Consider the following statements : A : Rishi is a judge. B : Rishi is honest. C : Rishi is not arro

Q: Wait, why can't I just negate the whole sentence by adding 'not' to the beginning?

That is a very common trap! In natural language, we often think negating a sentence means just saying 'It is not the case that...'. However, in formal logic, the negation of an implication $P \rightarrow Q$ has a specific structure. Think of it this way: the only time an 'If P, then Q' statement is false is when the condition $P$ is true, but the result $Q$ is false. Hence, the negation is $P \wedge \sim Q$.

Q: Is the order of $A$ and $C$ important in $(A \wedge C)$?

Great question! No, the order does not matter because the logical AND operator ($\wedge$) is commutative. This means $(A \wedge C)$ is logically equivalent to $(C \wedge A)$. You can arrange them in any order that helps you match the given options.

Q: Why is the negation of an implication $P \wedge \sim Q$ and not $\sim P \rightarrow \sim Q$?

That is a brilliant conceptual doubt. $\sim P \rightarrow \sim Q$ is actually the *inverse* of the original statement, not its negation. The negation must be a statement that is true whenever the original is false. If you use a truth table, you will see that $P \wedge \sim Q$ is the only expression that has the exact opposite truth values of $P \rightarrow Q$.

Question

Consider the following statements : \ $A$ : Rishi is a judge. \ $B$ : Rishi is honest. \ $C$ : Rishi is not arrogant. \ The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

Options:

(A) $B ightarrow (A \vee C)$

(B) $(\sim B) \wedge (A \wedge C)$

(C) $B ightarrow ((\sim A) \vee (\sim C))$

(D) $B ightarrow (A \wedge C)$

Sigma Percentile · Accepted Answer

### The Architecture of Logic: Unraveling the Negation

Logic is the silent, invisible skeleton upon which all of mathematics is built. It is the art of precise thinking.

Today, we are going to dissect a classic logical puzzle that often trips up students in the JEE Advanced. We are asked to find the negation of the statement: 'If Rishi is a judge and he is not arrogant, then he is honest.'

### Phase 1: The Building Blocks

Before we dive into the complex machinery of negation, we must simplify. Imagine you are an architect; you cannot build a skyscraper without first defining your materials.

Here, our materials are simple, atomic statements. Let us define them clearly:

- $A$: Rishi is a judge.
- $B$: Rishi is honest.
- $C$: Rishi is not arrogant.

By assigning these variables, we strip away the linguistic fluff and get to the core of the problem. We have three distinct facts, and now we can see how they interact.

### Phase 2: The Implication

Now, let us assemble these blocks. The statement says, 'If Rishi is a judge AND he is not arrogant, then he is honest.'

Look at the structure. We have a condition: 'Rishi is a judge AND he is not arrogant.' We represent this as $(A \wedge C)$.

This is our antecedent, or the 'If' part. Then, we have the conclusion: 'he is honest,' which is $B$. The entire sentence is an implication, which we write as:

$$ (A \wedge C) ightarrow B $$

This arrow, $ightarrow$, is the heart of the implication. It tells us that whenever the condition $(A \wedge C)$ is true, the conclusion $B$ must follow.

### Phase 3: The Negation Trap

Here is where the thrill begins. The question asks for the negation of this entire statement. We are looking for the logical opposite, which we write as:

$$ \sim ((A \wedge C) ightarrow B) $$

Many students instinctively try to distribute the negation symbol, perhaps writing $\sim(A \wedge C) ightarrow \sim B$. **Stop!** This is the most common trap in JEE logic.

You cannot simply distribute a negation over an implication. Instead, we must rely on a fundamental logical identity. The negation of an implication $P ightarrow Q$ is defined as:

$$ \sim (P ightarrow Q) \equiv P \wedge (\sim Q) $$

Think of this identity as a promise. The original statement $P ightarrow Q$ promises that if $P$ happens, $Q$ will follow. The only way to break that promise—to negate it—is for $P$ to happen, but for $Q$ to fail.

### Phase 4: The Final Assembly

Now, let us apply this identity to our specific problem. We set $P = (A \wedge C)$ and $Q = B$. Substituting these into our identity, we get:

$$ (A \wedge C) \wedge (\sim B) $$

We are almost there! We look at our options, and we might not see this exact order. But remember, the logical AND operator is commutative.

Just like $2 	imes 3 = 3 	imes 2$, the statement $X \wedge Y$ is the same as $Y \wedge X$. We can rearrange our expression to match the options provided:

$$ (\sim B) \wedge (A \wedge C) $$

And there it is. We have successfully navigated the logical maze. By breaking the sentence into atomic statements, identifying the implication, applying the correct negation identity, and using the commutative property, we have arrived at the solution.

**Logic is not just about memorizing rules; it is about understanding the structure of truth itself.** Keep this clarity with you, and no logical puzzle will ever be able to stand in your way.

Animated Solution for Mathematics - Basic Mathematics: Consider the following statements : \\ $A$ : Rishi is a judge. \\ $B$ : Rishi is honest. \\ $C$ : Rishi is not arrogant. \\ The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

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Visualized Solution

Identify Statement $A$

Identify Statement $C$

Identify Statement $B$

Build the Antecedent $(A \land C)$

Form the Implication $(A \land C) \to B$

Apply the Negation Operator

Recall the Logical Identity

Substitute $P$ and $Q$

Rearrange the Expression

Match with Options

The Sigma Insight: Theory of Indices

The Architecture of Logic

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Identify Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Identify Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Identify Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Build the Antecedent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A∧C)

Form the Implication .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A∧C)→B