Answer to Question #209047 in Discrete Mathematics for Ahmed Ali

Question

Find the simplest form for the following boolean expressions:

1) (A.B'.C')+(A'.B'.C')+(A'.B.C')+(A'.B'.C)

2) (A'.B.C)+(A'.B.C)+(A.B.C')+(A.B'.C')+(A'.B.C')+(A'.B'.C')

3) (A+B+C)(A+B'+C')(A+B+C')(A+B+C')

( ' = Not )

Answer 1

1) $A \cdot \overline B \cdot \overline C + \overline A \cdot \overline B \cdot \overline C + \overline A \cdot B \cdot \overline C + \overline A \cdot \overline B \cdot C = \overline A \cdot \overline B \left( {\overline C + C} \right) + A \cdot \overline B \cdot \overline C + \overline A \cdot B \cdot \overline C = \overline A \cdot \overline {B \cdot } 1 + A \cdot \overline B \cdot \overline C + \overline A \cdot B \cdot \overline C = \overline A \cdot \overline B + A \cdot \overline B \cdot \overline C + \overline A \cdot B \cdot \overline C = \overline A \left( {\overline B + B \cdot \overline C } \right) + A \cdot \overline B \cdot \overline C = \overline A \left( {\left( {\overline B + B} \right)\left( {\overline B + \overline C } \right)} \right) + A \cdot \overline B \cdot \overline C = \overline A \left( {\overline B + \overline C } \right) + A \cdot \overline B \cdot \overline C = \overline A \cdot \overline B + \overline A \cdot \overline C + A \cdot \overline B \cdot \overline C = \overline B \left( {\overline A + A \cdot \overline C } \right) + \overline A \cdot \overline C = \overline B \left( {\overline A + A} \right)\left( {\overline A + \overline C } \right) + \overline A \cdot \overline C = \overline B \left( {\overline A + \overline C } \right) + \overline A \cdot \overline C = \overline A \cdot \overline B + \overline B \cdot \overline C + \overline A \cdot \overline C$

Answer: $\overline A \cdot \overline B + \overline B \cdot \overline C + \overline A \cdot \overline C$

2) $\overline A \cdot B \cdot C + \overline A \cdot B \cdot C + AB\overline C + A \cdot \overline B \cdot \overline C + \overline A B\overline C + \overline A \cdot \overline B \cdot \overline C = \overline A \cdot B \cdot C + AB\overline C + A \cdot \overline B \cdot \overline C + \overline A B\overline C + \overline A \cdot \overline B \cdot \overline C = \overline B \cdot \overline C \left( {A + \overline A } \right) + \overline A \cdot B \cdot C + B\overline C \left( {A + \overline A } \right) = \overline B \cdot \overline C + \overline A \cdot B \cdot C + B\overline C = \overline C \left( {\overline B + B} \right) + \overline A \cdot B \cdot C = \overline C + \overline A \cdot B \cdot C = \left( {\overline C + \overline A \cdot B} \right)\left( {\overline C + C} \right) = \overline C + \overline A \cdot B$

Answer: $\overline C + \overline A \cdot B$

3)

$\left( {A + B + C} \right)\left( {A + \overline B + \overline C } \right)\left( {A + B + \overline C } \right)\left( {A + B + \overline C } \right) = \left( {A + B + C} \right)\left( {A + \overline B + \overline C } \right)\left( {A + B + \overline C } \right) = \left( {A + B + C} \right)\left( {A + \overline C } \right)\overline B B = \left( {A + B + C} \right)\left( {A + \overline C } \right) \cdot 0 = 0$

Answer: 0