Question

For three simple statements p, q, and r, p → (q ˅ r) is logically equivalent to

• A. (p ˅ q) → r
• B. (p → ∼q) ˄ (p → r)
• C. (p → q) ˅ (p → r)
• D. (p → q) ˄ (p → ∼r)

Answer 1

The Correct Option is C

Detailed Solution: Logical Equivalence of \( p \rightarrow (q \vee r) \)

Let’s dive into solving this logical equivalence problem step by step in a way that’s easy to understand, especially for students learning about propositional logic. The task is to determine which of the given options is logically equivalent to the expression \( p \rightarrow (q \vee r) \), where \( p \), \( q \), and \( r \) are simple statements. Logical equivalence means that the two expressions always have the same truth value regardless of the truth values of \( p \), \( q \), and \( r \). We’ll explore this using the definition of implication, De Morgan’s laws, and truth tables to ensure clarity.

Step 1: Understand the Given Expression

The expression is \( p \rightarrow (q \vee r) \). In logic, the implication \( p \rightarrow q \) is equivalent to \( \neg p \vee q \) (this is a fundamental rule). So, let’s rewrite \( p \rightarrow (q \vee r) \) using this equivalence:

• \( p \rightarrow (q \vee r) \equiv \neg p \vee (q \vee r) \)

According to the associative and commutative properties of disjunction (\( \vee \)), \( \neg p \vee (q \vee r) \) is the same as \( (\neg p \vee q) \vee r \). This gives us a starting point to compare with the options. However, let’s also consider the contrapositive and other logical manipulations to find the exact match.

Step 2: Analyze the Options

We have four options to evaluate:

• Option 1: \( (p \vee q) \rightarrow r \)
• Option 2: \( (p \rightarrow q) \wedge (p \rightarrow r) \)
• Option 3: \( (p \rightarrow q) \vee (p \rightarrow r) \)
• Option 4: \( (p \rightarrow q) \wedge (p \rightarrow r) \)

To determine the correct answer, we need to check which of these expressions is logically equivalent to \( p \rightarrow (q \vee r) \). Logical equivalence can be tested by constructing a truth table or by using logical identities. Since this is a detailed solution, let’s use both methods for a thorough understanding.

Step 3: Use Logical Identities

Let’s manipulate \( p \rightarrow (q \vee r) \) further. The implication \( p \rightarrow (q \vee r) \) can also be explored through its contrapositive, which is another form of equivalence. The contrapositive of \( p \rightarrow (q \vee r) \) is \( \neg (q \vee r) \rightarrow \neg p \). Using De Morgan’s law, \( \neg (q \vee r) \equiv \neg q \wedge \neg r \), so the contrapositive becomes:

• \( (\neg q \wedge \neg r) \rightarrow \neg p \)
• This is equivalent to \( \neg (\neg q \wedge \neg r) \vee \neg p \)
• Using De Morgan’s law again, \( \neg (\neg q \wedge \neg r) \equiv \neg \neg q \vee \neg \neg r \equiv q \vee r \)
• So, \( (q \vee r) \vee \neg p \), which is the same as \( \neg p \vee (q \vee r) \) (consistent with our initial rewrite).

This manipulation suggests we should compare the options by rewriting them. Let’s test each option:

• Option 1: \( (p \vee q) \rightarrow r 
  This is equivalent to \( \neg (p \vee q) \vee r \). Using De Morgan’s law, \( \neg (p \vee q) \equiv \neg p \wedge \neg q \), so \( (\neg p \wedge \neg q) \vee r \). This is different from \( \neg p \vee (q \vee r) \) because it includes \( \neg q \) and lacks the direct combination of \( q \vee r \).
• Option 2: \( (p \rightarrow q) \wedge (p \rightarrow r) \)
  \( p \rightarrow q \equiv \neg p \vee q \), and \( p \rightarrow r \equiv \neg p \vee r \). So, \( (\neg p \vee q) \wedge (\neg p \vee r) \). This is not the same as \( \neg p \vee (q \vee r) \) because the conjunction (\( \wedge \)) distributes differently and requires both conditions to hold, whereas \( \vee \) allows either.
• Option 3: \( (p \rightarrow q) \vee (p \rightarrow r) \)
  This is \( (\neg p \vee q) \vee (\neg p \vee r) \). Using the associative property, this simplifies to \( \neg p \vee q \vee r \), which is the same as \( \neg p \vee (q \vee r) \). This matches our rewritten form of \( p \rightarrow (q \vee r) \).
• Option 4: \( (p \rightarrow q) \wedge (p \rightarrow r) \)
  This is the same as Option 2 and does not match due to the conjunction.

Step 4: Verify with Truth Table

To confirm, let’s use a truth table for \( p \rightarrow (q \vee r) \) and \( (p \rightarrow q) \vee (p \rightarrow r) \). We’ll test all combinations of \( p \), \( q \), and \( r \) (8 rows):

- \( p \rightarrow (q \vee r) \) is T unless \( p \) is T and \( q \vee r \) is F (only the 4th row).
- \( (p \rightarrow q) \vee (p \rightarrow r) \) is T unless both \( p \rightarrow q \) and \( p \rightarrow r \) are F, which happens only when \( p \) is T and both \( q \) and \( r \) are F (4th row).
The truth values match in all cases, confirming logical equivalence.

Step 5: Conclusion

The expression \( p \rightarrow (q \vee r) \) is logically equivalent to Option 3: \( (p \rightarrow q) \vee (p \rightarrow r) \). This solution leverages logical identities and a truth table, making it a robust method for students to learn and apply in similar problems.