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The expression is:

[AB {C+(BD)'} + (AB)']CD

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1 Answer 1

Using De Morgan's laws,

$[AB \{C+(BD)'\} + (AB)']CD=[AB \{C+B'+D'\} + A'+B']CD=[ABC+ABD' + A'+B']CD$

Now, $AE+A'=AE+A'(1+E)=A'+AE+A'E=A'+E(A+A')=A'+E$ $\implies ABC+A'=BC+A'$

$[ABC+ABD'+ A'+B']CD$

$=[A'+(BC+B')+ABD']CD$

$=[A'+C+B'+ABD']CD $ (Using $AE+A'=A'+E$)

$=A'CD+CD+B'CD=CD(A'+1+B')=CD$

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Can you please define the 3rd last step of your answer...I could not really understand it. –  Ali Feb 8 '13 at 15:28
    
@Ali. observe that $A'=A'\cdot U=A'(U+E)=A'U+A'E=A'+A'E$ where $U$ is the universal set. –  lab bhattacharjee Feb 9 '13 at 15:14
    
Actually i am asking about the term (BC+B') how it comes....i still didn't understant.. –  Ali Feb 9 '13 at 15:33
    
@Ali, $ABC+A'=ABC+A'(U+BC)=ABC+A'BC+A'=(A+A')BC+A'=BC+A',B'$ was already there which is paired with $BC$ to reach $C+B'$ using the same formula. –  lab bhattacharjee Feb 9 '13 at 15:37

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