# Simplification of the boolean expression with XOR

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

A couple of useful principles are:

$$AB+AB'=A$$ (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 $$C'$$)

$$= B'+C'$$

• Thanks a lot!This was very helpful! – Jarvis Ferns Apr 2 at 16:21

Try using associativity to group terms with like elements. This should cause some tautology and let you remove unnecessary expressions.