Boolean algebra question.

Question

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.

Answer 1

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.

Answer 2

$X\overline{Y}+Y=X\overline{Y}+Y(X+\overline{X})=X(Y+\overline{Y})+Y\overline{X}=X(Y+\overline{Y})+Y\overline{X}+XY=X+Y$

Note: $A+A+...+A\equiv A;\qquad X+\overline{X}\equiv I$