Simplification of the boolean expression

Question

Simplify the following expressions to the simplest expression using De Morgan's theorem and Boolean algebra.

AB+(C+B')(AB+C')

=AB+ABC+CC'+ABB'+B'C'

=AB+CC'+A+B'C'

=A+CC'+B'C'

=A+B'C'

Answer 1

There is a mistake between the second and the third line: $B\overline{B}$ is a contradiction, and hence $A B \overline{B}$ can be dropped (the same for $C \overline{C}$).

\begin{align} AB+(C+\overline{B})(AB+\overline{C}) &= AB+ABC+C\overline{C}+AB\overline{B}+\overline{B}\,\overline{C} \\ &= AB + ABC + \overline{B}\,\overline{C} \\ &= AB + \overline{B}\, \overline{C} \end{align}

Answer 2

Is the given solution correct?

No. In the second to third line you substitute $\rm ABB'$ with $\rm A$.

However $\rm ABB' = 0$ . ($\rm BB'$ is a contradiction.)

The rest of your working is okay.   So correct the error, try again, and you should have it.