The output Y for the inputs A and B of circuit is given by Truth...

Output using DeMorgan's theorem can be written as,

$Y = \left(\left(A \cdot A \cdot B\right) \cdot \left(A \cdot B \cdot B\right)\right) = \left(A \cdot A \cdot B\right) + \left(A \cdot B \cdot B\right) = \left(A \cdot A \cdot B\right) + \left(A \cdot B \cdot B\right) = \left(A \cdot \left(A + B\right)\right) + \left(\left(A + B\right) \cdot B\right) = A \cdot A + A \cdot B + A \cdot B + B \cdot B = \left(A + B\right) \cdot \left(A + B\right)$

Which represents XOR gate.

Therefore, required truth table will be

\left[\begin{cases}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{cases}\right]