Let ⊕ denote the exclusive OR (XOR) operation. Let '1' and '0' denote the…
2014
Let ⊕ denote the exclusive OR (XOR) operation. Let '1' and '0' denote the binary constants. Consider the following Boolean expression for \(F\) over two variables \(P\) and \(Q\):
\(F(P,Q)=\left( \left(1 \oplus P \right) \oplus \left( P \oplus Q \right )\right ) \oplus \left(\left(P \oplus Q\right) \oplus \left(Q \oplus 0\right)\right)\)
The equivalent expression for \(F\) is
- A.
\(P+Q\) - B.
\(\overline{P+Q}\) - C.
\(P \oplus Q\) - D.
\(\overline {P \oplus Q}\)
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Correct answer: D
Key idea: use XOR associativity/commutativity and identities x ⊕ x = 0, x ⊕ 0 = x, 1 ⊕ x = ¬x.
Expand and group all XOR terms: (1 ⊕ P) ⊕ (P ⊕ Q) = 1 ⊕ P ⊕ P ⊕ Q
(P ⊕ Q) ⊕ (Q ⊕ 0) = P ⊕ Q ⊕ Q ⊕ 0
Combine the two results: 1 ⊕ P ⊕ P ⊕ Q ⊕ P ⊕ Q ⊕ Q ⊕ 0
Cancel pairs (x ⊕ x = 0) and simplify constants: this reduces to 1 ⊕ P ⊕ Q
Use 1 ⊕ (P ⊕ Q) = ¬(P ⊕ Q). Therefore F(P,Q) = ¬(P ⊕ Q).
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