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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'

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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}

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

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

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