)

    this adds 2 digit binary numbers. I used 3 half adders to get this. I dont know if it could be done more simply

    and multiplication can be done by just using the AND operator on every pair you get from this cross product \{a_0,a_1\} X \{b_0, b_1\} and then adding a 2 bit binary to a one bit binary (addition using the above method by using 0 for the leading digit of one 2 bit number).

    For ex

    11
    x 11
    -----------
    11 ----(save this digit without any operation and add 11 + 01 using the addition method above)
    11x (+)
    ------------
    1001
    Last edited by jakncoke; February 2nd 2013 at 04:35 AM.
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  3. #3
    Senior Member jakncoke's Avatar
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    Re: Boolean Algebra and proofs

    As for

    1)

    using the OR,AND,NOT def of XOR is the way to go

    for example
    \bar{x} \oplus y = \neg{} (\bar{x}  \text{ v }  y) ^  \neg(\bar{x} ^  y)= (take the negation inside)  \neg (x ^  \bar{y}) ^ (x v  \bar{y}) =  x \oplus \bar{y}


    2)
     x \oplus \bar{x} = (x v  \bar{x} ) ^  \neg (x ^  \bar{x} ) = 1 ^  \neg 0 = 1 ^ 1 = 1

    3)

     x \oplus 1 = (x v  1) ^  \neg (x ^  1) = 1 ^  \neg ( x ^  1)  = 1 ^  (\neg x v  0 ) = 1 ^ (  \neg x ) =  \neg x
    Last edited by jakncoke; February 2nd 2013 at 05:50 AM.
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  4. #4
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    Re: Boolean Algebra and proofs

    thank you, you are a beautiful human being.

    How would I solve this question?
    imgur: the simple image sharer

    I tried and got to:
    (A+B)[(A+C)' +C) [(B.A'+C')]


    This pic is also part of the "monster question" that I forgot to attach.
    Absolutely no idea how to take this, I shouldn't have taken an intro online course.. ><
    imgur: the simple image sharer
    Last edited by lunapt; February 2nd 2013 at 02:11 PM.
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  5. #5
    Senior Member jakncoke's Avatar
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    Re: Boolean Algebra and proofs

    Alrighty.

    So for that problem
    I rewrote is as  (A+B)(\bar{A}\bar{C}  + C) \neg{ (\bar{B} + AC)} Now i can rewrite the last term in parenthesis as  (B\bar{A} + \bar{C})
    So using commutativity and distributavity, i will distribute the  (A+B)(B\bar{A} + \bar{C}) = AB\bar{A} + BB\bar{A} + A\bar{C} + B\bar{C}
    This evaluates to since  (A\bar{A} = 0 and AA = A) 0 + B\bar{A} + A\bar{C} + B\bar{C} I will then distribute this term against the middle parenthetical term. to get  B\bar{A}\bar{A}C + B\bar{A}C + A\bar{C}\bar{A}\bar{C} + A\bar{C}C + B\bar{C}\bar{A}\bar{C} + B\bar{C}\bar{C} again note alot of terms cancel  (A\bar{A} = 0) so this equals  B\bar{A}C + B\bar{A}C + 0 + 0 + B\bar{A}\bar{C} + 0 = B\bar{A}C + B\bar{A}\bar{C} = B(\bar{A}C + \bar{A}\bar{C}) = B(\bar{A}(C + \bar{C})) = (Since (A + \bar{A} = 1)) = B(\bar{A} 1) = B(\bar{A}) = \bar{A}B
    Last edited by jakncoke; February 2nd 2013 at 02:54 PM.
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  6. #6
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    Re: Boolean Algebra and proofs

    thanks!!
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  7. #7
    Senior Member jakncoke's Avatar
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    Re: Boolean Algebra and proofs

    I was working on this, hopefully it might be of help




    Last edited by jakncoke; February 2nd 2013 at 04:56 PM.
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  8. #8
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    Re: Boolean Algebra and proofs

    Thanks, I'll def look into that.
    Are you able to help me with my last two questions? They got me stumped after I write the booleans

    imgur: the simple image sharer
    imgur: the simple image sharer



    For Question 1, I got:

    [(A+B).(C+D)] + [(C+D)' . (A+B)] + C
    C+(A+B+C+D) + (A+B+D) + (C+D)' + (C+D)+B
    A+B+C+(C+D)'

    I"m not sure how to simplify this further.

    Question 2 is really long and I would appreciate help. I got:
    A' + (D'+C') + A + (A' + C'+ D') + B' .D ' + (D+B)' = F

    I'm not even sure if this is correct, and these two questions are due as homework tonight.

    Any help would be much appreciated, thanks!
    Last edited by lunapt; February 2nd 2013 at 06:01 PM.
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  9. #9
    Senior Member jakncoke's Avatar
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    Re: Boolean Algebra and proofs

    Maybe I'm looking at it wrong but for 1 I got (A+B)(C+D)+~(C+D)(A+B)+C which simplify to A+B+C
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  10. #10
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    Re: Boolean Algebra and proofs

    could you please list the steps you did to get to a+b+c? I believe your answer is right, only our first expression is the same

    I tried distributing the (C+D) ' (A +B) and ended up with BABD +C as the final answer ><
    Last edited by lunapt; February 2nd 2013 at 06:35 PM.
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  11. #11
    Senior Member jakncoke's Avatar
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    Re: Boolean Algebra and proofs

    (a+b)(c+d)+~(c+d)(a+b)+c = (a+b)((c+d)+~(c+d))+c = (a+b)(1)+c= a+b+c
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  12. #12
    Senior Member jakncoke's Avatar
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    Re: Boolean Algebra and proofs

    As for the last one.

    The expression is ((~A + ~(C v D))A)((~ B \oplus D ) + (~BC)) = A~B~C~D
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