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how to solve this using rules of boolean algebra: 

2 Answers

Best answer
4 votes
4 votes

Given:

((ab'+ab)'c + a(b+ab'))'

((ab'+ab)'c + ab+aab')'

((ab'+ab)'c + ab+ab')'.......(a.a = a)

((ab'+ab)'c + a)'...............a(b+b')=a since b+b'=1

((a'+b)(a'+b')c + a)'...........DeMorgan's Law

(a'c+a'b'c+ba'c+bb'c+a)'

(a'c +a'c+a)'.....................a'cb+a'b'c=a'c

(a'c+a)'..................................a+a'b=a+b

(a+c)'

a'c'.................................Answer

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((ab'+abc)'+a(b+ab'))'

=((ab'+abc)'+(ab+ab'))'

=(ab'+abc).(ab+ab')'  [using De Morgan's law]

=(ab'+abc).((a'+b').(a'+b)) [using De Morgan's law]

=(ab'+abc)(a'+a'b+a'b')

=0

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