Using a suitable combination from a NOR, an OR and a NOT gate, draw circuits to obtain the truth table given below
(i) `|{:(A,B,y),(0,0,0),(0,1,0),(1,0,1),(1,1,0):}|` (ii) `|{:(A,B,y),(0,0,1),(0,1,1),(1,0,0),(1,1,1):}|`

Explanation

As in truth table (i), the Boolean expression `y=A+B` is not satisfied. Therefore, its circuit must consist of a NOR gate and a NOT gate. The combination of gates will be as shown in Fig.9.66.
In truth table (ii), the Boolean expression `y=A+B` is partly satisfied, therefore, its circuit must consist of OR gate and NOT gate. The combination of gates is shown in Fig.9.66.