Simplifying A XOR B XOR C: Is A Abar.B bar.C Correct?

Understanding the correct simplification of Boolean expressions is crucial in digital circuit design and logic gate optimization. In this article, we will explore whether it is correct to simplify the expression A ⊕ B ⊕ C as A Abar.B bar.C. We will break down the operations and verify the equivalence step by step.

Step 1: Understanding XOR

The XOR (exclusive OR) operation outputs true (1) if the number of true inputs is odd. This means that if any of the inputs is true, but not all of them, the output is true.

For three variables A, B, and C: A ⊕ B is true if either A or B is true, but not both. Extending this to three variables A ⊕ B ⊕ C will be true if an odd number of A, B, and C are true.

Step 2: Truth Table for A ⊕ B ⊕ C

Here is the truth table for A ⊕ B ⊕ C: ABCA ⊕ BA ⊕ B ⊕ C 00000 00101 01011 01100 10011 10100 11001 11110

From the truth table, we see that A ⊕ B ⊕ C is true for the following combinations:

0 0 1 0 1 0 1 0 0 1 1 0

Note that the output is 1 (true) only when an odd number of inputs are 1.

Step 3: Analyzing the Proposed Simplification

The proposed simplification is A Abar.B bar.C. Let's analyze each term:

A: This term is true when A 1. It covers the cases where A 1 and C can be 0 or 1, but it does not account for cases where A 0. This term is true when A 0, B 1, and C 0. Abar.B bar.C: This term is true when A 0, B 0, and C 1.

Now, let's check when the entire expression is true:

When A 1: The whole expression is true regardless of B and C. When A 0, B 1, C 0: The second term is true. When A 0, B 0, C 1: The third term is true.

Note that both the second and third terms cover the cases where the output of A ⊕ B ⊕ C is true. However, they do not account for the cases where A 1, B 1, and C 0.

Step 4: Comparison and Conclusion

The proposed simplification A Abar.B bar.C does not account for the case 1 1 0, where A ⊕ B ⊕ C 1 because A 1, B 1, and C 0. The output is true in the original expression but not in the simplification.

Since the simplified expression fails to capture the true output for certain input combinations, it is not correct to simplify A ⊕ B ⊕ C as A Abar.B bar.C.

Conclusion: The expression A ⊕ B ⊕ C cannot be simplified to A Abar.B bar.C. The correct simplification should include all combinations that yield an odd number of true values, which is not captured by the proposed simplification.