The given circuit has two inputs, A and B, each passing through a NOT gate, making the inputs Ā and B̄. These modified inputs are then fed into a NAND gate, which gives the output Y as follows:
1. **Output of NAND Gate:**
Y = (Ā ⋅ B̄).
2. **Apply De Morgan’s Law:**
Using De Morgan’s law:
Y = (Ā ⋅ B̄) = A + B.
Thus, the output Y behaves as an OR gate.
Answer: OR gate