Negation of the Boolean expression p ⇔ (q ⇒ p) is (A) (~p) ∧ q (B) p ∧ (~q) (C) (~p) ∨ (~q) (D) (~p)

Q: Wait, why can't we just use a truth table to solve this?

You absolutely can! Truth tables are a robust, foolproof method for any boolean expression. However, in competitive exams like the JEE, algebraic simplification is often faster and less prone to transcription errors. Mastering these logical identities builds your intuition, which is invaluable for more complex problems where truth tables become unwieldy.

Q: Why does $p \Leftrightarrow (\sim q \vee p)$ simplify to $p \vee q$?

Great question. When you expand the biconditional, you get two implications: $[p \Rightarrow (\sim q \vee p)] \wedge [(\sim q \vee p) \Rightarrow p]$. The first part, $p \Rightarrow (\sim q \vee p)$, is a tautology because if $p$ is true, the right side is automatically true. The second part, $(\sim q \vee p) \Rightarrow p$, simplifies to $p \vee q$. Since the whole expression is the AND of these two, it reduces to $p \vee q$.

Q: Is there a shortcut for negating a biconditional?

Yes! The negation of $A \Leftrightarrow B$ is $A \Leftrightarrow \sim B$ or $\sim A \Leftrightarrow B$. However, applying this directly to complex expressions can sometimes be trickier than simplifying first. When in doubt, simplify the expression to its minimal form before negating. It is the safest path to the correct answer.

Question

Negation of the Boolean expression $p \Leftrightarrow (q \Rightarrow p)$ is

Options:

(A) $(\sim p) \wedge q$

(B) $p \wedge (\sim q)$

(C) $(\sim p) \vee (\sim q)$

(D) $(\sim p) \wedge (\sim q)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the negation of the boolean expression $p \Leftrightarrow (q \Rightarrow p)$, we must systematically break down the logical operators. The goal is to simplify the expression into its most basic form before applying the negation.

### Phase 1: Deconstructing the Implication

Our first step is to simplify the inner bracket, $q \Rightarrow p$. Using the logical identity $A \Rightarrow B \equiv \sim A \vee B$, we transform the expression:

$$q \Rightarrow p \equiv \sim q \vee p$$

Substituting this back into the original expression, we obtain:

$$p \Leftrightarrow (\sim q \vee p)$$

### Phase 2: The Biconditional Bridge

Next, we address the biconditional operator. We utilize the definition $A \Leftrightarrow B \equiv (A \Rightarrow B) \wedge (B \Rightarrow A)$ to split the expression:

$$[p \Rightarrow (\sim q \vee p)] \wedge [(\sim q \vee p) \Rightarrow p]$$

This expansion allows us to analyze the two components independently.

### Phase 3: The Tautology Reveal

Let us evaluate the first part: $p \Rightarrow (\sim q \vee p)$. Applying the implication rule again:

$$\sim p \vee (\sim q \vee p) \equiv (\sim p \vee p) \vee \sim q$$

Since $\sim p \vee p$ is a tautology ($T$), the expression becomes $T \vee \sim q$, which simplifies to **True**.

Now, we evaluate the second part: $(\sim q \vee p) \Rightarrow p$. Applying the implication rule:

$$\sim(\sim q \vee p) \vee p$$

Using De Morgan's Law, this becomes $(q \wedge \sim p) \vee p$. Applying the distributive law, we get:

$$(q \vee p) \wedge (\sim p \vee p) \equiv (q \vee p) \wedge T \equiv p \vee q$$

### Final Calculation

Combining the two parts, the original expression simplifies to $T \wedge (p \vee q)$, which is equivalent to $p \vee q$.

Our final task is to find the negation of this result. By applying De Morgan's Law to $p \vee q$, we arrive at the final answer:

$$\sim (p \vee q) \equiv \mathbf{\sim p \wedge \sim q}$$

Animated Solution for Mathematics - Sets and Relations: Negation of the Boolean expression $p \Leftrightarrow (q \Rightarrow p)$ is

Select Answer:

Visualized Solution

Analyze the Expression

Simplify $q \Rightarrow p$

Substitute into Biconditional

Expand the Expression

Simplify First Bracket (Part 1)

Simplify First Bracket (Part 2)

Simplify Second Bracket (Part 1)

Simplify Second Bracket (Part 2)

Simplify Second Bracket (Part 3)

Combine Results

Apply Negation

Conclusion

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

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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; } Negation of the Boolean expression p⇔(q⇒p) is