The Boolean expression \(\left( {x + y} \right)\left( {x + \bar y
| The Boolean expression \(\left( {x + y} \right)\left( {x + \bar y} \right) + \overline {\left( {x\bar y} \right) + \bar x} \) simplifies to
A. x
B. y
C. xy
D. x + y
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
De Morgan’s law states that:
\(\overline {\left( {{A_1}\;.\;{A_2} \ldots \;{A_n}} \right)} \; = \left( {\overline {{A_1}} \; + \;\overline {{A_2}} \; + \; \ldots \; + \;\overline {{A_n}} } \right)\;\)
\(\overline {\left( {{A_1}\; + \;{A_2}\; + \; \ldots \; + \;{A_n}} \right)} \; = \left( {\overline {{A_1}} \;.\;\overline {{A_2}} \;.\;..\;\overline {{A_n}} } \right)\)
Analysis:
\(\left( {x + y} \right)\left( {x + \bar y} \right) + \overline {\left( {x\bar y} \right) + \bar x} \)
\(= xx + x\bar y + xy + y\bar y + \left( {\overline {x\bar y} \;.\overline{\overline x} } \right)\)
= x + xy̅ + xy + (x̅ + y) (x)
= x + xy
= x