(i) Using de Morgan's theorem
[(A′+B)C′]=(A′+B)′+C′′
Applying involution law
=(A′+B)′+C
Applying de Morgan's theorem
=A′′.B′+C
Applying involution law
=A.B′+C
(ii)
Applying de Morgan's theorem
=[(AB′+C)D′]′.E′′
Applying involution law
=[a(AB′+C)D′]′.E
Applying de Morgan's law
=[(AB′+C)′+D′′].E
Applying double negation
[(AB′+C)′+D].E
Applying de Morgan's theorem
=[(AB′)′.C′+D].E
=[(A′+B′′).C′+D].E
=[(A′+B).C′+D].E
=[A′C′+BC′+D].E
=E(D+C′).(B+D+A′)
Applying involution law
=[A+((BC′)′+D)]′′
Using involution law
=A+((BC′)′+D)
Applying de Morgan's theorem
=A+(B′+C′′)+D
Applying involution law
=A+B′+C+D
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