Simplification of the boolean expression with XOR

boolean-algebralogic

I need to simplify the following boolean expression

¬A¬B¬C + (B ⊕ C) + A¬B

I know B⊕C = ¬BC + B¬C

Then the expression will become

¬A¬B¬C +(¬BC + B¬C) +A¬B

However, I'm stuck on it and I don't know how to simplify it further. Can someone give me a hint or push me in the right direction? Thanks in advance

Best Answer

A couple of useful principles are:

Adjacency

$AB+AB'=A$ (I find the ' a little easier to work with than $\neg$)

Absorption

$A + AB = A$

Idempotence

$A + A = A$

So, starting with:

$A'B'C'+B'C+BC'+AB'$

(use Adjacency to rewrite $AB'$ as $AB'C+AB'C'$)

$= A'B'C'+B'C+BC'+AB'C+AB'C'$

($B'C$ absorbs $AB'C$)

$= A'B'C'+B'C+BC'+AB'C'$

(use Adjacency to rewrite $A'B'C'+AB'C'$ as $B'C'$)

$= B'C'+B'C+BC'$

(use Idempotence to make a copy of $B'C'$)

$= B'C'+B'C+B'C'+BC'$

(Use Adjancency to rewrite $B'C'+B'C$ as $B'$ and $B'C'+BC'$ as $C'$)

$= B'+C'$