Let, \(x_1⊕x_2⊕x_3⊕x_4 = 0\) where \(x_1, x_2, x_3, x_4\) are Boolean…
2016
Let, \(x_1⊕x_2⊕x_3⊕x_4 = 0\) where \(x_1, x_2, x_3, x_4\) are Boolean variables, and ⊕ is the XOR operator. Which one of the following must always be TRUE?
- A.
\(x_1x_2x_3x_4 = 0\) - B.
\(x_1x_3 +x_2 = 0\) - C.
\(\overline x_1 ⊕ \overline x_3 = \overline x_2 ⊕ \overline x_4\) - D.
\(x_1 +x_2 +x_3 +x_4 = 0\)
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Correct answer: C
Answer: The equality of complements ¬x1 ⊕ ¬x3 = ¬x2 ⊕ ¬x4 is always true under the given condition.
Reasoning:
Use the identity ¬a ⊕ ¬b = a ⊕ b (XOR of complements equals XOR of originals).
From x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0, group terms as (x1 ⊕ x3) ⊕ (x2 ⊕ x4) = 0, which implies x1 ⊕ x3 = x2 ⊕ x4.
Applying the complement identity to each side gives ¬x1 ⊕ ¬x3 = x1 ⊕ x3 and ¬x2 ⊕ ¬x4 = x2 ⊕ x4, so ¬x1 ⊕ ¬x3 = ¬x2 ⊕ ¬x4.
Why the other expressions may fail (quick counterexamples):
x1 x2 x3 x4 = 0: choose x1 = x2 = x3 = x4 = 1. Then x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0 but x1 x2 x3 x4 = 1.
(x1 x3) + x2 = 0 (interpreting + as OR): choose x1 = 1, x3 = 1, x2 = 0, x4 = 0. Then x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0 but (x1 x3) + x2 = 1.
x1 + x2 + x3 + x4 = 0 (OR): choose x1 = 1, x2 = 1, x3 = 0, x4 = 0. Then x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0 but the OR is 1.
Therefore, the equality of complements is the statement that must always be true.
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