Work forward gate by gate. The circuit has two identical halves feeding a final NAND, so simplify one half and the other follows by symmetry.

Step 1 — the upper half. The inverter gives \(\bar{A}\) and the NOR gives \(\overline{A+B}=\bar{A}\bar{B}\). The OR gate combines them:

\(Y_1=\bar{A}+\bar{A}\bar{B}\)

Apply the absorption law, \(X+XY=X\), with \(X=\bar{A}\):

\(Y_1=\bar{A}\)

Step 2 — the lower half. Identically, the inverter gives \(\bar{B}\), the NOR gives \(\overline{B+C}=\bar{B}\bar{C}\), and the OR gate produces

\(Y_2=\bar{B}+\bar{B}\bar{C}=\bar{B}\)

Notice that C has vanished completely — absorption swallowed the only branch that carried it. That is the key observation of the whole problem, and it immediately rules out options 1, 2 and 4, all of which contain C.

Step 3 — the output NAND.

\(Y=\overline{Y_1\cdot Y_2}=\overline{\bar{A}\cdot\bar{B}}\)

By De Morgan's theorem,

\(Y=\overline{\bar{A}}+\overline{\bar{B}}=A+B\)

Verify with a quick truth check. Take A = 0, B = 0, C = 1. Then \(Y_1=\bar{A}=1\) and \(Y_2=\bar{B}=1\), so the NAND gives 0 ✓, matching A + B = 0. Take A = 1, B = 0, C = 0: \(Y_1=0\), so the NAND gives 1 ✓, matching A + B = 1. Changing C never changes the output, confirming the redundancy.

The lesson. A NOR gate followed by an OR with one of its own inputs inverted is a redundant structure — the absorption law collapses it. Recognising \(X+XY=X\) and its dual \(X(X+Y)=X\) on sight is what turns a five-gate network into a single OR.

Hence, the output is A + B.