Want to check if my Boolean Algebra simplification is correct

Question

$(A+B)(B+\bar B)(\bar B+C)$

Distributive LAW

$(AB+A \bar B+B B+B \bar B)(\bar B+C)$

Distributive LAW

$(A B \bar B+A B C+A \bar B \bar B+A \bar B C+B B \bar B+B B C+B \bar B \bar B+B \bar B C)$

Idempotent Law {} + Complementary Law []

$(A [B \bar B]+A B C+A \{\bar B \bar B\}+A \bar B C+B [B \bar B]+C \{B B\}+\bar B[\bar B B]+C [\bar B B]$

$=A [0]+A B C+A \{\bar B\}+A \bar B C+B [0]+C \{B\}+\bar B [0]+C [0]$

Law of Intersection []

$A B C+A \bar B+A \bar B C+B C$

Law of Absorption{}

$\{A B C+B C\}+\{A \bar B C+A \bar B\}$

$=B C+A \bar B$

Answer 1

Yes.   Your steps are okay.

I would recommend using Complement as the first step.   It is not really faster, but perhaps cleaner.   But that's just my preference.

$$\begin{align}(A+B)(B+\bar B)(\bar B+C)~ = ~& (A+B)(1)(\bar B+C) & \textsf{Complement} \\ = ~& (A+B)(\bar B+C) & \textsf{Conjunctive Identity (Elim)} \\ = ~& A\bar B + A C + B\bar B+BC & \textsf{Distribution} \\ = ~& A\bar B + A C + 0+BC & \textsf{Contradiction} \\ = ~& A\bar B + A C + BC & \textsf{Disjunctive Identity (Elim)} \\ = ~& A\bar B + A(1) C + BC & \textsf{Conjunctive Identity (Intro)} \\ = ~& A\bar B + A(\bar B+B) C + BC & \textsf{Complement} \\ = ~& A\bar B+A\bar B C+ABC+BC & \textsf{Distribution} \\ = ~& A\bar B + BC & \textsf{Absorption} \end{align}$$

$\Box$