boolean algebra

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

Originally posted by avdk@Dec 13 2004, 06:00 PM
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

[post=4099]Quoted post[/post]​

Click to expand...

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