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The compound statement (prarrq)vv (p^^~q...

The compound statement `(prarrq)vv (p^^~q)` is logically equivalent to

A

`pharr q`

B

`p ^^ q`

C

tautology

D

contradiction

Text Solution

AI Generated Solution

The correct Answer is:
To determine the logical equivalence of the compound statement \( (p \rightarrow q) \lor (p \land \neg q) \), we can use a truth table to analyze the values of the components involved. Let's break it down step by step. ### Step 1: Identify the components We have two components in our statement: 1. \( p \rightarrow q \) (Implication) 2. \( p \land \neg q \) (Conjunction of \( p \) and the negation of \( q \)) ### Step 2: Create a truth table We will create a truth table with columns for \( p \), \( q \), \( p \rightarrow q \), \( \neg q \), \( p \land \neg q \), and finally \( (p \rightarrow q) \lor (p \land \neg q) \). | \( p \) | \( q \) | \( p \rightarrow q \) | \( \neg q \) | \( p \land \neg q \) | \( (p \rightarrow q) \lor (p \land \neg q) \) | |---------|---------|------------------------|---------------|-----------------------|-----------------------------------------------| | T | T | T | F | F | T | | T | F | F | T | T | T | | F | T | T | F | F | T | | F | F | T | T | F | T | ### Step 3: Fill in the truth values 1. **For \( p \rightarrow q \)**: - True when \( p \) is false or both \( p \) and \( q \) are true. - False only when \( p \) is true and \( q \) is false. 2. **For \( \neg q \)**: - True when \( q \) is false. - False when \( q \) is true. 3. **For \( p \land \neg q \)**: - True only when both \( p \) is true and \( q \) is false. 4. **For \( (p \rightarrow q) \lor (p \land \neg q) \)**: - True if at least one of the components \( p \rightarrow q \) or \( p \land \neg q \) is true. ### Step 4: Analyze the final column From the truth table, we can see that the final column \( (p \rightarrow q) \lor (p \land \neg q) \) is always true (T) for all combinations of \( p \) and \( q \). ### Conclusion Since the statement evaluates to true in all cases, it is a tautology. Therefore, the compound statement \( (p \rightarrow q) \lor (p \land \neg q) \) is logically equivalent to **True** (or a tautology).
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