Trying to simplify a Boolean expression

Simplify the following expression using Boolean Algebra: $$x = \bar{A} \bar{B} \bar{C} + \bar{A}BC + ABC + A \bar{B} \bar{C} + A \bar{B} C$$
\begin{align*} x &= \bar{A} \bar{B} \bar{C} + \bar{A}BC + ABC + A \bar{B} ( \bar{C} + C ) \\ x &= \bar{A} \bar{B} \bar{C} + \bar{A}BC + ABC + A \bar{B} \\ x &= \bar{A} \bar{B} \bar{C} + \bar{A}BC + A( BC + \bar{B} ) \\ x &= \bar{A} \bar{B} \bar{C} + \bar{A}BC + A( \bar{B} + C ) \\ x &= \bar{A} ( \bar{B} \bar{C} + BC ) + A( \bar{B} + C ) \\ \end{align*} Now, I feel I am stuck. The book's answer is: $$x = BC + \bar{B}(\bar{C} + A)$$.

Based upon @StainlessSteelRat comments I tired this: \begin{align*} x &= \bar{A} \bar{B} \bar{C} + \bar{A} B C + ABC + A \bar{B}\bar{C} + A \bar{B}C \\ x &= \bar{A} \bar{B} \bar{C} + A \bar{B}\bar{C} + \bar{A} B C + ABC + A\bar{B}C + ABC \\ x &= \bar{B} \bar{C} ( \bar{A} + A) + BC(\bar{A} + A) + AC(\bar{B} + B) \\ x &= \bar{B} \bar{C} + BC + AC \\ x &= \bar{B} \bar{C} + C(A + B ) \\ \end{align*} However, I still get the wrong answer.

• You can use the starting terms multiple times. Term 3 & Term 5. Your reduction on line 3 is incorrect. – StainlessSteelRat Sep 22 at 22:51
• @StainlessSteelRat I do not understand why this is wrong: $x = \bar{A} \bar{B} \bar{C} + \bar{A}BC + A( BC + \bar{B} )$ – Bob Sep 22 at 23:15
• Although, the 3rd line is correct mathematically, it is not the optimum. You are correct and wrong. Try reducing all 5 terms in the first step. And there is two optimum solutions. – StainlessSteelRat Sep 22 at 23:25
• Try to reduce each term with other terms. Repeat a term as required. Does your answer look like the answer? By doing two terms at a time, you have backed yourself into a corner. – StainlessSteelRat Sep 22 at 23:35
• No, DeMorgan's is not required. Read all my comments! – StainlessSteelRat Sep 22 at 23:36

Simplify the following expression using Boolean Algebra: $$x = \bar{A} \bar{B} \bar{C} + \bar{A}BC + ABC + A \bar{B} \bar{C} + A \bar{B} C$$
\begin{align*} x &= \bar{A} \bar{B} \bar{C} + (\bar{A}+A)BC + A \bar{B} ( \bar{C} + C ) \\ x &= \bar{A} \bar{B} \bar{C} + BC + A \bar{B} \\ x &= BC + \bar{B} ( A + \bar{A} \bar{C} ) \\ x &= BC + \bar{B} ( (A + A \bar{C}) + \bar{A} \bar{C} ) \\ x &= BC + \bar{B} ( A + (A + \bar{A}) \bar{C} ) \\ x &= BC + \bar{B} ( A + \bar{C} ) \end{align*}

The trick is to introduce $$\A \bar{C}\$$ in the 4th line.

• I do not understand how you went from $x = BC + \bar{B} ( (A + A C) + \bar{A} \bar{C} )$ to $x = BC + \bar{B} ( A + (A + \bar{A}) \bar{C} )$$– Bob Sep 23 at 1:02 • I do not understand how you went from line 4 to line 5. Where you suppose to introduce$A\bar{C}$instead of$AC\$ – Bob Sep 23 at 1:05
• Ok, I made a typeo - I forgot to add the bar on C. – le_top Sep 23 at 9:21

Actually your second solution as well as the textbook solution, both are right.

I put the expression for $$\X\$$ into K-map:

It gives me the reduced expression as: $$X=\overline B\overline C+A\overline B +BC$$ $$ie.,X=BC+\overline B(A+\overline C)$$

But I can round the 1's in this manner as well:

It gives me the reduced expression as: $$X=\overline B\overline C+AC +BC$$ $$ie.,X=\overline B\overline C+C(A+B)$$

First solution is what your textbook derived. Second solution is what you derived. Both reduced expressions are right.