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?

Answer 1

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:

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

Answer 2

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