Simplifying boolean algebra expression $(AB+AC)'+A'B'C$

Question

$$\eqalign{(AB+AC)'+A'B'C&=\overline{(AB+AC)}+\overline A \,\overline BC\\&=(\overline A+\overline B)(\overline A+\overline C)+\overline A\,\overline BC\\&=\overline A+\overline B\,\overline C+\overline A\,\overline BC\\&=\overline A+\overline B(C+C\overline B)}$$

this question i stuck at that, please show how to continuous simplify it?

Answer 1

Given

$$\eqalign{\overline{(AB+AC)} + \overline A \, \overline B \ C &=\overline{A(B+C)}+\overline A \,\overline B \, C\\ &=(\overline A +\overline B \, \overline C)+\overline A \, \overline B \, C\\ &=\overline A+\overline B\,\overline C+\overline A\,\overline BC \\ &=\overline A+\overline B(\overline C + \overline A C)\ \text {(wrong above)} }$$

$\ldots$ consider $$\eqalign {\overline C + \overline A \, C &= (\overline C + \overline A) \, (\overline C + C) \ \\ &= \cdots }$$

Answer 2

The last expression is wrong, it could read $\overline A+\overline B(\overline C+\overline AC)$, but not like this: $$\eqalign{(AB+AC)'+A'B'C&=\ldots \\ &=\overline A+\overline B\,\overline C+\overline A\,\overline BC \\ &=\overline A+\overline B\color{red}{(C+C\overline B)}}$$

Here are some equivalent expressions:

$$\eqalign{ (AB+AC)'+A'B'C & = (A(B+C))' + A'B'C \\ & = A' + (B+C)' + A'B'C \\ & = A' + B'C' + A'B'C \\ & = A'(1 + B'C) + B'C' \\ & = A' + B'C' \\ & = A' + (B+C)' \\ & = (A(B+C))' \\ & = (AB+AC)' }$$

It's up to you to choose which form is 'the simplest'.