Simplify Boolean Algebra

Discussion in 'Homework Help' started by azall122, Feb 8, 2008.

  1. azall122

    Thread Starter New Member

    Feb 8, 2008
    1
    0
    (a+b'=c')(a+b'c)=a=b'c

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

    (a'+a')'=a
     
  2. luck

    New Member

    Jan 25, 2008
    8
    0
    You Will want to clarify what you are asking in the future. (Also proof read your posts before you make them; it took me a few minutes to figure out some of the = signs were suppose to be + signs)

    (a+b'+c')(a+b'c)=a(a+b'+c')+b'c(a+b'+c')=a=b'c
    First I factor (a+b'+c') into a and b'c.
    From the first term we know if a is high then the expression will be high because we have a OR stuff.
    We also know that if b'c is high then b'c*b' will also be high and because it is OR'ed with stuff the expression will be high

    (a'+b')(a'+b)=a'(a'+b')+b(a'+b')=a'+a'b'+a'b+bb'=a'
    First I factor (a'+b') into a' and b.
    From the 4 terms I have we can factor an a' out to get a'(1+b'+b+0). Because of that 1 we know if a is low then the function will be high because we are OR'ing a' with stuff

    (a'+a')'=(a*a)=a
    This is straight up demorgan's. http://en.wikipedia.org/wiki/De_Morgan's_laws
     
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