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Help with understing how this boolean expression was sinplified

Discussion in 'Electronic Design' started by Maxim Vexler, Dec 22, 2004.

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  1. Maxim Vexler

    Maxim Vexler Guest

    Please help me understanding how this boolean expression was simplified

    1. ab + ab' + a'b =
    2. a(b+b') + a'b =
    3. a*1 + a'b =
    4. a + a'b + 0 =
    5. a + a'b + aa' =
    6. a + b(a + a') =
    7. a + b*1 =
    8. a + b

    I did a truth table on both function [1] & [8], they produce the same
    result so they are the same function, but I can't seem to understand
    using what theorem / postulate can move from line 5 to 6 ?

    Appreciate the help,
    Maxim.

    Example was taken from "Art Of Assembly (DOS-16bit), Chapter 02, Page 48"
     
  2. Maxim Vexler

    Maxim Vexler Guest

    My point exactly!

    I don't know based on what/how he did it, but check it yourself: both
    functions produce the same result (IE they are the same function).

    What I'm trying to figure out based on what boolean algebra theorem the
    expression was simplified?

    Look :
    P4: ab+ac=a(b+c)
    P5: a+a'=1, a*a'=0
    Th3: a+0=a
    Th4: a*1=a

    F=ab + ab' + a'b
    [by P4] = a(b+b') + a'b
    [by P5] = a*1 + a'b
    [by Th4] = a + a'b
    [by Th3] = a + a'b + 0
    [by P5] = a + a'b + aa'
    [by wtf?] a + b(a + a')
    [by P5] a + b*1 =
    [by Th4] = a + b
     
  3. Active8

    Active8 Guest

    since aa' = 0,
    then baa' = 0

    so you put the b there so you can use that, wtf?, associative
    property? cute.
     
  4. Active8

    Active8 Guest

    I'm wrong, well, I was right, but it doesn't justify the wtf? step.

    a'b + aa' =
    a'b + baa' =
    b(a' + aa') not b(a' + a) or b(a + a')

    don't be surprised if there was a misprint. Doesn't make sense,
    though. Somebody will be along to either straighten this out or
    confirm wtf?
     
  5. Can't help you, too much eggnog to focus...

    :)
     
  6. mark thomas

    mark thomas Guest

    The artist formerly known as Maxim Vexler <hq4ever (at) 012 (dot) net (dot)
    il> >" < wrote:

    | Please help me understanding how this boolean expression was simplified
    |
    | 1. ab + ab' + a'b =
    | 2. a(b+b') + a'b =
    | 3. a*1 + a'b =
    | 4. a + a'b + 0 =
    | 5. a + a'b + aa' =
    | 6. a + b(a + a') =
    | 7. a + b*1 =
    | 8. a + b
    |
    | I did a truth table on both function [1] & [8], they produce the same
    | result so they are the same function, but I can't seem to understand
    | using what theorem / postulate can move from line 5 to 6 ?
    |
    | Appreciate the help,
    | Maxim.
    |
    | Example was taken from "Art Of Assembly (DOS-16bit), Chapter 02, Page 48"

    I'll give you a hint:

    x' + xy
    = x'(1 + y) + xy
     
  7. Rich Grise

    Rich Grise Guest

    How does it work if you go,
    a + a'b + bb' = ?
    Cheers!
    Rich
     
  8. Ratch

    Ratch Guest

    "Maxim Vexler <hq4ever (at) 012 (dot) net (dot) il> >" <"Maxim Vexler
    There is mistake in step 4 and step 6. Using the absorption theorem:
    x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

    a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
    Ratch
     
  9. Ratch

    Ratch Guest

    hq4ever (at) 012 (dot) net (dot) il> wrote in message
    There is mistake in step 4 and step 6. Using the absorption theorem:
    x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

    a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
    Ratch
     
  10. Thaas

    Thaas Guest

    Step 6 should have been:

    6. a + a'(a + b) =

    Since x + x'y = x + y:

    7. a + (a + b) =

    Since x + x = x:

    8. a + b

    In the quoted text the author would have us believe between steps 5
    and 6 that:

    a'b + aa' = b(a + a')

    Which simplifies to:

    a'b = b

    A fallacy.
     
  11. Ratch

    Ratch Guest


    If one knows that the identity x + x'y = x + y is true, then a + a'b =
    a + b, and one does not need to go further. Ratch
     
  12. Thaas

    Thaas Guest

    Yeah, I just crawled back out of bed on that realization. Still, the
    fallacy remains. The author mistyped something between 5 and 6.
     
  13. Thaas

    Thaas Guest

    arrrgh! You're right, it was the use of aa' for 0 from step 4 to 5
    that threw the proof offcourse. Should've stayed in bed.
     
  14. Fred Bloggs

    Fred Bloggs Guest

    He must be relying on some fundamental results like a=a+ab and a=a+aa',
    which are fairly self-evident and require little reasoning beyond
    definition manipulation. Then 5) becomes: a+ ab + a'b + aa' when those
    substitutions are made, and 6) obviously follows. There is no error,
    typo, or illogic- it was a departure from one simple substitution per
    line that threw you.
     
  15. Fred Bloggs

    Fred Bloggs Guest

    Any term in the summation can be replicated - so replicate ab.

    This is rewritten as:

    2. ab + ab' + ab + a'b

    and then

    3. a(b+b') + (a+a')b

    and then

    4. a + b
    That book was written by a moron, throw it in the trash.
     
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