JEE Mains Maths · Mathematical Reasoning

Unknown Connectives and Missing Statements

Find which connective, or which of p, q, ∼p and ∼q, makes a statement a tautology or equal to a given one.

All Mathematical Reasoning notes JEE Mains Maths strategy

Why this matters

Eleven PYQs, all multiple choice. Eight ask which connective in place of a symbol such as Δ or ∇ makes a statement a tautology; three ask which of p, q, ∼p and ∼q to put in place of r. Two ideas cover the page.

Concept 1 of 2: Choosing the connective

The unknown symbol is one of

\wedge,\vee,\rightarrow,\leftrightarrow

, so there are at most four cases, or four pairs when there are two symbols. Test each case for a false row; one false row rules it out. A tautology usually needs a

\vee

or a

\rightarrow

that joins a statement with its own negation.

Definition

Test the options, not every case: each option fixes the symbols.
One false row rules a case out.
$A\wedge B$ is a tautology only if both $A$ and $B$ are.

The four connectives

p\wedge q,\quad p\vee q,\quad p\rightarrow q\equiv\sim p\vee q,\quad p\leftrightarrow q\equiv(p\rightarrow q)\wedge(q\rightarrow p)

Worked example

Let

\Delta,\nabla\in\{\wedge,\vee\}

. For how many pairs

(\Delta,\nabla)

(p\Delta q)\rightarrow(p\nabla q)

a tautology?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q151Moderate

Example 1 · Mathematical Reasoning · Unknown Connectives and Missing Statements

Let

\mathbf{\Delta,\nabla \in \{\land,\vee\}}

be such that

(p \rightarrow q)\Delta(p\nabla q)

is a tautology. Then

$\leftrightarrow$ is not $\rightarrow$

p\leftrightarrow q

is false at

p=F,\ q=T

, where

p\rightarrow q

is true. A case that works with

\rightarrow

can fail with

\leftrightarrow

, so test it separately.

Drill 7 more on choosing the connective

Concept 2 of 2: Choosing the missing statement

Here a letter

r

stands for one of

p,q,\sim p,\sim q

. Find the rows that could make the statement false, usually just one, and check each candidate there. A candidate that is true where it needs to be true, or false where it needs to be false, works.

Definition

Substitute each candidate for $r$ and test it.
For an implication, only rows where the antecedent is true and the consequent false matter.
When the question asks 'how many', test all four candidates.

Useful reductions

p\wedge\sim p\equiv F,\qquad F\rightarrow X\equiv T,\qquad X\rightarrow T\equiv T

Worked example

Let

r\in\{p,q,\sim p,\sim q\}

. For which

r

r\rightarrow(p\vee\sim q)

a tautology?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q167Moderate

Example 2 · Mathematical Reasoning · Unknown Connectives and Missing Statements

The number of values of

\mathbf{r \in \{ p,q, \sim p, \sim q\}}

for which

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

is a tautology, is:

Test every candidate

A 'how many values' question can have two or more answers. Stopping at the first

r

that works loses the count, so test all four.

Drill 2 more on choosing the missing statement

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 (2)

Choosing the connective
The four connectives
$p\wedge q,\quad p\vee q,\quad p\rightarrow q\equiv\sim p\vee q,\quad p\leftrightarrow q\equiv(p\rightarrow q)\wedge(q\rightarrow p)$
Choosing the missing statement
Useful reductions
$p\wedge\sim p\equiv F,\qquad F\rightarrow X\equiv T,\qquad X\rightarrow T\equiv T$

Watch out for (2)

$\leftrightarrow$ is not $\rightarrow$
→ Choosing the connective
Test every candidate
→ Choosing the missing statement

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