Why does become an AND (conjunction) instead of OR?

Because . By De Morgan's law, . Distributing gives .

JEE MainMathematics

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Mathematical Reasoning: Mathematics | JEE Main

Among the statements : (S1)

((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow(\mathrm{p} \Rightarrow \mathrm{r})

(S2)

((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))

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Aonly (S2) is a tautologyBonly (S1) is a tautologyCneither (S1) nor (S2) is a tautologyDboth (S1) and (S2) are tautologies

Step-by-step solutionView

Correct answer

Neither (S1) nor (S2) is a tautology because $(p \vee q) \Rightarrow r$ is logically equivalent to $(p \Rightarrow r) \wedge (q \Rightarrow r)$ , not $(p \Rightarrow r)$ alone or $(p \Rightarrow r) \vee (q \Rightarrow r)$ .

Option analysis

Why each option works or fails

A · only (S2) is a tautology

The student confuses the distribution of negation in De Morgan's laws or incorrectly distributes disjunction across implications, thinking $(p \vee q) \Rightarrow r$ distributes into a disjunction $\vee$ instead of a conjunction $\wedge$ . Remember that $(p \vee q) \Rightarrow r \equiv \neg(p \vee q) \vee r \equiv (\neg p \wedge \neg q) \vee r$ , which distributes to a conjunction of implications: $(p \Rightarrow r) \wedge (q \Rightarrow r)$ .

B · only (S1) is a tautology

The student assumes that $q$ has no effect on the validity of $(p \vee q) \Rightarrow r$ and reduces the premise prematurely to $p \Rightarrow r$ . Check truth assignments: if $p$ is false, $q$ is true, and $r$ is false, then $p \vee q \Rightarrow r$ is false while $p \Rightarrow r$ is true, disproving equivalence.

C · neither (S1) nor (S2) is a tautology

None. The student correctly applies logical equivalences or identifies counterexamples for both biconditionals. Correct: $(p \vee q) \Rightarrow r \equiv \neg(p \vee q) \vee r \equiv (\neg p \wedge \neg q) \vee r \equiv (\neg p \vee r) \wedge (\neg q \vee r) \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$ , making neither statement a tautology.

D · both (S1) and (S2) are tautologies

The student believes both reductions are valid logical simplifications, failing to apply De Morgan's laws correctly to the antecedent. Negating the antecedent $(p \vee q)$ yields $\neg p \wedge \neg q$ ; distributing $\vee r$ gives an 'AND' between the implications, so neither (S1) nor (S2) is equivalent.

Reviewed route

Solution

StepWorking

01given

Two statements: (S1) $((p \vee q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$ (S2) $((p \vee q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \vee (q \Rightarrow r))$

02goal

Determine whether (S1) and/or (S2) are tautologies.

03approach

Recall the standard equivalence: $(p \vee q) \Rightarrow r \equiv (\sim(p \vee q) \vee r) \equiv ((\sim p \wedge \sim q) \vee r) \equiv (\sim p \vee r) \wedge (\sim q \vee r) \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$ . Check if (S1) or (S2) match this, or find counterexamples (truth assignments giving False).

04execute

For (S1): We test the case $p = \mathrm{F}, q = \mathrm{T}, r = \mathrm{F}$ . Then $p \vee q = \mathrm{F} \vee \mathrm{T} = \mathrm{T}$ . Left side: $(p \vee q) \Rightarrow r \equiv \mathrm{T} \Rightarrow \mathrm{F} \equiv \mathrm{F}$ . Right side: $p \Rightarrow r \equiv \mathrm{F} \Rightarrow \mathrm{F} \equiv \mathrm{T}$ . The biconditional $\mathrm{F} \Leftrightarrow \mathrm{T}$ evaluates to $\mathrm{F}$ . Therefore, (S1) is not a tautology.

05execute

For (S2): Test the assignment $p = \mathrm{T}, q = \mathrm{F}, r = \mathrm{F}$ . Then $p \vee q = \mathrm{T} \vee \mathrm{F} = \mathrm{T}$ . Left side: $(p \vee q) \Rightarrow r \equiv \mathrm{T} \Rightarrow \mathrm{F} \equiv \mathrm{F}$ . Right side: $(p \Rightarrow r) \vee (q \Rightarrow r) \equiv (\mathrm{T} \Rightarrow \mathrm{F}) \vee (\mathrm{F} \Rightarrow \mathrm{F}) \equiv \mathrm{F} \vee \mathrm{T} \equiv \mathrm{T}$ . The biconditional $\mathrm{F} \Leftrightarrow \mathrm{T}$ evaluates to $\mathrm{F}$ . Therefore, (S2) is not a tautology.

✓verify

Since $(p \vee q) \Rightarrow r$ is logically equivalent to $(p \Rightarrow r) \wedge (q \Rightarrow r)$ , replacing the conjunction $\wedge$ with disjunction $\vee$ in (S2) produces a strictly weaker statement on the right, so the biconditional fails whenever one premise is true and the other false with $r$ false. Thus neither is a tautology.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does

(p \vee q) \Rightarrow r

become an AND (conjunction) instead of OR?

Because $(p \vee q) \Rightarrow r \equiv \sim(p \vee q) \vee r$ . By De Morgan's law, $\sim(p \vee q) \equiv \sim p \wedge \sim q$ . Distributing $\vee r$ gives $(\sim p \vee r) \wedge (\sim q \vee r) \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$ .

Answer

Neither (S1) nor (S2) is a tautology because $(p \vee q) \Rightarrow r$ is logically equivalent to $(p \Rightarrow r) \wedge (q \Rightarrow r)$ , not $(p \Rightarrow r)$ alone or $(p \Rightarrow r) \vee (q \Rightarrow r)$ .

Why each option works or fails

A: only (S2) is a tautology - The student confuses the distribution of negation in De Morgan's laws or incorrectly distributes disjunction across implications, thinking $((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))$ distributes into a disjunction $\vee$ instead of a conjunction $\wedge$ . Remember that $(p \vee q) \Rightarrow r \equiv \neg(p \vee q) \vee r \equiv (\neg p \wedge \neg q) \vee r$ , which distributes to a conjunction of implications: $(p \Rightarrow r) \wedge (q \Rightarrow r)$ .
B: only (S1) is a tautology - The student assumes that $q$ has no effect on the validity of $(p \vee q) \Rightarrow r$ and reduces the premise prematurely to $p \Rightarrow r$ . Check truth assignments: if $p$ is false, $q$ is true, and $r$ is false, then $p \vee q \Rightarrow r$ is false while $p \Rightarrow r$ is true, disproving equivalence.
C · correct: neither (S1) nor (S2) is a tautology - None. The student correctly applies logical equivalences or identifies counterexamples for both biconditionals. Correct: $(p \vee q) \Rightarrow r \equiv \neg(p \vee q) \vee r \equiv (\neg p \wedge \neg q) \vee r \equiv (\neg p \vee r) \wedge (\neg q \vee r) \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$ , making neither statement a tautology.
D: both (S1) and (S2) are tautologies - The student believes both reductions are valid logical simplifications, failing to apply De Morgan's laws correctly to the antecedent. Negating the antecedent $(p \vee q)$ yields $\neg p \wedge \neg q$ ; distributing $\vee r$ gives an 'AND' between the implications, so neither (S1) nor (S2) is equivalent.

Step-by-step solution

given: Two statements: (S1) $((p \vee q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$ (S2) $((p \vee q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \vee (q \Rightarrow r))$
goal: Determine whether (S1) and/or (S2) are tautologies.
approach: Recall the standard equivalence: $(p \vee q) \Rightarrow r \equiv (\sim(p \vee q) \vee r) \equiv ((\sim p \wedge \sim q) \vee r) \equiv (\sim p \vee r) \wedge (\sim q \vee r) \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$ . Check if (S1) or (S2) match this, or find counterexamples (truth assignments giving False).
execute: For (S1): We test the case $p = \mathrm{F}, q = \mathrm{T}, r = \mathrm{F}$ . Then $p \vee q = \mathrm{F} \vee \mathrm{T} = \mathrm{T}$ . Left side: $(p \vee q) \Rightarrow r \equiv \mathrm{T} \Rightarrow \mathrm{F} \equiv \mathrm{F}$ . Right side: $p \Rightarrow r \equiv \mathrm{F} \Rightarrow \mathrm{F} \equiv \mathrm{T}$ . The biconditional $\mathrm{F} \Leftrightarrow \mathrm{T}$ evaluates to $\mathrm{F}$ . Therefore, (S1) is not a tautology.
execute: For (S2): Test the assignment $p = \mathrm{T}, q = \mathrm{F}, r = \mathrm{F}$ . Then $p \vee q = \mathrm{T} \vee \mathrm{F} = \mathrm{T}$ . Left side: $(p \vee q) \Rightarrow r \equiv \mathrm{T} \Rightarrow \mathrm{F} \equiv \mathrm{F}$ . Right side: $(p \Rightarrow r) \vee (q \Rightarrow r) \equiv (\mathrm{T} \Rightarrow \mathrm{F}) \vee (\mathrm{F} \Rightarrow \mathrm{F}) \equiv \mathrm{F} \vee \mathrm{T} \equiv \mathrm{T}$ . The biconditional $\mathrm{F} \Leftrightarrow \mathrm{T}$ evaluates to $\mathrm{F}$ . Therefore, (S2) is not a tautology.
verify: Since $(p \vee q) \Rightarrow r$ is logically equivalent to $(p \Rightarrow r) \wedge (q \Rightarrow r)$ , replacing the conjunction $\wedge$ with disjunction $\vee$ in (S2) produces a strictly weaker statement on the right, so the biconditional fails whenever one premise is true and the other false with $r$ false. Thus neither is a tautology.

Shortcut:

When to use it: Fastest in competitive exams: find a single falsifying row for each biconditional.

given: Biconditionals $(A \Leftrightarrow B)$ and $(A \Leftrightarrow C)$ where $A = (p \vee q) \Rightarrow r$ , $B = (p \Rightarrow r)$ , $C = (p \Rightarrow r) \vee (q \Rightarrow r)$ .

approach: A biconditional is false whenever one side is False and the other is True. Set $r = \mathrm{F}$ to make implications easy to break.

execute: With $\mathrm{F}$ , pick $(p, q) = (\mathrm{F}, \mathrm{T})$ : $A = (\mathrm{F} \vee \mathrm{T}) \Rightarrow \mathrm{F} = \mathrm{F}$ , but $B = (\mathrm{F} \Rightarrow \mathrm{F}) = \mathrm{T}$ . Hence $A \Leftrightarrow B$ is $\mathrm{F}$ , so (S1) is not a tautology.

execute: With $r = \mathrm{F}$ , pick $(p, q) = (\mathrm{T}, \mathrm{F})$ : $A = (\mathrm{T} \vee \mathrm{F}) \Rightarrow \mathrm{F} = \mathrm{F}$ , but $C = (\mathrm{T} \Rightarrow \mathrm{F}) \vee (\mathrm{F} \Rightarrow \mathrm{F}) = \mathrm{F} \vee \mathrm{T} = \mathrm{T}$ . Hence $A \Leftrightarrow C$ is $\mathrm{F}$ , so (S2) is not a tautology.

verify: Both statements admit counterexamples, hence neither is a tautology.

Mathematical Reasoning: Mathematics | JEE Main

What feels right?

Neither (S1) nor (S2) is a tautology because $(p \vee q) \Rightarrow r$ is logically equivalent to $(p \Rightarrow r) \wedge (q \Rightarrow r)$ , not $(p \Rightarrow r)$ alone or $(p \Rightarrow r) \vee (q \Rightarrow r)$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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