Question

Let \( \oplus \) and \( \odot \) denote the Exclusive-OR and Exclusive-NOR operations respectively. Which one of the following is not correct?

• A. \( \overline{P} \oplus Q = P \odot Q \)
• B. \( \overline{P} \oplus Q = P \odot Q \)
• C. \( \overline{P} \oplus \overline{Q} = P \oplus Q \)
• D. \( (P \oplus \overline{P}) \oplus Q = (P \oplus P) \odot Q \)

Hint:

In Boolean algebra, XOR and XNOR have properties that can be used to simplify and compare expressions. Always remember the basic rules for these operations.

Answer 1

The Correct Option is D

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 \( \boxed{(d)} \).