Half-adders forming a full-adder — do we need any extra gate? True or False: Two half-adders can be combined to implement a full-adder with no additional gates required.

Introduction / Context:
Half-adders and full-adders are core arithmetic blocks. A well-known construction uses two half-adders to build a full-adder. However, designers must remember the role of the carry path when combining them. This question evaluates whether you know the exact gate requirements for that composite design.

Given Data / Assumptions:

• A half-adder outputs Sum = A XOR B and Carry = A AND B.
• A full-adder needs Sum = A XOR B XOR Cin and Cout = majority(A, B, Cin).
• Standard CMOS or TTL logic is assumed.

Concept / Approach:
To implement a full-adder from two half-adders: First half-adder computes S1 = A XOR B and C1 = A AND B. The second half-adder takes S1 and Cin to produce Sum = S1 XOR Cin and C2 = S1 AND Cin. The final carry-out is Cout = C1 OR C2. Note the need for an OR gate to combine the two carry signals.

Step-by-Step Solution:
Use Half-Adder 1: S1 = A XOR B; C1 = A AND B.Use Half-Adder 2: Sum = S1 XOR Cin; C2 = S1 AND Cin.Form Cout: Cout = C1 OR C2 → requires an additional OR gate beyond the two half-adders.

Verification / Alternative check:
Any textbook realization shows the extra OR gate for carry-out. Simulation or truth-table enumeration confirms that without the OR gate, Cout is not correct for all input combinations.

Why Other Options Are Wrong:

• True: Incorrect because at least one extra OR gate is required to produce the full-adder carry-out.

Common Pitfalls:

• Assuming the two half-adders alone are sufficient and forgetting the carry combination.
• Confusing the internal gates of an integrated “full-adder” IC with a discrete two half-adder construction.

Final Answer:
False