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Problem: Simplify the following expression using Boolean Algebra: $$ z = (B + \overline C)(\overline B + C) + \overline{ \overline A + B + \overline C} $$
Answer:
\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.

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

And now it is exactly the book answer.

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