The negation of compound statement p ∧ q ⇔ r ∧ q is equivalent to

p ∧ q ∨ q ∧ r ∧ ~ p ∧ q ∧ r

( p ∧ q ∧ r ) ∨ [ ( p ∧ q ) ∨ ( q ∧ r ) ]

~ p ∧ q ∧ r ∨ p ∧ q ∨ q ∧ r

The negation of compound statement p ∧ q ⇔ r ∧ q is equivalent to

p ∧ q ∨ q ∧ r ∧ ~ p ∧ q ∧ r

Negation of the statement: 5 is an integer or 5 is irrational is:

5 is not an integer and 5 is not irrational

5 is not an integer 5 is not irrational

5 is irrational or 5 is an integer

5 is an integer and 5 irrational

The negation of “ ∀ n ∈ N , n + 7 > 6 ” is ….

∃ n ∈ N , such that n + 7 ≤ 6

∃ n ∈ N , such that n + 7 ≥ 6

∃ n ∈ N , such that n + 7 < 6

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The negation of compound statement $(p \land q) \Leftrightarrow (r \land q)$ is equivalent to

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Important Questions on Mathematical Logic

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Mathematics>Algebra>Mathematical Logic>Negation of Statements

p \to (p \land ~ q)

is false, then the truth values of

p

and

q

are respectively

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Mathematics>Algebra>Mathematical Logic>Negation of Statements

Consider the following three statements:

$P : 5$ is a prime number

$Q : 7$ is a factor of $192$

$R : LCM$ of $5$ and $7$ is $35$

Then the truth value of which one of the following statements is true?

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

Negation of the statement:

\sqrt{5}

is an integer or

5

is irrational is:

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Mathematics>Algebra>Mathematical Logic>Negation of Statements

The statement

(p \to (q \to p)) \to (p \to (p \lor q))

is :

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The negation of the Boolean expression

~ s \lor (~ r \land s)

is equivalent to

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Mathematics>Algebra>Mathematical Logic>Negation of Statements

Let

p, q, r

be three statements such that the truth value of

(p \land q) \to (~ q \lor r)

F

. Then the truth values of

p, q, r

are respectively :

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Mathematics>Algebra>Mathematical Logic>Negation of Statements

For two statements

p

and

q

, the logical statement

(p \to q) \land (q \to ~ p)

is equivalent to

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The negation of the statement "For all real numbers

x

and

y, x + y = y + x

" is

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

p \to (~ p \lor ~ q)

is false, then the truth values of

p

and

q

are, respectively

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The Boolean expression

\sim (p \Rightarrow (\sim q))

is equivalent to

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Mathematics>Algebra>Mathematical Logic>Negation of Statements

The proposition

p \to ~ (p \land ~ q)

is equivalent to :

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The Boolean expression

~ (p \lor q) \lor (~ p \land q)

is equivalent to

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The Boolean expression

~ (p \lor q) \lor (~ p \land q)

is equivalent to

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The negation of the statement “All continuous functions are differentiable”

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

p : 4

is an even prime number,

q : 6

is a divisor of

12

and

r :

the

HCF

4

and

6

2

, then which of the following is correct?

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Mathematics>Algebra>Mathematical Logic>Negation of Statements

The statement

A \to (B \to A)

is equivalent to :

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The negation of the statement ''

72

is divisible by

2

and

3

'' is

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The negation of

“ \forall n \in N, n + 7 > 6 ”

is ….

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

For any two statement

p

and

q,

the negative of the expression

p \lor (~ p \land q)

EASY

Mathematics>Algebra>Mathematical Logic>Negation of Statements

The proposition

(~ p) \lor (p \land ~ q)

is equivalent to

The negation of compound statement p∧q⇔r∧q is equivalent to

Important Questions on Mathematical Logic

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The negation of compound statement $(p \land q) \Leftrightarrow (r \land q)$ is equivalent to