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UGC NET CSE | June 2009 | Part 2 | Question: 03

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Givne boolean expression is $(\bar x\bar yz+yz+xz)$

taking $z$ as common

$= z(\bar x\bar y+y+x)$

applying distributive law $(a+bc=(a+b)(a+c))$

$= z((y+\bar y)(\bar x+y)+x)$

$\because (y+\bar y=1)$

$=(x+\bar x)+y$

$= z(1+y)$

$\because (1+x=1)$

$=z.1=z$

So option $C$ is correct.
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