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Math Help - Trouble with de morgans law and boolean algebra

  1. #1
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    Trouble with de morgans law and boolean algebra

    So Ive attached an example of a couple of questions from a past exam paper. But, I have no idea how the answers can be achieved. If anyone can take a look, and explain step by step how to achieve each answer, that would be great, as I've tried and still dont get it :-|

    Thanks in advance

    Trouble with de morgans law and boolean algebra-de-morgans.png
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  2. #2
    Senior Member
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    Re: Trouble with de morgans law and boolean algebra

    use the following
    for first part B.(A+COMP(A))
    U KNOW A+COMP(A)=1
    AND B.1=B
    FOR SECOND PART TAKE OUT B AS COMMON
    U GET B(1+A)
    1+A=1 AND B.1=B
    FOR THIRD PART COMP(X+Y)=COMP(X)COMP(Y)
    HERE X=B AND Y=COMP(A)+COMP(B)
    SO COMP(X)COMP(Y) IS GIVEN FIND COMP(X+Y)
    COMP(X+Y)=COMP(B+COMP(A)+COMP(B))
    NOW B+COMP(B)=1 AND 1+COMP(A)=1
    ALSO COMP(1)=0
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