Simplifying a Boolean expression that has 3 variables in it

Question

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.

Answer 1

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.