The negation of the statement ~ p ∧ (p ∨ q) is : (1) ~ p ∨ q (2) p ∧ q (3) ~ p ∨ q (4) p ∨ ~ q

Q: Wait, why does the 'AND' operator change to an 'OR' operator when we apply the negation?

This is the essence of De Morgan's Law! Think of it this way: if you want to negate the condition 'I am hungry AND I am tired', the opposite is 'I am NOT hungry OR I am NOT tired'. If either one of those conditions is false, the original 'AND' statement is false. It is a fundamental rule of logic that negating a conjunction turns it into a disjunction.

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

You absolutely can! Truth tables are a fantastic, foolproof way to verify logical equivalences. However, in the JEE Advanced exam, time is your most precious resource. Algebraic manipulation using logical laws (like De Morgan's, Distributive, and Complement laws) is significantly faster and helps you develop a deeper intuition for the structure of logical statements.

Q: Why does $p \lor \sim p$ simplify to $T$ (Tautology)?

That is the Complement Law. Consider the statement 'It is raining OR it is not raining'. This statement is always true, regardless of the weather. In logic, a statement combined with its own negation using an 'OR' operator covers all possibilities, making it a Tautology, which we represent as $T$.

Question

The negation of the statement $~ p \land (p \lor q)$ is :

Options:

(A) ~ p \lor q

(B) p \land q

(C) ~ p \lor q

(D) p \lor \sim q

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the elegant world of Boolean Algebra. Today, we are going to dissect a problem that might seem like a simple exercise in symbols, but is actually a beautiful dance of logical laws.

We are tasked with finding the negation of the statement $\sim p \land (p \lor q)$. Our objective is to find the negation of the entire expression, which is represented as:

$$\sim [\sim p \land (p \lor q)]$$

### Applying De Morgan's Law

To find the complement of this logical region, we must apply the negation operator to the entire expression. We utilize **De Morgan's Law**, which states that $\sim (A \land B) \equiv \sim A \lor \sim B$.

Applying this to our outer bracket, the 'AND' operator flips to an 'OR', and the negation distributes:

$$\sim (\sim p) \lor \sim (p \lor q)$$

### Simplifying the Inner Terms

Now, let's simplify the two terms we have created. The first term is $\sim (\sim p)$, which simplifies to $p$ via the **Double Negation Law**.

For the second term, $\sim (p \lor q)$, we apply **De Morgan's Law** again for an 'OR' statement, where $\sim (A \lor B) \equiv \sim A \land \sim B$. This transforms the expression into:

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

### The Distributive Dance

We must now resolve the brackets using the **Distributive Law**, which states $A \lor (B \land C) \equiv (A \lor B) \land (A \lor C)$. Distributing the $p \lor$ across the terms inside the parenthesis yields:

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

### The Final Reveal

Look closely at the first bracket: $(p \lor \sim p)$. According to the **Complement Law**, a statement 'OR' its negation is always true, resulting in a Tautology ($T$):

$$T \land (p \lor \sim q)$$

Finally, we invoke the **Identity Law**, which states that the conjunction of a Tautology with any statement $X$ is simply $X$ ($T \land X \equiv X$). Therefore, the $T$ drops out, leaving us with our final, elegant result:

**$p \lor \sim q$**

Animated Solution for Mathematics - Sets and Relations: The negation of the statement $p \land (p \lor q)$ is :

Select Answer:

Visualized Solution

Identify the Statement $\sim p \land (p \lor q)$

Apply the Negation Operator

De Morgan's Law (Outer)

Double Negation Property

De Morgan's Law (Inner)

Distributive Law (Setup)

Distributive Law (Execution)

Complement Law (Tautology)

Identity Law (Simplification)

Final Result and Summary

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Applying De Morgan's Law

Simplifying the Inner Terms

The Distributive Dance

The Final Reveal

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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 negation of the statement p∧(p∨q) is :