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 $\neg$)


$A + AB = A$


$A + A = A$

So, starting with:


(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 $C'$)

$= B'+C'$

  • $\begingroup$ Thanks a lot!This was very helpful! $\endgroup$ – Jarvis Ferns Apr 2 '19 at 16:21

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


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