How do I go from $A(\overline BC+B)$ to $A(B+C)$? What definition should I use to get the final answer?
Would like an explanation and proof so I can learn rather than just memorise.
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$\displaystyle\bar BC+B=\bar BC+B\cdot1=\bar BC+B(1+C)$ as $1+C=1$
$\displaystyle\implies \bar BC+B=C(B+\bar B)+B=C\cdot1+B=C+B$
lab bhattacharjee
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Ahh thanks, makes sense. – NLed Dec 27 '13 at 16:25
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I'm confused, in what kind of algebra does $1+C = 1$ if we know nothing about $C$?? – Patrick Da Silva Dec 27 '13 at 16:38
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2@PatrickDaSilva, In Boolean algebra $C=1$ or $0$ and $0+1=1=1+1,$ right? – lab bhattacharjee Dec 27 '13 at 17:04
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@labbhattacharjee : ah, so I interpret $+$ as a union, not as a modulo $2$ thing. Okay – Patrick Da Silva Dec 27 '13 at 17:05
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@PatrickDaSilva, I maintained the notations used by OP. Sorry for the confusion – lab bhattacharjee Dec 27 '13 at 17:06
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@labbhattacharjee : It's an algebra, you can write $+$ as $+$ and $\cdot$ as $\cdot$! I just wasn't aware of what they meant. It's fine :) – Patrick Da Silva Dec 27 '13 at 17:18
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@PatrickDaSilva: Beware that a "Boolean algebra" is not an "algebra" (in the sense of a ring that is simultaneously a module). Using additive notation for the join operator can be confusing because Boolean algebras are in one-to-one correspondence with Boolean rings, and the ring addition (which is a modulo 2 thing) does not correspond to the Boolean-algebra join. – hmakholm left over Monica Dec 27 '13 at 18:54
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@HenningMakholm If this doesn't make a lengthy discussion, how do you perform this correspondence? I didn't manage to look it up. – Patrick Da Silva Dec 27 '13 at 19:04
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1@PatrickDaSilva: Given a Boolean algebra $(A,{\land},{\lor},{\neg},0,1)$ you get a Boolean ring by defining $x\cdot y = x\land y$ and $x+y = (x\lor y)\land \neg(x\land y)$. Conversely given a Boolean ring (i.e. a ring where $x^2=x$ for all $x$), you can make a Boolean algebra by defining $x\land y=xy$, $x\lor y=x+y-xy$, and $\neg x=1-x$. Wikipedia has a good discussion. – hmakholm left over Monica Dec 27 '13 at 19:17
