JEE Maths›Mathematical Reasoning›Visual Solution
Visual SolutionHard
Question
Consider the statement S=(p→(q→r))→((p→q)→(p→r))S = (p \to (q \to r)) \to ((p \to q) \to (p \to r)). The statement SS is:
(A)A tautology
(B)A contradiction
(C)True only when pp is false
(D)Equivalent to p∧q∧rp \wedge q \wedge r
Solution Path
Assume SS is false: hypothesis true, conclusion false. Chase truth values: p=T,r=F,q=Tp=T, r=F, q=T forces hypothesis to be false. Contradiction proves SS is a tautology.
01Question Setup
1/4
Determine whether S=(p→(q→r))→((p→q)→(p→r))S = (p \to (q \to r)) \to ((p \to q) \to (p \to r)) is a tautology, contradiction, or contingency.
Classify statement SS
02Proof by Contradiction
2/4
Assume SS is false. Then the hypothesis p→(q→r)p \to (q \to r) is true and the conclusion (p→q)→(p→r)(p \to q) \to (p \to r) is false. For the conclusion to be false: p→qp \to q is true and p→rp \to r is false.
p→rp \to r false ⇒p=T,r=F\Rightarrow p = T, r = F
03Reaching the ContradictionKEY INSIGHT
3/4
From p=Tp = T and p→qp \to q true: q=Tq = T. From q=T,r=Fq = T, r = F: q→r=Fq \to r = F. From p=Tp = T: p→(q→r)=T→F=Fp \to (q \to r) = T \to F = F. But we assumed this was true. Contradiction!
Hypothesis cannot be true when conclusion is false
04Final Answer
4/4
Since assuming SS is false leads to a contradiction, SS is always true. This is the distribution law for conditionals (also called the currying law).
Tautology\boxed{\text{Tautology}} - Answer (A)
Concepts from this question2 concepts unlocked
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Conditional Statement Equivalences

STANDARD

p -> q is logically equivalent to ~p OR q, and also to its contrapositive ~q -> ~p. The conditional is false only when p is true and q is false.

p→q≡∼p∨q≡∼q→∼pp \to q \equiv \sim p \vee q \equiv \sim q \to \sim p

Converts conditional problems into simpler OR statements. The equivalence p -> q = ~p OR q is the single most tested identity in this chapter.

Truth value problemsLogical equivalenceTautology identification
Practice (9 Qs) →

Negation Rules for Compound Statements

EASY

To negate a compound statement, apply De Morgan's laws: negate each component and swap AND with OR (and vice versa). For conditionals, ~(p -> q) = p AND ~q.

∼(p∧q)=∼p∨∼q,∼(p→q)=p∧∼q\sim(p \wedge q) = \sim p \vee \sim q, \quad \sim(p \to q) = p \wedge \sim q

Negation questions appear in almost every JEE Main paper on this topic. Getting De Morgan's wrong costs easy marks. The conditional negation formula is especially tricky since students default to ~p -> ~q.

Negation problemsStatement equivalenceContrapositive identification
Practice (10 Qs) →
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