I always get confused by $(p \Rightarrow q) \equiv (\sim p \vee q)$. Why is this true?

Think of it this way: an implication is only false when the premise ($p$) is true and the conclusion ($q$) is false. In all other cases, it is true. The expression $(\sim p \vee q)$ captures exactly that. If $p$ is false, the whole thing is true (because of $\sim p$). If $q$ is true, the whole thing is true. It is the exact same truth behavior!

JEE Main 2023 (12 April Shift 1)

LEVELBoard

Animated Solution for Mathematics - Sets and Relations: Among the two statements $(S 1) : (p \Rightarrow q) \land (p \land (\sim q))$ is a contradiction and $(S 2) : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))$ is a tautology

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Do not rush into truth tables immediately. Visualize the logic using a Venn diagram. For Statement 1, imagine two circles, p and q. The expression

(p \Rightarrow q) \land (p \land \sim q)

asks for the intersection of two specific regions. If those regions do not overlap, you have found a contradiction.

Hint 2

Hint 2: The Arsenal. You need two powerful tools from your logical toolkit. First, the Implication Law:

(p \Rightarrow q) \equiv (\sim p \lor q)

. Second, the Law of Excluded Middle:

(p \lor \sim p) \equiv t

(Tautology). Use these to simplify the expressions before you start drawing conclusions.

Hint 3

Hint 3: The Execution. For Statement 1, substitute the Implication Law into the expression and apply the Distributive Law:

A \land (B \lor C) \equiv (A \land B) \lor (A \land C)

. For Statement 2, group the terms in pairs and factor out the common variables to reveal the Tautology.

Visualized Solution

Introduction to Statement $S_{1}$

Given Statement $(S_{1}) : (p \Rightarrow q) \land (p \land \sim q)$
Objective: Verify if $(S_{1})$ is a Contradiction (always False).
We will use both algebraic laws and Venn diagrams for clarity.

Applying the Implication Law

Recall the Implication Law: $(p \Rightarrow q) \equiv (\sim p \lor q)$
Substitute this into $(S_{1})$ :
$(S_{1}) \equiv (\sim p \lor q) \land (p \land \sim q)$

Analyzing the Second Part of $S_{1}$

The second part of the statement is $(p \land \sim q)$ .
This represents the region strictly inside $p$ but outside $q$ .

Visualizing the Intersection

$(S_{1})$ asks for the intersection ( $\land$ ) of the two regions.
Blue region: $(\sim p \lor q)$
Red region: $(p \land \sim q)$
The intersection is empty $\to$ Contradiction.

Algebraic Proof using Distributive Law

Apply Distributive Law: $A \land (B \lor C) \equiv (A \land B) \lor (A \land C)$
Let $A = (p \land \sim q)$
$(S_{1}) \equiv (\sim p \land (p \land \sim q)) \lor (q \land (p \land \sim q))$

Identifying Internal Contradictions

Simplify the terms inside the brackets:
$(\sim p \land p) \equiv f$ (Falsehood)
$(q \land \sim q) \equiv f$ (Falsehood)
Substitute these values: $(S_{1}) \equiv (f \land \sim q) \lor (f \land p)$

Final Result for $S_{1}$

Since $(f \land anything) \equiv f$ :
$(S_{1}) \equiv f \lor f$
Final result: $(S_{1}) \equiv f$
Conclusion: $(S_{1})$ is a Contradiction. Statement 1 is True.

Introduction to Statement $S_{2}$

Given Statement $(S_{2}) : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))$
Objective: Verify if $(S_{2})$ is a Tautology (always True).

Grouping the First Half of $S_{2}$

Group the first two terms: $(p \land q) \lor ((\sim p) \land q)$
Visually: Intersection region (Green) $\cup$ Only $q$ region (Orange).
Factor out $q$ : $((p \lor \sim p) \land q)$

Grouping the Second Half of $S_{2}$

Group the last two terms: $(p \land \sim q) \lor ((\sim p) \land (\sim q))$
Visually: Only $p$ region (Purple) $\cup$ Outside both region (Yellow).
Factor out $\sim q$ : $((p \lor \sim p) \land \sim q)$

Applying the Law of Excluded Middle

Apply the Law of Excluded Middle: $(p \lor \sim p) \equiv t$ (Tautology)
Substitute $t$ into the expression:
$(S_{2}) \equiv (t \land q) \lor (t \land \sim q)$
Simplify using Identity Law: $(S_{2}) \equiv q \lor \sim q$

Final Result for $S_{2}$ and Conclusion

Apply Law of Excluded Middle again: $(q \lor \sim q) \equiv t$
Final result: $(S_{2}) \equiv t$ (Tautology)
Conclusion: $(S_{2})$ is a Tautology. Statement 2 is True.
Final Answer: Both are true (Option 4)

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The Sigma Insight: Types of Sets and Set Operations

Solution Diagram

Analyzing the Setup

We are tasked with evaluating the truth values of two logical statements,

(S_{1})

and

(S_{2})

. Our objective is to determine if

(S_{1})

is a contradiction and if

(S_{2})

is a tautology.

The Detective Work on Statement 1

We begin with the expression

(S_{1}) : (p \Rightarrow q) \land (p \land \sim q)

. Our goal is to determine if this expression is a contradiction, meaning it is always false.

The implication

(p \Rightarrow q)

is logically equivalent to

(\sim p \lor q)

. We can now rewrite the full expression as:

(S_{1}) \equiv (\sim p \lor q) \land (p \land \sim q)

Applying the Distributive Law, we distribute

(p \land \sim q)

over the bracket:

(S_{1}) \equiv (\sim p \land p \land \sim q) \lor (q \land p \land \sim q)

Observing the terms, we see that

(\sim p \land p)

is a contradiction (

f

), and

(q \land \sim q)

is also a contradiction (

f

). Thus, the expression simplifies to:

(S_{1}) \equiv (f \land \sim q) \lor (f \land p) \equiv f \lor f \equiv f

Since the result is always false,

(S_{1})

is indeed a contradiction. Therefore, the statement "

S_{1}

is a contradiction" is true.

The Tautology Hunt in Statement 2

Now, we examine

(S_{2}) : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))

. We group the terms to simplify the expression.

First, consider the first two terms:

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

. Factoring out

q

, we obtain:

(p \lor \sim p) \land q

By the Law of Excluded Middle,

(p \lor \sim p)

is always true (

t

). Thus, the expression simplifies to

(t \land q)

, which is simply

q

Next, consider the last two terms:

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

. Factoring out

\sim q

, we obtain:

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

Again,

(p \lor \sim p)

t

, so this simplifies to

(t \land \sim q)

, which is simply

\sim q

. Combining these results, we get:

(S_{2}) \equiv q \lor \sim q

By the Law of Excluded Middle,

q \lor \sim q

is always true (

t

). Therefore,

(S_{2})

is a tautology, and the statement "

S_{2}

is a tautology" is true.

Conclusion

Through rigorous algebraic manipulation and the application of logical laws, we have demonstrated that both statements are valid. We conclude that both $(S_{1})$ is a contradiction and $(S_{2})$ is a tautology are true statements.

Common Traps & FAQs

Wait, why do we reject the truth table method? Isn't it safer?

Truth tables are indeed safe, but they are the 'brute force' method of logic. In a high-stakes exam like JEE Advanced, time is your most precious resource. Learning to manipulate logical expressions algebraically or visualizing them via Venn diagrams is not just faster; it builds the mathematical intuition that separates a calculator from a mathematician. Use truth tables only as a last resort!

I always get confused by

(p \Rightarrow q) \equiv (\sim p \lor q)

. Why is this true?

Think of it this way: an implication is only false when the premise (

p

) is true and the conclusion (

q

) is false. In all other cases, it is true. The expression

(\sim p \lor q)

captures exactly that. If

p

is false, the whole thing is true (because of

\sim p

). If

q

is true, the whole thing is true. It is the exact same truth behavior!

How do I know when to use the Distributive Law versus just grouping terms?

Great question! Look at the structure. If you have a term like

(p \land \sim q)

being 'ANDed' with a bracket like

(\sim p \lor q)

, the Distributive Law is your best friend to break it open. If you have a long string of 'OR' operations, like in Statement 2, grouping and factoring is the most efficient way to simplify. Look for common factors, just like in standard algebra!

Animated Solution for Mathematics - Sets and Relations: Among the two statements $(S 1) : (p \Rightarrow q) \land (p \land (\sim q))$ is a contradiction and $(S 2) : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))$ is a tautology

Select Answer:

Visualized Solution

Introduction to Statement $S_{1}$

Applying the Implication Law

Analyzing the Second Part of $S_{1}$

Visualizing the Intersection

Algebraic Proof using Distributive Law

Identifying Internal Contradictions

Final Result for $S_{1}$

Introduction to Statement $S_{2}$

Grouping the First Half of $S_{2}$

Grouping the Second Half of $S_{2}$

Applying the Law of Excluded Middle

Final Result for $S_{2}$ and Conclusion

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Detective Work on Statement 1

The Tautology Hunt in Statement 2

Conclusion

Similar Questions

Among the statements: (S1) $(p \Rightarrow q) \lor (p \land (\sim q))$ is a tautology, (S2) $((\sim p) \Rightarrow (\sim q)) \land ((\sim p) \lor q)$ is a Contradiction. Then

Among the statements $(S 1) : (p \Rightarrow q) \lor ((\sim p) \land q)$ is a tautology, $(S 2) : (q \Rightarrow p) \Rightarrow ((\sim p) \land q)$ is a contradiction

Consider the two statements: (S1) : $(p \to q) \lor (\sim q \to p)$ is a tautology. (S2) : $(p \land \sim q) \land (\sim p \lor q)$ is a fallacy. Then :

Among the statements: (S1) $((p \lor q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$ , (S2) $((p \lor q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \lor (q \Rightarrow r))$

Consider Statement-1 : $(p \land \sim q) \land (\sim p \land q)$ is a fallacy. Statement-2 : $(p \to q) \leftrightarrow (\sim q \to\sim p)$ is a tautology.

Which of the following statement is a tautology?

Which of the following statements is a tautology ?

Which of the following statement is a tautology ?

Statement-1: $\sim (p \leftrightarrow\sim q)$ is equivalent to $p \leftrightarrow q$ . Statement-2: $\sim (p \leftrightarrow\sim q)$ is a tautology

The statement $(\sim (p \Leftrightarrow\sim q)) \land q$ is :

Select Answer:

Visualized Solution

Introduction to Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Applying the Implication Law

Analyzing the Second Part of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Visualizing the Intersection

Algebraic Proof using Distributive Law

Identifying Internal Contradictions

Final Result for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Introduction to Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Grouping the First Half of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Grouping the Second Half of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Applying the Law of Excluded Middle

Final Result for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​ and Conclusion

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Detective Work on Statement 1

The Tautology Hunt in Statement 2

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Among the statements: (S1) (p⇒q)∨(p∧(∼q)) is a tautology, (S2) ((∼p)⇒(∼q))∧((∼p)∨q) is a Contradiction. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Among the statements (S1):(p⇒q)∨((∼p)∧q) is a tautology, (S2):(q⇒p)⇒((∼p)∧q) is a contradiction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the two statements: (S1) : (p→q)∨(∼q→p) is a tautology. (S2) : (p∧∼q)∧(∼p∨q) is a fallacy. Then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Among the statements: (S1) ((p∨q)⇒r)⇔(p⇒r), (S2) ((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider Statement-1 : (p∧∼q)∧(∼p∧q) is a fallacy. Statement-2 : (p→q)↔(∼q→∼p) is a tautology.

Which of the following statement is a tautology?

Which of the following statements is a tautology ?

Which of the following statement is a tautology ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement-1: ∼(p↔∼q) is equivalent to p↔q. Statement-2: ∼(p↔∼q) is a tautology

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The statement (∼(p⇔∼q))∧q is :

Introduction to Statement $S_{1}$

Analyzing the Second Part of $S_{1}$

Final Result for $S_{1}$

Introduction to Statement $S_{2}$

Grouping the First Half of $S_{2}$

Grouping the Second Half of $S_{2}$

Final Result for $S_{2}$ and Conclusion

Among the statements: (S1) $(p \Rightarrow q) \lor (p \land (\sim q))$ is a tautology, (S2) $((\sim p) \Rightarrow (\sim q)) \land ((\sim p) \lor q)$ is a Contradiction. Then

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Among the statements: (S1) $((p \lor q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$ , (S2) $((p \lor q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \lor (q \Rightarrow r))$

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Statement-1: $\sim (p \leftrightarrow\sim q)$ is equivalent to $p \leftrightarrow q$ . Statement-2: $\sim (p \leftrightarrow\sim q)$ is a tautology

The statement $(\sim (p \Leftrightarrow\sim q)) \land q$ is :