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Answer: 1. AND gate

Key idea: Find the output of each first-stage gate, then apply the final NOR gate.

With the inputs of each gate joined, the NAND gives \(\overline{A}\) and the NOR gives \(\overline{B}\). The final gate is a NOR:

$$Y=\overline{\overline{A}+\overline{B}}=A\cdot B$$

(De Morgan's law). This is the AND function, option 1.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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