How do I simplify the Boolean function: ABC + B I've tried doing De Morgan's, but that doesn't do anything to simplify this. Can somebody show the steps on how to simplify this please?
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Can you simplify to $A'+ABC=A'+BC$? And then one step further. – Sarvesh Ravichandran Iyer Aug 09 '16 at 01:54
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@астонвіллаолофмэллбэрг How did you get that? I have no idea how – joe Aug 09 '16 at 01:56
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1Is ABC mean A and B and C?. Does B' mean not B? Does + mean and? Of or? – fleablood Aug 09 '16 at 02:03
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@fleablood Yes. That is standard convention for Boolean-algebra. $+$ is disjunction ("or"), and $\cdot$ is conjunction ("and"), while $x'$ is the negation of $x$. Also, $1$ is true, and $0$ is false. – Graham Kemp Aug 09 '16 at 02:15
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Usually, the previous notation means $(A \land B \land C)\lor \lnot A \lor \lnot C $ – JnxF Aug 09 '16 at 02:17
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@joe , астонвіллаолофмэллбэрг used distribution and complementation: $A' + AX = (A'+A)(A'+X) = 1(A'+X)$ – Graham Kemp Aug 09 '16 at 02:18
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@GrahamKemp thank you for the clarification. Unfortunately the question changed after my comment, but now the best way to do it is in fact, to go for karnaugh mapping. – Sarvesh Ravichandran Iyer Aug 09 '16 at 02:33
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@GrahamKemp How do you go from A′+AX to (A′+A)(A′+X) exactly? – joe Aug 09 '16 at 02:38
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@joe It is Distribution $\rm X+YZ = (X+Y)(X+Z)$ – Graham Kemp Aug 09 '16 at 05:11
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@GrahamKemp thanks – joe Aug 09 '16 at 17:35
2 Answers
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$ABC+\bar{A} +\bar{C} = ABC+ \bar{A}BC+ \bar{A} +\bar{C} =BC + \bar{A} +\bar{C}$.
$BC + \bar{A} +\bar{C}= BC + \bar{A} +\bar{C} + B\bar{C}= B + \bar{A} +\bar{C}$.
copper.hat
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how do you get from ABC + A' + C' to ABC + A'BC + A' + C'? what theorem do you use? – joe Aug 09 '16 at 04:04
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1It is easy to verify using a truth table that $X = X + XY$ for any $Y$. Hence $\bar{A} = \bar{A}+ \bar{A}BC$. – copper.hat Aug 09 '16 at 04:31
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DeMorgan's is the right way to go.
To get you started, it gives you: $$\begin{align}&\rm ABC+A'+C' \\[1ex]=~&\rm B(AC)+(AC)'\\[1ex] \ddots\end{align}$$
Now try distribution and simplify. (Alt: you could distribute and simplify twice.)
Graham Kemp
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@joe In Boolean-algebra disjunction($+$) distributes over conjunction($\cdot$). $$\rm X+Y\cdot Z = (X+Y)\cdot (X+Z)$$ – Graham Kemp Aug 09 '16 at 05:14