Simplification of: AB + A'C + BC in boolean algebra I am trying to understand the simplification of the boolean expression:

AB + A'C + BC

I know it simplifies to

A'C + BC

And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.
\begin{align*}
&amp;\mathrel{\phantom{=}}AB+A'C+BC\\
&amp;=AB+A'C+BC(A+A')          \quad \text{($A+A'=1$, Complementarity law)}\\
&amp;=AB+A'C+ABC+A'BC\\
&amp;=AB+ABC+A'C+ABC           \quad \text{(Associative law)}\\
&amp;=AB+A'C                 \quad   \text{(Absorption law)}
\end{align*}
a.b+a'.c+b.c
a.b+a'.c+b.c(a+a')     {Complementary Law}
a.b+a'.c+(a.b.c+a'.b.c)
(a.b+a.b.c)+(a'.c+a'.c.b)
a.b(1+c)+a'.c(1+b)       {As 1+c=1 and 1+b=1}
a.b+a'.c