# Problem with boolean algebra!

## 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.

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berkeman
Mentor
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.

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

awesome, thanks for the help guys!!