Thank you for reporting, we will resolve it shortly
Q.
The equivalent circuit is
The circuit represents
ManipalManipal 2015
Solution:
Output of $NOR$ gate
$y'=\overline{ A + B }$
Output of NAND gate
$y ''=\overline{( A + B ) \cdot( A + B )}$
$=\overline{( A + B )( A + B )}$
$=( A + B )( A + B )$
$A + B$
Output of $NOT$ gate
$y ''''=\overline{ A + B }$
$\therefore $ The Boolean expression is for NOR gate.