Simplify the following Boolean expression: $$F=A'B'C'+A'B'C+A'BC'+AB'C'+AB'C.$$
I have tried and am getting stuck at $A'+B'+C'+A'BC'$.
According to the K-Map it should get to $B'+A'C'$.
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$F$ is NOT equal to $$A'+B'+C'+A'BC'=A'+B'+C'+A'BC'=A′+B′+C′(1+A′B)=A′+B′+C′.$$ If $A$ and $B$ are true and $C$ is false then $F$ is false whereas $A′+B′+C′$ is true.
In order to simplify the given Boolean expression, note that $A'B'C'=A'B'C'+A'B'C'$ and \begin{align*} F&=(A'B'C'+A'B'C+AB'C'+AB'C)+(A'BC'+A'B'C')\\ &=B'(A'(C'+C)+A(C'+C))+A'C'(B+B')\\ &=B'(A'+A)+A'C'=B'+A'C'\\ \end{align*} where we used the fact that $A+A'=B+B'=C+C'=1$.
Robert Z
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I cant get how you get to the first step – Seffu Jul 11 '17 at 14:36
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@Seffu Note that $A'B'C'=A'B'C'+A'B'C'$. Is it clear now? – Robert Z Jul 11 '17 at 14:37
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is there a better editor for me o post the question as it is??? – Seffu Jul 11 '17 at 14:44
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Hi Robert would there be any difference if some of the bars are continous??? – Seffu Jul 11 '17 at 15:11
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F=A'+B'+C'+A'BC – Seffu Jul 12 '17 at 03:40
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@Seffu $F$ is NOT equal to $A'+B'+C'+A'BC'$. See my edited answer. – Robert Z Jul 12 '17 at 05:52
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I understood you.I am asking on how to simplify (A′+B′+C′+A′BC′) – Seffu Jul 12 '17 at 06:11
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@Seffu By the truth table $(A′+B′+C′+A′BC′)$ is true iff $A$,$B$,$C$ are all false hence it is equal to $A′+B′+C′$ or also $(ABC)'$. Another way: $$A′+B′+C′+A′BC′=A′+B′+C′(1+A′B)=A′+B′+C′.$$ – Robert Z Jul 12 '17 at 06:31
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@Seffu Any further doubts? BTW if you are new here take a few minutes for a tour: https://math.stackexchange.com/tour – Robert Z Jul 12 '17 at 06:38
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Thanks very much.I have just realized that the question have is different from what I wrote because its not possible to type double or triple bars here – Seffu Jul 12 '17 at 08:26
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@Seffu What do you mean? Something like $\overline{AB}$ or $\overline{ABC}$ or $\overline{\overline{ABC}}$? Have you a web-link to this exercise? – Robert Z Jul 12 '17 at 08:30
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yes something like that,how can I type those??? – Seffu Jul 12 '17 at 08:33
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The command \overline{A} between dollars gives $\overline{A}$. – Robert Z Jul 12 '17 at 08:39
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@Seffu If you are going to modify your problem I suggest you to ask a new question. Once again, I encourage you to see: https://math.stackexchange.com/tour – Robert Z Jul 12 '17 at 08:57
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There is a handy equivalence rule called Adjacency:
Adjacency
$PQ + PQ' = P$
'Adjacency' is of course named after the fact that it reflects exactly what you do in a K-map: you group together adjacent cells in order to get a simpler expression.
Another very useful principle is called Reduction:
Reduction
$P + P'Q = P + Q$ (that is, in the context of $P$, the term $P'Q$ reduces to just $Q$)
Applied to your problem:
$$A'B'C'+A'B'C+A'BC'+AB'C'+AB'C= \text{ (Adjacency) }$$
$$A'B'+A'BC'+AB'= \text{ (Adjacency) }$$
$$B'+A'BC'\text{ (Reduction) }$$
$$B'+A'C'$$
If you didn't know these two equivalences yet ... put them in your toolbox immediately!
And here is one more:
Absorption
$P+PQ=P$
$P(P+Q)=P$
Bram28
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@Seffu That will be A'+B'+C' by Absorption. That is, the A' term 'absorbs' the A'BC' term. I'll add Absorption to the post: you definitely need to know that one as well if you want to become skilled at Boolean algebra – Bram28 Jul 12 '17 at 11:17