JEE Mains Maths · Mathematical Reasoning

Truth Values, Tautologies and Contradictions

Decide when a compound statement is true or false: find the one row that makes an implication false, and test whether a statement is always true or always false.

All Mathematical Reasoning notes JEE Mains Maths strategy

Why this matters

Twenty-two PYQs, twenty of them multiple choice. Seven find the row that makes an implication false or count the rows where a statement is true, eight pick the tautology from four options, and seven judge two claims that a statement is a tautology or a contradiction. Three ideas cover the page.

Concept 1 of 3: The one row where an implication is false

An implication

A\rightarrow B

is false in exactly one case:

A

true and

B

false. So to find where a long implication fails, set the consequent false first; that usually fixes some letters at once. Then make the antecedent true with what is left. To count the true rows, count the false ones and subtract from

2^n

Definition

$A\rightarrow B$ is false only when $A=T$ and $B=F$ .
$A\vee B$ is false only when both are false; $A\wedge B$ is true only when both are true.
$A\leftrightarrow B$ (A if and only if B) is true exactly when $A$ and $B$ have the same truth value.
$\equiv$ reads 'is logically equivalent to': $P\equiv Q$ when $P$ and $Q$ have the same truth value in every row.
$n$ letters give $2^n$ rows, and true rows $=2^n-$ false rows.

When an implication is false

A\rightarrow B\equiv\sim A\vee B,\quad\text{false only at }A=T,\ B=F

Worked example

In how many of the eight rows is

(p\rightarrow q)\rightarrow(p\wedge r)

true?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q72Moderate

Example 1 · Mathematical Reasoning · Truth Values, Tautologies and Contradictions

p,q

and

r

three propositions, then which of the following combination of truth values of

p,q

and

r

makes the logical expression

\mathbf{\{(p \vee q) \land (( \sim p) \vee r)\} \rightarrow (( \sim q) \vee r)}

false?

A false antecedent makes it true

F\rightarrow B

is true whatever

B

is. A row with a false consequent is not a false row unless the antecedent is true there too. Only true-then-false breaks an implication.

Drill 6 more on the one row where an implication is false

Concept 2 of 3: Testing for a tautology

A tautology is true in every row. The fast test: rewrite each implication as

\sim A\vee B

and look for a letter and its negation joined by

\vee

. Since

p\vee\sim p

is always true, so is any disjunction that contains it. Or try to make the statement false: if the attempt forces a letter to be both true and false, there is no false row.

Definition

Tautology: true in every row. Contradiction (fallacy): false in every row.
$p\vee\sim p$ is a tautology; $p\wedge\sim p$ is a contradiction.
The negation of a tautology is a contradiction, and the other way round.
Two standard tautologies: $(p\wedge(p\rightarrow q))\rightarrow q$ and $((p\rightarrow q)\wedge\sim q)\rightarrow\sim p$ .

The tautology test

A\rightarrow B\equiv\sim A\vee B,\qquad p\vee\sim p\equiv T,\qquad p\wedge\sim p\equiv F

Worked example

(p\rightarrow q)\rightarrow(\sim q\rightarrow\sim p)

a tautology?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q159Moderate

Example 2 · Mathematical Reasoning · Truth Values, Tautologies and Contradictions

Which of the following statements is a tautology?

Neither is a third answer

A statement that is not a tautology need not be a contradiction.

p\rightarrow q

is true in three rows and false in one: it is neither.

Drill 7 more on testing for a tautology

Concept 3 of 3: Judging two claims, (S1) and (S2)

These questions make two claims, each saying a statement is a tautology or a contradiction. Judge each claim on its own: reduce the statement, then compare it with the claim. The claim 'tautology' fails as soon as one false row appears; the claim 'contradiction' fails as soon as one true row appears.

Definition

Reduce each statement with $A\rightarrow B\equiv\sim A\vee B$ and De Morgan's laws.
To refute 'tautology', find one false row. To refute 'contradiction', find one true row.
$X\vee\sim X$ is always true and $X\wedge\sim X$ always false, however long $X$ is.

A statement and its negation

X\vee\sim X\equiv T,\qquad X\wedge\sim X\equiv F

Worked example

(S1):

(p\wedge q)\rightarrow p

is a tautology. (S2):

(p\vee q)\wedge\sim p

is a contradiction. Which claims are correct?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q75Moderate

Example 3 · Mathematical Reasoning · Truth Values, Tautologies and Contradictions

(S1)

(p \Rightarrow q) \vee (p \land ( \sim q))

is a tautology (S2)

(( \sim p) \Rightarrow ( \sim q)) \land (( \sim p) \vee q)

is a contradiction. Then

Judge the claim, not only the statement

Each claim names a type. A statement that is a tautology makes the claim 'it is a contradiction' wrong. Settle what the statement is first, then compare it with the type the claim names.

Drill 6 more on judging two claims, (s1) and (s2)

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The one row where an implication is false
When an implication is false
$A\rightarrow B\equiv\sim A\vee B,\quad\text{false only at }A=T,\ B=F$
Testing for a tautology
The tautology test
$A\rightarrow B\equiv\sim A\vee B,\qquad p\vee\sim p\equiv T,\qquad p\wedge\sim p\equiv F$
Judging two claims, (S1) and (S2)
A statement and its negation
$X\vee\sim X\equiv T,\qquad X\wedge\sim X\equiv F$

Watch out for (3)

A false antecedent makes it true
→ The one row where an implication is false
Neither is a third answer
→ Testing for a tautology
Judge the claim, not only the statement
→ Judging two claims, (S1) and (S2)

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