Take the 2-minute tour ×
Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It's 100% free, no registration required.

When minimizing the full adder, I don't understand why $A(\bar{B}\bar{C} + BC)$ reduces to $A\overline{(B\oplus{C})}.$

$(\bar{B}\bar{C} + BC)\to (B\oplus{C})$ is partially decipherable, but why is $(B\oplus{C})$ inverted to $\overline{(B\oplus{C})}?$


Full adder simplification:

$ \bar{A}\bar{B}C + \bar{A}B\bar{C} + A\bar{B}\bar{C} + ABC \\ = \bar{A}(\bar{B}C + B\bar{C}) + A(\bar{B}\bar{C} + BC) \\ = \bar{A}(B\oplus{C}) + A(\overline{B\oplus{C}}) \\ = A\oplus{(B\oplus{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.

share|improve this question
    
Does $\overline{\vee}$ mean the exclusive-or symbol? –  Paresh Feb 17 '13 at 19:54
    
Yes, it's XOR. I think that I may have chosen the wrong symbol, remedying now. –  Delete Feb 17 '13 at 19:57
    
Yep, the wiki article says $\veebar$ to be a symbol, although I think $\oplus$ should be the most commonly used. –  Paresh Feb 17 '13 at 20:00
    
There. I changed $\veebar$ to $\oplus$ –  Delete Feb 17 '13 at 20:01
    
This may be a typo, but $(\bar{B}\bar{C} + BC)$ is not converted to $(B\oplus{C})$. Rather, $(\bar{B}C + B\bar{C})$ is converted. –  Paresh Feb 17 '13 at 20:06

2 Answers 2

up vote 4 down vote accepted

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 $(\bar{B}\bar{C} + BC)$.

share|improve this answer
    
Perfect, thank you. I'd upvote but I don't have the prerequisite rep points. –  Delete Feb 17 '13 at 20:15

Consider the truth table for $(\overline{B}\overline{C}+BC)$. It has 1's exactly when $B$ and $C$ get the same value, which is exactly when $(B\oplus C)$ gets 0, hence, $\overline{B\oplus C}$ is equivalent to $(\overline{B}\overline{C}+BC)$.

share|improve this answer
    
Perfect, thanks. I'll upvote when I have the prerequisite rep points. –  Delete Feb 17 '13 at 20:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.