JEE MainMathematics

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Testing Propositional Formulas for Tautology and Contradiction

Q: How do we instantly see that S1 is a tautology without writing all 4 rows?

Recall the negation of implication: ~(p ⇒ q) ≡ p wedge ~ q. Letting A = p ⇒ q, the expression is simply A vee ~ A, which is identically True.

(S1) (p ⇒ q) vee (p wedge (~ q)) is a tautology (S2) ((~ p) ⇒ (~ q)) wedge ((~ p) vee q) is a contradiction. Then

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What feels right?

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Aboth (S1) and (S2) are correctBonly (S1) is correctConly (S2) is correctDboth (S1) and (S2) are wrong

Step-by-step solutionView

Correct answer

Statement (S1) simplifies to $A \lor \sim A$ , making it a tautology, while (S2) simplifies to $q \lor \sim p$ , which is a contingency rather than a contradiction; thus only (S1) is correct.

Option analysis

Why each option works or fails

A · both (S1) and (S2) are correct

Believing that (S2) simplifies to false (a contradiction) because the implication antecedent $\sim p$ cancels out or contradicts the disjunction $\sim p \lor q$ . Express $((\sim p) \Rightarrow (\sim q))$ as $(p \lor \sim q)$ and $((\sim p) \lor q)$ as $(p \Rightarrow q)$ , or test truth values: when $p$ and $q$ are both false, (S2) evaluates to true, so it cannot be a contradiction.

B · only (S1) is correct

The student correctly applies propositional equivalences to identify (S1) as a tautology and (S2) as a contingency. Correct: $\sim (p \Rightarrow q) \equiv p \land \sim q$ , so (S1) has the form $A \lor \sim A \equiv T$ . For (S2), when $p = F$ and $q = F$ , $(T \Rightarrow T) \land (T \lor F) \equiv T \land T \equiv T$ , so it is not a contradiction.

C · only (S2) is correct

Failing to recognize that $p \land \sim q$ is the exact negation of $p \Rightarrow q$ , mistaking (S1) for a contingent statement while miscalculating (S2) as identically false. Recall the negation of conditional statement: $\sim (p \Rightarrow q) \equiv p \land \sim q$ . Therefore, $(p \Rightarrow q) \lor (p \land \sim q)$ is always true.

D · both (S1) and (S2) are wrong

Misapplying De Morgan's laws or conditional negation, incorrectly deducing that (S1) does not cover all truth assignments. Check truth values systematically or use algebraic duality: an expression of the form $X \lor \sim X$ is unconditionally a tautology by the law of excluded middle.

Reviewed route

Solution

StepWorking

01Given

Two statements: (S1) $(p \Rightarrow q) \lor (p \land \sim q)$ (S2) $((\sim p) \Rightarrow (\sim q)) \land ((\sim p) \lor q)$

02Goal

Determine whether S1 is a tautology (always True) and whether S2 is a contradiction (always False).

03Approach

Evaluate truth tables for both statements across all 4 combinations of $(p, q) \in {T, F}^{2}$ . Alternatively, use basic Boolean algebra properties.

04Execute

For statement S1: note that $p \land \sim q \equiv \sim (p \Rightarrow q)$ . Thus the expression is $(p \Rightarrow q) \lor \sim (p \Rightarrow q)$ . By the Law of Excluded Middle ( $A \lor \sim A \equiv T$ ), this statement evaluates to $T$ for all possible truth values of $p$ and $q$ . Therefore, S1 is a tautology. S1 is correct.

05Execute

For statement S2: let $p = T$ and $q = T$ . Then $\sim p = F$ and $\sim q = F$ . The first conjunct is $(\sim p \Rightarrow \sim q) = (F \Rightarrow F) = T$ . The second conjunct is $(\sim p \lor q) = (F \lor T) = T$ . Thus, S2 evaluates to $T \land T = T$ . Since S2 can be True, it is not a contradiction (a contradiction must be False for all assignments). Hence, S2 is incorrect.

✓Verify

Check S2 with truth table: when $(p, q) = (F, F)$ , $\sim p = T, \sim q = T ⟹ (T \Rightarrow T) \land (T \lor F) = T \land T = T$ . S2 has truth values $(T, F, F, T)$ , confirming it is a contingency, not a contradiction. Hence, only (S1) is correct.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

How do we instantly see that S1 is a tautology without writing all 4 rows?

Recall the negation of implication: $\sim (p \Rightarrow q) \equiv p \land \sim q$ . Letting $A = p \Rightarrow q$ , the expression is simply $A \lor \sim A$ , which is identically True.

Answer

Statement (S1) simplifies to $A \lor \sim A$ , making it a tautology, while (S2) simplifies to $q \lor \sim p$ , which is a contingency rather than a contradiction; thus only (S1) is correct.

Why each option works or fails

A: both (S1) and (S2) are correct - Believing that (S2) simplifies to false (a contradiction) because the implication antecedent $\sim p$ cancels out or contradicts the disjunction $\sim p \lor q$ . Express $((\sim p) \Rightarrow (\sim q))$ as $(p \lor \sim q)$ and $((\sim p) \lor q)$ as $(p \Rightarrow q)$ , or test truth values: when $p$ and $q$ are both false, (S2) evaluates to true, so it cannot be a contradiction.
B · correct: only (S1) is correct - The student correctly applies propositional equivalences to identify (S1) as a tautology and (S2) as a contingency. Correct: $\sim (p \Rightarrow q) \equiv p \land \sim q$ , so (S1) has the form $A \lor \sim A \equiv T$ . For (S2), when $p = F$ and $q = F$ , $(T \Rightarrow T) \land (T \lor F) \equiv T \land T \equiv T$ , so it is not a contradiction.
C: only (S2) is correct - Failing to recognize that $p \land \sim q$ is the exact negation of $p \Rightarrow q$ , mistaking (S1) for a contingent statement while miscalculating (S2) as identically false. Recall the negation of conditional statement: $\sim (p \Rightarrow q) \equiv p \land \sim q$ . Therefore, $(p \Rightarrow q) \lor (p \land \sim q)$ is always true.
D: both (S1) and (S2) are wrong - Misapplying De Morgan's laws or conditional negation, incorrectly deducing that (S1) does not cover all truth assignments. Check truth values systematically or use algebraic duality: an expression of the form $X \lor \sim X$ is unconditionally a tautology by the law of excluded middle.

Step-by-step solution

Given: Two statements: (S1) $(p \Rightarrow q) \lor (p \land \sim q)$ (S2) $((\sim p) \Rightarrow (\sim q)) \land ((\sim p) \lor q)$
Goal: Determine whether S1 is a tautology (always True) and whether S2 is a contradiction (always False).
Approach: Evaluate truth tables for both statements across all 4 combinations of $(p, q) \in {T, F}^{2}$ . Alternatively, use basic Boolean algebra properties.
Execute: For statement S1: note that $p \land \sim q \equiv \sim (p \Rightarrow q)$ . Thus the expression is $(p \Rightarrow q) \lor \sim (p \Rightarrow q)$ . By the Law of Excluded Middle ( $A \lor \sim A \equiv T$ ), this statement evaluates to $T$ for all possible truth values of $p$ and $q$ . Therefore, S1 is a tautology. S1 is correct.
Execute: For statement S2: let $p = T$ and $q = T$ . Then $\sim p = F$ and $\sim q = F$ . The first conjunct is $(\sim p \Rightarrow \sim q) = (F \Rightarrow F) = T$ . The second conjunct is $(\sim p \lor q) = (F \lor T) = T$ . Thus, S2 evaluates to $T \land T = T$ . Since S2 can be True, it is not a contradiction (a contradiction must be False for all assignments). Hence, S2 is incorrect.
Verify: Check S2 with truth table: when $(p, q) = (F, F)$ , $\sim p = T, \sim q = T ⟹ (T \Rightarrow T) \land (T \lor F) = T \land T = T$ . S2 has truth values $(T, F, F, T)$ , confirming it is a contingency, not a contradiction. Hence, only (S1) is correct.

Shortcut:

When to use it: Fastest when familiar with standard equivalences: $\sim (p \Rightarrow q) \equiv p \land \sim q$ and contrapositive laws.

Given: Expressions $(p \Rightarrow q) \lor (p \land \sim q)$ and $((\sim p) \Rightarrow (\sim q)) \land ((\sim p) \lor q)$ .

Goal: Verify if S1 is a tautology and S2 is a contradiction using propositional algebra.

Approach: Apply standard logical identities: $A \Rightarrow B \equiv \sim A \lor B$ and $\sim (A \Rightarrow B) \equiv A \land \sim B$ .

Execute: S1 is of the form $X \lor \sim X$ where $X = (p \Rightarrow q)$ , which simplifies directly to $T$ (tautology). For S2, $(\sim p \Rightarrow \sim q) \equiv p \lor \sim q$ . Thus S2 is $(p \lor \sim q) \land (\sim p \lor q) \equiv (p \Leftrightarrow q)$ . The biconditional $p \Leftrightarrow q$ is true when $p$ and $q$ have the same truth value, so it is neither a tautology nor a contradiction.

Verify: Since S1 is always $T$ , S1 is correct. Since S2 is $p \Leftrightarrow q$ , it is True for $(T, T)$ and $(F, F)$ , so it is not a contradiction. Hence, S2 is wrong. Only (S1) is correct.

Testing Propositional Formulas for Tautology and Contradiction

What feels right?

Statement (S1) simplifies to $A \lor \sim A$ , making it a tautology, while (S2) simplifies to $q \lor \sim p$ , which is a contingency rather than a contradiction; thus only (S1) is correct.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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