Boolean reduction of $AB+(B+C)BC$

Question

Reduce this Boolean expression $$AB+(B+C)BC$$ Please express step by step process to understand clearly.

Nah.. that's what those question marks indicate. I m totally clueless about what your problem is.. — Vidyanshu Mishra, Apr 19 '17 at 14:00
The expression appears to be fully reduced. What are you looking for? — lulu, Apr 19 '17 at 14:01
I mean, you can rewrite it as $AB+B^2C+BC^2$ or $B(A + BC + C^2)$ but I don't see why either or those would be described as more "reduced". — lulu, Apr 19 '17 at 14:02
@Thelonewolfisbackbaby... note the boolean-algebra tag. Dhani, it would be better to point out in the question that your variables come from a Boolean algebra, not, say, the real numbers! — Alex Kruckman, Apr 19 '17 at 14:03
@Dhani for you, does + denote "or" or "xor"? That is, do we have $x+x=x$ or $x+x=0$? — Alex Kruckman, Apr 19 '17 at 14:06

Bram28 · Answer 1 · 2017-04-19T14:13:43.267

4

Use:

Absorption

$P(P+Q) = P$

With that:

$$AB + (B + C)BC = AB + BC$$ ($B$ absorbs $B+C$)

And if you want, you can use Distribution to make that $B(A + C)$

edited Apr 19 '17 at 14:13

answered Apr 19 '17 at 14:08

Bram28

score 3 · Answer 2 · answered Apr 19 '17 at 14:04

3

\begin{align*} AB+(B+C)BC &= AB + B^2C + BC^2\\ & = AB + BC + BC\\ & = AB + BC \end{align*}

answered Apr 19 '17 at 14:04

Alex Kruckman

1

Sigh... did the downvoters miss the Boolean algebra tag too? – Alex Kruckman Apr 19 '17 at 14:10
Weird indeed ... – Bram28 Apr 19 '17 at 14:16

2 Answers2