The following configuration of gates is equivalent to
The following configuration of gates is equivalent to
NAND
$X o R$
$A N D$
$O R$
Solution
For the given logic gates combination,
$\begin{aligned}
& Y=\overline{\overline{\mathrm{A}+\mathrm{A}}+\overline{\mathrm{B}+\mathrm{B}}} \\
& =\overline{\overline{(\mathrm{A}+\mathrm{A})}} \overline{\overline{(\mathrm{B}+\mathrm{B})}}=\mathrm{A} \cdot \mathrm{~B}
\end{aligned}$
$\therefore$ The given combination is AND gate.
