Is it correct to say: Marking p → q as false whenever p is false?

p → q is false in exactly one case: p true and q false. With a false antecedent the conditional is true.

Is it correct to say: Writing ~(p ∧ q) ≡ ~p ∧ ~q?

De Morgan's law switches the connective: ~(p ∧ q) ≡ ~p ∨ ~q and ~(p ∨ q) ≡ ~p ∧ ~q.

Is it correct to say: Writing the negation of p → q as ~p → ~q?

~p → ~q is the inverse, not the negation. The negation is p ∧ ~q.

Is it correct to say: Treating the converse q → p as equivalent to p → q?

Only the contrapositive ~q → ~p is equivalent to p → q. Check with the row p = F, q = T.

Is it correct to say: Negating 'All students passed' as 'All students failed'?

The correct negation is 'Some student did not pass' (there exists a student who did not pass).

Is it correct to say: Finding the dual of a pattern that still contains → or changing ~ to something else?

Rewrite p → q as ~p ∨ q first. Then swap only ∧ ↔ ∨ and t ↔ c; keep every ~ as it is.

📄 Formula sheet (PDF)

Mathematical Logic

Statements, logical connectives, truth tables, tautology, logical equivalence, duality, negations and algebra of statements — Maharashtra HSC Maths Part I Ch 1

📐 18 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Build a truth table one column at a time — one column for every sub-statement — so a single slip does not spoil the final column.
→For a negation question, first write the sentence in symbols, apply the law (De Morgan, ~(p → q) ≡ p ∧ ~q), then translate back into words.
→Only the contrapositive is equivalent to p → q. The converse and the inverse are equivalent to each other, not to the original.
→To negate a quantified statement, swap 'for every' with 'there exists' and negate the condition: '<' becomes '≥', '=' becomes '≠'.
→When simplifying without a truth table, name the law used at every step (De Morgan, distributive, complement, identity).
→Remember that p → q is true whenever p is false. It is false in exactly one row: p true, q false.

📐 Formulas(18)

Negation

\sim p is T ⟺ p is F

Symbol	Meaning
$p$	A statement (truth value T or F)
$\sim p$	Negation of p, read 'not p'

Conjunction (and)

p \land q is T ⟺ p and q are both T

Symbol	Meaning
$p, q$	Component statements

Disjunction (or)

p \lor q is F ⟺ p and q are both F

Conditional (if … then)★ Board fav

p \to q is F ⟺ p is T and q is F

Biconditional (if and only if)

p \leftrightarrow q is T ⟺ p, q have the same truth value

Number of Rows in a Truth Table

rows = 2^{n}

Symbol	Meaning
$n$	Number of simple statements in the pattern

Tautology and Contradiction★ Board fav

p \lor \sim p \equiv T, p \land \sim p \equiv F

Conditional as a Disjunction

p \to q \equiv\sim p \lor q

Converse, Inverse and Contrapositive

Converse: q \to p, Inverse: \sim p \to\sim q, Contrapositive: \sim q \to\sim p

Contrapositive Equivalence★ Board fav

p \to q \equiv\sim q \to\sim p, q \to p \equiv\sim p \to\sim q

Biconditional as Two Conditionals

p \leftrightarrow q \equiv (p \to q) \land (q \to p)

Duality

Dual of p \lor (q \land r) is p \land (q \lor r)

De Morgan's Laws★ Board fav

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

Negation of a Conditional★ Board fav

\sim (p \to q) \equiv p \land \sim q

Negation of a Biconditional

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

Negation of Quantified Statements★ Board fav

\sim [\forall x, p (x)] \equiv \exists x such that \sim p (x), \sim [\exists x such that p (x)] \equiv \forall x, \sim p (x)

Symbol	Meaning
$\forall$	Universal quantifier: 'for all', 'for every'
$\exists$	Existential quantifier: 'there exists', 'for some'
$p (x)$	Open sentence in the variable x

Distributive Laws

p \land (q \lor r) \equiv (p \land q) \lor (p \land r), p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)

Identity and Involution Laws

p \land T \equiv p, p \lor F \equiv p, p \lor T \equiv T, p \land F \equiv F, \sim (\sim p) \equiv p

✏️ Solved Examples

1Solved Exampleeasy3 steps

If p: 5 + 3 = 8 and q: 4 is an odd number, find the truth value of (p ∧ ~q) → q.

Find the truth values of the simple statements. 5 + 3 = 8 is true; 4 is even, so q is false.

p = T, q = F, \sim q = T

2Solved Exampleboard3 steps

Using a truth table, examine whether [(p → q) ∧ p] → q is a tautology, contradiction or contingency.

Two simple statements, so 2² = 4 rows. Fill p → q first, then the conjunction with p, then the final conditional.

p T T F F q T F T F p \to q T F T T (p \to q) \land p T F F F [(p \to q) \land p] \to q T T T T

3Solved Exampleboard4 steps

Write the negation of: 'If it rains, then the match is cancelled and the players go home.'

Let p: It rains, q: The match is cancelled, r: The players go home. The statement is a conditional.

p \to (q \land r)

4Solved ExampleHOTS4 steps

Without using a truth table, show that ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Apply De Morgan's law to the first bracket.

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

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Marking p → q as false whenever p is false
✓p → q is false in exactly one case: p true and q false. With a false antecedent the conditional is true.
2
Writing ~(p ∧ q) ≡ ~p ∧ ~q
✓De Morgan's law switches the connective: ~(p ∧ q) ≡ ~p ∨ ~q and ~(p ∨ q) ≡ ~p ∧ ~q.
3
Writing the negation of p → q as ~p → ~q
✓~p → ~q is the inverse, not the negation. The negation is p ∧ ~q.
4
Treating the converse q → p as equivalent to p → q
✓Only the contrapositive ~q → ~p is equivalent to p → q. Check with the row p = F, q = T.
5
Negating 'All students passed' as 'All students failed'
✓The correct negation is 'Some student did not pass' (there exists a student who did not pass).
6
Finding the dual of a pattern that still contains → or changing ~ to something else
✓Rewrite p → q as ~p ∨ q first. Then swap only ∧ ↔ ∨ and t ↔ c; keep every ~ as it is.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
If p is true and q is false, find the truth value of (p ∨ q) → (p ∧ q).
$(p \lor q) \to (p \land q)$
Q2
Write the converse, inverse and contrapositive of: 'If a number is divisible by 6, then it is divisible by 3.'
Q3
Using a truth table, decide whether (p → q) ∧ (p ∧ ~q) is a tautology, contradiction or contingency.
$(p \to q) \land (p \land \sim q)$
Q4
Write the negation of: 'For every n ∈ N, n² + n is even.'
Q5
Write the dual of (p ∧ t) ∨ (q ∧ ~r), where t denotes a tautology.
Q6
Without a truth table, show that (p ∧ q) ∨ (p ∧ ~q) ≡ p.

📝 Notes

Mathematical Logic

This chapter turns English sentences into symbols so that their truth can be checked mechanically. Almost every question is one of three types: find a truth value, build a truth table, or write an equivalent form (negation, contrapositive, dual, simplified pattern).

From words to symbols

Start by naming the simple statements (p, q, r) and spotting the connective. 'And', 'but' and 'yet' give $\wedge$ ; 'or' gives $\vee$ ; 'if … then' and 'only if' give $\rightarrow$ ; 'if and only if' gives $\leftrightarrow$ . Only after the sentence is in symbols should you apply any law, and only at the end should you translate the result back into words.

Truth tables and equivalence

A pattern in n statements needs $2^n$ rows. Add one column for each sub-pattern, working from the inside of the brackets outwards. If the final column is all T the pattern is a tautology, all F a contradiction, otherwise a contingency. Two patterns are logically equivalent ( $\equiv$ ) when their final columns match row for row — this is how De Morgan's laws and $p \rightarrow q \equiv \sim q \rightarrow \sim p$ are proved.

Negations — the core skill

Learn these four results as a set:

$\sim (p \wedge q) \equiv \sim p \vee \sim q$
$\sim (p \vee q) \equiv \sim p \wedge \sim q$
$\sim (p \rightarrow q) \equiv p \wedge \sim q$
$\sim (p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (q \wedge \sim p)$

For quantified statements, 'for every' and 'there exists' swap and the condition is negated, so the negation of 'for every x, x² ≥ 0' is 'there exists x such that x² < 0'.

Simplifying without a truth table

Remove $\rightarrow$ with $p \rightarrow q \equiv \sim p \vee q$ , push negations inside with De Morgan's laws, take out a common statement with the distributive law, and finish with the complement and identity laws. Name each law beside its step.