Complex Boolean simplification

Question

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 $B'+A'C'$.

Answer 1

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

Answer 2

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

$P+PQ=P$

$P(P+Q)=P$