Boolean Algebra Simplification help

Discussion in 'Homework Help' started by PSU31, Jan 31, 2008.

  1. PSU31

    Thread Starter New Member

    Jan 31, 2008
    2
    0
    I'm trying to simplify this expression.

    a'b'c'+a'bc'+a'bc+ab'c+abc'+abc

    so far I have

    (a'b'c'+abc)+ (a'bc'+abc)+(a'bc+abc)+ (ab'c+abc)+ (abc'+abc)

    b(a'+a)(c'+c) + bc(a'+a) + ac(b'+b) + ab(c'+c)

    b + bc + ac + ab

    I'm stuck here and the final solution is 3 terms with 5 literals.

    Thank you for your help.
     
  2. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    I'm stuck too, how did I pass my digital logic class? I did do a K-map and found the solution to be : = b + a'c' + ac

    I need to refresh my boolean reduction, anyone help???

    Steve
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    a'b'c'+a'bc' +a'bc +ab'c +abc'+abc

    (next line, we know what is...)

    a'(b'c'+bc'+bc)+a(b'c+bc'+bc)

    (so, look for what is not!)

    (abc'+a'bc)'

    (b'(ac'+a'c))'

    b + a'c + ac'

    Make sense?

    Scubasteve, I don't want to even tell you how long ago that was, and how I had to reach back for this one ;)

    I double-checked using a truth table
    abc
    000
    001---
    010
    011
    100---
    101
    110
    111
    The entries with the dashes were not in the original statement. Then I realized I'd messed it up the first time around LOL

    One of my early work assignments was checking a 64Kbit Boolean table that had been burned into a EPROM by reading the waveform from a digital analyzer scope. That was a looooonnnnnggggg string of 1's and 0's. I thought my eyes were going to bleed to death by the time I was done.
     
  4. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    hi Wookie,

    Sadly enough, this was last semester for me! But, at least I got good at 5-bit K-maps :)

    I think there is a problem with your solution, I checked it over, it doesn't seem to hold the same logic as the original problem.

    I checked over = B + A'C' + AC and it holds. I didn't prove via boolean operations though, just via k-mapping it out..

    Steve

    Yours is B + A'C + AC'
    Code ( (Unknown Language)):
    1.  
    2. abc                                =                        should be
    3. 000                                     0                          1
    4. 001              1                     1                           0
    5. 010       1                            1                           1
    6. 011       1      1                     1                           1
    7. 100                      1              1                           0
    8. 101                                      0                           1
    9. 110       1              1              1                           1
    10. 111       1                             1                            1
     
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