Statement (P ⇒ Q) ∧ (R ⇒ Q) is logically equivalent to (1) (P ⇒ R) ∨ (Q ⇒ R) (2) (P ∧ R) ⇒ Q (3) (P

Q: Why can't we just use a truth table to solve this?

Truth tables are a perfectly valid and reliable method in logic! However, in the context of JEE Advanced, time is your most precious resource. Algebraic manipulation using logical laws like the Implication Rule and Distributive Law is significantly faster and demonstrates a deeper mastery of the underlying structure of logic.

Q: Is it always valid to factor out a term like that in logic?

Yes, absolutely! The Distributive Law in logic, $(X \wedge Y) \vee Z \equiv (X \vee Z) \wedge (Y \vee Z)$, works exactly like factoring in algebra. When you see a common term like $\vee Q$ in both brackets, you can treat it as a common factor and pull it out, just as you would with $ax + bx = (a+b)x$.

Q: Why does $\sim P \wedge \sim R$ become $\sim(P \vee R)$?

This is the essence of De Morgan's Law. Think of it this way: if it is not the case that $P$ is true, AND it is not the case that $R$ is true, then it must be that neither $P$ nor $R$ is true. Therefore, the negation of the disjunction $(P \vee R)$ is equivalent to the conjunction of the negations $(\sim P \wedge \sim R)$.

Question

Statement $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$ is logically equivalent to

Options:

(A) $(P \Rightarrow R) \vee (Q \Rightarrow R)$

(B) $(P \wedge R) \Rightarrow Q$

(C) $(P \Rightarrow R) \wedge (Q \Rightarrow R)$

(D) $(P \vee R) \Rightarrow Q$

Sigma Percentile · Accepted Answer

### Analyzing the Logical Structure

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a problem; we are peeling back the layers of logical structure to reveal a beautiful, hidden symmetry.

We are given the statement $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$. At first glance, it might look like a dry collection of symbols, but let us look closer.

We have two separate paths, $P$ and $R$, both leading to the same destination, $Q$. In the world of logic, this is a powerful setup. Our goal is to find an equivalent expression by translating these implications into a language that is easier to manipulate.

### Phase 1: The Implication Rule

The first step in our journey is to simplify the conditional statements. We use the fundamental identity:

$$P \Rightarrow Q \equiv \sim P \vee Q$$

This rule is our Swiss Army knife in logic. It allows us to break down the rigid structure of an implication and turn it into a flexible disjunction.

By applying this to both parts of our given statement, we transform $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$ into:

$$(\sim P \vee Q) \wedge (\sim R \vee Q)$$

Suddenly, the problem feels less like a rigid condition and more like a set of choices.

### Phase 2: The Art of Factoring

Now, look at the expression we have created: $(\sim P \vee Q) \wedge (\sim R \vee Q)$. Do you see the commonality? Both brackets end with $\vee Q$.

In algebra, if you had $(x + q)(y + q)$, you might be tempted to expand it. But here, we want to simplify. We can use the **Distributive Law** in reverse.

Just as we can factor out a common term in an algebraic equation, we can factor out the $\vee Q$ here. When we pull $\vee Q$ out, we are left with:

$$(\sim P \wedge \sim R) \vee Q$$

It is elegant, isn't it? We have successfully consolidated our logical structure.

### Phase 3: De Morgan's Law and the Final Synthesis

We are almost at the finish line. We have the term $(\sim P \wedge \sim R) \vee Q$.

The part inside the bracket, $\sim P \wedge \sim R$, is a classic application of **De Morgan's Law**. It tells us that the conjunction of negations is the same as the negation of a disjunction.

So, $\sim P \wedge \sim R$ becomes $\sim(P \vee R)$. Now, substitute this back into our expression:

$$\sim(P \vee R) \vee Q$$

### The Final Revelation

Look at what we have: $\sim(P \vee R) \vee Q$. This is the exact inverse of the implication rule we started with!

If $\sim X \vee Y$ is equivalent to $X \Rightarrow Y$, then by letting $X = (P \vee R)$ and $Y = Q$, we arrive at our final, simplified form:

$$(P \vee R) \Rightarrow Q$$

We have taken a complex compound statement and distilled it down to its core essence. It is not just about getting the right answer; it is about seeing the underlying harmony of the logical operators. Keep practicing this, and soon, you will see these patterns everywhere!

Animated Solution for Mathematics - Sets and Relations: Statement $(P \Rightarrow Q) \land (R \Rightarrow Q)$ is logically equivalent to

Select Answer:

Visualized Solution

Analyzing the Statement

The Implication Rule

Applying the Implication Rule

Identifying the Common Term

The Distributive Law

De Morgan's Law

Reversing the Implication Rule

Matching with Options

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Logical Structure

Phase 1

Phase 2

Phase 3

The Final Revelation

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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; } Statement (P⇒Q)∧(R⇒Q) is logically equivalent to