# Simplification of the boolean expression

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'

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}

• Can I ask the reason of the downvote? – Taroccoesbrocco Apr 5 at 10:37
• Thanks a lot!This was very helpful! – Jarvis Ferns Apr 9 at 20:31

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.

• Thanks a lot!This was very helpful! – Jarvis Ferns Apr 9 at 21:52