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.
$\endgroup$ 27 Answers
$\begingroup$\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*}
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 4 $\begingroup$\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}
$\endgroup$ $\begingroup$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..$\begin{align*}&= AB+A′C+BC\\ &= AB+A′C+BC(A+A′)\\ &= AB+A′C+ABC+A′BC\\ &= AB+ABC+A′C+A′BC\\ &= AB(1+C)+A′C(1+B)\\ &= AB+A′C\end{align*}$
$\endgroup$ $\begingroup$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
=AB+A'C+BC =AB.A'C+BC(A+A') =AB+A'C+ABC+A'BC =AB+ABC+A'C+ABC =AB+A'C
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