TITLE: Boolean algebra question.
QUESTION [5 upvotes]: Is there a way to show that $$A\bar{B}C\bar{D}+D=A\bar{B}C+D$$ using the rules of boolean algebra? I  tried several methods such as expanding D with $$D(D+\bar{D})$$ or adding $$D\bar{D}$$ to the equation but nothing worked. From the Karnaugh map it is evident however that the $$\bar{D}$$ term doesn't matter. I just can't come up with a boolean algebra solution for the proof.

REPLY [3 votes]: How about:
$\bar{A}B + A = \overline{\overline{\bar{A}B + A}} = \overline{(A + \bar{B})\bar{A}} = \overline{\bar{A}\bar{B}} = A + B$
There's probably a less messy way of doing this.