Let *, □ ∈ {∧, ∨} be such that the Boolean expression (p * ∼ q) ⇒ (p □ q) is a tautology. Then :
(1)

Q: Wait, why do we reject the negative root here? Or rather, why can't we just use Boolean algebra laws like De Morgan's to simplify this?

That is a fantastic question! You absolutely *can* use Boolean algebra laws. You could simplify $(p * \sim q) \Rightarrow (p \square q)$ into $\sim(p * \sim q) \vee (p \square q)$ and check if it simplifies to a constant 'True'. However, in the heat of a JEE exam, truth tables are often less prone to algebraic slips. They provide a visual, systematic 'proof' that is harder to mess up under pressure.

Q: Is there a shortcut to avoid checking all 4 rows for every single option?

Yes! Focus on the 'killer' condition: $A \Rightarrow B$ is False only when $A$ is True and $B$ is False. Instead of filling the whole table, just look for a scenario where the antecedent $(p * \sim q)$ is True. If you find one, immediately check if the consequent $(p \square q)$ is also True. If it is, you are safe for that row. If the consequent is False, you have found a counter-example, and you can immediately discard that option.

Q: What if I get a False in the final column? Does that mean I made a calculation error?

Not necessarily! If you get a False in the final column, it simply means that the specific combination of operators you tested does not form a tautology. It doesn't mean your math is wrong; it means your hypothesis (the choice of operators) was incorrect. Simply move to the next option with confidence. In JEE, elimination is just as powerful as direct calculation.

Question

Let $*, \square \in \{\wedge, \vee\}$ be such that the Boolean expression $(p * \sim q) \Rightarrow (p \square q)$ is a tautology. Then :

Options:

(A) $* = \vee, \square = \vee$

(B) $* = \wedge, \square = \wedge$

(C) $* = \wedge, \square = \vee$

(D) $* = \vee, \square = \wedge$

Sigma Percentile · Accepted Answer

### The Architecture of Truth: Mastering Logical Tautologies

Welcome, future engineers! Today, we are not just solving a problem; we are exploring the very foundation of mathematical reasoning. In the world of JEE Advanced, logic is not just about symbols; it is about the structure of truth itself.

We are tasked with finding the hidden operators in the expression $(p * \sim q) \Rightarrow (p \square q)$ such that it becomes a tautology. Let us embark on this journey.

### Phase 1: The Golden Rule of Implication

Before we touch the operators, we must respect the 'Implication' operator, denoted by $\Rightarrow$. Many students stumble here. They treat it like an algebraic equation, but it is a logical flow.

The statement $A \Rightarrow B$ is a promise. It says, 'If $A$ happens, then $B$ must happen.' The only way to break this promise—to make the statement False—is to have $A$ occur (True) while $B$ fails to occur (False).

If $A$ is False, the promise is vacuously true. Keep this in your heart: the only 'danger zone' is $T \Rightarrow F$. If we can avoid this, we have a tautology.

### Phase 2: The Map of Possibilities

Since we have two variables, $p$ and $q$, we have $2^2 = 4$ possible worlds. We must map these out. We list them as $(T, T), (T, F), (F, T),$ and $(F, F)$.

We also need the negation of $q$, which is $\sim q$. This is simply the inversion of $q$. If $q$ is True, $\sim q$ is False. If $q$ is False, $\sim q$ is True.

It is a simple flip, but do not rush it. A single sign error here cascades into the final answer.

### Phase 3: Testing the Hypothesis

We test the hypothesis where $* = \wedge$ and $\square = \vee$. Our expression becomes $(p \wedge \sim q) \Rightarrow (p \vee q)$.

Let us evaluate the antecedent $(p \wedge \sim q)$. The 'AND' operator is strict; it demands both inputs be True. Scanning our four rows, we find this only happens when $p$ is True and $\sim q$ is True (which means $q$ is False).

So, the antecedent values are: False, True, False, False.

Now, we evaluate the consequent $(p \vee q)$. The 'OR' operator is generous; it only fails if both are False. Looking at our table, this only happens in the final row where both $p$ and $q$ are False.

Thus, the consequent values are: True, True, True, False.

### The Grand Finale

Now, we check the implication $(p \wedge \sim q) \Rightarrow (p \vee q)$. We look for the 'danger zone' ($T \Rightarrow F$):

- Row 1: $F \Rightarrow T$ (Safe)
- Row 2: $T \Rightarrow T$ (Safe)
- Row 3: $F \Rightarrow T$ (Safe)
- Row 4: $F \Rightarrow F$ (Safe)

Every single row results in True! We have successfully navigated the logic and proven that for these operators, the expression is a **tautology**.

You see, logic is not about memorizing tables; it is about understanding the flow of truth. Keep this clarity, and no logical problem will ever stand in your way.

Animated Solution for Mathematics - Sets and Relations: Let $, □ \in {\land, \lor}$ be such that the Boolean expression $(p \sim q) \Rightarrow (p □ q)$ is a tautology. Then :

Select Answer:

Visualized Solution

Understanding the Problem

The Implication Rule

Setting Up the Truth Table

Calculating $\sim q$

Testing Option 3: $* = \land, □ = \lor$

Evaluating the Antecedent: $p \land \sim q$

Evaluating the Consequent: $p \lor q$

The Final Implication

Conclusion: A Tautology

The Sigma Insight: Types of Sets and Set Operations

The Architecture of Truth

Phase 1

Phase 2

Phase 3

The Grand Finale

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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; } Let ∗,□∈{∧,∨} be such that the Boolean expression (p∗∼q)⇒(p□q) is a tautology. Then :