A digital logic circuit has three inputs A , B , and C . The circuit consists of four logic gates connected as follows: Gate 1 is a NOR gate whose inputs are A and B . Gate 2 is a NAND gate whose inputs are B and C . Gate 3 is an AND gate whose inputs are the outputs of Gate 1 and Gate 2. Gate 4 is a NOT gate whose input is the output of Gate 3. If the final output of the circuit is taken from Gate 4 and denoted by Y

Step-by-step solution

Let us find the Boolean expression for the output of each gate step by step. The output of Gate 1 (NOR gate) is G₁ = A + B . The output of Gate 2 (NAND gate) is G₂ = B C . The output of Gate 3 (AND gate) is G₃ = G₁ G₂ = A + B B C . The final output Y from Gate 4 (NOT gate) is Y = G₃ = A + B B C . Applying De Morgan's laws ( X Y = X + Y ): Y = A + B + B C Y = (A + B) + (B C) Using the Boolean absorption law or factoring out B : Y = A + B(1 + C) Since 1 + C = 1 , we get: Y = A + B The final output Y is 1 whenever A =