boolean algebra

Discussion in 'Homework Help' started by avdk, Dec 13, 2004.

  1. avdk

    Thread Starter New Member

    Dec 13, 2004
    1
    0
    hey... i was going over my midterm.. studying for the final today at 6. i came up to this question and couldnt reduce it...

    Z = ( B' + C + D' )( A' + B' + C' )( B + C' + D )

    PLEASE HELP

    thanx
     
  2. dragan733

    Senior Member

    Dec 12, 2004
    152
    0
    Z=(A'B'+B'B'+B'C'+A'C+B'C+C'C'+A'D'+B'D'+C'D')(B+C'+D)
    Z=(A'B'+B'+B'C'+A'C+B'C+C'+A'D'+B'D'+C'D')(B+C'+D)
    Z=(B'(A'+1+C'+C+D')+A'C+C'(1+D')+A'D')(B+C'+D)
    Z=(B'+A'C+C'+A'D')(B+C'+D)
    A'C+C'=A'+C'
    Z=(B'+A'+C'+A'D')(B+C'+D)
    Z=(B'+A'(1+D')+C')(B+C'+D)
    Z=(B'+A'+C')(B+C'+D)
    Z=BB'+B'C'+B'D+A'B+A'C'+A'D+BC'+CC'+C'D
    Z=B'C'+B'D+A'B+A'C'+A'D+BC'+C'D
    Z=C'(B'+B+A')+D(B'+C')+A'B
    Z=C'+B'D+C'D+A'B
    Z=C'(1+D)+B'D+A'B
    Z=C'+B'D+A'B
     
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