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: Build the truth table for the two propositions \(p\) and \(q\) over the four cases \(TT, TF, FT, FF\).

Step 2: Evaluate \((p \vee q) \wedge (\neg p \vee \neg q)\). At \(p = T, q = T\): \(p \vee q = T\) but \(\neg p \vee \neg q = F \vee F = F\), so the conjunction is \(F\). Since the expression is false for at least one assignment, it is not a tautology, so statement (I) is false. This expression is in fact exactly the exclusive-or \(p \oplus q\), true only when \(p\) and \(q\) differ.

Step 3: Evaluate \((p \vee q) \wedge (\neg p \wedge \neg q)\). The factor \(\neg p \wedge \neg q\) is true only when \(p = F\) and \(q = F\); but then \(p \vee q = F\), making the whole conjunction \(F\). For every other assignment \(\neg p \wedge \neg q\) is already \(F\), so the conjunction is \(F\) again. The expression is false for all four assignments, so it is a contradiction, and statement (II) is true.

Step 4: Only statement (II) is true.

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