Given that x + y = y (with y = 1), F_1 = xy' + y', F_2 = x' + y and F_3 = xy', then value of F_1, F_2…

From $x + y = y$, multiplying both sides by $y'$ gives $xy' = 0$, so $F_3 = xy' = 0$, and by De Morgan's law $F_2 = x' + y = (xy')' = 1$. With $y = 1$, $y' = 0$, which yields $F_1 = xy' + y' = 0 + 0 = 0$, giving the triple 0 , 1 , 0.