Question

If the truth value of the statement pattern \([p \wedge \sim r] \to \sim r \wedge q\) is False, then which of the following has truth value False?

• A. \((p \vee r) \to \sim r\)
• B. \((r \vee q) \to \sim p\)
• C. \(\sim (p \vee q) \to \sim r\)
• D. \(\sim (r \vee q) \to \sim p\)

Hint:

An implication is false only in one case: \[ \text{True} \to \text{False} \] Use this rule first to extract truth values quickly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
An implication \(X \to Y\) is False only when \(X\) is True and \(Y\) is False.
Step 2: Key Formula or Approach:
For \([p \wedge \sim r] \to [\sim r \wedge q]\) to be False:
1. \(p \wedge \sim r \equiv \text{True}\)
2. \(\sim r \wedge q \equiv \text{False}\)
Step 3: Detailed Explanation:
From (1), \(p \equiv \text{True}\) and \(\sim r \equiv \text{True} \implies r \equiv \text{False}\).
From (2), since \(\sim r \equiv \text{True}\), for the conjunction to be False, \(q\) must be \(\text{False}\).
So: \(p = T, q = F, r = F\).
Testing options:
(A) \((T \vee F) \to T \equiv T \to T \equiv \text{True}\).
(B) \((F \vee F) \to F \equiv F \to F \equiv \text{True}\).
(C) \(\sim (T \vee F) \to T \equiv \sim T \to T \equiv F \to T \equiv \text{True}\).
(D) \(\sim (F \vee F) \to \sim T \equiv \sim F \to F \equiv T \to F \equiv \text{False}\).
Step 4: Final Answer:
The statement in option (D) is False.