Semiconductor Electronics question for JEE Main (JEE Main 2019 (12 Jan, Shift 1)), with solution

Answer: (A) $A\overline{B}$

First NAND: $X = \overline{AB}$.

Upper NAND: $Y_1 = \overline{A\cdot\overline{AB}} = \overline{A(\overline A + \overline B)} = \overline{A\overline B} = \overline A + B$.

OR gate: $Y_2 = \overline{AB} + B = \overline A + \overline B + B = 1$.

Final NAND: $Y = \overline{Y_1\cdot Y_2} = \overline{\overline A + B} = A\overline B$.

Solution by Sreeraj P, M.Sc Physics