Correct Answer - Option 4 : 0
Concept:
De Morgan’s law states that:
\(\overline {\left( {{A_1}.{A_2} \ldots {A_n}} \right)} = \left( {\overline {{A_1}} + \overline {{A_2}} + \ldots + \overline {{A_n}} } \right)\)
Also,
\(\overline {\left( {{A_1} + {A_2}+ \ldots + {A_n}} \right)} = \left( {\overline {{A_1}} .\overline {{A_2}} ...\overline {{A_n}} } \right)\)
Analysis:
Given
\({\rm{F}} = \overline {\left( {{\rm{a}} + {\rm{̅ b}} + {\rm{c}} + {\rm{̅ d}}} \right) + \left( {{\rm{b}} + {\rm{̅ c}}} \right)}\)
Applying the De-Morgans theorem in the above function F
\(\begin{array}{l} = \overline {\left( {{\rm{a}} + {\rm{̅ b}} + {\rm{c}} + {\rm{̅ d}}} \right)} \cdot \overline {\left( {{\rm{b}} + {\rm{̅ c}}} \right)} \\ \end{array}\)
= a̅.b.c̅.d.b̅. c
As, b.b̅ = c.c̅ = 0
∴ F = a̅.b.c̅.d.b̅. c = 0
Hence option (4) is correct
