Problem with boolean algebra!

  • Thread starter mr_coffee
  • Start date
  • #1
1,629
1

Main Question or Discussion Point

Hello everyone i'm having problems proving this:
note: A' will stand for complemented;

AB + BC'D' + A'BC + C'D = B + C'D
B(A+C'D')
B(A+C'+D')
BA + BC' + BD' + A'BC + C'D
BA + BD' + A'BC + C'(B+D)
B(A + D' + A'C) + C'(B+D)
now i'm stuck, i don't see how that is going to work out. Any help would be great.
 

Answers and Replies

  • #2
berkeman
Mentor
56,455
6,369
It works on a Karnaugh map. Try mapping it out to help you with the algebra. Look for combinations on the map that you can use as tricks in the algebraic manipulations.
 
  • #3
133
0
OK this is gonna be long but i but a lot off steps for explanation so you can skip some
steps

AB+BC'D'+A'BC+C'D
B(A+A'C)+BC'D'+C'D
B((A+A').(A+C))+BC'D'+C'D
B((1).(A+C))+BC'D'+C'D
AB+BC+BC'D'+C'D
B(C+C'D')+AB+C'D
B((C+C').(C+D'))+AB+C'D
BC+BD'+AB+C'D
BC.(D+D')+BD'+AB+C'D
BCD+BCD'+BD'+AB+C'D
BCD+BD'+AB+C'D
D(C'+BC)+BD'+AB
D((C'+C).(C'+B))+BD'+AB
D((1).(C'+B))+BD'+AB
CD'+BD+BD'+AB
C'D+B+AB
B(A+1)+C'D
B+C'D = RHS
 
  • #4
1,629
1
awesome, thanks for the help guys!!
 
  • #5
133
0
Your welcome
 

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