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 $B'+A'C'$.


$F$ 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 $A+A'=B+B'=C+C'=1$.

  • $\begingroup$ I cant get how you get to the first step $\endgroup$ – Seffu Jul 11 '17 at 14:36
  • $\begingroup$ @Seffu Note that $A'B'C'=A'B'C'+A'B'C'$. Is it clear now? $\endgroup$ – Robert Z Jul 11 '17 at 14:37
  • $\begingroup$ is there a better editor for me o post the question as it is??? $\endgroup$ – Seffu Jul 11 '17 at 14:44
  • $\begingroup$ Hi Robert would there be any difference if some of the bars are continous??? $\endgroup$ – Seffu Jul 11 '17 at 15:11
  • $\begingroup$ F=A'+B'+C'+A'BC $\endgroup$ – Seffu Jul 12 '17 at 3:40

There is a handy equivalence rule called 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:


$P + P'Q = P + Q$ (that is, in the context of $P$, the term $P'Q$ reduces to just $Q$)

Applied to your problem:

$$A'B'C'+A'B'C+A'BC'+AB'C'+AB'C= \text{ (Adjacency) }$$

$$A'B'+A'BC'+AB'= \text{ (Adjacency) }$$

$$B'+A'BC'\text{ (Reduction) }$$


If you didn't know these two equivalences yet ... put them in your toolbox immediately!

And here is one more:




  • $\begingroup$ What will A'+B'+C'+A'BC' give me? $\endgroup$ – Seffu Jul 12 '17 at 3:39
  • $\begingroup$ @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 $\endgroup$ – Bram28 Jul 12 '17 at 11:17

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