1

Need to solve by Boolean algebra. I tried but didn't solve it and the answer on K-map is $\;A'C'D+B'CD+ACD'+A'BC'$.

The question is $\;ABCD'+AB'CD'+A'BC'D+AB'D'+A'B'CD+A'C'D+AB'CD$.

Angelo
  • 12,328
Omar Mamdouh Zake
  • 13

1 Answers1

1

HINT: Use the Adjacency principle: $PQ + PQ'=P$

For example: $ABCD'+AB'CD'=ACD'$

If for some reason you are not given Adjacency as a basic rule (and many textbooks don't which I never understand, since it's such a useful principle!), you can always derive it as follows from the rules that I am sure are given to you:

$PQ+PQ' \overset{\text{Distribution}}{=} P(Q + Q') \overset{\text{Complement}}{=} P\!\cdot\!1\overset{\text{Identity}}{=} P$

By the way, when you look at a K-map, you'll understand why this principle is called Adjacency: this is exactly the algebraic equivalent of the process of grouping together adjacent cells in a K-map

Angelo
  • 12,328
Bram28
  • 100,612
  • 6
  • 70
  • 118
  • I know this rule I tried with it but only works with ACD' Other didn't work as k map – Omar Mamdouh Zake Jun 05 '21 at 14:08
  • If I used it as u Wright so I can't repeat the equation with other one u used ABCD'+AB'CD' CAn I use one of them agine with other equation like ABCD'+AB'D' – Omar Mamdouh Zake Jun 05 '21 at 14:12
  • Adjacency goes both ways. So, for example, you can always take something like $ABC$ and reqrite it as $ABCD + ABCD'$. So, for example, the $AB'D'$ that occurs in the prompt does not occur in the answer, so you'll need to break that one into two. If you relate each term to the cell or grouping in the K-map for both the prompt and the answer, then you'll need to see how to break things down and then group together again. Also please note that by Idempotene you have $P + P = P$, so you can always add or remove duplicate terms as needed – Bram28 Jun 05 '21 at 14:25
  • Thx for help I solve it ☺️☺️ – Omar Mamdouh Zake Jun 05 '21 at 16:29
  • Do u know app or something transform circuit to NaND only or NOR only – Omar Mamdouh Zake Jun 05 '21 at 16:30