(Solved): Convert this Truth table to its Logic Circuit. Correct diagram of the circuit will get thumbs up. Th ...

\begin{tabular}{|c|c|c|c|c|} \hline \( \mathrm{A} \) & \( \mathrm{B} \) & \( \mathrm{C} \) & \( \mathrm{D} \) & \( \mathrm{Y} \) \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 & 1 \\ \hline 0 & 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 0 & 0 \\ \hline 0 & 1 & 0 & 1 & 1 \\ \hline 0 & 1 & 1 & 0 & 1 \\ \hline 0 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 0 & 1 & 1 \\ \hline 1 & 0 & 1 & 0 & 1 \\ \hline 1 & 0 & 1 & 1 & 1 \\ \hline 1 & 1 & 0 & 0 & 0 \\ \hline 1 & 1 & 0 & 1 & 1 \\ \hline 1 & 1 & 1 & 0 & 1 \\ \hline 1 & 1 & 1 & 1 & 1 \\ \hline \end{tabular}