The logical statement [~(∼p ∨ q) ∨ (p ∧ r)] ∧ (∼q ∧ r) is equivalent to: (1) (p ∧ r) ∧ ∼q (2) (∼p ∧

Q: Why does the OR operator flip to an AND when we apply De Morgan's Law?

Think of it as a logical inversion. When you negate a statement like $p \lor q$, you are saying 'it is not the case that either $p$ or $q$ is true'. This implies that both $p$ must be false AND $q$ must be false. Hence, the OR flips to an AND. It is a fundamental rule of logic that ensures the truth values remain consistent.

Q: Can I just use a truth table for this? It feels safer.

You absolutely can, and it will give you the correct answer! However, in a high-pressure exam like JEE Advanced, truth tables are time-consuming and prone to simple calculation errors. Mastering these laws is like learning a shortcut—it is faster, more elegant, and builds your logical intuition for more complex problems.

Q: How did you know that $(\sim q \lor r) \land (\sim q \land r)$ simplifies to $(\sim q \land r)$?

This is a classic application of the Absorption Law. If you have a statement $X$ and a statement $Y$, and you know that $X$ implies $Y$, then $X \land Y$ is just $X$. Here, $(\sim q \land r)$ is a stronger, more specific condition than $(\sim q \lor r)$. Therefore, the intersection of the two is just the stronger condition.

Question

The logical statement $[\sim(\sim p \lor q) \lor (p \land r)] \land (\sim q \land r)$ is equivalent to:

Options:

(A) $(p \land r) \land \sim q$

(B) $(\sim p \land \sim q) \land r$

(C) $\sim p \lor r$

(D) $(p \land \sim q) \lor r$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to face a problem that looks like a tangled mess of symbols. You might see $[\sim(\sim p \lor q) \lor (p \land r)] \land (\sim q \land r)$ and feel a sudden urge to skip it.

But I want you to take a deep breath. In the world of JEE Advanced, intimidation is the first trap. Our goal is to dismantle this 'monster' piece by piece, using the elegant tools of mathematical logic.

### Phase 1: The De Morgan's Surgery

Let us start by focusing on the most complex-looking part: $\sim(\sim p \lor q)$. This is a classic setup for De Morgan's Law.

Remember, when you distribute a negation inside a bracket, the OR operator flips to an AND, and the individual terms are negated. So, the negation of $\sim p$ becomes $p$, the OR becomes an AND, and the negation of $q$ becomes $\sim q$.

Just like that, the term transforms into $p \land \sim q$. It is clean, it is simple, and it is the first step toward victory.

### Phase 2: The Distributive Dance

Now, let us substitute this back into our expression. Our Part A now looks like $[(p \land \sim q) \lor (p \land r)]$.

Do you see the beauty here? We have a common factor of $p \land$ on both sides of the OR operator. This is exactly like factoring out a variable in algebra.

By applying the Distributive Law in reverse, we can pull out the $p \land$ and group the remaining terms:
$$p \land (\sim q \lor r)$$
We have successfully tamed the first half of our expression.

### Phase 3: The Logical Collapse

Now, we bring back Part B: $(\sim q \land r)$. Our full expression is now $[p \land (\sim q \lor r)] \land (\sim q \land r)$.

Since all our operators are now ANDs, we can use the Associative Law to regroup the terms in whatever way makes our lives easier. Let us group the bracket from Part A with Part B:
$$p \land [(\sim q \lor r) \land (\sim q \land r)]$$

Look closely at the bracketed section: $(\sim q \lor r) \land (\sim q \land r)$. This is a beautiful logical identity.

If $(\sim q \land r)$ is true, then $(\sim q \lor r)$ must also be true. Because the first term is a subset of the second, their intersection is simply the first term itself. The entire bracket collapses into $(\sim q \land r)$.

### Conclusion: The Final Victory

We are at the finish line! We are left with $p \land (\sim q \land r)$.

Since all operators are ANDs, we can drop the brackets and rearrange the terms using the Associative and Commutative laws to get $(p \land r) \land \sim q$. This matches our first option perfectly.

You see? What looked like a monster was just a series of small, logical steps. Keep practicing these laws, and you will find that even the most intimidating problems have a simple, **elegant soul**.

Animated Solution for Mathematics - Sets and Relations: The logical statement $[\sim (\sim p \lor q) \lor (p \land r)] \land (\sim q \land r)$ is equivalent to:

Select Answer:

Visualized Solution

Analyze the Given Expression

Apply De Morgan's Law

Substitute back into Part A

Apply Distributive Law

Combine Part A and Part B

Simplify the Grouped Term

Final Simplification

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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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 logical statement [∼(∼p∨q)∨(p∧r)]∧(∼q∧r) is equivalent to: