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.
 A: \begin{align*}
&\mathrel{\phantom{=}}AB+A'C+BC\\
&=AB+A'C+BC(A+A')          \quad \text{($A+A'=1$, Complementarity law)}\\
&=AB+A'C+ABC+A'BC\\
&=AB+ABC+A'C+ABC           \quad \text{(Associative law)}\\
&=AB+A'C                 \quad   \text{(Absorption law)}
\end{align*}
A: The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.
A: \begin{align}
F & = AB + A'C + BC \\
 & = AB + A'C + BC(A+A') \\ 
 & = AB + A'C + ABC + A'BC \\
 & = AB + ABC + A'C + A'BC \\ 
 & = AB (1 + C) + A'C (1 + B) \\
 & = AB + A'C
\end{align}
A: In this way , this can be simplified
                 LHS = AB+A'C+BC   
                 =  AB+A'C+BC (A+A')                  [ A+A'=1 ]
                  =  AB+A'C+ABC+A'BC
                  = AB+ABC+A'C+A'BC
                  =  AB (1+C)+A'C (1+B)
                  = AB+A'C                           [  1+C=1 ]
                                        =RHS..

A: $\begin{align*}&= AB+A′C+BC\\
&= AB+A′C+BC(A+A′)\\
&= AB+A′C+ABC+A′BC\\
&= AB+ABC+A′C+A′BC\\
&= AB(1+C)+A′C(1+B)\\
&= AB+A′C\end{align*}$
A: 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  
A: =AB+A'C+BC
=AB.A'C+BC(A+A')
=AB+A'C+ABC+A'BC
=AB+ABC+A'C+ABC
=AB+A'C
