Steps to simplify this boolean expression

Question

How do you simplify: ~A*B+A*~C+B*~C to A * ~C + B * ~A

I tried the distributive law but I end up going in circles.

Answer 1

This equivalence is well known and called the Consensus Theorem.

It can be proven as follows:

$$A'B + AC' + BC' \overset{Adjacency}{=}A'B + AC' + ABC' +A'BC' \overset{Absorption (2x)}{=}A'B + AC' $$

Answer 2

\begin{align} \bar{a}\,b+a\,\bar{c}+b\,\bar{c}&= \bar{a}\,b\,(c+\bar{c})+a\,\bar{c}\,(b+\bar{b})+b\,\bar{c}\,(a+\bar{a}) \\ &= \bar{a}\,b\,c+\bar{a}\,b\,\bar{c}+a\,\bar{c}\,b+a\,\bar{c}\,\bar{b} \\ &= \bar{a}\,b\,(c+\bar{c})+a\,\bar{c}\,(b+\bar{b}) \\ &= \bar{a}\,b+a\,\bar{c} . \end{align}

Edit

This simplification can be demonstrated with the unit-distance cube: