[PSD6] Exercise 2.13 #147

matypist · 2022-12-29T22:26:08Z

matypist
Dec 29, 2022
Maintainer

Elia-Belli · 2022-12-30T16:50:43Z

(a) $AC + \bar A \bar B C$
$C(A+ \bar A \bar B) = C((A+\bar A)(A+ \bar B)) = C(A+\bar B) = AC + C\bar B$
(b) $\bar A \bar B+ \bar A B \bar C + \bar A C$
$\bar A (\bar B + B \bar C + C) = \bar A ((B+\bar B)(\bar B + \bar C)+C) = \bar A (\bar B + \bar C+C) = \bar A$
(c) $\bar A \bar B \bar C \bar D + A\bar B \bar C + A \bar B C \bar D + ABD + \bar A \bar B C \bar D + B \bar C D + \bar A$
$\bar A + (ABD + AB\bar C D + \bar A B \bar C D) + A\bar B \bar C + A \bar B C \bar D=$
$\bar A + ABD + A\bar B (\bar C + C \bar D) =$
$\bar A + ABD + A\bar B ((C + \bar C)(\bar C + \bar D)) = \bar A + ABD + A\bar B \bar C + A \bar B \bar D =$
$\bar A + BD + \bar B \bar C + \bar B \bar D$

0 replies

This comment has been hidden.

Sign in to view