Question

The Boolean expression $(p \implies q) \wedge (q \implies \sim p)$ is equivalent to :

• A. $p$
• B. $q$
• C. $\sim p$
• D. $\sim q$

Hint:

Truth tables are definitive but algebraic simplification is faster. For this expression, if $p$ is True, the second part $(q \implies \text{False})$ forces $q$ to be False, but then $(p \implies q)$ becomes $(\text{True} \implies \text{False})$, which is False. Thus if $p$ is True, the result is False. This behavior matches $\sim p$.

Answer 1

The Correct Option is C

To solve the given problem, we need to simplify the Boolean expression \((p \implies q) \wedge (q \implies \sim p)\). Let's break down each part of the expression:

1. Understand the implication in logical expressions:
   • The implication \(p \implies q\) is logically equivalent to \(\sim p \vee q\), which means "not \(p\) or \(q\)".
   • Similarly, \(q \implies \sim p\) is equivalent to \(\sim q \vee \sim p\), meaning "not \(q\) or not \(p\)".
2. Substitute these equivalents back into the expression:
   • The expression \((p \implies q) \wedge (q \implies \sim p)\) becomes \((\sim p \vee q) \wedge (\sim q \vee \sim p)\).
3. Simplify the expression using Boolean algebra:
   • We can use the distribution law to distribute \(\wedge\) over \(\vee\):
   • \((\sim p \vee q) \wedge (\sim q \vee \sim p)\) simplifies to \((\sim p \wedge \sim q) \vee (q \wedge \sim p)\).
   • Here, \(\sim p\) is common in both parts and can be factored out:
   • \(\sim p \wedge (\sim q \vee q)\). The expression \((\sim q \vee q)\) is a tautology because it always evaluates to true regardless of the values of \(q\).
   • Therefore, the expression simplifies to \(\sim p\).

As a result, the Boolean expression \((p \implies q) \wedge (q \implies \sim p)\) is equivalent to \(\sim p\).

Thus, the correct answer is: \(\sim p\).