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Math Help - Simplification of boolean algebra

  1. #1
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    Post Simplification of boolean algebra

    PLEASE LOOK AT THIS TABLE


    TABLE 1. POSTULATES OF BOOLEAN ALGEBRA
    Basic Identity
    (A) x+0 = x
    (B) x . 1 = x
    Commutative Property
    (A) x+y = y+x
    (B) xy = yx
    Distributive Property
    (A) x (y+z) = xy + xz
    (B) x+yz = (x+y) (x+z)
    Basic Identity
    (A) x+x’ = 1
    (B) x . x’ = 0

    TABLE 2. THEOREMS OF BOOLEAN ALGEBRA
    Theorem 1: Identity
    (A) x+x = x
    (B) x.x = x
    Theorem 2: Identity
    (A) x+1 = 1
    (B) x.0 = 0
    Theorem 3: Involution
    (x’)’= x

    Theorem 4: Associative
    (A) x+(y+z) = (x+y)+z
    (B) x(yz) = (xy)z
    Theorem 5: DeMorgan
    (A) (x+y)’= x’y’
    (B) (xy)’ = x’+y’
    Theorem 6: Absorption
    (A) x+xy = x
    (B) x(x+y) = x

    Now look at this sample problem and solution:

    SIMPLIFY ABC+A’B’C+A’BC+ABC’+A’B’C’
    SOLUTION
    REASON
    =ABC+A’BC+A’B’C+A’B’C’+ABC’
    Commutative (a)
    =BC (A+A’) + A’B’ (C+C’) + ABC’
    Distributive (a)
    =BC (1) + A’B’ (1) + ABC’
    Basic Identity (a)
    =BC+ A’B’+ABC’
    Basic Identity (b)
    =A’B’ + BC +ABC’
    Commutative (a)
    =A’B’+B (C+AC’)
    Distributive (a)
    =A’B’{B’+C’ (A’+C)}
    DeMorgan (a,b)
    =A’B’ (B’+C’A+C’C)
    Distributive (a)
    =A’B’ (B’+A’C+CC’)
    Commutative (b)
    =A’B’ (B’+A’C’+0)
    Basic Identity (b)
    =A’B’ (B’+A’C’)
    Basic Identity (a)
    =A’B’+B (A+C)
    DeMorgan (a,b)

    Now, can you simplify this Boolean equation using Postulates and theorems of Boolean Algebra into 3 terms only?

    WXYZ + W’XYZ + WX’YZ + WXYZ’ + W’X’YZ + WXY’Z’ + W’X’Y’Z +W’X’Y’Z’


    Thanks!
    Last edited by mr fantastic; January 10th 2009 at 01:20 AM. Reason: Removed email address from post
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  2. #2
    Lord of certain Rings
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    I will suggest an easier method. Write the Karnaugh map for that expression, then simplify it to the three terms you require. Do you know how to write the Karnaugh map for logic expressions?
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  3. #3
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    simplification of boolean algebra

    What is required is to simplify it using postulates and theorem but not by using Karnaugh map... Thanks
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  4. #4
    nvv
    nvv is offline
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    Hi, Urfreiend. And what is a problem? There is a sample, so simply do like in the sample

    Here is in my attempt. I obtain 4 terms, but maybe You'll find mistake in my solution and obtain correct solutions with 3 terms only

    WXYZ + WXYZ + WXYZ + WXYZ + WXYZ + WXYZ + WXYZ +WXYZ =

    |
    WXYZ + WXYZ = Distributive Property = XYZ(W+W')= Basic Identity=XYZ(1) =XYZ|
    |WXYZ +WXYZ =Distributive Property =WXY(Z+Z')= Basic Identity=WXY(1)=WXY|

    =
    XYZ + WXYZ + WXYZ + WXYZ + WXYZ + WXY =

    |
    XYZ + WXYZ = Distributive Property = YZ(X+WX)=Distributive Property =YZ((X+W)(X+X'))=Basic Identity=YZ((X+W)(1))=YZ((X+W))=YZX+YZW|


    =
    YZX+YZW+ WXYZ + WXYZ + WXYZ + WXY =

    |
    YZW + WXYZ=(from this place I woun't write POSTULATES and THEOREMS names)=YZ(W+WX)=YZW+YZX'|

    =
    YZX+YZW+YZX'+WXYZ +WXYZ + WXY =

    |
    YZX+YZX'=YZ|

    YZ+YZW+WXYZ +WXYZ + WXY =

    |
    YZW+WXYZ=YW(Z+XZ')=YWZ+YWX|

    =
    YZ+YWZ+YWX+WXYZ + WXY =

    |
    YWX+WXYZ=WX(Y+Y'Z')=WXY+WXZ'|

    =
    YZ+YWZ+WXY+WXZ+WXY =

    |
    YZ+YWZ=YZ(1+W)=YZ(1)=YZ|

    =YZ+
    WXY+WXZ+WXY =...
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  5. #5
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    Anyone

    Anyone who can answer this problem is great! Please use the theorem and postulates to simplify it
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