Simplifying Boolean Algebra issue

Question

I'm having trouble simplifying a Boolean equation using Boolean algebra, I have the answer and have attempted the question but I keep getting stuck. I would appreciate it if someone can let me know what I am doing wrong.

The original equation is: $$Y = ABCD + \bar{A}B + AC + \bar{B}D + \bar{A}\bar{C}D$$

My work is as follows: \begin{align*} Y &= ABCD + \bar{A}B + AC + \bar{B}D + \bar{A}\bar{C}D\\ Y &= AC(BD + 1)+\bar{A}B + \bar{B}D + \bar{A}\bar{C}D\\ Y &= AC+\bar{A}B + \bar{B}D + \bar{A}\bar{C}D\\ \end{align*}

However this is as far as I can simplify it, which cant be correct.

I've tried multiplying by $\bar{A}+A$, also done the same for $B$,$C$ and $D$, and I've also tried multiple combinations, But still no luck.

I know the answer is: $Y = AC + \bar{A}B + \bar{B}D$

But I cannot figure out how to get rid of the $\bar{A}\bar{C}D$ term, does anyone have any idea where I went wrong?

Answer 1

Well, !A.!C.D = !A.B.!C.D + !A.!B.!C.D

Then, !A.B + !A.B.!C.D = !A.B

Also, !B.D + !A.!B.!C.D = !B.D

And that's how you can get rid of !A.!C.D.

Answer 2

For future work you might also look up Veitch diagrams or Karnaugh maps as simple tools to help reduce the number of truth statements.

http://mathworld.wolfram.com/VeitchDiagram.html

http://www.mathematik.uni-marburg.de/~thormae/lectures/ti1/code/karnaughmap/