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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.

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

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Perfect, thank you. I'd upvote but I don't have the prerequisite rep points. – Tyler 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 $(\overline{B}\overline{C}+BC)$.

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Perfect, thanks. I'll upvote when I have the prerequisite rep points. – Tyler Feb 17 '13 at 20:15

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