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

$((p\wedge q)\Rightarrow (r\vee q))\wedge ((p\wedge r)\Rightarrow q)$
We know, $p\Rightarrow q$ is equivalent to
$\sim p\vee q$
$(\sim (p\wedge q)\vee (r\vee q))\wedge (\sim (p\wedge r))\vee q))$
$\Rightarrow (\sim p\vee \sim q\vee r\vee q)\wedge (\sim p\vee \sim r\vee q)$
$\Rightarrow (\sim p\vee r\vee t)\wedge (\sim p\vee \sim r\vee q)$
$\Rightarrow (t)\wedge (\sim p\vee \sim r\vee q)$
For this to be tautology, $(\sim p\vee \sim r\vee q)$ must be always true which follows for $r=\sim p$ or $r=q$.
So, the correct option is (B) : 2

Answer 2

Step 1: We are given the expression: \[ \left( (p \land q) \Leftrightarrow (r \vee q) \right) \land \left( (p \land r) \Leftrightarrow q \right). \] Step 2: We know that \( p \Leftrightarrow q \) is equivalent to \( \neg p \vee q \). \[ \left( (p \land q) \Leftrightarrow (r \vee q) \right) \quad \text{and} \quad \left( (p \land r) \Leftrightarrow q \right). \] Now, simplify the first part: \[ \left( (p \land q) \Leftrightarrow (r \vee q) \right) = \neg (p \land q) \vee (r \vee q). \] Step 3: For the second part: \[ \left( (p \land r) \Leftrightarrow q \right) = \neg (p \land r) \vee q. \] Step 4: We must find the values of \( r \) for which the expression is always true. For this to be a tautology, we must have: \[ \left( \neg (p \land q) \vee (r \vee q) \right) \land \left( \neg (p \land r) \vee q \right) \quad \text{is true for all cases}. \] Step 5: After solving the logical expression, we find that there are only 2 values of \( r \) that make this expression a tautology. Hence, the number of values of \( r \) is 2.