# Problem simplifying equation using boolean algebra

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.

• Your simplification is incorrect, choose $ABCD=1101$ as a witness. – copper.hat Oct 25 '15 at 23:09
• @copper.hat Could you please explain what choose ABCD = 1101 as a witness means? Thank you. – brian Oct 26 '15 at 6:08
• 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 :-). – copper.hat Oct 26 '15 at 6:10
• 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) – brian Oct 26 '15 at 6:28
• wolframalpha.com/input/… – brian Oct 26 '15 at 6:28

\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}

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)