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how can I simplify this sop expression using Boolean algebra ? :

$A'BC'D'+A'BC'D+A'BCD'+A'BCD+AB'C'D+AB'CD'+ABC'D'+ABC'D$

I have to use Boolean algebra rules only (no K map); the answer should be this:

$A'B+BC'+AC'D+AB'CD'.$

Thanks!

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HINT

This equivalence principle will be your friend:

Adjacency

$PQ + PQ' = P$

If you're not allowed to use Adjacency in $1$ step, here is a derivation of Adjacency in terms of more basic equivalence principles:

$$P Q + (P Q') \overset{Distribution}=$$

$$P (Q + Q') \overset{Complement}=$$

$$P 1 \overset{Identity}=$$

$$P$$

To apply Adjacency, note that $P$ and $Q$ can be any complex expressions, so in this case, where every terms has $4$ variables, just look for two terms that are the same for $3$ of the variables, but differ in the fourth. For example, the first two terms are the same except for the $D$ variable, so those can be combined:

$A'BC'D'+A'BC'D=A'BC'$

You can also combine the first and seventh terms:

$A'BC'D'+ABC'D'=BC'D'$

To do both of those, you would need to 'reuse' the first term, but you can get as many copies as you want by:

Idempotence

$P + P = P$

So, for example, focusing on the first, second, and seventh term:

$$A'BC'D'+A'BC'D+ABC'D'\overset{Idempotence}=$$

$$A'BC'D'+A'BC'D'+A'BC'D+ABC'D'\overset{Commutation}=$$

$$A'BC'D'+A'BC'D+A'BC'D'+ABC'D'\overset{Adjacency \ x \ 2}=$$

$$A'BC'+BC'D'$$

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  • $\begingroup$ Thank you, I have tried to apply the rules, but I'm still not getting the correct answer, I would really appreciate it if you could post the solution steps $\endgroup$ – Abdalrahman Abouzaied May 19 '18 at 4:35
  • $\begingroup$ @AbdalrahmanAbouzaied I am wondering if maybe you were struggling with the fact that you have to use the same term multiple times, so I added some more hints to my Answer. I really would like you to be able to find this by yourself though, so I won't give the whole derivation. $\endgroup$ – Bram28 May 19 '18 at 13:22

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