Question

The statement $(p \wedge(\sim q)) \Rightarrow(p \Rightarrow(\sim q))$ is

• A. equivalent to $p \vee q$
• B. equivalent to $(-p) \vee(-q)$
• C. a contradiction
• D. a tautology

Answer 1

The Correct Option is D

To determine whether the statement \((p \wedge(\sim q)) \Rightarrow(p \Rightarrow(\sim q))\)is a tautology, we can simplify the statement using logical equivalences.

1. Let's break down the statement: \((p \wedge \sim q) \Rightarrow (p \Rightarrow \sim q)\). This can be rewritten using the implication definition: \(\sim(p \wedge \sim q) \vee (p \Rightarrow \sim q)\).
2. Simplifying further: - From the implication \(p \Rightarrow \sim q = \sim p \vee \sim q\). So, the expression becomes: \(\sim(p \wedge \sim q) \vee (\sim p \vee \sim q)\).
3. The expression \(\sim(p \wedge \sim q)\) becomes \(\sim p \vee q\) using De Morgan's laws.
4. Substitute back, and we have: \((\sim p \vee q) \vee (\sim p \vee \sim q)\).
5. Applying the associative and distributive laws simplifies this to: \(\sim p \vee q \vee \sim q\).
6. Since \(q \vee \sim q\) is a tautology (any statement or its negation is always true), the entire expression reduces to: \(\sim p \vee \text{true}\), which is always true regardless of the truth value of \(p\).

Therefore, the given statement is a tautology, as it is always true for all truth values of \(p\) and \(q\).