Help with a boolean algebra simplification.

Question

I know this is annoying and probably easy for some poeple, but this is giving me a hard time, so i thought Id ask here.

~ is NOT; + is OR;

(~A+~B+~C)(~A+B+C)(A+~B+~C)(A+~B+C)(A+B+~C)(A+B+C)

I used a Karnaugh-Veitch Map to get a solution to check if im right later, and got to A(~B+~C)(B+C).

How can i solve this using only boolean algebra?

Answer 1

(Please allow me to use $P'$ instead of $\sim P$)

Use:

Adjacency

$PQ + PQ'=P$

$(P+Q)(P+Q')=P$

Idempotence

$P+P=P$

$PP=P$

With those:

$$(A'+B'+C')(A'+B+C)(A+B'+C')(A+B'+C)(A+B+C')(A+B+C)$$

$$\overset{\text{Idempotence x 2}}{=}$$

$$(A'+B'+C')(A'+B+C)(A+B'+C')(A+B'+C')(A+B'+C)(A+B+C')(A+B+C)(A+B+C)$$

$$\overset{\text{Commutation}}{=}$$

$$(A+B+C)(A+B+C')(A+B'+C)(A+B'+C')(A+B'+C')(A'+B'+C')(A+B+C)(A'+B+C)$$

$$\overset{\text{Adjacency x 4}}{=}$$

$$(A+B)(A+B')(B'+C')(B+C)$$

$$\overset{\text{Adjacency}}{=}$$

$$A(B'+C')(B+C)$$