Problem Detail: When minimizing the full adder, I don’t understand why $A(bar{B}bar{C} + BC)$ reduces to $Aoverline{(Boplus{C})}.$ $(bar{B}bar{C} + BC)to (Boplus{C})$ is partially decipherable, but why is $(Boplus{C})$ inverted to $overline{(Boplus{C})}?$ Full adder simplification: $ bar{A}bar{B}C + bar{A}Bbar{C} + Abar{B}bar{C} + ABC = bar{A}(bar{B}C + Bbar{C}) + A(bar{B}bar{C} + BC) = bar{A}(Boplus{C}) + A(overline{Boplus{C}}) = Aoplus{(Boplus{C})} $
Could you help me out?
PS: I hope that this is the correct subforum of StackExchange to ask this (perhaps Electrical Engineering is the proper venue). I couldn’t find appropriate tags on either site.
Asked By : Tyler
Answered By : Paresh
Hint 1: Intuitively, $oplus$ implies exactly one of the two inputs is $1$ ($B$ and $C$ here). Whereas, $(bar{B}bar{C} + BC)$ implies both inputs are $0$ or both are $1$. Hint 2: Start from $overline{(B oplus C)}$, expand it, use De’Morgan’s laws, simplify and you should reach $(bar{B}bar{C} + BC)$.
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Question Source : http://cs.stackexchange.com/questions/9866
