Question

For the propositions p and q, consider the following statements:
(I) \((p \vee q) \wedge (\neg p \vee \neg q)\) is a tautology.
(II) \((p \vee q) \wedge (\neg p \wedge \neg q)\) is a contradiction.
Choose the correct answer:

• A. Only (I) is true
• B. Only (II) is true
• C. Both (I) and (II) are true
• D. Both (I) and (II) are false

Hint:

Simplify each expression using De Morgan's laws before checking whether it is always true or always false.

Answer 1

The Correct Option is B

Step 1: Simplify \((p \vee q) \wedge (\neg p \wedge \neg q)\) using De Morgan's law: \(\neg p \wedge \neg q \equiv \neg (p \vee q)\). So the expression becomes \((p \vee q) \wedge \neg(p \vee q)\), which has the form \(A \wedge \neg A\) with \(A = p \vee q\). By the law of non-contradiction, \(A \wedge \neg A\) is identically false for every truth value of \(A\). Hence statement (II) is a contradiction, so it is true.

Step 2: Simplify \((p \vee q) \wedge (\neg p \vee \neg q)\) using De Morgan's law on the second factor: \(\neg p \vee \neg q \equiv \neg (p \wedge q)\). So the expression is \((p \vee q) \wedge \neg (p \wedge q)\), the standard identity for exclusive-or, \(p \oplus q\). Since \(p \oplus q\) is false whenever \(p\) and \(q\) share the same truth value, for example \(p = q = T\) gives \(p \oplus q = F\), the expression is not always true, so it is not a tautology. Statement (I) is false.

Step 3: Only statement (II) is true.

\[\boxed{\text{Only (II) is true}}\]