Let \( \oplus \) and \( \odot \) denote the Exclusive-OR and Exclusive-NOR operations respectively. Which one of the following is not correct?
Show Hint
- \( \overline{P} \oplus Q = P \odot Q \)
- \( \overline{P} \oplus Q = P \odot Q \)
- \( \overline{P} \oplus \overline{Q} = P \oplus Q \)
- \( (P \oplus \overline{P}) \oplus Q = (P \oplus P) \odot Q \)
The Correct Option is D
Solution and Explanation
Let's examine the operations:
1. \( \oplus \) (Exclusive-OR) is defined as:
- \( P \oplus Q = (P \land \overline{Q}) \lor (\overline{P} \land Q) \)
2. \( \odot \) (Exclusive-NOR) is defined as:
- \( P \odot Q = \overline{P \oplus Q} \) Now, let's evaluate each option:
- Option (a): \( \overline{P} \oplus Q = P \odot Q \) is valid because it is one of the properties of the Exclusive-OR and Exclusive-NOR gates.
- Option (b): \( \overline{P} \oplus Q = P \odot Q \) is valid, as this is true by definition of XOR and XNOR gates.
- Option (c): \( \overline{P} \oplus \overline{Q} = P \oplus Q \) is valid. This is a property of XOR.
- Option (d): \( (P \oplus \overline{P}) \oplus Q = (P \oplus P) \odot Q \) is incorrect because the left-hand side is 1 (since \( P \oplus \overline{P} = 1 \)) and the right-hand side simplifies to 0 (since \( P \oplus P = 0 \)).
Thus, the correct answer is \( \boxed{(d)} \).
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