Boolean algebra equation: ABC OR A'B'C'

Question

I have also tried with K-map but i get a different result:

      CD
     |00 | 01 | 11 | 10|
     +------------------
AB 00| 1 |  0 |  0 |  0|
   --+---+----+----+----
   01| 0 |  0 |  0 |  0|
   --+---+----+----+----
   11| 1 |  1 |  1 |  1|
   --+---+----+----+----
   10| 1 |  1 |  0 |  1|
   ---------------------

The equation that i have to solve is this:

Y = ABCD'+ A(BCD)' + (A+B+C+D)'

The steps that i follow are:

1. modify middle term as: A(B'+C'+D')
2. modify last term as: A'B'C'D'

*Rewrite equation after replacements:

Y = ABCD' + A'B'C'D'+ AB' + AC' + AD' 
Y = D'(ABC + A'B'C') + AB' + AC' + AD'

The answer should be:

Y = B'C'D' + A(B'+C'+D')

I think the answer is 100% right (i have it on the back of my book and also tested on BoolCalculator) If you want to test it on that site, copy and paste this equation there:

(A * B * C * ~D) + (A * (~B + ~C + ~D)) + (~A * ~B * ~C * ~D)

I am stuck after rewriting the equation. I'm certainly missing something.

ABC + A'B'C' can be optimized more?

Bram28 · Accepted Answer · 2019-10-21T19:18:41.243

2

Let's start where you have:

$ABCD' + A'B'C'D'+ AB' + AC' + AD' $

By

Absorption

$P + PQ=P$

we can get rid of the $ABCD'$ term, as it gets absorbed by $AD'$

So, that leaves:

$A'B'C'D'+ AB' + AC' + AD' $

OK, by multiple instances of

Adjacency

$PQ+PQ'=P$

we get that:

$AD'=ACD'+AC'D'=ABCD'+AB'CD'+ABC'D'+AB'C'D'$

And so we get:

$A'B'C'D'+ AB' + AC' + ABCD'+AB'CD'+ABC'D'+AB'C'D'$

By

Idempotence

$P+P=P$

we can duplicate $AB'C'D'$:

$A'B'C'D'+ AB' + AC' + ABCD'+AB'CD'+ABC'D'+AB'C'D'+AB'C'D'$

(and note: now we have that $ABCD'$ back ... but this is exactly why the $ABCD'$ term is absorbed by $AD'$)

But now we use Adjacency again:

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

to get:

$B'C'D'+ AB' + AC' + ABCD'+AB'CD'+ABC'D'+AB'C'D'+AB'C'D'$

and now we use $AD'=ABCD'+AB'CD'+ABC'D'+AB'C'D'$ again to get:

$B'C'D'+ AB' + AC' + AD'$

which is the same as:

$B'C'D'+ A(B' +C' + D')$

edited Oct 21 '19 at 19:18

answered Oct 21 '19 at 18:25

Bram28

100,612
6
70
118

its a three-way XNOR, which might be what they're looking for – Legorhin Oct 21 '19 at 18:28
@Legorhin No. The XNOR is true iff an even number of the variables are true. So, XNOR(ABC)=F – Bram28 Oct 21 '19 at 18:32
ok I thought XNOR was always an equality check – Legorhin Oct 21 '19 at 18:34
Ok, so i rewrite the question as it should have been from the beginning. This is the context why i was asking about simplifying that equation! Thanks for the fast answers – Dragos Makovei Oct 21 '19 at 18:38
@DragosMakovei Just rewrote my Answer – Bram28 Oct 21 '19 at 19:05
Thank you very much. Could you still get to this final form even if you didn't know the answer? I mean, have u look from the end to the beginning? – Dragos Makovei Oct 21 '19 at 19:31
@DragosMakovei Well, study my Answer and try and understand what I did ... if you have questions let me know. – Bram28 Oct 21 '19 at 19:32
No, the answer is very clear and brief. I still have problems while taking the general rules and applying them for 3-4 members terms. – Dragos Makovei Oct 21 '19 at 19:35
@Bram28 i would appreciate karnaugh-map for this equation. The weird thing is that i still get the wrong answer with this map. I will load it in my question. – Dragos Makovei Oct 21 '19 at 19:43
@DragosMakovei The bottom half of the K-map you posted corresponds to $A(B'+C'+D')$, because that half correspond to $A$, and the only $0$ in that half is when you have $BCD$, meaning that all the true ones correspond to $A(ABC)'$. – Bram28 Oct 21 '19 at 20:13
@DragosMakovei Also, the $1$ in the top left corner forms a group of two with the $1$ in the lower left corner. This corresponds to $B'C'D'$ – Bram28 Oct 21 '19 at 20:13

score 0 · Answer 2 · answered Oct 22 '19 at 07:53

0

And the way you can optimize the boolean equation using boolean algebra rules is explained very well by Bram28

answered Oct 22 '19 at 07:53

Dragos Makovei

25

Boolean algebra equation: ABC OR A'B'C'

2 Answers2