Statements and logical connectives
The sentence "It rained today" can be called true or false, but the command "Open the door!" is neither. In mathematics we use only sentences that are true or false. Building on class 11 : the "and / or" of sets and Venn diagrams now applies to statements.
A statement is a sentence that is either true or false, never both. True = T, false = F is its truth value. Words that join statements are connectives: and (∧), or (∨), not (∼), if … then (→), if and only if (↔).
Rule 1: Recognise a statement — Questions, commands, exclamations and "x + 2 = 5" (x unknown) are not statements; "7 is a prime number" is a statement (true).
Rule 2: The ∧ and ∨ connectives — p ∧ q is true only when both are true. p ∨ q is false only when both are false.
Rule 3: The → and ↔ connectives — p → q is false only when p is true and q is false. p ↔ q is true when both have the same value.
In the figure : p ∧ q is true only when both are true; p ∨ q is false only when both are false.
- Easy : p: "12 is even" (T), q: "12 is a multiple of 5" (F): p ∧ q = F, p ∨ q = T.
- Medium : p: "7 < 3" (F), q: "9 is even" (F): p → q = T (a false "if" part makes the whole true).
- Exam-type : "2 + 2 = 5 ↔ Mumbai is in Maharashtra": F ↔ T = F.
p → q is false in just one row: (T, F).
Treating the rows (F, T) and (F, F) of p → q as F; they are T.
Check p → q by writing it as ∼p ∨ q. Memory clue : "A promise is false only if it is broken" (T → F).
"x > 3" is not a statement; it has no truth value until x is given.
CET-style example : Which of these is a statement? (A) 5 + 3 = 8 (B) x + 3 = 8 (C) Where are you going? (D) Close the door
Answer : (A). Only (A) can be called true or false. Time-saving method : Drop questions, commands and sentences with unknowns at once.
Spot the difference
| सत्य कधी ? | True when |
|---|---|
| दोन्ही T | both T |
| किमान एक T | at least one T |
| (T, F) सोडून | except (T, F) |
| दोन्ही सारखे | both the same |
Practice questions
- 1Is "15 is a prime number" a statement? What is its truth value?
- 2If p = T and q = F, what is p ∧ q?
- 3If p = F and q = T, what is p ∨ q?
- 4If p = T and q = F, find p → q and q → p.
- 5If p = F and q = F, what is p ↔ q?
- 6Find the truth value of "If 3 > 5 then 4 is even".
- 7If p = T, q = F, r = T, find (p ∨ q) ∧ r.
- 8If p = T, q = T, r = F, find p → (q ∧ r).
Show answers (try them yourself first)
- 1Yes; F (15 = 3 × 5).
- 2F.
- 3T.
- 4p → q = F; q → p = T.
- 5T (both the same).
- 6F → T = T.
- 7(T ∨ F) = T; T ∧ T = T.
- 8q ∧ r = F; T → F = F.
Statement patterns and logical equivalence
"If it rains I will take an umbrella" and "It did not rain or I will take an umbrella" are true or false in exactly the same situations, so they are equivalent.
A combination of p, q, r joined by connectives is a statement pattern. With n statements the truth table has 2ⁿ rows. If the final columns of two patterns are identical, they are logically equivalent, written ≡.
Rule 1: Build the truth table — List all T/F combinations of p, q (and r) first; then fill one column at a time from the innermost bracket outwards.
Rule 2: Key equivalences — p → q ≡ ∼p ∨ q; ∼(p → q) ≡ p ∧ ∼q; ∼(p ∧ q) ≡ ∼p ∨ ∼q; ∼(p ∨ q) ≡ ∼p ∧ ∼q; p ↔ q ≡ (p → q) ∧ (q → p).
In the figure : The columns of p → q and ∼p ∨ q match, so they are equivalent.
How is the formula obtained? : p → q fails only at (T, F); ∼p ∨ q also fails only at p = T, q = F (there ∼p = F and q = F). The two agree in all four rows.
- Easy : Table of (p ∧ q): T, F, F, F. Then ∼(p ∧ q): F, T, T, T.
- Medium : p ∧ ∼q over rows (TT, TF, FT, FF): F, T, F, F, the same as ∼(p → q).
- Exam-type : (p → q) ∧ (q → p): T, F, F, T, which is exactly the table of p ↔ q.
With three statements there are 2³ = 8 rows.
Losing a combination by changing the row order; always keep TT, TF, FT, FF.
To test equivalence, look for one row that differs; if you find one they are not equivalent. Memory clue : "If-then = not-or" (p → q = ∼p ∨ q).
Do not think ∼(p → q) ≡ ∼p → ∼q; the correct form is p ∧ ∼q.
CET-style example : Which is equivalent to ∼(p → q)? (A) ∼p ∨ q (B) p ∧ ∼q (C) ∼p ∧ q (D) p ∨ ∼q
Answer : (B). p → q fails only at (T, F), so its negation is true only there: p ∧ ∼q. Time-saving method : Check only the (T, F) row for the four options.
Spot the difference
| समतुल्य | Equivalent to |
|---|---|
| ∼p ∨ q | ∼p ∨ q |
| ∼p ∨ ∼q | ∼p ∨ ∼q |
| ∼p ∧ ∼q | ∼p ∧ ∼q |
| p ∧ ∼q | p ∧ ∼q |
Practice questions
- 1How many rows does the truth table of two statements p, q have?
- 2Write the last column of p ∧ ∼q over (TT, TF, FT, FF).
- 3Write the truth values of ∼p ∨ q over (TT, TF, FT, FF).
- 4Write an equivalent form of ∼(p ∨ q).
- 5Write the table of p ↔ q over (TT, TF, FT, FF).
- 6Write the negation of "If x = 2 then x² = 4".
- 7Show that p → q ≡ ∼q → ∼p.
- 8Why are (p → q) ∧ (q → p) and p ↔ q equivalent?
Show answers (try them yourself first)
- 12² = 4.
- 2F, T, F, F.
- 3T, F, T, T.
- 4∼p ∧ ∼q.
- 5T, F, F, T.
- 6x = 2 and x² ≠ 4 (p ∧ ∼q).
- 7Table of ∼q → ∼p: T, F, T, T, the same as p → q.
- 8Both give T, F, F, T over the four rows.
Tautology, contradiction and contingency
"Tomorrow it will rain or it will not" is always true; "He was and was not at school today" is always false; "He was at school today" depends on the situation.
If the final column is all T the pattern is a tautology. If it is all F it is a contradiction. If it has both T and F it is a contingency.
Rule 1: Full table — Fill every row and look at the last column: all T, all F or mixed.
Rule 2: Short method — Assume the pattern is F and assign values to its parts; if you hit a conflict, it is a tautology.
In the figure : The column of p ∨ ∼p is all T and that of p ∧ ∼p is all F.
How is the formula obtained? : For (p ∧ q) → p to be F we would need p ∧ q = T and p = F; but p ∧ q = T forces p = T. Conflict, so it is a tautology.
- Easy : p ∨ ∼p is a tautology; p ∧ ∼p is a contradiction.
- Medium : (p ∧ q) → p is a tautology (short method above).
- Exam-type : [p ∧ (p → q)] → q is a tautology; p → q is a contingency (T, F, T, T).
The negation of a tautology is a contradiction, and vice versa.
Judging from one row; a tautology needs every row to be T.
Assume the pattern is F and look for a conflict. Memory clue : "All T = tautology, all F = contradiction".
p → q is not a tautology; it is a contingency.
CET-style example : Which is a tautology? (A) p ∧ ∼p (B) p → q (C) p ∨ ∼p (D) p ∧ q
Answer : (C). T in every row. Time-saving method : Try only p = T and p = F in each option.
Spot the difference
| शेवटचा स्तंभ | Last column |
|---|---|
| सर्व T | all T |
| सर्व F | all F |
| T आणि F दोन्ही | both T and F |
Practice questions
- 1What type is p ∨ ∼p?
- 2What type is p ∧ ∼p?
- 3What type is p ∧ q?
- 4What type is (p ∧ q) → p?
- 5What type is ∼(p ∨ ∼p)?
- 6What type is p → (p ∨ q)?
- 7What type is [(p → q) ∧ (q → r)] → (p → r)? (short method)
- 8What type is (p → q) ↔ (∼p ∨ q)?
Show answers (try them yourself first)
- 1Tautology.
- 2Contradiction.
- 3Contingency (T, F, F, F).
- 4Tautology.
- 5Contradiction (negation of a tautology).
- 6Tautology; if p = T then p ∨ q = T.
- 7Tautology: assuming F gives p = T, r = F; p → q = T ⇒ q = T; q → r = T ⇒ r = T, a conflict.
- 8Tautology (both sides are equivalent).
Quantifiers and their negations
To disprove "All children in the class are clever" one child who is not clever is enough. That is exactly how the negation of ∀ and ∃ works.
The universal quantifier ∀ means "for every". The existential quantifier ∃ means "for at least one". A quantified statement uses one of them, e.g. ∀ x ∈ N, x > 0.
Rule 1: Negation rule — ∼(∀x, p(x)) ≡ ∃x, ∼p(x); ∼(∃x, p(x)) ≡ ∀x, ∼p(x). Switch the quantifier and negate the inside.
Rule 2: Truth value — One counter-example makes a ∀ statement false; one example makes a ∃ statement true.
- Easy : ∀ x ∈ N, x > 0: true. Negation: ∃ x ∈ N, x ≤ 0 (false).
- Medium : ∃ x ∈ R, x² = −1: false (a square is never negative). Negation: ∀ x ∈ R, x² ≠ −1 (true).
- Exam-type : "Some primes are even" is true (2); negation: "All primes are odd", which is false.
Of a quantified statement and its negation exactly one is true.
Taking the negation of "all" as "none"; the correct negation is "at least one is not".
Negation = swap ∀ and ∃ + negate the inner statement. Memory clue : "All and some swap places".
"Some" means "at least one", not "exactly one".
CET-style example : The negation of "∀ x ∈ N, x + 3 > 5" is: (A) ∃ x ∈ N, x + 3 ≤ 5 (B) ∀ x ∈ N, x + 3 ≤ 5 (C) ∃ x ∈ N, x + 3 > 5 (D) ∀ x ∈ N, x + 3 < 5
Answer : (A). ∀ becomes ∃ and > becomes ≤. Time-saving method : Two changes: the quantifier and the sign.
Spot the difference
| अर्थ / नकार | Meaning / negation |
|---|---|
| प्रत्येकासाठी ; नकार ∃ | for every; negation is ∃ |
| किमान एकासाठी ; नकार ∀ | for at least one; negation is ∀ |
Practice questions
- 1Truth value of ∀ x ∈ N, x + 1 > x?
- 2Truth value of ∃ x ∈ N, x + 5 = 3?
- 3Read "∀ x ∈ R, x² ≥ 0" and give its truth value.
- 4Write the negation of "∃ x ∈ N, x < 1".
- 5Write the negation of "All birds can fly".
- 6Is "Some even numbers are prime" true?
- 7Truth value of ∀ x ∈ N, (x² > x), with reason.
- 8Write ∼[∀ x ∈ N, ∃ y ∈ N, y > x].
Show answers (try them yourself first)
- 1T.
- 2F (x = −2 is not in N).
- 3For every real x, x² ≥ 0; T.
- 4∀ x ∈ N, x ≥ 1.
- 5At least one bird cannot fly.
- 6Yes; 2.
- 7F; for x = 1, 1 > 1 is false (counter-example).
- 8∃ x ∈ N, ∀ y ∈ N, y ≤ x (the original is T, so this is F).
Duality, converse, inverse and contrapositive
From "If it rains the road gets wet", the reverse "If the road is wet it rained" need not be true (the road may have been washed). But "If the road is not wet it did not rain" is always true.
For p → q: the converse is q → p; the inverse is ∼p → ∼q; the contrapositive is ∼q → ∼p. The dual of a pattern is obtained by swapping ∧ ↔ ∨ and T ↔ F, leaving ∼ unchanged.
Rule 1: Equivalent pairs — p → q ≡ ∼q → ∼p (contrapositive); converse ≡ inverse. p → q is not equivalent to its converse.
Rule 2: Find the dual — The dual of p ∧ (q ∨ r) is p ∨ (q ∧ r); the dual of p ∨ t is p ∧ c. The dual of the dual is the original.
Rule 3: Laws of statements — Double negation ∼(∼p) ≡ p; De Morgan ∼(p ∧ q) ≡ ∼p ∨ ∼q; distributive p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r); absorption p ∨ (p ∧ q) ≡ p.
- Easy : "If x = 3 then x² = 9": converse "If x² = 9 then x = 3" (false because of x = −3).
- Medium : Contrapositive: "If x² ≠ 9 then x ≠ 3", true exactly as the original.
- Exam-type : Dual of (p ∨ q) ∧ r: (p ∧ q) ∨ r; ∼(p ∨ ∼q) ≡ ∼p ∧ q (De Morgan).
The contrapositive always has the same truth value as the original.
Treating the converse as equivalent to the original.
When p → q is hard to prove, prove its contrapositive ∼q → ∼p. Memory clue : "Reverse + negate = contrapositive"; "only reverse = converse"; "only negate = inverse".
While dualising do not touch ∼; change only ∧, ∨, t, c.
CET-style example : The contrapositive of p → q is: (A) q → p (B) ∼p → ∼q (C) ∼q → ∼p (D) ∼(p → q)
Answer : (C). Time-saving method : "Reverse and negate both" gives the contrapositive.
Spot the difference
| रूप (p → q साठी) | Form (for p → q) |
|---|---|
| q → p | q → p |
| ∼p → ∼q | ∼p → ∼q |
| ∼q → ∼p | ∼q → ∼p |
Practice questions
- 1Write the converse of p → q.
- 2Write the inverse of p → q.
- 3Write the dual of p ∨ q.
- 4Write the contrapositive of "If a number is divisible by 4 then it is even".
- 5Simplify ∼(p ∧ ∼q) by De Morgan.
- 6Write the dual of (p ∧ q) ∨ (p ∧ r).
- 7Prove p ∨ (p ∧ q) ≡ p by a table.
- 8Show by an example that the converse of "If x > 2 then x² > 4" is false.
Show answers (try them yourself first)
- 1q → p.
- 2∼p → ∼q.
- 3p ∧ q.
- 4If a number is not even then it is not divisible by 4.
- 5∼p ∨ q (≡ p → q).
- 6(p ∨ q) ∧ (p ∨ r).
- 7p ∧ q: T, F, F, F; p ∨ (p ∧ q): T, T, F, F = p.
- 8Converse: x² > 4 ⇒ x > 2. For x = −3, 9 > 4 but −3 > 2 is false.
Switching circuits
A staircase light that can be switched on from the top or the bottom is a parallel connection (or). A door that opens only if both locks are opened is a series connection (and).
A closed switch (current flows) = 1 = T; an open switch = 0 = F. A series connection is p ∧ q; a parallel connection is p ∨ q; the opposite switch p′ is ∼p (when p is closed, p′ is open).
Rule 1: Circuit to statement — Write ∧ for series, ∨ for parallel and group with brackets.
Rule 2: Simplify — Use the laws: p ∧ p′ = 0; p ∨ p′ = 1; p ∧ (p ∨ q) = p.
In the figure : Series means and, parallel means or.
How is the formula obtained? : (p ∧ q) ∨ (p ∧ q′) = p ∧ (q ∨ q′) = p ∧ 1 = p, so both q switches can be removed and only p kept.
- Easy : p and q in series: the lamp glows only if both are closed (1 ∧ 1 = 1).
- Medium : p, q in parallel and r in series: (p ∨ q) ∧ r; with p = 0, q = 1, r = 1 the output is 1.
- Exam-type : p ∨ (p′ ∧ q) = (p ∨ p′) ∧ (p ∨ q) = 1 ∧ (p ∨ q) = p ∨ q.
Series is like multiplication (and); parallel is like addition (or).
Taking p and p′ as closed together; they are always opposite.
Write 0 and 1 straight away for p ∧ p′ and p ∨ p′. Memory clue : "Series = queue = and", "Parallel = options = or".
Do not treat the opposite switch p′ as an independent statement.
CET-style example : Which single switch replaces (p ∧ q) ∨ (p ∧ q′)? (A) p (B) q (C) p ∨ q (D) p ∧ q
Answer : (A). p ∧ (q ∨ q′) = p. Time-saving method : Take p outside the bracket.
Spot the difference
| विधान | Statement |
|---|---|
| p ∧ q | p ∧ q |
| p ∨ q | p ∨ q |
| ∼p | ∼p |
Practice questions
- 1Two switches in series; p = 1, q = 0. Will the lamp glow?
- 2Two switches in parallel; p = 1, q = 0. Will the lamp glow?
- 3What is p ∧ p′?
- 4Find (p ∨ q) ∧ p′ for p = 1.
- 5Simplify p ∧ (p ∨ q).
- 6Simplify (p ∧ q) ∨ (p ∧ q′).
- 7Simplify p ∨ (p′ ∧ q).
- 8Simplify (p ∨ q) ∧ (p′ ∨ q).
Show answers (try them yourself first)
- 1No (1 ∧ 0 = 0).
- 2Yes (1 ∨ 0 = 1).
- 30.
- 4p′ = 0, so the value is 0.
- 5p (absorption).
- 6p.
- 7p ∨ q.
- 8q ∨ (p ∧ p′) = q ∨ 0 = q.