The statement (p ∧ q) ⇒ (p ∧ r) is equivalent to : (A) q ⇒ (p ∧ r) (B) p ⇒ (q ∧ r) (C) (p ∧ r) ⇒ (p

Q: Wait, why can't I just cancel $p$ from both sides of the implication?

This is a classic trap! Logical statements do not behave like algebraic equations. You cannot 'divide' or 'cancel' variables across an implication. Logic follows specific laws like De Morgan's, Distributive, and Implication laws. Always transform the expression into a standard form using these laws before attempting any simplification.

Q: Why do we convert the implication to $\sim A \vee B$?

Implications are notoriously hard to manipulate because they are not associative or commutative in the same way as AND/OR operators. By converting $A \Rightarrow B$ into $\sim A \vee B$, we move into the realm of disjunctions and conjunctions, where we have a powerful toolkit of laws (like Distributive and Associative laws) to simplify the expression systematically.

Q: How do I know when to use the Distributive Law versus just rearranging terms?

Great question! Use the Distributive Law when you see a pattern like $X \vee (Y \wedge Z)$, which allows you to break the bracket into $(X \vee Y) \wedge (X \vee Z)$. This is often the key to creating a term that simplifies via the Complement Law (like $\sim p \vee p \equiv T$). If you don't see a clear path to a tautology, try rearranging terms first to see if you can group negations together.

Question

The statement $(p \wedge q) \Rightarrow (p \wedge r)$ is equivalent to :

Options:

(A) $q \Rightarrow (p \wedge r)$

(B) $p \Rightarrow (q \wedge r)$

(C) $(p \wedge r) \Rightarrow (p \wedge q)$

(D) $(p \wedge q) \Rightarrow r$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The logical statement provided is $(p \wedge q) \Rightarrow (p \wedge r)$. Our objective is to simplify this expression into an equivalent, more elegant form.

The implication operator $\Rightarrow$ is often difficult to manipulate directly. We utilize the **Implication Law**, which states that $A \Rightarrow B \equiv \sim A \vee B$.

Applying this law to our expression, we obtain:
$$\sim(p \wedge q) \vee (p \wedge r)$$

### Applying De Morgan's Law

Next, we address the negation acting on the conjunction. According to **De Morgan's Law**, $\sim(p \wedge q) \equiv \sim p \vee \sim q$.

Substituting this into our expression, we get:
$$(\sim p \vee \sim q) \vee (p \wedge r)$$

### Rearranging and Distributing

Since the OR operator ($\vee$) is both associative and commutative, we can rearrange the terms to facilitate simplification:
$$(\sim p \vee (p \wedge r)) \vee \sim q$$

Now, we focus on the bracketed term $(\sim p \vee (p \wedge r))$. We apply the **Distributive Law** to expand this:
$$((\sim p \vee p) \wedge (\sim p \vee r)) \vee \sim q$$

### Simplifying via Complement and Identity Laws

Observe the term $(\sim p \vee p)$. By the **Complement Law**, any statement ORed with its negation is always True ($T$):
$$(T \wedge (\sim p \vee r)) \vee \sim q$$

According to the **Identity Law**, $T \wedge X \equiv X$. Therefore, the $T$ term effectively vanishes:
$$(\sim p \vee r) \vee \sim q$$

### Reaching the Final Conclusion

We rearrange the terms one final time to group the negations:
$$(\sim p \vee \sim q) \vee r$$

Recognizing that $(\sim p \vee \sim q)$ is equivalent to $\sim(p \wedge q)$ via **De Morgan's Law**, we rewrite the expression:
$$\sim(p \wedge q) \vee r$$

Finally, applying the **Implication Law** in reverse, where $\sim A \vee B \equiv A \Rightarrow B$, we arrive at the simplified result:
$$(p \wedge q) \Rightarrow r$$

The equivalent form of the original statement is **$(p \wedge q) \Rightarrow r$**.

Animated Solution for Mathematics - Sets and Relations: The statement $(p \land q) \Rightarrow (p \land r)$ is equivalent to :

Select Answer:

Visualized Solution

Given Statement

Implication Law

Applying the Law

De Morgan's Law

Substituting De Morgan's Result

Rearranging Terms

Distributive Law

Complement Law

Identity Law

Grouping Terms

Reversing De Morgan's Law

Final Implication Form

Conclusion

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Applying De Morgan's Law

Rearranging and Distributing

Simplifying via Complement and Identity Laws

Reaching the Final Conclusion

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The statement (p∧q)⇒(p∧r) is equivalent to :