# Simplifying a Boolean expression that has 3 variables in it

Problem: Simplify the following expression using Boolean Algebra: $$z = (B + \overline C)(\overline B + C) + \overline{ \overline A + B + \overline C}$$
\begin{align*} z &= (B + \overline C)(\overline B + C) + A \overline{B} C \\ z &= \overline C \, \overline B + BC + A \overline{B} C \\ z &= \overline B \, \overline C + C ( B + A \overline B ) \\ z &= \overline B \, \overline C + C ( A + B ) \\ z &= AC + BC + \overline B \, \overline C \\ \end{align*} However, the book gets: $$BC + \overline B ( \overline C + A )$$ I believe both answers are right but I would like to know how to get the book's answer.

• At first glance, one thing is you didn't convert AND to OR for ABC in your answer's first line. – TonyM Sep 24 '19 at 13:35
• The answers are not equivalent, and yours has an error between the 4th and 5th line. – Kevin Kruse Sep 24 '19 at 13:53
• Consider ABC = 101 and check both expressions. – Eugene Sh. Sep 24 '19 at 14:00
• Yes, now these are equivalent – Eugene Sh. Sep 24 '19 at 14:10
• Didn't you ask the same question yesterday – Meenie Leis Sep 24 '19 at 17:47

Take your answer: $$z = AC + BC + \overline B \, \overline C$$
Now make the following transformation: $$z = AC(B+\overline B) + BC + \overline B \, \overline C$$ Expand: $$z = ABC+A\overline B C + BC + \overline B \, \overline C$$ Use the redundancy rule $$\X+XY=X\$$ on the first and third terms:
$$z = BC + A\overline B C + \overline B \overline C =$$ $$= BC + \overline B(AC+\overline C)$$
Use another rule: $$\X+\overline XY=X + Y\$$ on the expression in parentheses: $$z = BC + \overline B(A+\overline C)$$