0
$\begingroup$

I have this boolean equation:

A'.B'.C'.D' + A'.B.C'.D' + B'.C'.D + B.C'.D

Using a Karnaugh map I find I can simplify the above to:

C'.(A' + D)

I have used Wolfram Alfa website to confirm this simplification is correct.

I would like to try simplifying the original statement using Boolean algebra but get stuck along the way:

A'.B'.C'.D' + A'.B.C'.D' + B'.C'.D + B.C'.D

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

C'.D'.(A' + 1)  + C'.D.1

I would be grateful if someone could point out where I have gone wrong. Thank you.

$\endgroup$
  • $\begingroup$ Your simplification is incorrect, choose $ABCD=1101$ as a witness. $\endgroup$ – copper.hat Oct 25 '15 at 23:09
  • $\begingroup$ @copper.hat Could you please explain what choose ABCD = 1101 as a witness means? Thank you. $\endgroup$ – brian Oct 26 '15 at 6:08
  • $\begingroup$ Set $A,B,D$ to true and $C$ to false and evaluate the equation and the 'simplification'. One expression will evaluate to true and one to false, thereby 'witnessing' that they are not the same :-). $\endgroup$ – copper.hat Oct 26 '15 at 6:10
  • $\begingroup$ Sorry, just realised I had mistyped the algebraic statements in what I had posted here. However, my problem still stands in that I am making a mistake in my simplification as I am unable to get to C' . (A' + D) $\endgroup$ – brian Oct 26 '15 at 6:28
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – brian Oct 26 '15 at 6:28
0
$\begingroup$

\begin{eqnarray} \bar{A} \bar{B} \bar{C} \bar{D} + \bar{A} {B} \bar{C} \bar{D} +\bar{B} \bar{C} {D} + {B} \bar{C} {D} &=& B (\bar{A} \bar{C} \bar{D} + \bar{C}) + \bar{B} (\bar{A} \bar{C} \bar{D} + \bar{C} {D}) \\ &=& \bar{A} \bar{C} \bar{D} + \bar{C} {D} \\ &=& \bar{C} (\bar{A} \bar{D} + {D} ) \\ &=& \bar{C} (\bar{A} \bar{D} + \bar{A} D + {D} ) \\ &=& \bar{C} (\bar{A} + {D} ) \\ \end{eqnarray}

$\endgroup$
-1
$\begingroup$

A'.B'.C'.D' + A'.B.C'.D' + B'.C'.D + B.C'.D

=c'.(A'.B'.D' + A'.B.D' + B'.D + B.D)

=c'.(A'.D'(B' + B) + D.(B' + B))

=c'.(A'.D'(1) + D.(1)) can write as:

=c'.(A'.D' + D.(1 + A'))

=c'.(A'.D' + D + D.A')

=c'.(A'.D' + A'.D + D)

=c'.(A'.(D' + D) + D)

=c'.(A'.(1) + D)

=c'.(A' + D)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.