The Boolean expression $(p \implies q) \wedge (q \implies \sim p)$ is equivalent to :
Show 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 $\sim p$.
Step 1: Understanding the Concept:
Mathematical logic uses symbolic representations for logical connectors. The implication \(A \implies B\) is logically equivalent to \(\sim A \vee B\). We can simplify expressions using distributive laws and properties of basic connectors. Step 3: Detailed Explanation:
The expression is \((p \implies q) \wedge (q \implies \sim p)\).
Step 1: Rewrite implications using the equivalent OR form:
\[ (\sim p \vee q) \wedge (\sim q \vee \sim p) \]
Step 2: Recognize the common term \(\sim p\) in both brackets and apply the Distributive Law in reverse (\(A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)\)):
\[ \sim p \vee (q \wedge \sim q) \]
Step 3: Simplify the term in the parenthesis. Since \(q \wedge \sim q\) is a contradiction (always False):
\[ \sim p \vee F \]
Step 4: Any statement ORed with False is just the statement itself:
\[ \sim p \] Step 4: Final Answer:
The expression is equivalent to \(\sim p\).
