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Boolean algebra question

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  • 28-01-2009 1:09pm
    #1
    Owen
    Registered Users Posts: 12,683 ✭✭✭✭


    My head is wrecked, I've got the following statement :

    X=A'BCD + AB'CD + ABC'D + ABCD' + ABCD

    I've reduced this to :

    X=CD(A'B+AB') + AB(C+D)

    Can this be reduced any further using boolean algebra? I've been googling lots, but most examples are only using ABC, not ABCD, and I'm stumped.


Comments

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    #2
    nialo
    Registered Users Posts: 610 ✭✭✭nialo


    i think you can reduce it to CD + AB(C +D) as A'B + AB' is 1.

    or

    CD(A'B + AB') + AB(C'D + CD') + ABCD
    CD(1) + AB(1) + ABCD


  • Options
    #3
    Owen
    Registered Users Posts: 12,683 ✭✭✭✭Owen


    nialo wrote: »
    A'B + AB' is 1.

    Oh, well, that's where I've been screwing up! Thanks Nialo :)


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    #4
    nialo
    Registered Users Posts: 610 ✭✭✭nialo


    Dont take that as golden. :) but i think its right. in simpler terms A + A' is 1 from what ive read. so would stand to reason it applys to AB as well.


  • Options
    #5
    nobodythere
    Closed Accounts Posts: 2,349 ✭✭✭nobodythere 


    Karnaugh Map:
   CD: 00  01  11  10
AB
00     
01              1
11          1   1   1
10              1

Best reduction is
ABC + ABD + BCD + ACD
= BC(A+D) + AD(B+C)


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    #6
    Cpt Tremendous
    Registered Users Posts: 80 ✭✭Cpt Tremendous


    Divide by zero


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