Identify the logic gate given in the circuit :

The correct answer is OR gate.

Looking at the circuit diagram carefully, the following components are visible:

- Input A is fed into a NOT gate (inverter — shown as a triangle with a small circle/bubble at the output).
- Input B is fed into another NOT gate (inverter).
- The outputs of both NOT gates (A̅ and B̅) are then fed as inputs into a NAND gate (AND gate symbol with a bubble at its output).
- The final output is Y.

So the Boolean expression for the output Y can be written as:

Y=\overline{\mathrm{A̅}}\cdot \overline{\mathrm{B̅}}

In cleaner notation, the output of the NAND gate applied to A̅ and B̅ is:

Y=\stackrel{-}{\stackrel{-}{A}·\stackrel{-}{B}}

That is: Y = NOT( NOT(A) · NOT(B) )

Step 1: Apply the inner NOT gates.

The two NOT gates produce A̅ (NOT A) and B̅ (NOT B).

Step 2: Apply the NAND operation on A̅ and B̅.

A NAND gate gives the complement of (A AND B). So applied to A̅ and B̅:

Y=\stackrel{-}{\stackrel{-}{A}·\stackrel{-}{B}}

Step 3: Apply De Morgan's Theorem.

De Morgan's theorem states:

\stackrel{-}{X·Z}=\stackrel{-}{X}+\stackrel{-}{Z}

Applying this to our expression (where X = A̅ and Z = B̅):

Y=\stackrel{-}{\stackrel{-}{A}}+\stackrel{-}{\stackrel{-}{B}}

Step 4: Apply the double negation law (NOT NOT X = X).

The double complement of any variable returns the variable itself:

\stackrel{-}{\stackrel{-}{A}}=A\text{and}\stackrel{-}{\stackrel{-}{B}}=B

Therefore:

Y=A+B

Conclusion: The output Y = A + B, which is exactly the Boolean expression for an OR gate.

This circuit — two NOT gates feeding into a NAND gate — is a well-known implementation of an OR gate using only NAND-based logic, consistent with De Morgan's equivalence theorem. The circuit is therefore equivalent to an OR gate.

Truth Table Verification:

The output column matches exactly the truth table of an OR gate, confirming the answer is the OR gate.