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Prove that

$\bar{A}B + AC + BC = \bar{A}B + AC$

with the help of boolean algebraic manipulations. I have no clue from where to start, how should I tackle these type of questions?

Or

$$ \left(\neg A \wedge B\right)\vee \left(A \wedge C\right)\vee\left(B \wedge C\right) = \left(\neg A \wedge B\right)\vee \left(A \wedge C\right)$$

Jared
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Soham
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  • You haven't written anything in any kind of boolean algebra that I understand. It looks more like set algebra, where $\bar{A}$ means the complement of the set $A$ (i.e. every value that is not in the set $A$). I also don't know how to interpret juxtaposition, i.e. $\bar{A}B$, and I don't know how to interpret addition, i.e. $+$. Does juxtaposition mean an intersection? And addition ($+$) mean union? Or what? – Jared Sep 21 '14 at 07:06
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    A and B are boolean variables, the $\bar A$ denotes NOT A, AB denotes A AND B and A+B denotes A OR B – Soham Sep 21 '14 at 07:08

2 Answers2

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HINT Since you want to get rid of the term $BC$, maybe it makes sense to do something to it : $$\begin{align}\bar{A}B + AC + BC &= \bar{A}B + AC + (A+\bar{A})BC\\&=\bar{A}B + AC + ABC+\bar{A}BC\end{align} $$

Group first and last terms
Group middle two terms

AgentS
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  • thanks! could you explain the motivation for this problem? – Soham Sep 21 '14 at 07:18
  • I think simplification is the main motivation for most boolean algebra problems. I'm not gonna lie here, its not that obvious why/how one comes up with anding $A+\bar{A}$ to $BC$; there might be other ways to approach this problem but since we know that we need to kick out $BC$, anding A+A' looks like a good start :) – AgentS Sep 21 '14 at 07:24
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Just to start from the obvious, if $A$ is false, then clearly either $B$ or $B$ and $C$ must be true. Right away, this means that when $A$ is false, $B$ cannot be false. So this suggests $\neg A \wedge B$. Now, if $A$ is true then either $C$ is true or $C$ and $B$ is true--either way $C$ is true, therefore it's clear that this is equivalent to $(A \wedge B)\vee (\neg A \wedge C)$.

Jared
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