The following logic diagram is equivalent to ______ gate.

Concept

A NOR gate outputs 1 only when all its inputs are 0, i.e. its output is the complement of the OR of its inputs: (X + Y)'. NOR is a universal gate, so any logic function can be built from NOR gates alone. Two algebraic tools reduce such a network: De Morgan's law, which rewrites a complemented sum as a product of complements, (X + Y)' = X'·Y'; and the complement law together with distribution, X'·(X + Y) = X'X + X'Y = X'Y, since X'X = 0. The standard equivalence (XNOR) function is the one whose output is 1 exactly when the two inputs are the same; its Boolean expression is AB + A'B'.

Applying it to this circuit

Label the inputs A (top) and B (bottom) and trace the four NOR gates left to right. Note that the small open circle at the very end of the output line is just the output terminal/node marker, not an extra inverting bubble: there is exactly one inverting bubble on the final gate.

1. First (left) NOR: inputs A and B give G1 = (A + B)'.

2. Top-middle NOR: inputs A and G1 give G2 = (A + G1)'.

3. Bottom-middle NOR: inputs G1 and B give G3 = (G1 + B)'.

4. Final (right) NOR: inputs G2 and G3 give the output O = (G2 + G3)'.

Now reduce each intermediate output with those two identities, one gate at a time:

1. Top-middle gate: G2 = (A + G1)' = A'·G1' by De Morgan's law.

2. Since G1' = ((A + B)')' = A + B, this gives G2 = A'(A + B) = A'A + A'B = A'B, because A'A = 0.

3. Bottom-middle gate: G3 = (G1 + B)' = G1'·B' = (A + B)B' = AB' + BB' = AB', because BB' = 0.

4. Final gate: O = (G2 + G3)' = (A'B + AB')'.

5. A'B + AB' is the non-equivalence (difference) function, so its complement is the equivalence function: O = (A'B)'·(AB')' = (A + B')(A' + B) = AA' + AB + A'B' + BB' = AB + A'B'.

So the network's output is O = AB + A'B' — it is 1 exactly when A and B carry the same value.

Cross-check by truth table

Evaluating all four input combinations confirms the output is 1 only when A and B are equal:

The output column is 1 for A=B (00 and 11) and 0 when the inputs differ (01 and 10). That is precisely the equivalence function AB + A'B', so the circuit is equivalent to an XNOR gate.