Question

The number of values of $r \in\{p, q, \sim p, \sim q\}$ for which $((p \wedge q) \Rightarrow(r \vee q)) \wedge((p \wedge r) \Rightarrow q)$ is a tautology, is :

• A. 1
• B. 2
• C. 4
• D. 3

Hint:

A tautology is a logical expression that is always true, regardless of the truth values of its components. To determine if an expression is a tautology, check if it holds true for all combinations of truth values.

Answer 1

The Correct Option is B

To solve the given problem, we need to determine the number of values of \( r \in \{ p, q, \sim p, \sim q \} \) for which the logical expression \(((p \wedge q) \Rightarrow (r \vee q)) \wedge ((p \wedge r) \Rightarrow q)\) is a tautology, meaning it is true for all possible truth values of \( p, q, \) and \( r \).

Let's break down the expression:

1. \((p \wedge q) \Rightarrow (r \vee q)\): This implication is true unless the antecedent \((p \wedge q)\) is true and the consequent \((r \vee q)\) is false. If \( p \wedge q \) is true, then both \( p \) and \( q \) must be true. For \((r \vee q)\) to be false, both \( r \) and \( q \) must be false, which contradicts \( q \) being true. Thus, this part is always true.
2. \((p \wedge r) \Rightarrow q\): This implication is true unless the antecedent \((p \wedge r)\) is true and the consequent \(q\) is false. So, if both \( p \) and \( r \) are true, \( q \) must also be true.

Now consider the truth values for \( r \) from the set \(\{ p, q, \sim p, \sim q \}\):

• Case 1: \( r = p \)
  For \((p \wedge r) = (p \wedge p) = p\), and since we assumed \( p \) is true, \( q \) must also be true, satisfying the conditions and making the expression a tautology.
• Case 2: \( r = q \)
  Here, \((p \wedge r) = (p \wedge q)\). If both \( p \) and \( q \) are true, the implications hold, making the expression a tautology.
• Case 3: \( r = \sim p \)
  If \( r = \sim p \), then for \((p \wedge r)\) to be true, \( p \) must be true, but \( r \) is false (since \( r = \sim p \)), which contradicts, thus not a tautology.
• Case 4: \( r = \sim q \)
  If \( r = \sim q \), the conjunction with \( p \) and the requirement \( q \) must also be true create contradictions similar to the previous case, failing to make the expression a tautology.

Therefore, the number of values for \( r \) that make the expression a tautology is \(2\). Thus, the correct answer is 2.