Simplify the following expression using Boolean algebra ABCD

Simplify the following expression using Boolean algebra ABCD + AB(CD)bar + A bar ABCD bar + ABC BAR D + AB bar CD + A bar BCD + (AB) bar CD A(BC) bar + (AB) bar C + ABC bar ABC + ABC bar D + AD bar + AB bar C bar D + A bar B ABC bar D bar + A bar B C bar D + ABCD + C bar D bar (AB) bar(CD) bar + AB(CD) bar + ABD + ABC + A bar

Solution

a) ABCD + AB(CD)\' + A\' = AB(CD+(CD)\') + A\' = AB(1) + A\' = AB + A\' = A\' + B

b) ABCD\' + ABC\'D + AB\'CD + A\'BCD + (AB)\'CD = ABCD\' + ABC\'D + AB\'CD + A\'BCD + (A\' + B\')CD

= ABCD\' + ABC\'D + AB\'CD + A\'BCD + A\'CD + B\'CD = ABCD\' + ABC\'D + B\'CD(A+1) + A\'CD(B+1)

= ABCD\' + ABC\'D + CD(A\'+B\') = ABCD\' + ABC\'D +(AB)\'CD = AB(CD\'+C\'D)+(AB)\'CD

c) A(BC)\' + (AB)\'C + ABC\' = A(B\'+C\')+(A\'+B\')C+ABC\' = AB\'+AC\'+A\'C+B\'C+ABC\' = AB\'+AC\'(1+B)+A\'C+B\'C

=AB\'+AC\'+A\'C+B\'C = A(BC)\'+(AB)\'C

d) ABC + ABC\'D + AD\' + AB\'C\'D + A\'B = B(AC+A\')+ ABC\'D + AD\' + AB\'C\'D = B(A\'+C)+ ABC\'D + AD\' + AB\'C\'D

=A\'B+BC+ ABC\'D + AD\' + AB\'C\'D = A\'B+BC+AC\'D(B+B\')+AD\' = A\'B+BC+AC\'D+AD\'

e) ABC\'D\' + A\'BC\'D + ABCD + C\'D\' = C\'D\'(AB+1)+BD(A\'C\'+AC) = C\'D\' + BD(A\'C\'+AC)

f) (AB)\'(CD)\'+AB(CD)\'+ABD+ABC+A\' = ((AB)\'+AB)(CD)\'+ABD+ABC+A\' = (1)(CD)\'+ABD+ABC+A\'

= C\'+D\'+ABD+ABC+A\' = C\'+D\'+AB(C+D)+A\' = (CD)\' + A\' + B(C+D)