Mathematical Logic — MHT-CET Maths Formulas and Traps

Statements, Connectives and Truth Tables

The Conditional and Vacuous Truth

The conditional and its disjunction form

p \to q \;\text{ agrees with }\; \sim p \vee q \;\text{ in every row} \qquad p \to q = F \;\text{ only when } p = T,\; q = F \qquad F \to q = T

$p$ the antecedent (hypothesis)
$q$ the consequent (conclusion)
$F \to q$ vacuously true — a false antecedent makes the whole conditional true

Building the Full Truth Table

Number of rows in a truth table

\text{rows} = 2^{n} \qquad n = 2 \Rightarrow 4 \text{ rows}, \qquad n = 3 \Rightarrow 8 \text{ rows}

$n$ the number of DISTINCT statement letters, not the number of connectives

The Five Logical Connectives

Connective	Symbol	Read as	Value
Negation	$\sim p$	not p	Flips: $\sim T = F$ , $\sim F = T$
Conjunction	$p \wedge q$	p and q	T only when both p and q are T
Disjunction	$p \vee q$	p or q	F only when both p and q are F Inclusive OR: 'p or q' is TRUE when both hold. Everyday English often means the exclusive one; logic never does.
Conditional	$p \to q$	if p then q	F only when p is T and q is F The single most-tested row in the chapter. A conditional with a FALSE antecedent is TRUE, whatever the consequent says.
Biconditional	$p \leftrightarrow q$	p if and only if q	T when p and q have the SAME value

Every question in this chapter is this table applied repeatedly. Learn the Value column and the rest is bookkeeping.

Reading the Last Column: Equivalence, Tautology, Contradiction, Contingency

Reading	Last column looks like	Example
Tautology	All T	$p \vee \sim p$
Contradiction	All F	$p \wedge \sim p$
Contingency	Mixed — some T, some F	$p \vee q$ The default case. Most statement patterns are contingencies; the exam asks you to spot the ones that are not.
Logically equivalent $A \equiv B$	Two patterns whose columns match in EVERY row	$p \to q$ and $\sim p \vee q$ ONE disagreeing row destroys equivalence; agreeing in one row proves nothing. The asymmetry is the whole test.

Every 'which of the following is a tautology / is equivalent to' question on this paper is one of these four readings. Subtopic 5 gives you the algebra that reaches them without writing the table out.

Common traps

Treating a false sentence as 'not a statement'

A statement only has to HAVE a truth value, not to be true. '2 + 2 = 5' is a perfectly good statement whose truth value is F. The test is whether the sentence makes a definite claim, never whether the claim is correct.

Reading 'or' as exclusive

In logic

p \vee q

is TRUE when p and q are both true. A student who reads 'or' as 'one or the other but not both' will mark the both-true row F and get every disjunction question wrong by exactly one row.

Calling a conditional false because its parts are false

'If 3 + 2 = 7, then the earth is flat' has a false antecedent AND a false consequent, and the whole statement is true. Students reject it because both halves are nonsense. Only the

T \to F

pattern makes a conditional false — check the PATTERN, not the plausibility.

Chained stems that hide a false antecedent

A recurring MHT-CET stem gives statements (A) and (B), then (C) = 'If both (A) and (B) are true, then …'. If either (A) or (B) is false, (C)'s antecedent is false and (C) is automatically true — however absurd its consequent. Evaluate (A) and (B) first; the answer to (C) usually falls out with no work at all.

Resolving the negation last instead of first

\sim(p \wedge q)

and

\sim p \wedge q

are different statements. The tilde binds only as far as its bracket reaches, so decide what the

\sim

is sitting on BEFORE you substitute anything.

Counting connectives instead of letters

The row count is

2^{n}

where

n

is the number of DISTINCT letters.

[p \to (q \wedge \sim p)] \vee [(p \vee \sim q) \wedge p]

has six connectives and only two letters, so it needs 4 rows, not 64.

Matching the last column in the wrong row order

Options are quoted as a bare string like 'TTFT'. That only matches your table if you filled the input columns in the standard

TT,\,TF,\,FT,\,FF

order. Writing the rows in a different order gives a correct table and the wrong option.

Reading $\equiv$ as another name for $\leftrightarrow$

They are different kinds of object.

p \leftrightarrow q

is a statement pattern with its own truth table, true in some rows and false in others.

A \equiv B

is a verdict on two whole tables, and it is either right or wrong — it has no rows. The bridge between them is that

A \equiv B

holds exactly when the pattern

A \leftrightarrow B

is a tautology.

Doing the logic correctly on a misjudged claim

These questions are graded on the mathematics, not the logic. If you record 'there are 26 primes below 100' as true, every connective afterwards is applied flawlessly to the wrong input and the answer is wrong. Settle each letter before reading the options.

Assuming a 'for all' claim is true because it works for small n

A universally quantified claim fails on ONE counterexample. '

10n - 3

is prime whenever

n

is not divisible by 3' survives

n = 1, 2, 4

and dies at

n = 8

, where

77 = 7 \times 11

. Hunt for the counterexample rather than confirming the pattern.

Finding Truth Values of Component Statements

Learn this subtopic in the notes

The Forced Row of a False Conditional

The one false row of a conditional

X \to Y = F \iff X = T \;\text{ and }\; Y = F \qquad (A \wedge B) = T \Rightarrow A = B = T \qquad (A \vee B) = F \Rightarrow A = B = F

$X$ the antecedent — forced TRUE
$Y$ the consequent — forced FALSE

Forced Values from a Biconditional

The matching rule

X \leftrightarrow Y = T \iff X = Y \qquad X \leftrightarrow Y = F \iff X \neq Y

$X, Y$ the two sides, which may themselves be compound

Common traps

Trying to work backwards from a TRUE conditional

X \to Y = T

is satisfied by three of the four rows, so it forces nothing on its own. If a stem says a conditional is true, the information you need is somewhere else in the stem — look for a second given, not for a forced row.

Hunting for p when p is not determined

In 'q is false and

(p \wedge q) \leftrightarrow r

is true',

p

never gets pinned down and does not need to be — the answer depends only on

r

. Students lose time trying to force a value that the stem deliberately leaves open. If a letter cancels out, move on.

Starting with the given that forces least

Both givens are true statements about the same letters, but they are not equally useful. Beginning with a TRUE conditional leaves you enumerating three rows; beginning with the FALSE one settles two letters at a stroke. Scan the givens and start with a false conditional, a true conjunction or a false disjunction.

Evaluating a consequent you never needed

If an option's antecedent works out false, the option is true and the consequent is irrelevant — however elaborate it looks. Checking the antecedent first turns several of these questions into a single glance.

Reporting the truth values when the question asked for an option

These stems often end 'then which of the following is true?' rather than 'find p, q and r'. Deducing the letters correctly and then answering the wrong question is a common and entirely avoidable loss.

Negation of Statements and Quantifiers

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De Morgan Laws for And and Or

De Morgan laws

\sim(p \wedge q) \equiv\; \sim p \vee \sim q \qquad \sim(p \vee q) \equiv\; \sim p \wedge \sim q \qquad \sim(\sim p) \equiv p

$\sim$ negation — distributes inwards, flipping the connective as it goes

Negating a Conditional

Negation of an implication

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

$p$ the antecedent — asserted, not negated
$\sim q$ the consequent — negated

Negating a Biconditional

Negation of a biconditional

\sim(p \leftrightarrow q) \equiv (p \wedge \sim q) \vee (\sim p \wedge q) \qquad \sim(p \leftrightarrow \sim q) \equiv p \leftrightarrow q

$\leftrightarrow$ biconditional — true when the sides agree

Negating Quantified Statements

Quantifier negation

\sim(\forall x,\, p(x)) \equiv \exists x,\; \sim p(x) \qquad \sim(\exists x,\, p(x)) \equiv \forall x,\; \sim p(x) \qquad \sim(x \geq M) \equiv (x < M)

$\forall$ for all / for every
$\exists$ there exists / for some

Common traps

Negating both parts but keeping the connective

\sim(p \wedge q)

is NOT

\sim p \wedge \sim q

. The connective must flip. This is the most common single error in the chapter, and the distractor list always contains the unflipped version.

Negating an implication as another implication

The tempting wrong answer is

\sim p \to \sim q

, which negates both parts and keeps the arrow. It is not the negation — it is the inverse, and it is not even equivalent to the original. The negation has no arrow at all.

Negating both sides of a biconditional

\sim p \leftrightarrow \sim q

is equivalent to

p \leftrightarrow q

itself — if two things always agree, so do their denials. It is therefore the exact opposite of the negation, and it appears in the option list every time this is asked.

Assigning a negative statement to a letter

If you set

p

= 'the triangle is NOT isosceles', every subsequent negation needs an extra cancellation and the bookkeeping collapses. Always let the letters stand for the positive forms and carry the

\sim

explicitly.

Flipping the quantifier but leaving the predicate alone

The negation of 'for all x, p(x)' is 'there exists x with not p(x)' — the inner claim must be negated too. Options that flip only the quantifier are the standard distractor and look convincing at a glance.

Forgetting that the inequality changes

Negating

x \geq M

gives

x < M

, not

x \leq M

and not

x > M

. When the predicate is an inequality, the boundary moves to the other side, so check the equality case explicitly.

Converse, Inverse and Contrapositive

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Only the Contrapositive Shares the Truth Value

The equivalence pairs

p \to q \equiv\; \sim q \to \sim p \qquad q \to p \equiv\; \sim p \to \sim q \qquad p \to q \not\equiv q \to p

$\equiv$ logically equivalent — identical last column

Converting to Conditional Form First

Conditional law, used in reverse

\sim p \vee q \equiv p \to q \qquad p \vee q \equiv\; \sim p \to q \qquad p \to q \equiv\; \sim q \to \sim p

$\vee$ the disjunction hiding a conditional

Stacked Operations: Negation of a Contrapositive, Contrapositive of an Inverse

The chain that collapses

\text{contrapositive}(\text{inverse}(p \to q)) \equiv q \to p \qquad \sim(\text{contrapositive}(p \to q)) \equiv\; \sim q \wedge p

inverse $\sim p \to \sim q$
contrapositiveswap and negate

Necessary and Sufficient Condition Language

Equivalent phrasings of one conditional

p \to q \;\equiv\; (p \text{ only if } q) \;\equiv\; (q \text{ necessary for } p) \;\equiv\; (p \text{ sufficient for } q) \;\equiv\; \sim q \to \sim p

necessarythe CONSEQUENT — it must hold for the antecedent to
sufficientthe ANTECEDENT — it is enough to guarantee the consequent

The Three Relatives of a Conditional

Form	Symbolic	Built by	Equivalent to original?
Original	$p \to q$	—	Yes, trivially
Converse	$q \to p$	Swap the two parts	No
Inverse	$\sim p \to \sim q$	Negate both parts, keep the order	No
Contrapositive	$\sim q \to \sim p$	Swap and negate both	Yes — always The only equivalent relative, and the one the paper asks about most. A statement and its contrapositive always share a truth value.
Converse and inverse	$q \to p$ and $\sim p \to \sim q$	Each is the contrapositive of the other	Equivalent to EACH OTHER, not to the original

Memorise the last column. Most option lists contain all three relatives, so knowing the forms is not enough — you must know which one is being asked for.

Common traps

Offering the converse where the contrapositive was asked

Both are single-step transformations of the same statement and both appear in the options every time. Read the question word again before you commit: 'converse' swaps, 'contrapositive' swaps AND negates.

Assuming the converse follows from the original

'If a number is a multiple of 9 then it is a multiple of 3' is true; its converse is false (take 6). A true conditional says nothing whatever about its converse, and stems are built around exactly this gap.

Negating the wrong part when converting an OR

p \vee q

becomes

\sim p \to q

, not

p \to q

. The letter that moves into the antecedent position picks up a negation — check it against the law rather than by feel.

Applying the second operation to the original statement

'The negation of the contrapositive' does not mean 'the negation, and also the contrapositive, of the original'. Form the contrapositive, write it down, and treat THAT as the new statement. Skipping the written intermediate is where these go wrong.

Reading 'p only if q' as 'if q then p'

The word order tempts you to put q first, but 'p only if q' is

p \to q

— q is the NECESSARY condition, so it sits as the consequent. 'If q then p' would be the converse, and it is the distractor supplied.

Swapping necessary and sufficient

Necessary is the consequent; sufficient is the antecedent. A quick anchor: being a multiple of 3 is NECESSARY for being a multiple of 9, and being a multiple of 9 is SUFFICIENT for being a multiple of 3.

Logical Equivalence and Algebra of Statements

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Why Equivalence Licenses Substitution

Equivalence and its test

A \equiv B \iff (A \leftrightarrow B) \text{ is a tautology} \iff A \text{ and } B \text{ share every row}

$\equiv$ logically equivalent — a claim about all rows
$\leftrightarrow$ a connective, evaluated row by row

Classifying a Pattern Without Building Its Table

Classification by last column

\text{tautology} \Rightarrow \text{all } T \qquad \text{contradiction} \Rightarrow \text{all } F \qquad \text{contingency} \Rightarrow \text{mixed}

$p \vee \sim p$ the standard tautology
$p \wedge \sim p$ the standard contradiction

Finding the Statement That Makes a Pattern a Tautology

Tautology condition and modus tollens

(X \to r) \text{ is a tautology} \iff X = T \Rightarrow r = T \qquad (p \to q) \wedge \sim q \equiv\; \sim p \wedge \sim q

$X$ the antecedent — simplify this before anything else
$r$ the statement being solved for

The Dual of a Statement Pattern

Dual versus negation

\text{dual}(p \wedge \sim q) = p \vee \sim q \qquad \sim(p \wedge \sim q) \equiv\; \sim p \vee q

dualswap the connectives only — letters untouched
$\sim$ negation — swaps connectives AND negates every letter

The Algebra of Statements

Law	With AND	With OR
Commutative	$p \wedge q \equiv q \wedge p$	$p \vee q \equiv q \vee p$
Associative	$(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)$	$(p \vee q) \vee r \equiv p \vee (q \vee r)$
Distributive	$p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$	$p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)$ Both directions are legal here, unlike ordinary arithmetic where only one distribution holds.
Identity	$p \wedge T \equiv p$	$p \vee F \equiv p$
Domination	$p \wedge F \equiv F$	$p \vee T \equiv T$
Complement	$p \wedge \sim p \equiv F$	$p \vee \sim p \equiv T$ The engine of most simplifications: spot a letter meeting its own negation and a whole branch collapses to F or T.
Idempotent	$p \wedge p \equiv p$	$p \vee p \equiv p$
Absorption	$p \wedge (p \vee q) \equiv p$	$p \vee (p \wedge q) \equiv p$ The whole bracket vanishes. Worth memorising by shape: a letter outside meeting itself inside swallows the rest.
De Morgan	$\sim(p \wedge q) \equiv \;\sim p \vee \sim q$	$\sim(p \vee q) \equiv \;\sim p \wedge \sim q$
Conditional	$\sim(p \to q) \equiv p \wedge \sim q$	$p \to q \equiv \;\sim p \vee q$ Always apply this first. The other laws cannot see through an arrow.

Ten laws cover every simplification the paper sets. Complement, distributive and absorption account for most of the work.

Common traps

Checking one row and declaring equivalence

Agreeing in a row proves nothing; equivalence needs EVERY row. The asymmetry is worth holding onto: one row can disprove equivalence but never establish it.

Distributing only one way

In logic, OR distributes over AND as well:

p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)

. Students trained on ordinary algebra expect only

p \wedge (q \vee r)

to expand and miss the other half.

Simplifying around an arrow instead of clearing it

De Morgan, distribution and absorption are stated for

\wedge

and

\vee

only. Leaving a

\to

in place and trying to distribute across it produces confident nonsense. Clear every arrow first.

Calling a contingency a tautology after checking two rows

A pattern that comes out T in the rows you happened to try may still be F elsewhere. Either simplify all the way to a bare

T

, or check every row — a partial table cannot establish a tautology.

Looking for an r that is true everywhere

r

only has to hold in the rows where the ANTECEDENT is true — it may be false elsewhere without harming the tautology. Hunting for a universally true

r

rules out the correct option, which is usually a plain

\sim p

or

\sim q

.

Negating the letters when asked for a dual

The dual of

p \wedge \sim q

is

p \vee \sim q

, NOT

\sim p \vee q

. The second is the negation. Duals touch connectives and constants only, and the two operations produce different statements that both appear in the option list.

Switching Circuits

Learn this subtopic in the notes

Series is And, Parallel is Or

The translation rule

\text{series} \;\longrightarrow\; p \wedge q \qquad \text{parallel} \;\longrightarrow\; p \vee q \qquad S_1' \;\longrightarrow\; \sim p

$p$ the switch $S_1$ is closed
$S_1'$ the complementary switch — closed exactly when $S_1$ is open

Common traps

Reading a branch before finishing it

A branch often contains its own series run before rejoining the parallel node. Resolve each branch completely into a single expression FIRST, then join the branches — reading left to right across the whole picture mixes the levels up.

Giving a primed switch its own letter

S_1'

is not a third switch. It is

\sim p

, and treating it as a new letter destroys every simplification — the whole point of these circuits is that

p

meeting

\sim p

collapses a branch.

Trying to simplify by staring at the picture

Redrawing a circuit by eye, without writing the expression down, is where these questions are lost. The simplification is algebraic; the drawing is only the last step, and it follows mechanically from the simplified expression.

Deleting a repeated switch instead of factoring it

In

(p \wedge q) \vee (p \wedge r)

,

S_1

genuinely appears on both branches and both copies matter. Factoring gives

p \wedge (q \vee r)

, which is a different arrangement — not the result of rubbing one copy out.

Matching circuits by how they look

Two circuits drawn with the same number of switches in a similar arrangement can behave differently, and two that look nothing alike can be equivalent. The pictures are not the evidence — the simplified expressions are.

Mathematical Logic formulas

Statements, Connectives and Truth Tables

The Conditional and Vacuous Truth

Building the Full Truth Table

The Five Logical Connectives

Reading the Last Column: Equivalence, Tautology, Contradiction, Contingency

Common traps

Finding Truth Values of Component Statements

The Forced Row of a False Conditional

Forced Values from a Biconditional

Common traps

Negation of Statements and Quantifiers

De Morgan Laws for And and Or

Negating a Conditional

Negating a Biconditional

Negating Quantified Statements

Common traps

Converse, Inverse and Contrapositive

Only the Contrapositive Shares the Truth Value

Converting to Conditional Form First

Stacked Operations: Negation of a Contrapositive, Contrapositive of an Inverse

Necessary and Sufficient Condition Language

The Three Relatives of a Conditional

Common traps

Logical Equivalence and Algebra of Statements

Why Equivalence Licenses Substitution

Classifying a Pattern Without Building Its Table

Finding the Statement That Makes a Pattern a Tautology

The Dual of a Statement Pattern

The Algebra of Statements

Common traps

Switching Circuits

Series is And, Parallel is Or

Common traps

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