1. **State the problem:** We have a logic circuit with inputs A, B, and C. - A and B go into a NAND gate. - C goes into a NOT gate. - The outputs of the NAND gate, the original B, and the NOT gate feed into an OR gate. - The OR gate output is Y. 2. **Find the output expression Y:** - NAND gate output: $$\text{NAND}(A,B) = \overline{A \cdot B}$$ - NOT gate output: $$\text{NOT}(C) = \overline{C}$$ - Inputs to OR gate: $$\overline{A \cdot B}, B, \overline{C}$$ - OR gate output: $$Y = \overline{A \cdot B} + B + \overline{C}$$ 3. **Simplify the expression:** - Note that $$\overline{A \cdot B} + B = \overline{A \cdot B} + B$$ - Since $$B + \overline{A \cdot B} = B + \overline{A} + \overline{B}$$ (using De Morgan's law on NAND) - But $$B + \overline{B} = 1$$, so $$B + \overline{A \cdot B} = 1 + \overline{A} = 1$$ - Therefore, $$Y = 1 + \overline{C} = 1$$ - So the output Y is always 1 regardless of inputs. 4. **Evaluate Y for A=1, B=0, C=0:** - Using the original expression: $$\overline{A \cdot B} = \overline{1 \cdot 0} = \overline{0} = 1$$ $$\overline{C} = \overline{0} = 1$$ - So, $$Y = 1 + 0 + 1 = 1$$ **Final answers:** - (a) $$Y = \overline{A \cdot B} + B + \overline{C}$$ which simplifies to always 1. - (b) For A=1, B=0, C=0, $$Y=1$$.